1
INTEGERS
Content Items
Learning Outcomes
1.0 Introduction
The learner is able to
l
l
understand the process of multiplication and
division of integers.
1.1 Multiplication of integers
solve the problems related to multiplication and
division of integers.
1.3 Properties of integers
1.2 Division of integers
l
verify and explain the properties of integers.
l
solve the real-life problems related to integers.
l
1.4 BODMAS rule
simplify and solve the numerical expressions by
using BODMAS rule.
1.0 Introduction :
We have learnt the ordering of integers, addition and
subtraction of integers in the previous class. We know
that the collection of natural numbers {1,2,3,4,5...},
zero {0} and negative numbers {-1, -2, -3, -4, -5...} are
Integers. We use the letter ‘Z’ to represent integers.There
are several situations in our daily life where we use
integers. Look at the following situations and explain
your observations regarding integers.
+4000m
+ 100 C
+
+ 100 C
hotter than
00C
Sea level
–
–+ 100 C
colder than
00C
–900m
– 100 C
Height above the sea level is positive
Depth below the sea level is negative
Temperature above 00C is positive
Temperature below 00C is negative
VII Class Mathematics
1
Integers
n<ó‘«jáT+
|ŸPsÁ’dŸ+K«\T
nuó„«dŸq |˜Ÿ*Ԑ\T
$wŸjáÖ+Xæ\T
nuó²«dŸÅ£”&ƒT,
l
l
|Ÿ]#ájáTeTT
1.1 |ŸPsÁ’dŸ+K«\ >·TD¿±sÁ+
1.2 |ŸPsÁ’dŸ+K«\ uó²>±VŸäsÁ+
1.3 |ŸPsÁ’dŸ+K«\ <óŠsˆ\T
1.4 BODMAS “jáTeT+
1.0
|ŸPsÁd’ +Ÿ K«\ >·TD¿±sÁ+ eT]jáTT uó²>·VäŸ s\qT #ûd $<ó‘H“•
ne>±VŸ²q #ûdŸTÅ£”+{²&ƒT.
|ŸPsÁ’dŸ+K«\ >·TD¿±sÁ+ eT]jáTT uó²>·VŸäs\Å£” dŸ+‹+~ó+ºq
dŸeTdŸ«\T kÍ~ókÍï&ƒT.
l
|ŸPsÁd’ +Ÿ K«\ <ósŠ ˆ\qT dŸ] #áÖ&ƒ>\· &ƒT eT]jáTT $e]+#á>\· &ƒT.
l
|ŸPsÁd’ +Ÿ K«\Å£” dŸ+‹+~ó+ºq “Ôá«J$Ôá dŸeTdŸ«\qT kÍ~ókÍï&Tƒ .
l
dŸ+U²« dŸeÖkÍ\qT BODMAS “jáTeT+ €<ó‘sÁ+>± dŸÖ¿¡Œ ˆ¿£]kÍï&Tƒ .
1.0 |Ÿ]#ájáTeTT :
|ŸPsÁd’ +Ÿ K«\ ç¿£eT+, dŸ+¿£\q+ eT]jáTT e«e¿£\q+\qT eTT+<ŠT Ôás>Á Ü· ýË
HûsÁTÌÅ£”H•eTT. dŸVŸ²È dŸ+K«\T {1,2,3,4,5...}, dŸTH• {0} eT]jáTT
sÁTDdŸ+K«\qT {-1, -2, -3, -4, -5...} ¿£*|¾ |ŸPsÁ’dŸ+K«\T n+{²sÁ“
Ôî \ TdŸ T Å£ ” H•eTT. |Ÿ P sÁ ’ d Ÿ + K«\qT dŸ Ö º+#á ³ Å£ ” ‘Z’ nHû n¿£ Œ s Á +
–|ŸjîÖÐkÍïsÁT. eTq <îÕq+~q J$Ôá+ýË eTq+ |ŸPs’+¿±\qT $$<óŠ
dŸ+<ŠsÒÛ\ýË –|ŸjîÖÐkÍïeTT. ç¿ì+~ dŸ+<ŠsÒÛ\qT >·eT“+º |ŸPsÁ’dŸ+K«\
>·T]+º MT |Ÿ]o\q\qT $e]+#áTeTT.
|ŸPsÁ’dŸ+K«\T
+4000m
+100
+
00C ¿£H•
100C mÅ£”Øe
dŸeTTç<Š eT³¼eTT
–
–00C ¿£H•
–900m
100C ÔáŔ£ Øe
–100
–cþ’ç>·Ôá 0 C ¿£+fñ mÅ£”Øe –+fñ <óŠHÔሿ£+
–cþ’ç>·Ôá 0 C ¿£+fñ ÔáŔ£ Øe –+fñ ‹TTD²Ôሿ£+.
dŸeTTç<ŠeT³¼eTT qT+& mÔáTïýË –+fñ <óŠHÔሿ£+
dŸeTTç<ŠeT³¼eTT qT+& ýËÔáTýË –+fñ ‹TTD²Ôሿ£+
0
0
7e ÔásÁ>·Ü >·DìÔá+
2
|ŸPsÁ’ dŸ+K«\T
In this class we will learn multiplication, division of integers and properties of integers. Let us
recall what we have learnt in the previous class through following exercise.
1.
Represent the following statements with suitable integer.
i)
Sneha deposited `2000 in her savings account.
ii)
A submarine is in the depth of 350 feet in the sea level.
iii) The height of Mount Everest is 8848 m above the sea level.
iv) Temperature of 14 degrees below 00 C.
2.
Fill the missing integers on the number line.
3.
Represent the following additions and subtractions on number line.
i) 7 + 4
ii) –8 + 3
iii) 9 – 11
iv) 13 – 5
Write the following integers in descending order and ascending order.
i) –9, –1, 0, –10, –6
ii) –6, 6, –9, 5, 10, –3
iii) –15, –20, –35, 0, 2
Calculate the following.
i) –2 + 3
ii) –6 + (–2)
iii) 8 – (–6)
iv) –9 + 4
v) –23 – (–30)
vi) 50 – 153
vii) 71 + (–10) –8
viii) –30 + 58 –38
4.
5.
–7
–3
0
4
6
Siachen Glacier
6.
The temperature in Siachen at 5 a.m. was 10 degrees
below 00C. Six hours later, it had increased to 140C.
What is the temperature at 11 a.m.?
7.
A fish was at 16 feet below the surface of sea and went down
another 17 feet. What is its position now from the surface of sea?
8.
A green grocer had a profit of `250 on Monday, a loss of `120 on Tuesday and loss of `180
on Wednesday. Find total profit or loss after 3 days?
9.
Fill the boxes in first
diagram by doing
addition of two
integers and in
second diagram by
doing subtraction of
two integers.
–1+(–4)
(i)
VII Class Mathematics
(ii)
3
Integers
‡ Ôás>Á Ü· q+<ŠT |ŸPsÁd’ +Ÿ K«\ jîTT¿£Ø >·TD¿±sÁ+, uó²>·VäŸ sÁ+ eT]jáTT y{ì <ósŠ ˆ\qT >·T]+º eTq+ Ôî\TdŸTÅ£”+<‘+.
eTT+<ŠT>± ç¿ì+~ ÔásÁ>·ÜýË HûsÁTÌÅ£”q• n+Xæ\qT ç¿ì+~ nuó²«dŸ+ €<ó‘sÁ+>± eTq+ |ŸÚq]ÇeTsÁô #ûdŸTÅ£”+<‘+.
|ŸÚq]ÇeTsÁô nuó²«dŸ+
ç¿ì+~ y¿±«\qT dŸÂsÕq |ŸPsÁ’dŸ+K« –|ŸjîÖÐ+º sjáT+&.
i) d•VŸ² Ôáq bõ<ŠT|ŸÚ U²ÔýË `2000 ÈeT #ûd¾q~.
ii) Èý²+ÔásZ$T dŸeTTç<Š eT³¼eTT qT+& 350 n&ƒT>·T\ ýËÔáTýË –+~.
iii) meÂsdŸT¼ ¥KsÁ+ dŸeTTç<Š eT³¼eTT qT+& 8848 MT³sÁ¢ mÔáTïýË –+~.
iv) 00 C ¿£H• 14 &ç^\ ÔáÅ£”Øe –cþ’ç>·Ôá –+~.
2. ç¿ì+~ dŸ+K« ¹sK™|Õ ýñ“ |ŸPsÁ’dŸ+K«\qT >·T]ï+#á+&.
1.
–7
3.
7+4
ii) –8 + 3
–9, –1, 0, –10, –6
iii) 9 – 11
ii)
ç¿ì+~ y{ì“ ýÉ¿ìØ+#áTeTT.
i) –2 + 3
iv) –9 + 4
vii) 71 + (–10) –8
6.
4
6
iv) 13 – 5
ç¿ì+~ |ŸPsÁ’dŸ+K«\qT nesÃVŸ²D eT]jáTT €sÃVŸ²D ç¿£eT+ýË sjáT+&.
i)
5.
0
ç¿ì+~ |ŸPsÁ’dŸ+K«\ dŸ+¿£\qeTT, e«e¿£\qeTT\qT dŸ+U²«¹sK™|Õ dŸÖº+#á+&.
i)
4.
–3
–6, 6, –9, 5, 10, –3
ii) –6 + (–2)
v) –23 – (–30)
viii) –30 + 58 –38
iii) 8 – (–6)
vi) 50 – 153
d¾jáÖºHŽ e<ŠÝ –<ŠjáT+ ×<ŠT >·+³\Å£” –cþ’ç>·Ôá 0 C ¿£H•10 C
ÔáÅ£”Øe –+~. €sÁT >·+³\ ÔásÇÔá n~ 140C ™|]Ðq~. nsTTq
–<ŠjTá + 11 >·+³\Å£” –cþ’ç>·Ôá m+Ôá –+³T+~?
0
iii) –15,–20, –35, 0, 2
d¾jáÖºHŽ V¾²eTúq<Š+
0
7. ÿ¿£ #û|Ÿ dŸeTTç<Š –|Ÿ]Ôá\+ qT+& 16 n&ƒT>·T\ ýËÔáTýË –+&, eTsÁý²
17 n&ƒT>·T\ ¿ì+<ŠÅ£” yî[ß+~. ç|ŸdŸTïÔá+ dŸeTTç<Š eT³¼+ qT+& #û|Ÿ k͜q+
@$T{ì?
8. ÿ¿£ €Å£” Å£LsÁ\ y«bÍ] kþeTysÁ+ H&ƒT `250 ý²uó+„ , eT+>·Þyø sÁ+ H&ƒT `120 qwŸ+¼ eT]jáTT ‹T<óyŠ sÁ+
H&ƒT `180 qwŸ+¼ bõ+<‘&ƒT. eTÖ&ƒT sÃE\ ÔásÁTyÔá nÔᓿì eºÌq yîTTÔáï+ ý²uó+„ ýñ<‘ qwŸ+¼ m+Ôá?
9.
|Ÿ ¿ £ Ø |Ÿ { ²\ýË ™|fÉ ¼ \ qT
yî T T<Š { ì |Ÿ ³ +ýË Â s +&ƒ T
|Ÿ P sÁ ’ d Ÿ + K«\ jî T T¿£ Ø
dŸ + ¿£ \ q+ €<ó ‘ sÁ + >±
eT]já T T  s +&à |Ÿ ³ +ýË
 s +&ƒ T |Ÿ P sÁ ’ d Ÿ + K«\
e«e¿£ \ q+ €<ó ‘ sÁ + >±
|ŸP]ï#ûjáTTeTT.
–1+(–4)
(i)
7e ÔásÁ>·Ü >·DìÔá+
(ii)
4
|ŸPsÁ’ dŸ+K«\T
1.1
Let us learn the multiplication of integers.
Multiplication of integers :
Tulasi madam conducted a quiz in her class room. Each correct answer carried 5 marks
and each wrong answer carried - 2 marks. Sufiya answered 4 questions correctly and
John answered 4 questions wrongly in first round. Let us calculate their marks.
Sufiya’s marks
5 + 5 + 5 + 5 = 20
John’s marks
(–2) + (–2) + (–2) + (–2) = –8
4 times of 5 = 20
4 times of –2 = –8
4 × 5 = 20
4 × (–2) = –8
We can represent this using number line also.
–2
–10
–9
–8
–7
–2
–2
–6
–5
–4
–3
–2
–2
–1
0
1
another example, 6 × (–5)
6 × (–5) = –30
Let us find 6 × (–5) in another way. Put minus sign (–) before the product of 6 and 5 we get
–30.
6 × (–5) = – (6 × 5) = –30
few more examples : 3 × (–2) = – (3 × 2) = –6
7 × (–4) = – (7 × 4) = ____
2 × (–6) =_____ = ____
Find the values of (i) 4 × (–8) (ii) 5 × (–20) (iii) 7 × (–8) (iv) 10 × (–9)
Observe the following pattern,
3 × 4 = 12
2×4=8
1× 4 = 4
0×4=0
–1 × 4 = –4
–2 × 4 = –8
–3 × 4 = ___
–4 × 4 = ___
What is your observation?
How you answered, –4 × 4 = –16?
1. Prepare a pattern to find (–3) × 5 starting from 4 × 5
2. Prepare a pattern to find (–7) × 3 starting from 5 × 3
VII Class Mathematics
5
Integers
1.1
|ŸPsÁ’dŸ+K«\ >·TD¿±sÁ+ >·T]+º eTq+ Ôî\TdŸTÅ£”+<‘eTT.
|ŸPsÁd
’ +Ÿ K«\ >·TD¿±sÁ+
>·DìÔá –bÍ<ó‘«jáTT“ ÔáT\d¾ yûT&ƒyŽT Ôáq ÔásÁ>·ÜýË ¿ìÇCÙ “sÁÇV¾²+#sÁT. ç|ŸÜ dŸÂsÕq dŸeÖ<ó‘qeTTqÅ£” 5
eÖsÁTØ\T eT]jáTT ç|ŸÜ Ôá|Ÿð dŸeÖ<ó‘qeTTqÅ£” `2 eÖsÁTØ\T. yîTT<Š{ì sš+&ƒTýË dŸT|˜¾jáÖ 4 ç|ŸXø•\Å£”
dŸ]>±Z dŸeÖ<ó‘q+ ‚ºÌ+~. C²HŽ 4 ç|ŸXø•\Å£” Ôá|Ÿð>± dŸeÖ<ó‘q+ ‚#Ì&ƒT y] eÖsÁTØ\qT eTq+
ýÉ¿ìØ<‘eTT.
C²HŽ eÖsÁTØ\T
dŸT|˜¾jáÖ eÖsÁTØ\T
(–2) + (–2) + (–2) + (–2) = –8
–2 Å£” 4 Âs³T¢ = –8
4 × (–2) = –8
5 + 5 + 5 + 5 = 20
5 Å£” 4 Âs³T¢ = 20
4 × 5 = 20
B““ dŸ+U²« ¹sK™|Õ Å£L&† dŸÖº+#áe#áTÌ.
–2
–10
–9
–8
–7
–2
–2
–6
–4
–5
–3
–2
–2
–1
0
1
eTs=¿£ –<‘VŸ²sÁD #áÖ<‘Ý+, 6 × (–5)
6 × (–5) = (–30)
jîTT¿£Ø $\TeqT eTs=¿£ |Ÿ<ŠÆÜýË ¿£qT>=+<‘+ 6 eT]jáTT 5\ \‹ÝeTT sd¾ <‘“ eTT+<ŠT
‹TTD²Ôሿ£ (–) dŸ+È㠖+#áTeTT. n|Ÿð&ƒT eTqÅ£” (–30) dŸeÖ<ó‘qeTòÔáT+~.
6 × (–5)
6 × (–5) = – (6 × 5) = –30
eT]¿=“• –<‘VŸ²sÁD\T 3 × (–2) = – (3 × 2) = –6
7 × (–4) = – (7 × 4) = ____
2 × (–6) =_____ = ____
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
ç¿ì+~ y““ ýÉ¿Øì +#áTeTT (i) 4 × (–8) (ii) 5 × (–20) (iii) 7 × (–8) (iv) 10 × (–9)
ç¿ì+~ neT]¿£qT |Ÿ]o*+#áTeTT.
3 × 4 = 12
2×4=8
1× 4 = 4
0×4=0
–1 × 4 = –4
–2 × 4 = –8
–3 × 4 = ___
–4 × 4 = ___
nHûÇw¾<‘Ý+
MTsÁT @$T >·eT“+#sÁT?
–4 × 4 = –16 n“ mý² #î|ŸÎ>·*>±sÁT?
1. 4 × 5 Ôà çbÍsÁ+_ó+º (–3) × 5 $\Te ¿£qT>=qT³Å£” neT]¿£qT sjáTTeTT.
2. 5 × 3 Ôà çbÍsÁ+_ó+º (–7) × 3 $\Te ¿£qT>=qT³Å£” neT]¿£qT sjáTTeTT.
7e ÔásÁ>·Ü >·DìÔá+
6
|ŸPsÁ’ dŸ+K«\T
Find in similar way,
(–6) × 7 = ____
(–2) × 5 = ______
(–3) × 6 = ____
(–4) × 5 = ______
Thus, to multiply the integers having different signs put the negative sign before the product.
For any two integers ‘a’ and ‘b’, (–a) × b = – (a × b) = a × (–b)
Find the values of
(i) (– 6) × 5
(ii) (–15) × 2
(iii) (–12) × 8
(iv) (–10) × 6
Multiplication of two negative integers :
Observe the following pattern,
–2 × 3 = –6
–2 × 2 = –4
–2 × 1 = –2
1. Prepare a pattern to find (–5) × (–4) starting from (–5) × 3
–2 × 0 = 0
2. Prepare a pattern to find (–7) × (–2) starting from (–7) × 5
–2 × (–1) = 2
–2 × (–2) = 4
–2 × (–3) = 6
What is your observation?
Based on this observation complete the following
–2 × (–4) =____
–2 × (–5) = ______
So, the product of two negative integer is always positive integer.
For example, (–9) × (–2) = 9 × 2 = 18
(–10) × (–2) = 10 × 2 = 20
Find the values of
(i) (– 8) × (–7) (ii) (–15) × (–6) (iii) (–12) × (–2) (iv) (–10) × (–9)
Here we will use the tokens which has green colour on one side and red
colour on other side.
Let green side of token represents a positive and red side of token represents a negative.
Positive
Negative
Instructions: To multiply two integers,

Arrange the 2nd Integer as columns and 1st Integer as rows.

If 1st Integer is negative then flip the tokens after arrangement.

Count the tokens, that number will be answer.
VII Class Mathematics
7
Integers
n<û $<óŠ+>± ç¿ì+~ y{ì“ ýÉ¿ìØ+#áTeTT.
(–6) × 7 = ____ (–2) × 5 = ______
(–3) × 6 = ____ (–4) × 5 = ______
n+<ŠTe\q. $_óq• dŸ+Èã\T ¿£*Ð –q• |ŸPsÁ’ dŸ+K«\qT >·TDì+#á{²“¿ì y{ì \‹ÆeTT eTT+<ŠT ‹TTD²Ôሿ£ dŸ+Èã (–) qT
–+#áTeTT.
a, b \T @yîH
Õ  s +&ƒT |ŸPsÁ’ dŸ+K«\T nsTTq, (–a) × b = – (a × b)= a × (–b)
¿ì+~ y{ì“ ýÉ¿ìØ+#áTeTT.
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
(i) (– 6) × 5
(ii) (–15) × 2 (iii) (–12) × 8 (iv) (–10) × 6
Âs+&ƒT |ŸPsÁ’ dŸ+K«\ \‹ÝeTT :
¿ì+~ neT]¿£qT |Ÿ]o*+#áTeTT.
–2 × 3 = –6
–2 × 2 = –4
nHûÇw¾<‘Ý+
–2 × 1 = –2
1. (–5) × 3Ôà çbÍsÁ+_ó+º (–5) × (–4) $\Te ¿£qT>=qT³Å£” neT]¿£qT sjáTTeTT.
–2 × 0 = 0
2. (–7) × 5 Ôà çbÍsÁ+_ó+º (–7) × (–2) $\Te ¿£qT>=qT³Å£” neT]¿£qT sjáTTeTT.
–2 × (–1) = 2
–2 × (–2) = 4
–2 × (–3) = 6
™|Õ neT]¿£ qT+& @$T >·eT“+#sÁT?
‡ |Ÿ]o\q €<ó‘sÁ+>± ¿ì+~ y{ì“ |ŸP]ï #ûjáT+&.
–2 × (–4) =____
–2 × (–5) = ______
¿±‹{ì¼ Âs+&ƒT ‹TTD |ŸPsÁ’ dŸ+K«\ \‹ÝeTT m\¢|Ÿð&ƒÖ <óŠq |ŸPsÁ’ dŸ+K« neÚÔáT+~.
–<‘VŸ²sÁDÅ£”, (–9) × (–2) = 9 × 2 = 18
(–10) × (–2) = 10 × 2 = 20
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
‚~ #û<‘Ý+!
¿£Ôá«+
¿ì+~ y{ì“ ýÉ¿ìØ+#áTeTT
(i) (– 8) × (–7) (ii) (–15) × (–6) (iii) (–12) × (–2) (iv) (–10) × (–9)
‡ ¿£Ôá«eTTýË ÿ¿£ yîÕ|ŸÚ €Å£” |Ÿ#áÌsÁ+>·T eT]jáTT eTsà yîÕ|ŸÚ msÁT|ŸÚ sÁ+>·T >·\ {Ë¿q¢qT
–|ŸjîÖÐkÍï+.
{Ë¿q¢ ™|Õ yîÕ|ŸÚ €Å£”|Ÿ#áÌ sÁ+>·TýË –+fñ n~ <óŠq dŸ+K«qT, msÁT|ŸÚ sÁ+>·TýË –+fñ n~ ‹TTD
dŸ+K«qT dŸÖºdŸTï+<Š“ nqTÅ£”+<‘eTT.
<óŠq dŸ+K«
‹TTD dŸ+K«
dŸÖ#áq\T : Âs+&ƒT |ŸPsÁ’dŸ+K«\qT >·TDì+#\+fñ,
 Âs+&ƒe |ŸPsÁ’dŸ+K«qT “\TeÚ esÁTdŸýË yîTT<Š{ì |ŸPsÁ’dŸ+K«qT n&ƒT¦esÁdŸ\ýË neTsÌ*.
 yîTT<Š{ì |ŸPsÁ’dŸ+K« ‹TTD dŸ+K« nsTTÔû, eºÌq neT]¿£ýË {Ë¿HŽ\qT çÜ|¾Î neTsÌ*.
 {Ë¿q¢qT ýÉ¿ìØ+ºq³¢sTTÔû € dŸ+K« dŸeÖ<ó‘qeTT neÚÔáT+~.
7e ÔásÁ>·Ü >·DìÔá+
8
|ŸPsÁ’ dŸ+K«\T
We remember that 2 × 5 means add 5 for 2 times
Case (i): Positive integer  Positive integer
Eg: (+2) ´ (+5) means 2 times of 5
 (+2) ´ (+5) = 10
Case (ii): Positive integer  Negative integer
 4 ´ (– 6) = – 24
Eg: 4 ´ (– 6) means 4 times of (– 6)
Case (iii): Negative integer ´ Positive integer
Eg: (–8) ´ 5
(–8) ´ 5 = –40
flip the tokens
Case (iv): Negative integer ´ Negative integer
Eg: (–2) ´ (–7)
(–2) ´ (– 7) = 14
flip the tokens
So, we can generalise the multiplication of two integers as follows
1.
If the signs of two integers are same then the product is positive integer.
2.
If the signs of two integers are different then the product is negative integer.
The idea of the products involving negative numbers was
first considered by Brahmagupta of India in the seventh
century AD. It is described in his book Brahmasphuta
Siddantham. He makes the definition such as negative
times negative is positive to give a single general method
for formulizing problems involving a number and its
square, and for finding their solutions.
VII Class Mathematics
9
Integers
2 × 5 nq>± 5Å£” 2 Âs³T¢ n“ >·TsÁTï#ûdŸTÅ£”+<‘eTT.
dŸ+<ŠsÒÁ eÛ TT (i) : <óŠq |ŸPsÁ’ dŸ+K«  <óŠq |ŸPsÁ’ dŸ+K«
–<‘ : (+2) ´ (+5) nq>± 5Å£” 2 Âs³T¢
 (+2) ´ (+5) = 10
dŸ+<ŠsÒÁ eÛ TT (ii) : <óŠq |ŸPsÁ’ dŸ+K«  ‹TTD |ŸPsÁ’ dŸ+K«
–<‘ : 4 ´ (– 6) nq>± – 6 Å£” 4 Âs³T¢
 4 ´ (– 6) = – 24
dŸ+<ŠsÒÁ eÛ TT (iii) : ‹TTD |ŸPsÁ’ dŸ+K«  <óŠq |ŸPsÁ’ dŸ+K«
–<‘ : (–8) ´ 5
{Ë¿HŽ\qT çÜ|¾Î neTsÌ*
(–8) ´ 5 = –40
dŸ+<ŠsÁÒÛeTT (iv): ‹TTD |ŸPsÁ’ dŸ+K« ´ ‹TTD |ŸPsÁ’ dŸ+K«
–<‘ : (–2) ´ (–7)
{Ë¿HŽ\qT
çÜ|¾Î neTsÌ*
(–2) ´ (– 7) = 14
¿±eÚq, Âs+&ƒT |ŸPsÁ’ dŸ+K«\ >·TD¿±s“• ¿ì+~ $<óŠ+>± kÍ<ó‘sÁD¡¿£]+#áe#áTÌ.
1. Âs+&ƒT |ŸPsÁ’ dŸ+K«\T ÿ¹¿ >·TsÁTï ¿£*Ð –+fñ y{ì \‹ÝeTT <óŠq |ŸPsÁ’ dŸ+K« neÚÔáT+~.
2. s +&ƒT |ŸPsÁ’ dŸ+K«\T yûsÁT yûsÁT >·TsÁTï\T ¿£*Ð –+fñ y{ì \‹ÝeTT ‹TTD |ŸPsÁ’ dŸ+K« neÚÔáT+~.
#]çÔá¿£ n+XøeTT
ç¿¡.Xø. 7e XøԐ‹ÝeTTýËHû uó²sÁrjáT >·DìÔá XægyûÔáï ç‹VŸ²ˆ>·T|ŸÚï&ƒT Ôáq
|ŸÚdŸï¿£yîT®q »ç‹VŸ²ˆdŸTγ d¾<‘Æ+ÔáeTTµ q+<ŠT ‹TTDdŸ+K«\ \‹ÆeTT\qT
yîTT<Š{ìkÍ]>± ç|ŸkÍï$+#&ƒT. ‚ÔáqT ÿ¿£ dŸ+K« eT]jáTT y{ì esZ\ÔÃ
Å£L&q dŸeTdŸ«\ dŸÖçr¿£sÁDÅ£” ÿ¿£ kÍ<ó‘sÁD |Ÿ<ŠÆÜ ‚eÇ&†“¿ì eT]jáTT
y{ì kÍ<óqŠ ¿£qT>=qT³Å£” ‹TTDdŸ+K«qT ‹TTDdŸ+K«Ôà >·TDì+ºq \‹ÆeTT
<óŠqdŸ+K« n“ “sÁǺ+#&ƒT.
7e ÔásÁ>·Ü >·DìÔá+
10
|ŸPsÁ’ dŸ+K«\T
Fill the grid by multiplying each number in the first column with
each number in the first row and answer the following questions.

–5
–5 –4 –3 –2 –1
0
25
1
2
3
4
5
-15
–4
-20
–3
9
–2
1. Write your observations from
the table.
2. What happens when an integer
multiplied with (–1)?
3. When will we get product of
two integers is zero?
0
–1
0
0
1
2
3
4
-8
5
20
Example 1: A sump is full of water, when the motor started pumping the level of water decreasing
2 inches per minute then what is the level of water from the ground level after
20 minutes?
Solution:
The change in level of water per minute = –2 inches (decreasing 2 inches)
The level of water after 20 minutes
= 20 × (–2)
= –40 inches
So, the level of water in sump is 40 inches depth from
the ground level.
Do you know?
1 inch = 2.54cm
Example 2 : An elevator begins from 20m above the ground. It descends into a mine shaft at the
rate of 6m per minute. what will be its position after 15 minutes?
Solution :
Since the elevator is going down, so the distance covered
by it will be represented by a negative integer.
Change in position of the elevator in one minute = –6m
Change in position of the elevator in 15 minutes = 15 × (–6) = –90m
So, the final position of the elevator = 20 + (–90) = – 70m
 The elevator is at 70m below the ground level.
VII Class Mathematics
Integers
‚~ #û<‘Ý+!
ç¿ì+~ |Ÿ{켿£ýË yîTT<Š{ì “\TeÚ esÁdŸýË ç|ŸÜdŸ+K«qT yîTT<Š{ì n&ƒT¦esÁTdŸý˓ ç|ŸÜ dŸ+K«ÔÃ
>·TDìdŸÖï |Ÿ{켿£qT |ŸP]+#áTeTT. ‚eNj&q ç|ŸXø•\Å£” dŸeÖ<ó‘H\T ‚eÇ+&.
¿£Ôá«+

–5
–5 –4 –3 –2 –1
0
25
1
2
3
4
-15
–4
-20
–3
9
–2
0
–1
0
5
0
1. |Ÿ{¿¼ì £ qT+& MTsÁT @$T >·eT“+#sÃ
sjáT+&
2. |ŸPsÁ’dŸ+K«qT (–1) #û >·TD¿±sÁ+
#ûd¾q|ŸÚ&ƒT @eTeÚÔáT+~?
3. Âs+&ƒT |ŸPsÁ’dŸ+K«\ \‹ÆeTT dŸTH•
m|Ÿð&ƒT neÚÔáT+~?
1
2
3
4
-8
5
20
–<‘VŸ²sÁD 1: ú{ì >·T+³ (dŸ+|t) |ŸP]ï>± ú{ìÔà “+&eÚ+~. yîÖ{²sÁTÔà ú{ì“ ÔÃ&ƒ&ƒ+ e\q ú{ì k͜sTT
“eTTc͓¿ì Âs+&ƒT n+>·TÞ²\ #=|ŸÚÎq ÔáÐZq 20 “eTTcÍ\ ÔásÁTyÔá Hû\eT³¼eTT qT+& úsÁT m+Ôá
<ŠÖsÁeTTýË –+³T+~?
kÍ<óŠq :
ÿ¿£ “eTTwŸeTTýË ú{ì eT³¼eTTýË eÖsÁTÎ
= -2 n+>·TÞ²\T (2 n+>·TÞ²\T ÔáÐZq~)
20 “eTTcÍ\ ÔásÁTyÔá ú{ì eT³¼eTTýË eÖsÁTÎ = 20 × (-2)
MTÅ£” Ôî\TkÍ?
= –40 n+>·TÞ²\T
¿±eÚq, ú{ì >·T+³ýË úsÁT Hû\ eT³¼eTT qT+& 40 n+>·TÞ²\ ýËÔáTýË –+&ƒTqT. 1 n+>·TÞø+ R
2.54 ™d+.MT
–<‘VŸ²sÁD 2: uó„Ö$T qT+º 20MT³sÁ¢ mÔáTï qT+º ÿ¿£ *|˜t¼ çbÍsÁ+uó„eTsTT+~.n~ >·“ ýË|Ÿ*¿ì “eTTc͓¿ì 6
MT³sÁ¢ #=|ŸÚÎq ¿ì+~¿ì yî[ßq, 15 “eTTcÍ\ ÔásÁTyÔá <‘“ k͜qeTT ¿£qT>=qTeTT.
kÍ<óŠq :
*|˜t¼ ¿ì+~¿ì yîÞø—ïq•~ ¿±eÚq n~ yîÞâß <ŠÖs“• sÁTD |ŸPsÁ’ dŸ+K«Ôà dŸÖºkÍïeTT.
ÿ¿£ “$TwŸ+ýË *|˜t¼ jîTT¿£Ø k͜qeTTýË eÖsÁTÎ =-6 MT.
15 “eTTcÍ\ýË *|˜t¼ jîTT¿£Ø k͜qeTTýË eÖsÁTÎ =15 × (-6) R -90MT.
15 “eTTcÍ\ ÔásÁTyÔá *|˜t¼ jîTT¿£Ø k͜qeTT R 20 G (-90) R -70MT.
*|˜t¼ Hû\ eT³¼eTT qT+& 70 MT³sÁT¢ ýËÔáTýË –+&ƒTqT.
7e ÔásÁ>·Ü >·DìÔá+
12
|ŸPsÁ’ dŸ+K«\T
Example 3 :In a test, (+5) marks are given for every correct answer and (–3) marks are given
for every incorrect answer. Lakshmi gets 45 correct and 15 incorrect answers. What
is her score?
Solution : Marks given for one correct answer = 5
Marks obtained for 45 correct answers = 45 × 5 = 225
Marks given for one incorrect answer = –3
Marks obtained for 15 incorrect answers = 15 × (–3) = –45
 Lakshmi’s score = 225 + (–45) = 180.
Exercise - 1.1
1.
2.
Multiply the following
i)
5×7
(–9) × (6)
iii)
(9) × (–4)
iv)
(8) × (–7)
v)
(–124) × (–1) vi)
(–12) × (–7)
vii)
(–63) × 7
viii)
7 × (–15)
2 × (–5) or 3 × (–4)
ii)
(–6) × (–7) or (–8) × 5
iii) (–6) × 10 or (–3) × (–21)
iv)
9 × (–11) or 6 × (–16)
Which is greater?
i)
v)
3.
(–8) × (–5) or (–9) × (–4)
Write the pair of integers whose product gives,
i)
4.
ii)
A negative integer
ii)
A positive integer
iii) Zero
A frog is slipping into a well from upper surface at a rate
of 3 meters per minute, after 5 minutes what is the position
of the frog in the well?
5.
During the summer, the level of water in a pond decreases
by 5 inches every week due to evaporation. What is the
change in the level of the water over a period of 6 weeks?
6.
A shop keeper earns a profit of `5 on one note book and loss
of `3 on one pen by selling in the month of July. He sells 1500
books and 1500 pens. Find out what is his profit or loss?
7.
A cement company earns a profit of `8 per bag of white cement and a loss of `6 per bag of
grey cement by selling.The company sells 2,000 bags of white cement and 3,000 bags of
grey cement in a month. Find out what is its profit or loss?
VII Class Mathematics
13
Integers
–<‘VŸ²sÁD 3 : ÿ¿£ |Ÿ¯¿£ŒýË sjáT‹&q dŸÂsÕq dŸeÖ<ó‘H“¿ì (G5) eÖsÁTØ\T, Ôá|ŸÚÎ nsTTq dŸeÖ<ó‘H“¿ì (-3)
eÖsÁTØ\T ¹¿{²sTT+#á&ƒ+ È]Ð+~. \¿ìë sd¾q dŸeÖ<ó‘H\ýË 45 dŸÂsÕq$, 15 Ôá|Ÿð nsTTq €yîTÅ£” eºÌq
eÖsÁTØ\T m“•?
kÍ<óŠq : ÿ¿£ dŸÂsÕq dŸeÖ<ó‘H“¿ì ¿=sÁÅ£” ‚eNj&ƒ¦ eÖsÁTØ\T R 5
45 dŸÂsÕq dŸeÖ<ó‘H\Å£” eÖsÁTØ\T R 45 × 5 R 225
ÿ¿£ Ôá|ŸÚÎ dŸeÖ<ó‘H“¿ì ‚eNj&ƒ¦ eÖsÁTØ\T R -3
15 Ôá|ŸÚÎ dŸeÖ<ó‘H\¿ì ‚eNj&ƒ¦ eÖsÁTØ\T R 15 × (-3) R -45
\¿ìë¿ì eºÌq eÖsÁTØ\T R 225 G (-45) R 180.
nuó²«dŸ+ - 1.1
1. ¿ì+~ y{ì“ >·TDì+#á+&.
i)
5×7
ii)
(–9) × (6)
iii)
(9) × (–4)
iv)
(8) × (–7)
v)
(–124) × (–1)
vi)
(–12) × (–7)
vii)
(–63) × 7
viii)
7 × (–15)
2. ¿ì+~ y{ìýË ™|<ŠÝ~ @~?
i) 2 × (–5) ýñ<‘ 3 × (–4)
ii)
(–6) × (–7) ýñ<‘ (–8) × 5
iii) (–6) × 10 ýñ<‘ (–3) × (–21)
iv)
9 × (–11) ýñ<‘ 6 × (–16)
v) (–8) × (–5) ýñ<‘ (–9) × (–4)
3. \‹ÝeTT
i) ÿ¿£ <óŠq |ŸPsÁ’ dŸ+K« ii) ÿ¿£ sÁTD |ŸPsÁ’ dŸ+K« iii) dŸTH• n>·Tq³T¢ |ŸPsÁ’ dŸ+K«\ ÈÔá\qTsjáTTeTT.
4. ÿ¿£ ¿£|ŸÎ “eTTc͓¿ì 3 MT³sÁ¢ e+ÔáTq u²$ ™|Õ –|Ÿ]Ôá\+ qT+& ýË|Ÿ*¿ì
C²sÁTÔáTq•~. ¿£|ŸÎ 5 “eTTcÍ\ ÔásÁTyÔá u²$ýË @ k͜qeTTýË –+³T+~?
5. yûdŸ$ýË ÿ¿£ ¿=\qTýË ú{ì eT³¼eTT u²wÓÎ uó„eq+ e\q ÿ¿£ sÃE¿ì 5
n+>·TÞ²\ #=|ŸÚÎq Ôá>·TZÔáTq•~. ú{ì k͜sTT d¾œsÁ |Ÿ]eÖDeTTýË
Ôá>·TZ#áTq•#Ã, 6 ys\ ÔásÁTyÔá ¿=\qTýË ú{ì eT³¼eTTýË eÖsÁTÎ m+Ôá
–+&ƒTqT?
6. ÿ¿£ <ŠT¿±D<‘sÁT&ƒT ÿ¿=Ø¿£Ø |ŸÚdŸ¿ï +£ neTˆ&ƒ+ e\q `5\T ý²u󲓕, ÿ¿=Ø¿£Ø
™|qT• neTˆ&ƒ+ e\q `3\T qwŸ¼eTT bõ+<ŠTqT. pýÉÕ Hî\ýË nÔáqT
1500 |ŸÚdŸï¿±\T eT]jáTT 1500 ™|qT•\T n$Tˆq yîTTÔáï+ MT<Š nÔᓿì
eºÌq ý²uó„eTT ýñ<‘ qwŸ¼eTTqT ¿£qT>=qTeTT.
7. ÿ¿£ d¾yîT+³T ¿£+™|ú ÿ¿=Ø¿£Ø Ôî\T|ŸÚ sÁ+>·T ‹kÍï d¾yîT+³T ™|Õ `8 ý²uó„+, ‹Ö&<Š sÁ+>·T ‹kÍï d¾yîT+³T ™|Õ
`6 qwŸ¼eTTÔà n$Tˆ+~. ÿ¿£ Hî\ýË 2,000 ‹kÍï\ Ôî\T|ŸÚ d¾yîT+³T 3,000 ‹kÍï\ ‹Ö&<Š sÁ+>·T d¾yîT+³T
n$Tˆq d¾yîT+³T ¿£+™|ú¿ì € Hî\ýË eºÌq ý²uó„eTT ýñ<‘ qwŸ¼eTT ¿£qT>=qTeTT.
7e ÔásÁ>·Ü >·DìÔá+
14
|ŸPsÁ’ dŸ+K«\T
8.
Fill in the blanks with suitable integer to make the statement true.
i)
(–4) × ___ = –20
iv) ___ × (–9) = 45
ii)
___ × 5 = –35
iii) (–6) × ____ = 48
v)
___ × 7 = – 42
vi)
8 × ______= –8
1.2 Division of Integers :
Division is the inverse operation of multiplication. Let us see an example, take two nonzero integers 6 × 4 = 24. So, 24 ÷ 4 = 6 and 24 ÷ 6 = 4
We can say for each multiplication statement there are two division statements.
Observe the following.
Multiplication statement
Division statements
5 × 3 = 15
15 ÷ 3 = 5
15 ÷ 5 = 3
6 × (–2) = –12
(–12) ÷ 6
= ––––––
Do you know?
Division by zero
is not defined
(–12) ÷ (–2) = ––––––
(–10) × 2 = –20
(–20) ÷ (–10) = ––––––
––––––––––––––––––
(–5) × (–6) = 30
––––––––––––––––––
––––––––––––––––––
What do you observe from the table?
Thus, we can conclude as follows :
12 ÷ 3 = 4
Positive integer divided by positive integer, the quotient is positive.
(–12) ÷ 3 = –4
Negative integer divided by positive integer, the quotient is negative.
12 ÷ (–3) = –4
Positive integer divided by negative integer, the quotient is negative.
(–12) ÷ (–3) = 4
Negative integer divided by negative integer, the quotient is positive.
Division of two integers follows the same rules of multiplication.
If the signs are same then the quotient is positive.
If the signs are different then the quotient is negative.
Fill the following table:
S.No.
VII Class Mathematics
Integer 1 ÷ Integer 2
1
(+25) ÷ (+5)
2
42 ÷ (– 6)
3
(–75) ÷ 15
4
(–27) ÷ (–3)
15
Quotient
5
Integers
8. ¿ì+~ U²°\qT dŸÂsÕq |ŸPsÁ’ dŸ+K«#û |ŸP]+#áTeTT.
i)
(–4) × ___ = –20
iv) ___ × (–9) = 45
ii)
___ × 5 = –35 iii)
(–6) × ____ = 48
v)
___ × 7 = – 42 vi)
8 × ______= –8
1.2 |ŸPsÁ’ dŸ+K«\ uó²>·VŸäsÁeTT :
uó²>·VŸäsÁ+ nHû~ >·TD¿±sÁ+ jîTT¿£Ø $ýËeT |Ÿ]ç¿ìjáT n“ eTqÅ£” Ôî\TdŸT. –<‘VŸ²sÁDÅ£” Âs+&ƒT XøSHû«ÔásÁ
|ŸPsÁ’ dŸ+K«\qT rdŸTÅ£”+<‘+. 6 × 4 = 24 ¿±eÚq 24 ÷ 4 = 6 eT]jáTT 24 ÷ 6 = 4
ç|ŸÜ >·TD¿±s“¿ì Âs+&ƒT uó²>·VŸäsÁ y¿±«\T –+{²jáT“ #î|ŸÎe#áTÌ.
¿ì+~ |Ÿ{켿£qT |Ÿ]o*+º $TÐ*q U²°\qT |ŸP]+#á+&.
>·TD¿±sÁ y¿£«eTT
5 × 3 = 15
uó²>·VŸäsÁ y¿±«\T
15 ÷ 3 = 5
15 ÷ 5 = 3
6 × (–2) = –12
(–12) ÷ 6
= ––––––
(–12) ÷ (–2) = ––––––
(–10) × 2 = –20
(–20) ÷ (–10) = ––––––
––––––––––––––––––
(–5) × (–6) = 30
MTÅ£” Ôî\TkÍ?
dŸTH•Ôà uó²>·VŸäsÁeTT
“sÁǺ+#áýñeTT
––––––––––––––––––
––––––––––––––––––
™|Õ |Ÿ{켿£ qT+& MTsÁT @$T >·eT“+#sÁT?
¿ì+<Š $<óŠeTT>± kÍ<ó‘sÁD¡¿£]+e#áTÌ.
12 ÷ 3 = 4
<óŠq |ŸPsÁ’dŸ+K«qT <óŠq |ŸPsÁ’dŸ+K«Ôà uó²Ð+ºq uó²>·|˜Ÿ\+ <óŠHÔሿ£+.
(–12) ÷ 3 = –4
‹TTD |ŸPsÁ’dŸ+K«qT <óŠq |ŸPsÁ’dŸ+K«Ôà uó²Ð+ºq uó²>·|˜Ÿ\+ ‹TTD²Ôሿ£+.
12 ÷ (–3) = –4
<óŠq |ŸPsÁ’dŸ+K«qT ‹TTD |ŸPsÁ’dŸ+K«Ôà uó²Ð+ºq uó²>·|˜Ÿ\+ ‹TTD²Ôሿ£+.
(–12) ÷ (–3) = 4
‹TTD |ŸPsÁ’dŸ+K«qT ‹TTD |ŸPsÁ’dŸ+K«Ôà uó²Ð+ºq uó²>·|˜Ÿ\+ <óŠHÔሿ£+.
Âs+&ƒT |ŸPsÁ’dŸ+K«\ uó²>·VŸäsÁeTT Å£L&† >·TsÁTï\ |ŸsÁeTT>± >·TD¿±sÁeTT jîTT¿£Ø “jáTeÖýHû bÍ{ì+#áTqT.
ÿ¹¿ >·TsÁTï >·\ Âs+&ƒT |ŸPsÁ’ dŸ+K«\ uó²>·|˜Ÿ\eTT <óŠHÔሿ£+
yûsÁT yûsÁT >·TsÁTï>·\ Âs+&ƒT |ŸPsÁ’ dŸ+K«\ uó²>·|˜Ÿ\eTT ‹TTD²Ôሿ£+.
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
ç¿ì+~ |Ÿ{켿£qT |ŸP]ï #ûjáTTeTT.
ç¿£.dŸ+
7e ÔásÁ>·Ü >·DìÔá+
1e |ŸPsÁ’ dŸ+K« ÷ 2e |ŸPsÁ’ dŸ+K«
1
(+25) ÷ (+5)
2
42 ÷ (– 6)
3
(–75) ÷ 15
4
(–27) ÷ (–3)
16
uó²>·|˜Ÿ\eTT
5
|ŸPsÁ’ dŸ+K«\T
Example 4 : A borewell machine drills down 72 feet per hour from surface of the earth. If the
water is at 360 feet down from surface of earth, after how many hours it will touch
the water layer?
Solution :
Depth of drilling in one hour = –72 feet
Depth of water layer from surface of earth = –360 feet
Number of hours required = –360 ÷ (–72) = 5
Hence, the borewell machine will touch water layer at 5 hours of drilling.
Example 5 : In a test, (+4) marks are given for every correct answer and (–2) marks are given
for every incorrect answer. Sai answered all the questions and scored 26 marks
from 8 correct answers. How many incorrect answers had Sai attempted?
Solution :
Marks given for one correct answer = 4
So, marks given for 8 correct answers = 4 × 8 = 32
Sai score = 26
Marks obtained for incorrect answers = 26 – 32 = –6
Marks given for one incorrect answer = (–2)
 Number of incorrect answers = (–6) ÷ (–2) = 3
Example 6 : Shop keeper Yaseen earns a profit of `20 per bag of Sonamasoori rice sold and
loss of `12 per bag of Hamsa rice. In one week he gets neither profit nor loss, if he
solds1440 Sonamasoori rice bags. How many Hamsa rice bags did he sell?
Solution :
In the given problem there is neither profit or loss.
So, Profit earned + loss incurred = 0
Profit earned = – loss incurred
Profit earned by selling one Sonamasoori rice bag = `20
Profit earned by selling 1440 Sonamasoori rice bag = 1440 × 20 = `28800
Loss incurred by selling one Hamsa rice bag = ` 12 which we denoted by –12
Loss incurred by selling one Hamsa rice bags = ` –28800
Total number of Hamsa rice bags sold = (–28800) ÷ (–12) = 2400 bags
The fish in the pond below, carry some
numbers. Choose any 4 pairs and carry out
four multiplications with those numbers. Now,
choose four other pairs and carry out divisions
with those numbers.
For example
(i) (–10) × 6 = –60
VII Class Mathematics
(ii) (–36) ÷ 6 = –6
17
Integers
–<‘VŸ²sÁD 4 : uó„Ö –|Ÿ]Ôá\+ qT+º ÿ¿£ uËsYyîýÙ jáT+çÔá+ ç|ŸÜ>·+³Å£” 72 n&ƒT>·T\ ýËÔáTqT çÔáeÇ>·\<ŠT. uó„Ö
–|Ÿ]Ôá\+ qT+º 360 n&ƒT>·T\ ýËÔáTýË –q• ú{ìbõsÁqT #ûsÁT³Å£” € jáT+çԐ“¿ì m+Ôá
dŸeTjáT+ |Ÿ&ƒTÔáT+~?
kÍ<óŠq:
ÿ¿£ >·+³ýË çÔáeÚÇ ýËÔáT R -72 n&ƒT>·T\T
–|Ÿ]Ôá\eTT qT+& ú{ìbõsÁ >·\ <ŠÖsÁeTT R -360
n&ƒT>·T\T
ú{ì“ #ûsÁT³Å£” |Ÿ³T¼ dŸeTjáTeTT R -360 ÷ (-72) R 5
¿±eÚq, uËsYyîýÙ jáT+çÔá+ ú{ìbõsÁqT #ûsÁT³Å£” 5 >·+³\
dŸeTjáT+ |Ÿ&ƒTÔáT+~.
–<‘VŸ²sÁD 5 : ÿ¿£ |Ÿ¯¿£ŒýË, ç|ŸÜ dŸÂsÕq dŸeÖ<ó‘H“¿ì (G4) eÖsÁTØ\T, eT]jáTT ç|ŸÜ Ôá|ŸÚÎ dŸeÖ<ó‘H“¿ì (-2)
eÖsÁTØ\T ‚eNj&ƒÔsTT. kÍsTT n“• ç|ŸXø•\Å£” dŸeÖ<ó‘q+ ‚#ÌsÁT. kÍsTT çyd¾q 8 dŸÂsÕq
dŸeÖ<ó‘H\ <‘Çs 26 eÖsÁTØ\T kÍ~ó+#&ƒT. kÍsTT sd¾q Ôá|ŸÚÎ dŸeÖ<ó‘H\T m“•?
kÍ<óŠq:
i) ÿ¿=Ø¿£Ø dŸÂsÕq dŸeÖ<ó‘H“¿ì eÖsÁTØ\T R 4
8 dŸÂsÕq dŸeÖ<ó‘H\Å£” eÖsÁTØ\T R 4 × 8 R 32
kÍsTT¿ì eºÌq eÖsÁTØ\T R 26
Ôá|ŸÚÎ dŸeÖ<ó‘H\Å£” ‚eNj&q eÖsÁTØ\T R 26 - 32 R -6
ÿ¿=Ø¿£Ø Ôá|ڟ Î dŸeÖ<ó‘H“¿ì eÖsÁTØ\T R (-2)
 kÍsTT sd¾q Ôá|ŸÚÎ dŸeÖ<ó‘H\T R (-6) ÷ (-2) R 3
–<‘VŸ²sÁD 6 : <ŠT¿±D<‘sÁT&ƒT jáÖd¾HŽ ÿ¿£ kþH eTdŸÖ] _jáT«|ŸÚ ‹kÍï ™|Õ `20 ý²uó„eTTÔÃeT]jáTT VŸ²+dŸ
kÍ<óŠq :
‚~ #û<‘Ý+!
¿£Ôá«+
_jáT«|ŸÚ ‹kÍï ™|Õ `12 qwŸ¼eTTÔÃneֈ&ƒT. ÿ¿£ Hî\ýË1440 kþH eTdŸÖ] _jáT«|ŸÚ ‹kÍï\T
n$TˆH ý²uóe„ TT ¿±ú, qwŸe¼ TT ¿±ú sýñ<TŠ . nsTTq € Hî\ýË m“• VŸ²+dŸ _jáT«|ŸÚ ‹kÍï\T
neֈ&ƒT?
‚ºÌq dŸeTdŸ«ýË ý²uó„eTT ¿±ú, qwŸ¼eTT ¿±ú ýñ<ŠT.
¿±eÚq, eºÌq ý²uó„eTT G eºÌq qwŸ¼eTT R 0
eºÌq ý²uó„eTT R - eºÌq qwŸ¼eTT
ÿ¿£ kþH eTdŸÖ] _jáT«|ŸÚ ‹kÍï ™|Õ e#áTÌ ý²uóe„ TT R `20
1440 kþH eTdŸÖ] _jáT«|ŸÚ ‹kÍï\ ™|Õ e#áTÌ ý²uó„eTT R 1440 × 20 R `28800
VŸ²+dŸ _jáT«|ŸÚ ‹kÍï\™|Õ e#áTÌ qwŸ¼eTT R -`28800
ÿ¿£ VŸ²+dŸ _jáT«|ŸÚ ‹kÍï ™|Õ e#áTÌ qwŸ¼eTT R `12, B““ eTq+ -12>± dŸÖºkÍï+.
€ Hî\ýË n$Tˆq VŸ²+dŸ _jáT«|ŸÚ ‹kÍï\ dŸ+K« R (-28800) ÷ (-12) R 2400 ‹kÍï\T.
|Ÿ¿Ø£ ¿=\qTý˓ #û|\Ÿ ™|Õ ¿=“• dŸ+K«\T –q•$. @yû“
4 ÈÔá\ dŸ+K«\qT mqT•¿=“ 4 >·TD¿±sÁ y¿±«\T
sjáTTeTT. ÔásÁTyÔá 4 ÈÔá\ dŸ+K«\qT mqT•¿=“
4 uó²>·VŸäsÁ y¿±«\T sjáTTeTT.
–<‘VŸ²sÁD:
(i) (–10) × 6 = –60
7e ÔásÁ>·Ü >·DìÔá+
(ii) –36 ÷ 6 = –6
18
|ŸPsÁ’ dŸ+K«\T
Exercise - 1.2
1.
Calculate the following.
i)
(–96) ÷ 16
ii)
98 ÷ (–49)
iii) (–51) ÷ 17
iv) 38 ÷ (–19)
v)
(–80) ÷ 20
vi) (–150) ÷ (–25)
vii) (–600) ÷ 60
viii) (–54) ÷ 9
ix) 130 ÷ 65
x)
(–315) ÷ (–315)
2.
3.
The product of two integers is –165. If one number is –15, Find the other integer.
Because of covid-19 a company lockdown for 6 months and
got loss of `1,32,000 in the year 2020. Find the average loss
of each month.
4.
The temperature at 12 noon was 10°C above zero. If it
decreases at the rate of 2°C per hour until midnight, at what
time would the temperature be 8°C below zero? What would
be the temperature at mid-night?
5.
A green grocer earns a profit of `7 per kg of tomato and got
loss of `4 per kg of brinjal by selling. On Monday he gets
neither profit or loss, if he sold 68 kgs of tomato. How many
kgs of brinjal did he sell?
6.
In a test, + 3 marks are given for every correct answer and –1 mark are given for every
incorrect answer. Sona attempted all the questions and scored +20 marks though she got 10
correct answers.
i)
How many incorrect answers she attempted?
ii) How many questions were given in the test?
7.
Write 5 pairs of integers (a, b) such that a ÷ b = –4.
(Example:(12, –3) because 12÷ (–3) = – 4)
Jasvi says her favorite number through puzzle.
Find it.
–15
¸3
‘Î’ ‘Ï’
× (–2)
–20
¸ (–2)
+4
–5
×4
VII Class Mathematics
Do you know?
¸ (–4)
19
These symbols are used to indicate
whether the given element belongs to
the given collection or not.
0 W (0 belongs to whole numbers)
0 N (0 does not belongs to natural
numbers)
–3 Z (–3 belongs to integers)
Integers
nuó²«dŸ+ - 1.2
1. ç¿ì+~ y{ì“ ýÉ¿ìØ+#á+&.
i)
(–96) ÷ 16
ii) 98 ÷ (–49)
iii) (–51) ÷ 17
iv) 38 ÷ (–19)
v)
(–80) ÷ 20
vi) (–150) ÷ (–25)
vii) (–600) ÷ 60
viii) (–54) ÷ 9
ix) 130 ÷ 65
x)
(–315) ÷ (–315)
2. Âs+&ƒT |ŸPsÁ’ dŸ+K«\ \‹ÝeTT -165 eT]jáTT n+<ŠTýË ÿ¿£ dŸ+K« -15 nsTTq Âs+&ƒe dŸ+K« m+Ôá?
3. 2020 dŸ+. ýË ¿Ã$&Ž-19 ý²¿ù &êHŽ e\q ÿ¿£ ¿£+™|ú 6 Hî\\ýË
`1,32,000 qwŸ¼bþsTTq, Hî\dŸ] dŸsdŸ] qwŸ¼eTT ¿£qT>=qTeTT.
4. eT<ó‘«VŸ²•+ 12 >·+³\Å£” –cþ’ç>·Ôá 00 ™|Õq 10°C n“ >·T]ï+#á‹&q~.
–cþ’ç>·Ôá ç|ŸÜ >·+³Å£” 20C #=|Ÿðq nsÁÆsçÜ esÁÅ£” Ôá>·TZÔáT+~. @
dŸeTjá֓¿ì 00 ¿£H• 80C ÔáŔ£ Øe>± –+³T+~. nsÁsÆ çÜ –cþ’ç>·Ôá m+Ôá?
5. ÿ¿£ Å£LsÁ>±jáT\ y«bÍ] ÿ¿£ ¿ì.ç>±. ³yîÖ{² ™|Õ `7 ý²uó„eTTÔÃ, ÿ¿£
¿ì.ç>±. e+¿±jáT\ ™|Õ `4 qwŸ¼eTTÔà neֈ&ƒT. nÔáqT kþeTysÁeTT 68
¿ì.ç>±. \ ³eÖ{²\T n$TˆH ý²uóe„ TT ¿±ú qwŸe¼ TT ¿±ú sýñ<TŠ . nsTTq
nÔáqT € sÃE m“• ¿ì.ç>±.\ e+¿±jáT\T neֈ&ƒT?
6. ÿ¿£ |Ÿ¯¿£ýŒ Ë, ç|ŸÜ dŸs qÕ dŸeÖ<ó‘H“¿ì (+3) eÖsÁTØ\T, eT]jáTT ç|ŸÜ Ôá|ڟ Πţ” (-1) eÖsÁTØ\T ‚eNj&ƒÔsTT.
kþH n“• ç|ŸX•ø \Å£” dŸeÖ<ó‘H\T çyjáT>±, n+<ŠTýË 10 dŸ]jîT® q$ eT]jáTT €yîT 20 eÖsÁTØ\T bõ+~q~.
i) €yîT sd¾q Ôá|ŸÚÎ dŸeÖ<ó‘H\T m“•?
ii) |Ÿ¯¿£ŒýË ‚eNj&q yîTTÔáï+ ç|ŸXø•\T m“•?
7. a ÷ b = –4 n>·Tq³T¢ 5 |ŸPsÁ’ dŸ+K«\ ÈÔá\T (a, b) çyjáTTeTT.
(–<‘VŸ²sÁD: (12, -3) m+<ŠT¿£q>± 12 ÷ (-3) = - 4).
|Ÿ›ýÙ fÉ®yŽT
鴂 Ôáq ‚wŸ¼yîT®q dŸ+K«qT ÿ¿£ |Ÿ›ýÙ sÁÖ|ŸeTTýË #î|¾Îq~.
MTsÁT € dŸ+K«qT ¿£qT>=q+&.
–15
¸3
× (–2)
–20
¸ (–2)
+4
–5
×4
7e ÔásÁ>·Ü >·DìÔá+
MTÅ£” Ôî\TkÍ ‘Î’ ‘Ï’
ÿ¿£ dŸ+K«, ‚ºÌq dŸeTT<‘já֓¿ì #î+~q<‘? ýñ<‘?
n“ Ôî*jáTCñjTá T³Å£” ‡ >·TsÁT\ï T –|ŸjÖî ÐkÍïeTT.
0 W (0 |ŸPs’+¿±\ dŸeTT<‘já֓¿ì #î+<ŠTqT)
0 N (0 dŸV²Ÿ È dŸ+U²« dŸeTT<‘já֓¿ì #î+<Š<TŠ )
–3 Z (-3 |ŸPsÁ’ dŸ+K«\ dŸeTT<‘já֓¿ì #î+<ŠTqT).
¸ (–4)
20
|ŸPsÁ’ dŸ+K«\T
1.3 Properties of Integers :
In previous class we learnt the properties of whole numbers. Here we will learn the properties
of integers. Let us learn properties of integers under four fundamental operations
simultaneously.
i) Closure property: Observe the following and complete the tables.
Addition
Multiplication
Statement
Conclusion
–5  3= –15
The product is an integer
–3  (–2) =
0  (–3) =
7  (–6) =
The sum is an integer
The product of any two integers
The sum of any two integers also an integer.
is also an integer.
For any two integers a and b, a + b is also
For any two integers a and b, a ´ b is also
an integer (a, b Î Z then a + b ÎZ)
an integer (a, b Î Z then a ´ b ÎZ)
\ Integers are closed under addition and multiplication.
Statement
–5 + 3= –2
–3 + (–2) =
0 + (–3) =
7 + (–6) =
Conclusion
The sum is an integer
Can you find at least one pair of integers whose sum or
product is not an integer?
Subtraction
Statement
–7 –8 = –15
–5 – (–3) =
0 – (–3) =
9 – (–6) =
Division
Conclusion
The difference is an integer
The difference is an integer
The difference of any two integers is an integer.
Statement
Conclusion
The quotient is not an integer
9 ÷ 10 = 0.9
The quotient is an integer
–6 ÷ (–2) =3
0 ÷ (–3) =
7 ÷ (–6) =
In general, the quotient of any two
integers need not an integer.
For any two integers a and b (b ¹ 0), a ÷ b need
not an integer (a, b Î Z then a ÷ b ÎZ)
Integers need not closed under division
For any two integers a and b, a – b is also
an integer (a, b Î Z then a – b ÎZ)
\ Integers are closed under subtraction
ii) Commutative Law :
Observe the following and complete the tables.
Addition
Multiplication
Statement 1 Statement 2 Conclusion
Statement 1 Statement 2
Conclusion
–6  3 = –3
3  (–6) = –3
(–9) + 3 =
3  (–9) =
(–9)  3 =
(–5) + (–6) =
–6  (–5) =
(–5)  (–6) =
–6 + 3 = –3
3 + (–6) = –3
3 + (–9) =
–6 + (–5) =
–3 + 2 =
–6 + 3=3 + (–6)
–6  3 = 3  (–6)
–3  2 =
4 + (–5) =
4  (–5) =
For any two integers a and b, a + b = b + a For any two integers a and b, a ´ b = b ´ a
\ Integers are commutative under additon and multiplication.
VII Class Mathematics
21
Integers
1.3 |ŸPsÁ’ dŸ+K«\ <óŠsˆ\T:
ç¿ì + ~ Ôá s Á > · Ü ýË |Ÿ P s’ + ¿±\ <ó Š s ˆ\T Hû s Á T ÌÅ£ ” H•eTT. ‚|Ÿ Ú &ƒ T eTqeTT |Ÿ P sÁ ’ d Ÿ + K«\ <ó Š s ˆ\qT
HûsÁTÌÅ£”+<‘eTT.H\T>·T #áÔáT]Ç<óŠ |Ÿç¿ìjáT\™|Õ |ŸPsÁ’ dŸ+K«\ <óŠsˆ\qT ÿ¹¿ kÍ] |Ÿ]o*<‘Ý+.
i) dŸ+eÔá <ós
Š ˆÁ eTT : ¿ì+~ |Ÿ{켿£\qT |Ÿ]o*+º $TÐ*q U²°\T |ŸP]+#áTeTT.
>·TD¿±sÁ+
dŸ+¿£\q+
ç|Ÿe#áq+
kÍs+Xø+
ç|Ÿe#áq+
kÍs+Xø+
–5 + 3= –2
yîTTÔáï+ ÿ¿£ |ŸPsÁ’ dŸ+K«
–5  3= –15
\‹ÝeTT ÿ¿£ |ŸPsÁ’ dŸ+K«
–3 + (–2) =
–3  (–2) =
0 + (–3) =
7 + (–6) =
yîTTÔáï+ ÿ¿£ |ŸPsÁ’ dŸ+K«
@yû“ Âs+&ƒT |ŸPsÁ’ dŸ+K«\ jîTT¿£Ø yîTTÔáï+ Å£L&† ÿ¿£ |ŸPsÁ’
dŸ+K« neÚÔáT+~.
a eT]jáTT b, \T @yîHÕ  s +&ƒT |ŸPsÁd’ +Ÿ K«\T nsTTq
a + b Å£L&† |ŸPsÁ’ dŸ+U« (a, b Î Z nsTTq a + b Î Z)
\ |ŸPsÁ’
€ý˺<‘Ý+!
ç|Ÿe#áq+
–7 –8 = –15
–5 – (–3) =
0 – (–3) =
9 – (–6) =
0  (–3) =
7  (–6) =
@yû“ Âs+&ƒT |ŸPsÁ’ dŸ+K«\ jîTT¿£Ø \‹ÝeTT Å£L&†
ÿ¿£ |ŸPsÁ’ dŸ+K« neÚÔáT+~.
a eT]jáTT b, \T @yîH
Õ  s +&ƒT |ŸPsÁ’ dŸ+K«\T nsTTq
a ´ b Å£L&† |ŸPsÁ’ dŸ+U«. (a, b Î Z nsTTq a ´ b Î Z)
dŸ+K«\T dŸ+¿£\qeTT, >·TD¿±sÁeTT <Šcͼ« dŸ+eÔá <óŠsˆ“• bÍ{ìkÍïsTT.
yîTTÔáïeTT ýñ<‘ \‹ÝeTT |ŸPsÁ’ dŸ+K« ¿±“ dŸ+K« n>·Tq³T¢ ¿£údŸ+ Âs+&ƒT |ŸPsÁ’
dŸ+K«\ ÈÔá #î|ŸÎ>·\eÖ?
e«e¿£\q+
kÍs+Xø+
ç|Ÿe#áq+
uó<ñ eŠ TT ÿ¿£ |ŸPsÁ’ dŸ+K«
uó<ñ eŠ TT ÿ¿£ |ŸPsÁ’ dŸ+K«
@yû“ s +&ƒT |ŸPsÁ’ dŸ+K«\ jîTT¿£Ø uó<ñ eŠ TT Å£L&†
ÿ¿£ |ŸPsÁ’ dŸ+K« neÚÔáT+~.
a eT]jáTT b, \T @yîH
Õ  s +&ƒT |ŸPsÁ’ dŸ+K«\T nsTTq
a – b Å£L&† |ŸPsÁ’ dŸ+U«. (a, b Î Z nsTTq a – b Î Z)
\ |ŸPsÁ’ dŸ+K«\T e«e¿£\qeTT <Šcͼ« dŸ+eÔá
ii) $“eTjáT (d¾Ô
œ «á +Ôás)Á H«jáTeTT:
9 ÷ 10 = 0.9
–6 ÷ (–2) =3
0 ÷ (–3) =
7 ÷ (–6) =
uó²>·VŸäsÁ+
kÍs+Xø+
uó²>·|\Ÿ˜ + ÿ¿£ |ŸPsÁ’ dŸ+K« ¿±<ŠT
kÍ<ó‘sÁD+>±, s +&ƒT |ŸPsÁ’ dŸ+K«\ uó²>·|\
Ÿ˜ + ÿ¿£ |ŸPsÁ’
dŸ+K« ¿±e\d¾q nedŸs+Á ýñ<TŠ .
a eT]jáTT b (b ¹ 0) \T @yîH
Õ  s +&ƒT |ŸPsÁ’ dŸ+K«\T nsTTq
a ÷ b ÿ¿£ |ŸPsÁ’ dŸ+K« ¿±<ŠT. (a, b Î Z nsTTq a ÷ b ÏZ)
<óŠsˆ“• bÍ{ìkÍïsTT.¿±ú uó²>·VŸäsÁeTT <Šcͼ« dŸ+eÔá <óŠsˆ“• bÍ{ì+#áeÚ.
¿ì+~ |Ÿ{켿£\qT |Ÿ]o*+º, |ŸP]ï#ûjáTTeTT.
dŸ+¿£\q+
ç|Ÿe#áq+ 1
ç|Ÿe#áq+ 2
kÍs+Xø+
–6  3 = –3
3  (–6) = –3
–6 + 3 = –3
3 + (–6) = –3
3 + (–9) =
(–9) + 3 =
3  (–9) =
(–9)  3 =
–6 + (–5) =
(–5) + (–6) =
–6  (–5) =
(–5)  (–6) =
–3 + 2 =
–6 + 3=3 + (–6)
ç|Ÿe#áq+ 1
>·TD¿±sÁ+
ç|Ÿe#áq+ 2
kÍs+Xø+
–6  3 = 3  (–6)
–3  2 =
4 + (–5) =
4  (–5) =
a eT]jáTT b \T @yîHÕ  s +&ƒT |ŸPsÁ’ dŸ+K«\T nsTTq, a + b = b + a a eT]jáTT b \T @yîHÕ  s +&ƒT |ŸPsÁ’ dŸ+K«\T nsTTq, a ´ b = b ´ a
\ |ŸPsÁ’ dŸ+K«\T dŸ+¿£\qeTT eT]jáTT >·TD¿±sÁeTT <Šcͼ« $“eTjáT (d¾Ô
œ «á +Ôás)Á H«jáTeTT bÍ{ìkÍïsTT.
7e ÔásÁ>·Ü >·DìÔá+
22
|ŸPsÁ’ dŸ+K«\T
Subtraction
Statement 1 Statement 2 Conclusion
–6 – 3 = –9
3 – (–6) = 9
5 – (–9) =
(–9) –5 =
–6 – 3  3 + (–6)
For any two integers a and b, a – b ¹ b – a
Statement 1
Division
Statement 2 Conclusion
2  10 = 0.2
3  (–6) = –3
2  10  10  2
(–9)  3 =
3  (–9) = – 1/3
In general, for any two integers a,b (b ¹ 0), a ¸ b ¹ b ¸ a
\ Integers are not commutative under subtraction and division.
iii) Associative Law :
Observe the following and complete the tables.
Addition
Statement 1 Statement 2 Conclusion
Multiplication
Statement 1 Statement 2 Conclusion
(–6 + 3) + 2
–6 + (3 +2)= (–6 + 3) + 2=
(–6  3)  2
–6  (3  2)=
(–6  3)  2=
= –3 + 2
–6 + 5
= –18  2
–6  6
–6  (3  2)
= –1
= –1
= – 36
= –36
–6 + (3 + 2)
[3+ (–9)] + (–5)] 3 + [(–9) +(–5)]
[3  (–9)]  (–5)] 3  [(–9)  (–5)]
=
=
=
=
=
=
=
=
[–6 + (–5)] + 6 –6 + [(–5) + 6]
[–6  (–5)]  6
–6  [(–5)  6]
=
=
=
=
=
=
=
=
For any three integers a, b and c,
(a + b) +c = a + (b + c)
For any three integers a, b and c,
(a ´ b) ´ c = a ´ (b ´ c)
\ Integers are associative under additon and multiplication.
Subtraction
Statement 1 Statement 2 Conclusion
Division
Statement 1 Statement 2 Conclusion
(–6 – 3) –2
–6 – (3 – 2)
(–6 – 3) –2
(–18 ÷ 6) ÷ 3
–18 ÷ (6 ÷ 3)
= –9 –2
= –6 –1
–6 – (3 –2)
= –3 ÷ 3
= –18 ÷ 2
= –11
= –7
= –1
= –9
[3– (–9)] –(–5)
3 – [(–9) –(–5)]
[16÷ (–4)] ÷ (–2) 16 ÷ [(–4) ÷ (–2)]
=
=
=
=
=
=
=
=
(–6 ÷ 3) ÷ 2  –6 ÷ (3÷2)
In general, for any three integers a, b and c,
(a ÷ b) ÷ c ¹ a ÷ (b ÷ c)
In general, for any three integers a, b and c,
(a – b) – c ¹ a – (b – c)
\ Integers are not associative under subtraction and division.
VII Class Mathematics
23
Integers
e«e¿£\q+
ç|Ÿe#áq+ 2
kÍs+Xø+
ç|Ÿe#áq+ 1
–6 – 3 = –9
3 – (–6) = 9
5 – (–9) =
(–9) –5 =
–6 – 3  3 + (–6)
a eT]jáTT b \T @yîH
Õ  s +&ƒT |ŸPsÁ’ dŸ+K«\T
ç|Ÿe#áq+ 1
2  10 = 0.2
3  (–9) = – 1/3
iii)
kÍs+Xø+
3  (–6) = –3
(–9)  3 =
2  10  10  2
kÍ<ó‘sÁD+>±, a eT]jáTT b (b ¹ 0) \T
@yîHÕ  s +&ƒT |ŸPsÁ’ dŸ+K«\T nsTTq a ¸ b ¹ b ¸ a
nsTTq a – b ¹ b – a
\ |ŸPsÁ’
uó²>·VŸäsÁ+
ç|Ÿe#áq+ 2
dŸ+K«\T e«e¿£\qeTT eT]jáTT uó²>·VäŸ sÁeTT\ <Šcͼ« $“eTjáT (d¾Ôœ «á +Ôás)Á H«jáTeTTqT bÍ{ì+#áeÚ.
dŸVŸ²#ásÁ H«jáTeTT : ¿ì+~ |Ÿ{켿£\qT |Ÿ]o*+º, |ŸP]ï#ûjáTTeTT.
ç|Ÿe#áq+ 1
>·TD¿±sÁ+
ç|Ÿe#áq+ 2
kÍs+Xø+
(–6 + 3) + 2=
(–6  3)  2
–6  (3  2)=
(–6  3)  2=
–6 + (3 + 2)
= –18  2
–6  6
–6  (3  2)
= – 36
= –36
ç|Ÿe#áq+ 1
dŸ+¿£\q+
ç|Ÿe#áq+ 2
kÍs+Xø+
(–6 + 3) + 2
–6 + (3 +2)=
= –3 + 2
–6 + 5
= –1
= –1
[3+ (–9)] + (–5)] 3 + [(–9) +(–5)]
[3  (–9)]  (–5)] 3  [(–9)  (–5)]
=
=
=
=
=
=
=
=
[–6 + (–5)] + 6
–6 + [(–5) + 6]
[–6  (–5)]  6
–6  [(–5)  6]
=
=
=
=
=
=
=
=
a, b eT]jáTT c \T @yîH
Õ  s +&ƒT |ŸPsÁ’ dŸ+K«\T nsTTq
(a + b) + c = a + (b + c)
\ |ŸPsÁ’
a, b eT]jáTT c \T @yîH
Õ  eTÖ&ƒT |ŸPsÁ’ dŸ+K«\T nsTTq
(a ´ b) ´ c = a ´ (b ´ c)
dŸ+K«\T dŸ+¿£\qeTT eT]jáTT >·TD¿±sÁeTT\ <Šcͼ« dŸV²Ÿ #ásÁ H«jáTeTT bÍ{ìkÍïsTT.
ç|Ÿe#áq+ 1
e«e¿£\q+
ç|Ÿe#áq+ 2
kÍs+Xø+
ç|Ÿe#áq+ 1
(–6 – 3) –2
–6 – (3 – 2)
(–6 – 3) –2
(–18 ÷ 6) ÷ 3
–18 ÷ (6 ÷ 3)
= –9 –2
= –6 –1
–6 – (3 –2)
= –3 ÷ 3
= –18 ÷ 2
= –11
= –7
= –1
= –9
[3– (–9)] –(–5)
3 – [(–9) –(–5)]
[16÷ (–4)] ÷ (–2) 16 ÷ [(–4) ÷ (–2)]
=
=
=
=
=
=
=
=
uó²>·VŸäsÁ+
ç|Ÿe#áq+ 2
kÍs+Xø+
(–6 ÷ 3) ÷ 2  –6 ÷ (3÷2)
kÍ<ó‘sÁD+>±, a, b eT]jáTT c, \T @yîHÕ  eTÖ&ƒT
kÍ<ó‘sÁD+>±, a, b eT]jáTT c, \T @yîHÕ  eTÖ&ƒT
|ŸPsÁ’ dŸ+K«\T nsTTq (a – b) – c ¹ a – (b – c)
|ŸPsÁ’ dŸ+K«\T nsTTq (a ÷ b) ÷ c ¹ a ÷ (b ÷ c)
\ |ŸPsÁ’ dŸ+K«\T e«e¿£\qeTT eT]jáTT uó²>·VäŸ sÁeTT\ <Šcͼ« dŸV²Ÿ #ásÁ H«jáTeTTqT bÍ{ì+#áeÚ.
7e ÔásÁ>·Ü >·DìÔá+
24
|ŸPsÁ’ dŸ+K«\T
iv) Identity property:
Addition
3 + ___ = 3
Multiplication
3×1=3
0 + (–3) = ___
___ × (–3) = –3
–2 + ___ = –2
–2 × ___= –2
___ + 5 = 5
___ × 5 = 5
–6 + ___ = –6
–6 × ___ = –6
For any integer a, a + 0 = 0 + a = a
For any integer a, a × 1 = 1 × a = a
\ 1 is multiplicative identity .
\ Zero is the additive identity
v)
Additive inverse property :
What should be added to –3 to get additive identity 0?
Observe the following, 4 + (–4) = 0
(–5) + 5 = __
(–6) + __ = 0
In each pair given above, one integer is called the additive inverse of the other integer.
For any integer ‘a’, there exists an integer (–a) such that a + (–a) = (–a) + a = 0.
Both the integers a and –a are called ‘additive inverse’ of each other.
We get additive inverse of an integer a when we multiply (–1) to a.
Example 7 : Find the additive inverses of (+2) and (–3).
Solution :
Additive inverse of +2 = – (+2) = –2.
Additive inverse of –3 = – (–3) = + 3.
Write the additive inverses of 5, –8, 1 and 0
What should be multiplied by 6 to get multiplicative identity 1?
Is it exist in integers?
vi) Distributive law :
Let us take three integers –2, 1 and 3,
i) –2 × (1 + 3) = [(–2) × 1] + [(–2) × 3]
–2 × 4 = –2 + (–6)
–8 = –8
Let us take three integers –1, 3 and –5,
ii) –1 × [3 + (–5)] = [(–1) × 3] + [(–1) × (–5)]
–1 × (–2) = –3 + (+5)
2=2
You will find that in each case, the left-hand side is equal to the right-hand side.
For any integers a, b and c, a × (b + c) = a × b + a × c
Thus, multiplication distributes over addition of integers.
VII Class Mathematics
25
Integers
iv) ÔáÔàá eT <ós
Š ˆÁ eTT :
dŸ+¿£\q+
a @<û“
>·TD¿±sÁ+
3 + ___ = 3
3×1=3
0 + (–3) = ___
___ × (–3) = –3
–2 + ___ = –2
–2 × ___= –2
___ + 5 = 5
___ × 5 = 5
–6 + ___ = –6
dŸ+K« nsTTq a + 0 = 0 + a = a
–6 × ___ = –6
dŸ+K« nsTTq a × 1 = 1 × a = a
|ŸPsÁ’
a @<û“ |ŸPsÁ’
\ |ŸPsÁ’ dŸ+K«\Å£” dŸ+¿£\q ÔáÔáàeÖ+XøeTT »0µ
\ |ŸPsÁ’ dŸ+K«\Å£” >·TD¿±sÁ ÔáÔàá eÖ+XøeTT »1µ
v) dŸ+¿£\q $ýËeT H«jáTeTT:
-3 Å£” m+Ôá ¿£*|¾q dŸ+¿£\q ÔáÔáàeÖ+XøeTT »0µ e#áTÌqT?
4 + (–4) = 0
¿ì+~ y{ì“ |Ÿ]o*+#á+&,
(–5) + 5 = __
(–6) + __ = 0
™|Õ ÈÔá\ýË ç|ŸÜ dŸ+K«qT Âs+&ƒe dŸ+K«Å£” dŸ+¿£\q $ýËeTeTT n+{²sÁT.
ç|ŸÜ |ŸPsÁ’ dŸ+K« a ¿ì a + (–a) = (–a) + a = 0 n>·Tq³T¢ –a |ŸPsÁ’ dŸ+K« –+³T+~.
a eT]jáTT »–aµ \T |ŸsÁdŸÎsÁeTT dŸ+¿£\q $ýËeÖ\T neÚԐsTT.
a nHû |ŸPsÁ’ dŸ+K«qT »-1µ #û >·TDì+#áT³ <‘Çs <‘“ dŸ+¿£\q $ýËeTeTT bõ+<Še#áTÌ.
–<‘VŸ²sÁD 7 : (G2) eT]jáTT (-3) \ jîTT¿£Ø dŸ+¿£\q $ýËeÖ\qT sjáTTeTT.
kÍ<óŠq :
G2 jîTT¿£Ø dŸ+¿£\q $ýËeTeTT R - (G2) R -2
-3 jîTT¿£Ø dŸ+¿£\q $ýËeTeTT R -(-3) R G3
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
5, -8, 1, 0 \ jîTT¿£Ø dŸ+¿£\q $ýËeÖ\qT sjáTTeTT.
€ý˺<‘Ý+!
vi)
6 qT @ dŸ+K«Ôà >·TDì+ºq >·TD¿±sÁ ÔáÔáàeÖ+XøeTT »1µ edŸTï+~?
n~ |ŸPsÁ’ dŸ+K«jûTH?
$uó²>· H«jáTeTT:
eTÖ&ƒT |ŸPsÁ’ dŸ+K«\T -2, 1 eT]jáTT 3 rdŸTÅ£”+<‘+.
i)
ii)
–2 × (1 + 3) = [(–2) × 1] + [(–2) × 3]
–2 × 4 = –2 + (–6)
–8 = –8
eTsà eTÖ&ƒT |ŸPsÁ’ dŸ+K«\T -1, 3 eT]jáTT -5 rdŸTÅ£”+<‘+.
–1 × [3 + (–5)] = [(–1) × 3] + [(–1) × (–5)]
–1 × (–2) = –3 + (+5)
2=2
ç|ŸÜ dŸ+<ŠsÁÒÛeTTýËqÖ m&ƒeT #ûÜ yîÕ|ŸÚ –q• $\Te, Å£”& #ûÜ yîÕ|ŸÚ –q• $\TeÅ£” dŸeÖqeTT.
a, b eT]jáTT c \T
@yû“ eTÖ&ƒT |ŸPsÁ’ dŸ+K«\T nsTTq, a × (b + c) = a × b + a × c
¿±eÚq, |ŸPsÁ’ dŸ+K«\T dŸ+¿£\qeTT ™|Õ >·TD¿±sÁeTT $uó²>· H«jáTeTTqT bÍ{ìkÍïsTT.
7e ÔásÁ>·Ü >·DìÔá+
26
|ŸPsÁ’ dŸ+K«\T
Verify –3 × [ (–4) – 2] = [(–3) × (–4)] – [(–3) × 2]. Is multiplication
distribute over subtraction of integers? Write your conclusion.
Let us see multiplication of two or more negative numbers.
Observe the following :
(–1) × (–1) = + 1
(–1) × (–1) × (–1) = –1
(–1) × (–1) × (–1) × (–1) = + 1
(–1) × (–1) × (–1) × (–1) × (–1) = –1
This means if negative integer multiplied even number of times, the product is positive integer.
If negative integer multiplied odd number of times, the product is negative integer.
Example 8 : Multiply the following using associative law.
Solution :
i)
–25 × (–4) × 2 × (– 8)
i)
–25 × (–4) × 2 × (–8) = [ –25 × (–4)] × 2 × (–8)
ii)
(–20) × (–2) × (–5) × 7
= [100 × 2] × (–8)= 200 × (–8)= –1600
ii) (–20) × (–2) × (–5) × 7 = (–20) × [(–2) × (–5)] × 7
= [(–20) × 10] × 7 = –200 × 7 = –1400
Example 9 : Are (–42) × (–7) and (–7) × (–42) equal? Write the law.
Solution :
(–42) × (–7) = + 294
(–7) × (–42) = +294
(–42) × (–7) = (–7) × (–42)
It is multiplicative commutative law.
Example 10 :Simplify 26 × (–48) + (–48) × (–36) using suitable laws.
Solution :
26 × (–48) + (–48) × (–36) = (–48) × 26+ (–48) × (–36) (Commutative law)
= (–48) × [26+ (–36)] (Distributive law)
= (–48) × (–10)
= 480
Exercise - 1.3
1.
Identify the laws in the following statements :
i)
–3 + 5 = 5 + (–3)
ii) –2 × 1 = 1 × (–2) = –2
iii) [(–5) × 2)] × 3 = (–5) × [(2 × 3)]
iv) 18 × [7 + (–3)] = [18 × 7] + [18 × (–3)]
v)
vi) –3 + 0 = 0 + (–3) = –3
–5 × 6 = –30
VII Class Mathematics
27
Integers
dŸ]#áÖ&ƒTeTT.
|ŸPsÁ’dŸ+K«\T e«e¿£\qeTT ™|Õ >·TD¿±sÁeTT $uó²>· H«jáTeTTqT bÍ{ìkÍïjáÖ?
MTsÁT @$T |Ÿ]o*+#sÁT?
Âs+&ƒT ýñ<‘ n+Ôá¿£+fñ mÅ£”Øe ‹TTD dŸ+K«\ >·TD¿±sÁeTT >·T]+º Ôî\TdŸTÅ£”+<‘eTT.
¿ì+~ neT]¿£qT |Ÿ]o*+#áTeTT.
€ý˺<‘Ý+!
–3 × [ (–4) – 2] = [(–3) × (–4)] – [(–3) × 2]qT
(–1) × (–1) = + 1
(–1) × (–1) × (–1) = –1
(–1) × (–1) × (–1) × (–1) = + 1
(–1) × (–1) × (–1) × (–1) × (–1) = –1
™|Õ y{ì“ |Ÿ]o*+ºq, >·TDì+ºq ‹TTD |ŸPsÁ’ dŸ+K«\ dŸ+K« dŸ]dŸ+K« nsTTq y{ì\‹ÝeTT
<óq
Š |ŸPsÁ’ dŸ+K« n>·TqT. ný²¹> >·TDì+ºq ‹TTD |ŸPsÁ’ dŸ+K«\ dŸ+K« uñd¾ dŸ+K« nsTTq y{ì
\‹ÝeTT ‹TTD |ŸPsÁ’ dŸ+K« n>·TqT.
–<‘VŸ²sÁD 8 : ç¿ì+~ y{ì“ dŸVŸ²#ásÁ H«jáTeTT€<ó‘sÁ+>± ¿£qT>=qTeTT.
kÍ<óŠq:
i)
i)
–25 × (–4) × 2 × (– 8)
ii) (–20) × (–2) × (–5) × 7
–25 × (–4) × 2 × (–8) = [ –25 × (–4)] × 2 × (–8)
= [100 × 2] × (–8)= 200 × (–8)= –1600
ii) (–20) × (–2) × (–5) × 7 = (–20) × [(–2) × (–5)] × 7
= [(–20) × 10] × 7 = –200 × 7 = –1400
–<‘VŸ²sÁD 9 : (-42) × (-7) eT]jáTT (-7) × (-42) dŸeÖqeÖ? ‚~ @ H«jáTeTT?
kÍ<óŠq :
(-42) × (-7) = + 294
(-7) × (-42) = + 294
(-42) × (-7) = (-7) × (-42)
‚~ >·TD¿±sÁ $“eTjáT (d¾œÔá«+ÔásÁ) H«jáTeTT.
–<‘VŸ²sÁD 10 : 26 × (-48) + (-48) × (-36) qT ÔáÐq H«jáÖ\qT|ŸjîÖÐ+º dŸÖ¿¡Œˆ¿£]+#áTeTT.
kÍ<óŠq :
26 × (-48) + (-48) × (-36) = (-48) × 26 + (-48) × (-36) ($“eTjáT H«jáTeTT)
= (–48) × [26+ (–36)] ($uó²>· H«jáTeTT)
= (–48) × (–10)
= 480
nuó²«dŸ+ - 1.3
1. ¿ì+~ y{ìýË –q• <óŠsˆ\qT >·T]ï+º sjáT+&.
i)
–3 + 5 = 5 + (–3)
ii) –2 × 1 = 1 × (–2) = –2
iii) [(–5) × 2)] × 3 = (–5) × [(2 × 3)]
iv) 18 × [7 + (–3)] = [18 × 7] + [18 × (–3)]
v)
vi) –3 + 0 = 0 + (–3) = –3
–5 × 6 = –30
7e ÔásÁ>·Ü >·DìÔá+
28
|ŸPsÁ’ dŸ+K«\T
2.
3.
4.
5.
6.
7.
What will be the sign of the product of the following :
i) 24 times of negative integer.
ii) 35 times of negative integer.
Write the numbers in the blanks by using appropriate law.
i) –3 + ____ = –3
ii) 2 × (–3) = (–3) × ___
iii) –6 + [3 + (–2)] = [(–6) +____ ] +____ iv) –6 × ___ = –6
v) 5 × [(–6) + 9] = ____ × (–6) + 5 × ____
State true or false. Give the reasons.
i) 2 is the multiplicative identity of –2.
ii) Integers are commutative under subtraction.
iii) For any two integers a and b, a × b = b × a
iv) The division of integer by zero is not defined.
v) 6 + (–6) = (–6) + 6 = 0 indicates additive identity property.
Simplify the following using suitable laws.
i) –11 × (–25) × (–4)
ii) 3 × (–18) + 3 × (–32)
Are the integers are associative under subtraction? Explain by an example.
Verify [(–5) × 2)] × 3 = (–5) × [(2 × 3)]
1.4 BODMAS Rule :
Salma madam gave following problem to her students.
Find the value of 2 + 3 × 4
Two students Sarayu and Sanvi solved the problem in the following way.
Sarayu
Sanvi
2 + 3 × 4= 2 + 12 = 14
2 + 3 × 4 = 5 × 4 = 20
Who is correct?
To simplify arithmetic expressions, which involve various operations like brackets,
multiplication, subtraction, addition, etc., a particular sequence of operations has to be followed.
In the above problem Sanvi is correct because in arithmetic
operations, multiplication should be done first before addition
is taken up.
The hierarchy of arithmetic operations to be followed is
given by a rule called BODMAS rule.
To simplify an arithmetic expressions, we should perform
the operation in order as shown in figure. After brackets O in
the BODMAS rule that stands for ‘of ’ which means
multiplication. After O, the next operation is D it stands for
division. This is followed by M which stands for
multiplication. After multiplication, A that stands for addition
will be performed. Then, S meant for subtraction is performed.
VII Class Mathematics
29
Integers
2. ¿ì+~ dŸ+<ŠsÒÛ\ýË \‹ÝeTT jîTT¿£Ø dŸ+Èã (>·TsÁTï)qT sjáT+&.
i) 24 ‹TTD |ŸPsÁ’ dŸ+K«\ \‹ÝeTT
ii) 35 ‹TTD |ŸPsÁ’ dŸ+K«\ \‹ÝeTT
3. ¿ì+~ U²°\ q+<ŠT dŸÂsÕq |ŸPsÁ’ dŸ+K«\qT ÔáÐq H«jáÖ\ €<ó‘sÁ+>± |ŸP]+#áTeTT.
i) –3 + ____ = –3
iii) –6 + [3 + (–2)] = [(–6) +____ ] +____
v) 5 × [(–6) + 9] = ____ × (–6) + 5 × ____
ii)
iv)
2 × (–3) = (–3) × ___
–6 × ___ = –6
4. ¿ì+~ y¿±«\T dŸÔá«eÖ? ndŸÔá«eÖ? ¿±sÁD²\T Ôî\T|ŸÚeTT.
i) 2 jîTT¿£Ø >·TD¿±sÁ ÔáÔáàeÖ+XøeTT -2.
ii) |ŸPsÁ’ dŸ+K«\T e«e¿£\qeTT <Šcͼ« $“eTjáT H«jáTeTT bÍ{ìkÍïsTT.
iii) a eT]jáTT b \T @yîÕH Âs+&ƒT |ŸPsÁ’ dŸ+K«\T nsTTq a × b = b × a.
iv) dŸTH•Ôà |ŸPsÁ’dŸ+K«\ uó²>·VŸäsÁeTT “sÁǺ+#á‹&ƒ<ŠT.
v) 6 + (–6) = (–6) + 6 = 0 nqTq~ dŸ+¿£\q ÔáÔáàeT <óŠsÁˆeTTqT dŸÖº+#áTqT.
5. ¿ì+~ y{ì“ ÔáÐq H«jáÖ\qT|ŸjîÖÐ+º dŸÖ¿¡ë¿£]+#áTeTT.
i)
–11 × (–25) × (–4)
ii)
3 × (–18) + 3 × (–32)
6. |ŸPsÁ’ dŸ+K«\T e«e¿£\qeTT <Šcͼ« dŸVŸ²#ásÁ H«jáTeTTqT bÍ{ìkÍïjáÖ?–<‘VŸ²sÁD <‘Çs $e]+#áTeTT.
7. [(–5) × 2)] × 3 = (–5) × [(2 × 3)] dŸ]#áÖ&ƒTeTT.
1.4 BODMAS “jáTeTeTT :
Ôáq $<‘«sÁTœ\Å£” dŸý²ˆ yûT&ƒ+ ¿ì+~ dŸeTdŸ«qT ‚#ÌsÁT.
2 + 3 × 4 jîTT¿£Ø $\Te ¿£qT>=qTeTT?
™|Õ dŸeTdŸ«qT dŸsÁjáTT eT]jáTT Xæ“Ç nqT ‚<ŠÝsÁT $<‘«sÁTœ\T ¿ì+~ $<óŠ+>± kÍ~ó+#sÁT.
dŸsÁjáTT
Xæ“Ç
2 + 3 × 4= 2 + 12 = 14
2 + 3 × 4 = 5 × 4 = 20
™|Õ ‚<ŠÝ]ýË mesÁT dŸÂsÕq dŸeÖ<ó‘q$T#ÌsÁT?
$$<óŠ sÁ¿±\ n+¿£>·DìÔá |Ÿ]ç¿ìjáT\T nsTTq dŸ+¿£\qeTT,
e«e¿£\qeTT, >·TD¿±sÁeTT, uó²>·VäŸ sÁeTT eT]jáTT çu²Â¿³¢Ôà ţL&q dŸ+U²«
dŸeÖkÍ\qT dŸÖ¿¡ë¿£]+#á&†“¿ì ÿ¿£ ç¿£eT|Ÿ<ŠÆÜýË |Ÿ]ç¿ìjáT\qT
|ŸP]ï#j
û Öá *à –+³T+~.™|Õ –<‘VŸ²sÁDýË Xæ“Ç dŸs qÕ dŸeÖ<ó‘q$TºÌ+~.
m+<ŠT¿£q>± n+¿£>D
· Ôì á |Ÿ]ç¿ìjTá \ýË dŸ+¿£\qeTT ¿£H• eTT+<ŠT >·TD¿±sÁ
|Ÿ]ç¿ìjáT #ûjáÖ*.
n+¿£>D
· Ôì á |Ÿ]ç¿ìjTá \qT |ŸP]ï #ûjÖá *àq ç¿£e֓• Ôî*| “jáTeTyûT
BODMAS “jáTeTeTT.
dŸ+U²« dŸeÖkÍ\qT dŸÖ¿¡ˆŒ ¿£]+#á&†“¿ì|³
Ÿ eTTýË #áÖ|¾q $<óeŠ TT>±
ç¿£eT |Ÿ<ŠÝÜýË |Ÿ]ç¿ìjáT\qT |ŸP]ï#ûjáÖ*à –+³T+~. BODMASýË
yîTT<Š³ çu²Â¿{Ùà (B), ÔásÁTyÔá –q• »Oµ nqTq~ ‘of’ (>·TD¿±sÁeTT)qT
Ôî*jáTCñjáTTqT. ÔásÁTyÔá D uó²>·VŸäsÁeTTqT, M >·TD¿±sÁeTT, A
dŸ+¿£\qeTTqT, S e«e¿£\qeTTqT Ôî*jáTCñjáTTqT.‡ ç¿£eTeTTýË
|Ÿ]ç¿ìjáT\T |ŸP]ï #ûjáÖ*.
7e ÔásÁ>·Ü >·DìÔá+
30
dŸ+U²« dŸeÖkÍ\qT
dŸÖ¿¡Œ ˆ¿£sÁD
çu²Â¿{Ùà
|Ÿ]ç¿ìjáT\T
B
$dŸTØ\+
çu²Â¿{Ùà
O
€|˜t
D
uó²>±VŸäsÁ+
M
>·TD¿±sÁ+
A
dŸ+¿£\q+
S
e«e¿£\q+
( ) kÍ<ó‘sÁD çu²Â¿{Ùà
{}
¿£¯¢ çu²Â¿{Ùà
[ ] #áÔTá sÁçdŸ çu²Â¿{Ùà
|ŸPsÁ’ dŸ+K«\T
There are four types of brackets:
1.
Vinculum: This is represented by a bar on the top of the numbers.
For example: 2 + 3 – 4 + 3. Here, the figures under the vinculum have to be calculated as
4 + 3 first and the ‘minus’ sign before 4 is applicable to 7. Thus, the given expression is
equal to 2 + 3 – 7 which is equal to –2.
2.
Simple Bracket: This is represented by ( )
3.
Curly Bracket: This is represented by { }
4.
Square Bracket: This is represented by [ ]
The brackets in an expression have to be opened in the order of vinculum, simple brackets,
curly brackets and square brackets, i.e., [{( )}] to be opened from inside to outwards.
Within every bracket also we should follow order of BODMAS rule.
Example 11 : Simplify 3 × 2 + 8 ÷ 4
Solution :
3×2+8÷4
(Division)
=3×2+2
(Multiplication)
=6+2
(Addition)
=8
(i) 5 × 6 – 6
Simplify the following.
(ii) 24 ÷ 3 × 3 – 30 (iii) 5 × 5 – 5 ÷ 5 + 5
Example 12 :Simplify 7 × 6 – 8 – 4
Solution :
7×6–8–4
(Vinculum)
=7×6–4
(Multiplication)
= 42 – 4
(Subtraction)
= 38
Example 13 :Simplify 18 + 64 ÷ 4 {26 – (14 – 7 – 3)}
Solution :
18 + 64 ÷ 4{26 – (14 – 4)}
(Vinculum)
= 18 + 64 ÷ 4{26 – (14 – 4)}
(Simple bracket)
= 18 + 64 ÷ 4{26 – 10}
(Curly bracket)
= 18 + 64 ÷ 4 {16}
(Of )
= 18 + 64 ÷ 64
(Division)
= 18 + 1
(Addition)
= 19
VII Class Mathematics
31
Integers
çu²Â¿³T¢ H\T>·T sÁ¿±\T>± –+{²sTT.
1. $qTØ\+ çu²Â¿{Ù : B““ dŸ+K«\™|Õ ^Ôá –+#á&ƒ+ <‘Çs dŸÖºkÍïeTT.
–<‘VŸ²sÁD: 2 + 3 – 4 + 3. ‚¿£Ø&ƒ $qTØ\+ çu²Â¿{Ù ¿ì+<Š –q• 4 + 3 “ yîTT<Š³ ýÉ¿Øì +#*. nq>± 4
eTT+<ŠT eÚq• ‹TTD >·TsÁTï 7 Å£” e]ï+#áTqT. ¿±eÚq ‚ºÌq dŸ+U²« dŸeÖdŸ|ŸÚ $\Te 2 + 3 - 7Å£”
dŸeÖqeTT. ‚~ -2 Å£” dŸeÖqeTT neÚÔáT+~.
2. kÍ<ó‘sÁD çu²Â¿{Ù: B““ ( ) Ôà dŸÖºkÍïeTT.
3. ¿£¯¢ çu²Â¿{Ù: B““ { } Ôà dŸÖºkÍïeTT.
4. #áÔáTsÁçdŸ çu²Â¿{Ù: B““ [ ] Ôà dŸÖºkÍïeTT.
dŸ+U²« dŸeÖkÍ\ dŸÖ¿¡ë¿£sÁDýË çu²Â¿{ÙàqT $qTØ\+ çu²Â¿{Ù, kÍ<ó‘sÁD çu²Â¿{Ù, ¿£¯¢ çu²Â¿{Ù, #áÔáTsÁçdŸ
çu²Â¿{Ù ç¿£eTeTTýË |ŸP]ï #ûjáÖ*. nq>± [{( )}] \qT ýË|Ÿ* qT+& ‹jáT³Å£” ç¿£eTeTTýË |ŸP]ï#ûjáÖ*.
ç|ŸÜ çu²Â¿{Ù q+<ŠT Å£L&† BODMAS “jáTeTeTT bÍ{ì+#*.
–<‘VŸ²sÁD 11 : 3 × 2 + 8 ÷ 4 dŸÖ¿¡ë¿£]+#áTeTT.
kÍ<óŠq :
3×2+8÷4
(uó²>·VŸäsÁeTT)
=3×2+2
(>·TD¿±sÁeTT)
=6+2
(dŸ+¿£\qeTT)
=8
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
(i) 5 × 6 – 6
dŸÖ¿¡Œ ˆ¿£]+#áTeTT:
(ii) 24 ÷ 3 × 3 – 30 (iii) 5 × 5 – 5 ÷ 5 + 5
–<‘VŸ²sÁD 12 : 7 × 6 – 8 – 4 dŸÖ¿¡Œ ˆ¿£]+#áTeTT
kÍ<óŠq :
7×6–8–4
($qTØ\+)
=7×6–4
= 42 – 4
(>·TD¿±sÁeTT)
(e«e¿£\qeTT)
= 38
–<‘VŸ²sÁD 13 : 18 + 64 ÷ 4 {26 – (14 – 7 – 3)} dŸÖ¿¡Œ ˆ¿£]+#áTeTT
kÍ<óŠq :
18 + 64 ÷ 4 {26 – (14 – 4)}
($qTØ\+)
(kÍ<ó‘sÁD çu²Â¿{Ù)
(¿£¯¢ çu²Â¿{Ù)
(€|˜t)
(uó²>·VŸäsÁeTT)
(dŸ+¿£\qeTT)
= 18 + 64 ÷ 4 {26 – (14 – 4)}
= 18 + 64 ÷ 4 {26 – 10}
= 18 + 64 ÷ 4 {16}
= 18 + 64 ÷ 64
= 18 + 1
= 19
7e ÔásÁ>·Ü >·DìÔá+
32
|ŸPsÁ’ dŸ+K«\T
Exercise - 1.4
1.
Simplify the following:
i)
6 × 9 – 6 ÷3
iii) 80 – 56 ÷ 8 × 9
v)
2.
iv)
15 ÷ 5 + 17 – 30
ii)
{12 – 14 – 8 + 7} – 15
iv)
{6 of 145 ÷ (3 + 2)} ÷2 – 4 of 20
8+8–8×8÷8
8 × 3 – 13 – 7
iii) 16 – (4 + 18 ÷ 6 – 7 – 5) × 5
v)
25 + [14 – 18 + {12 of 5 – (– 4 + 14)}]
Identify the true statements.
i)
48 ÷ 6 – 4 = 24
iii) –11 + 3 × 7 = –56
4.
12 ÷ 4 – 8 + 5
Simplify the following:
i)
3.
ii)
ii)
–18 + 12 ÷ 3 = –14
iv)
2020 ÷ 20 – 100 = 1
Fill the blanks with +, –, ×, ÷ to make the statement true.
i)
9 __ 3 __ 6 = –3
ii)
–6 __ 12 __ 6 = –4
iii)
–15__3 __6 = –30
Absolute value :
Look at the picture, at what level does the
ship sails on the sea?
How can we represent the height of the
temple using integer? How can we represent
the depth of the fish from the sea level using
integer?
Though they represent with different
integers, their distance from the sea level is same.
In this case we need absolute value.
Numerical value of an integer without
considering it’s sign is the absolute value of
that integer.
The absolute value of a number is always
non negative. Absolute value of a is denoted by
| a |. Read it as modulus of a.
30m
20m
15m
10m
0m
–10m
–15m
–20m
Examples: i) | –50 | = 50 ii) | 60 | = 60
iii) | –701 | = 701
iv) | 0 | = 0
v) If x > 5, then |x – 5| = x – 5 (why?)
vi) If x < 5, then |x – 5| = 5 – x (why?)
So, we can generalise,
If x > 0, then | x | = x
If x < 0, then | x | = –x (why?)
If x = 0, then | x | = 0
VII Class Mathematics
33
Integers
nuó²«dŸ+ - 1.4
1. ¿ì+~ y{ì“ dŸÖ¿¡Œ ˆ¿£]+#áTeTT.
i)
6 × 9 – 6 ÷3
iii) 80 – 56 ÷ 8 × 9
v)
ii)
12 ÷ 4 – 8 + 5
iv)
15 ÷ 5 + 17 – 30
ii)
{12 – 14 – 8 + 7} – 15
iv)
{6 of 145 ÷ (3 + 2)} ÷2 – 4 of 20
8+8–8×8÷8
2. ¿ì+~ y{ì“ dŸÖ¿¡Œ ˆ¿£]+#áTeTT.
i)
8 × 3 – 13 – 7
iii) 16 – (4 + 18 ÷ 6 – 7 – 5) × 5
v)
25 + [14 – 18 + {12 of 5 – (– 4 + 14)}]
3. ¿ì+~ y{ìýË dŸÔá« y¿±«\qT >·T]ï+#áTeTT.
i)
48 ÷ 6 – 4 = 24
iii) –11 + 3 × 7 = –56
ii)
–18 + 12 ÷ 3 = –14
iv)
2020 ÷ 20 – 100 = 1
4. ¿ì+~ y¿±«\T dŸÔá«eTjûT« $<óŠeTT>± U²°\qT +, –, ×, ÷ \Ôà |ŸP]+#áTeTT.
i)
9 __ 3 __ 6 = –3
ii)
–6 __ 12 __ 6 = –4
iii)
–15__3 __6 = –30
|ŸseÁ T eTÖ\«eTT:
|Ÿ³eTTqT >·eT“+#áTeTT. dŸeTTç<Š –|Ÿ]Ôá\eTTqT+& m+Ôá
mÔáTýï Ë z&ƒ ç|ŸjÖá “dŸTqï •~?
|Ÿ³eTTý˓ dŸeTTç<Š –|Ÿ]Ôá\eTT qT+& <ûy\jáTeTT
mÔáTqï T |ŸPsÁ’ dŸ+K«Ôà mý² dŸÖºkÍïeTT?
dŸeTTç<Š –|Ÿ]Ôá\eTT qT+& #û|Ÿ ‡<ŠTÔáTq• ýËÔáTqT |ŸPsÁ’
dŸ+K«Ôà mý² dŸÖºkÍïeTT?
™|Õ s +&ƒT yûsTÁ yûsTÁ |ŸPsÁ’ dŸ+K«\Ôà dŸÖº+ºq|ŸÎ{ì¿¡
dŸeTTç<Š –|Ÿ]Ôá\eTT qT+& n$ dŸeÖq <ŠÖsÁeTTýË –q•$.
‚ý²+{ì dŸ+<ŠsÒ\ýË |ŸseÁ TeTÖ\«eTT nedŸseÁ TòÔáT+~.
30m
ÿ¿£ |ŸPsÁ’ dŸ+K«jîTT¿£Ø dŸ+U²« $\Te (<‘“ >·TsÁTïqT
|Ÿ]>·DqýË¿ì rdŸT¿ÃÅ£”+&†)qT <‘“ jîTT¿£Ø |ŸseÁ T eTÖ\«eTT
n+{²sÁT.
ÿ¿£ dŸ+K« jîTT¿£Ø |ŸseÁ T eTÖ\«eTT m\¢|ڟ Î&ƒÖ sÁTD²Ôሿ£+
¿±<ŠT.a jîTT¿£Ø |ŸseÁ T eTÖ\«eTTqT |a| Ôà dŸÖºkÍïeTT.B““
a jîTT¿£Ø yîÖ&ƒT«\dt n“ #á<TŠ eÚԐ+.
–<‘VŸ²sÁD
20m
15m
10m
0m
–10m
–15m
–20m
i) | –50 | = 50 ii) | 60 | = 60
iii) | –701 | = 701
iv) | 0 | = 0
v) x > 5 nsTTq, |x – 5| = x – 5 (m+<ŠTÅ£”?) vi) x < 5 nsTTq, |x – 5| = 5 – x (m+<ŠTÅ£”?)
¿±eÚq B““ eTq+ ç¿ì+~ $<óŠ+>± kÍ<ó‘sÁD¡¿£]+#áe#áTÌ.
x > 0 nsTTq | x | = x
x < 0 nsTTq | x | = –x (m+<ŠTÅ£”?)
x = 0 nsTTq | x | = 0
7e ÔásÁ>·Ü >·DìÔá+
34
|ŸPsÁ’ dŸ+K«\T
If | x | = 15 then what will be the value of x? Discuss
1.
2.
Calculate the following.
i)
8 × (–1)
ii) (–2) × 175
iii)
(–3) × (–40)
v)
(–7) ÷ (–1)
vi) (–12) ÷ (+6)
vii) (–49) ÷ (–7)
(–7) × _____ = 21
iv) _____ × (–11) = 132
v)
vi) (42) ÷___= –6
(–25) ÷______=1
viii) _____÷ (–9) = 16
Write all the possible pairs of integers that give a product of –50.
4.
Sankar, a fruit vendor sells 100kg of oranges and 75 kg of
pomegranates. If he makes a profit of ` 11 per one kg of
pomegranates and loss of `8 per one kg oranges, what will be his
overall profit or loss?
5.
Bhargavi lost 5700 calories in the month of June using yoga. If the
calory loss is uniform, calculate the loss of calories per day?
6.
Simplify 625 × (–35) + 625 × 30 using suitable law.
7.
Simplify the following using BODMAS.
i)
12 – 36 ÷ 3
ii)
6 × (–7) + (–3) ÷ 3
iii)
38 – {35 – (36 – 34 – 37)}
Write the absolute values of following numbers.
i) –700
ii) 150
iii) –150
v) If p < 10, then |p –10|



(+63) ÷ (–9)
iii) _____ × (– 9) = –72
3.

viii)
ii) 7 × _____ = –42
vii) _____÷ (–15) = 6

(–24) × (–7)
Replace the blank with an integer to make it a true statement.
i)
8.
iv)
iv) –35
vi) If y > 7, then |7 – y|
The collection of natural numbers (1,2,3,4,5 ...) zero(0) and negative
numbers (-1, -2, -3, -4, -5 ...) are integers.
The product of two positive integers (two negative integers) is positive
integer.
The product of one positive integer and one negative integer is negative integer.
The quotient of two positive integers (or two negative integers) is positive.
The quotient of one positive integer and one negative integer is negative.
VII Class Mathematics
35
Integers
nHûÇw¾<‘Ý+
| x | =15
nsTTq x jîTT¿£Ø $\Te @eTeÚÔáT+~? #á]Ì+#á+&.
jáT֓{Ù nuó²«dŸ+
1. ¿ì+~ y{ì“ ýÉ¿ìØ+#áTeTT.
i)
8 × (–1)
ii) (–2) × 175
iii)
(–3) × (–40)
v)
(–7) ÷ (–1)
vi) (–12) ÷ (+6)
vii) (–49) ÷ (–7)
iv)
(–24) × (–7)
viii)
(+63) ÷ (–9)
2. ¿ì+~ y¿±«\T dŸÔá«eTjûT« $<óŠeTT>± U²°\qT |ŸPsÁ’ dŸ+K«\Ôà |ŸP]+#áTeTT.
i)
(–7) × _____ = 21
ii) 7 × _____ = –42
iii) _____ × (– 9) = –72
iv) _____ × (–11) = 132
v)
vi) (42) ÷___= –6
(–25) ÷______=1
vii) _____÷ (–15) = 6
viii) _____÷ (–9) = 16
3. \‹ÝeTT -50 njûT« $<óŠeTT>± MýÉÕq“• |ŸPsÁ’ dŸ+K«\ ÈÔá\T sjáT+&.
4. |Ÿ+&ƒ¢ y«bÍ] Xø+¿£sY, 100 ¿ì.ç>±. ¿£eTý² |Ÿ+&ƒT¢, 75 ¿ì.ç>±. <‘“eTˆ |Ÿ+&ƒT¢
neֈ&ƒT. nÔáqT ÿ¿£ ¿ì.ç>±. <‘“eTˆ |Ÿ+&ƒ¢ ™|Õ `11 ý²u󲓕, ÿ¿£ ¿ì.ç>±.
¿£eTý² |Ÿ+&ƒ¢ ™|Õ `8 qcͼ“• bõ+~q yîTTÔáïeTT MT<Š nÔᓿì m+Ôá ý²uó„eTT
ýñ<‘ qwŸ¼eTT?
5. uó²sÁZ$ pHŽ Hî\ýË jîÖ>· <‘Çs 5700 ¹¿\¯\qT ÔáÐZ+#sÁT. ¹¿\¯\ Ôá>·TZ<Š\
d¾œsÁ+>± eÚq•, € Hî\ýË sÃEy] dŸsdŸ] ¹¿\¯\ Ôá>·TZ<Š\ m+Ôá?
6.
7.
625 × (–35) + 625 × 30 qT
ÔáÐq H«jáÖ\qT|ŸjîÖÐ+º dŸÖ¿¡ë¿£]+#áTeTT.
BODMAS qT –|ŸjîÖÐ+º dŸÖ¿¡Œ ˆ¿£]+#áTeTT.
i)
12 – 36 ÷ 3
ii)
6 × (–7) + (–3) ÷ 3
iii)
38 – {35 – (36 – 34 – 37)}
8. ¿ì+~ dŸ+K«\Å£” |ŸsÁeT eTÖ\« $\TeqT ¿£qT>=qTeTT.
i)
–700
v)
p < 10 nsTTq, |p –10|
ii) 150
iii) –150
iv) –35
vi) y > 7 nsTTq, |7 – y|
>·TsÁTï+#áT¿Ãy*àq
n+Xæ\T





dŸV²Ÿ È dŸ+K«\T (1,2,3,4,5 ...), dŸTq• (0) eT]jáTT sÁTD dŸ+K«\qT (-1, -2, -3,
-4, -5 ...) ¿£*|¾ |ŸPsÁ’dŸ+K«\T n+{²sÁT.
Âs+&ƒT <óŠq |ŸPsÁ’ dŸ+K«\ ýñ<‘ Âs+&ƒT ‹TTD |ŸPsÁ’ dŸ+K«\ \‹ÝeTT ÿ¿£ <óŠq |ŸPsÁ’ dŸ+K«.
ÿ¿£ <óŠq |ŸPsÁ’ dŸ+K«qT ‹TTD |ŸPsÁ’ dŸ+K«#û >±ú ýñ<‘ ‹TTD |ŸPsÁ’ dŸ+K«qT <óŠq |ŸPsÁ’
dŸ+K«Ôà >·TDì+ºq \‹ÝeTT ‹TTD |ŸPsÁ’ dŸ+K« neÚÔáT+~.
ÿ¿£ <óŠq |ŸPsÁ’ dŸ+K«qT eTs=¿£ <óŠq |ŸPsÁ’ dŸ+K«#û ýñ<‘ ÿ¿£ ‹TTD |ŸPsÁ’ dŸ+K«qT
eTs=¿£ ‹TTD |ŸPsÁ’ dŸ+K«Ôà uó²Ð+ºq uó²>·|˜Ÿ\+ ÿ¿£ <óŠq dŸ+K«.
ÿ¿£ <óŠq |ŸPsÁ’ dŸ+K«qT ‹TTD |ŸPsÁ’ dŸ+K«#û >±ú ýñ<‘ ‹TTD |ŸPsÁ’ dŸ+K«qT <óŠq |ŸPsÁ’
dŸ+K«Ôà uó²Ð+ºq uó²>·|˜Ÿ\+ ‹TTD |ŸPsÁ’ dŸ+K« neÚÔáT+~.
7e ÔásÁ>·Ü >·DìÔá+
36
|ŸPsÁ’ dŸ+K«\T
Property
Name
Closure
Commutative
Associative
Identity
Distributive



Addition
Properties of Integers
Operations
Multiplication
Subtraction
a + b Z
a – b Z
a  b Z
Division
If a, b, c are nonzero integers
a  b  Z
a+b=b+a
a–b  b–a
a  b=b  a
a b ba
(a + b) + c
= a + (b + c)
a+0=a
0+a=a
a × (b + c)
= ab + ac
(a – b) – c
 a – (b – c)
If a, b, c  Z
Not applicable
a × (b – c)
= ab – ac
(a  b)  c
= a  (b  c)
a×1=a
1×a=a
Not applicable
(a  b)  c
 a  (b  c)
Not applicable
Not applicable
To simplify arithmetic expressions, we must follow the BODMAS rule (Brackets, Of,
Division, Multiplication, Addition, Subtraction).
The brackets in an expression have to be opened in the order of vinculum, simple brackets,
curly brackets and square brackets, i.e., [{( )}] to be opened from inside to outwards.
Numerical value of an integer without considering its sign is the absolute value of that
integer. Absolute value never be negative.
Number Series - 1
A number series can be considered as a collection of numbers in which all the terms are
formed according to some particular rule or all the terms follow a particular pattern. The relation
of any term to its proceeding term will be same throughout the series. we have to find out the
rule in which the terms of the series areselected and depending on that rule we have to find out
the missing number.
Some of the series can be in the following types.
1.
Prime Series : In which the terms are the prime numbers in order.
Ex(i): 2, 3, 5, 7, 11, 13, ___.
Here the terms of the series are the prime numbers in order. The Prime number after 13 is 17.
So, the answer is 17.
Ex (ii): 2, 5, 11,17,23, ___.
This series is by taking the alternative Prime numbers, after 23 the prime numbers are 29
and 31. So, the answer is 31.
VII Class Mathematics
37
Integers
<ósŠ ˆÁ eTT |sÁT
dŸ+¿£\q+
|ŸPsÁ’ dŸ+K«\ <ósŠ ˆ\T
|Ÿ]ç¿ìjáT\T
>·TD¿±sÁ+
e«e¿£\q+
‚¿£Ø&ƒ a, b, c  Z
uó²>·VŸäsÁ+
a, b, c \T XøSHû«ÔásÁ
|ŸPsÁ’ dŸ+K«\T
a – b Z
a  b Z
a  b  Z
d¾œÔá«+ÔásÁ H«jáTeTT a + b = b + a
a–b=b–a
a  b=b  a
a b  ba
(a + b) + c
= a + (b + c)
a+0=a
0+a=a
a × (b + c)
= ab + ac
(a – b) – c
= a – (b – c)
dŸ+eÔá <óŠsÁˆeTT
dŸVŸ²#ásÁ H«jáTeTT
ÔáÔáàeT <óŠsÁˆeTT
$uó²>· H«jáTeTT



a + b Z
e]ï+#á<ŠT
a × (b – c)
= ab – ac
(a  b)  c
= a  (b  c)
a×1=a
1×a=a
(a  b)  c
 a  (b  c)
e]ï+#á<ŠT
e]ï+#á<ŠT
e]ï+#á<ŠT
n+¿£>·DìÔá |Ÿ]ç¿ìjáT\qT |ŸP]ï #ûjáÖ*àq ç¿£e֓• Ôî*| “jáTeTyûT BODMAS “jáTeTeTT.(çu²Â¿{Ùà,
€|˜t, uó²>·VŸäsÁeTT, >·TD¿±sÁeTT, dŸ+¿£\qeTT, e«e¿£\qeTT).
dŸ+U²« dŸeÖkÍ\ dŸÖ¿¡Œ ˆ¿£sD
Á ýË çu²Â¿{ÙàqT $qTØ\+ çu²Â¿{Ù, kÍ<ó‘sÁD çu²Â¿{Ù, ¿£¯¢ çu²Â¿{Ù, #áÔTá sÁçdŸ
çu²Â¿{Ù, ç¿£eTeTTýË |ŸP]ï#ûjáÖ*. nq>± [{( )}] \qT ýË|Ÿ*qT+& ‹jáT³Å£” ç¿£eTeTTýË |ŸP]ï#ûjáÖ*.
ÿ¿£ |ŸPsÁ’ dŸ+K« jîTT¿£Ø dŸ+U²« $\Te (<‘“ >·TsÁTïqT |Ÿ]>·Dq ýË¿ì rdŸT¿ÃÅ£”+&†)qT <‘“ jîTT¿£Ø |ŸsÁeT
eTÖ\«eTT n+{²sÁT. ÿ¿£ dŸ+K« jîTT¿£Ø |ŸsÁeT eTÖ\«eTT m\¢|ŸÚÎ&ƒÖ sÁTD²Ôሿ£+ ¿±<ŠT.
dŸ+U²« çXâDT\T ` 1
ÿ¿£ ç|ŸÔû«¿£ “jáTeTeTT ýñ<‘ ç|ŸÔû«¿£ neT]¿£ €<ó‘sÁ+>± @sÁÎ&q dŸ+K«\ dŸeÖVŸäsÁyûT dŸ+U²« çXâDT\T >±
|Ÿ]>·Dì+#áe#áTÌ. ÿ¿£ |Ÿ<ŠeTTqÅ£” <‘“ eTT+<ŠT |Ÿ<ŠeTTqÅ£” eT<óŠ« dŸ+‹+<óŠ+ @ $<óŠ+>± –+<à € çXâDì ý˓
n“• |Ÿ<‘\ eT<óŠ« n<û $<óŠ+>± –+&ƒTqT. eTq+ çXâDýì ˓ |Ÿ<‘\T @ “jáTeT+ ç|Ÿ¿±sÁ+ @sÁÎ&†¦jîÖ
¿£qT>=“, € “jáTeT+ €<ó‘sÁ+>± eTqÅ£” Ôî*jáT“ dŸ+K«qT ¿£qT>=H*.
¿=“• sÁ¿±\ çXâDT\T ¿ì+<Š ‚eNj&†¦sTT.
1. ç|Ÿ<‘ó q dŸ+K«\T çXâD:ì ‚+<ŠTýË ç|Ÿ<ó‘q dŸ+K«\T ç¿£eT+ýË –+&ƒTqT.
–<‘ (i) : 2, 3, 5, 7, 11, 13 ___.
‡ çXâDìýË esÁTdŸ ç|Ÿ<ó‘q dŸ+K«\T –H•sTT.13 ÔásÇÔá e#áTÌ ç|Ÿ<ó‘q dŸ+K« 17.
¿±eÚq, dŸeÖ<ó‘qeTT 17.
–<‘ (ii): 2, 5, 11, 17, 23 ___.
‡ çXâDýì Ë ç|ŸÔ«e֕jáT(ÿ¿£{ì e~* eTs=¿£{)ì ç|Ÿ<‘ó q dŸ+K«\T –H•sTT.23 ÔásÇÔá e#áTÌ ç|Ÿ<‘ó q dŸ+K«\T
29, 31. ¿±eÚq, dŸeÖ<ó‘qeTT 31.
7e ÔásÁ>·Ü >·DìÔá+
38
|ŸPsÁ’ dŸ+K«\T
2.
Addition Series : Each number in the series obtained by adding a specific number or series
of numbers to previous number.
Ex i) 7, 10, 13, 16, 19, 22, _____
So, the answer is 22 + 3 = 25.
Ex ii) 10, 14, 19, 25, 32, ____
So, the answer is 32 + 8 = 40
Ex iii) 5, 7, 10, 15, 22, ___
Each number in the series obtained by adding consecutive primes. So, the answer is 22 + 11 = 33.
3.
Fibonacci Series: Every third number can be the sum of the preceding two numbers.
Ex i) 3, 5, 8, 13, 21,__
Here starting from third number,
3 + 5 = 8,
5 + 8 = 13,
8 + 13 = 21, So, the answer is 13 + 21 = 34
Ex ii) 6, 10, 16, 26, 42, ___
Here starting from third number, 6 + 10 = 16, 10 + 16 = 26, 16 + 26 = 42, So, the answer is
26 + 42 = 68
Practice problems.
1)
12, 19, 26, 33, 40, 47, __
a)
57
b)
54
c) 52
d) 50
2)
2, 13, 24, 35, 46, 57, __
a)
65
b)
67
c)
68
d) 72
3)
61, 67, 71, 73, 79, ___
a)
89
b)
87
c) 85
d) 83
4)
3, 7, 13, 21, 31, ___
a)
43
b)
48
c) 51
d) 53
5)
8, 12, 20, 32, 52, 84, ___
a)
111
b)
126
c) 136
d) 174
6)
23, 28, 38, 53, 73, 98, ___
a)
121
b)
128
c) 135
d) 146
7)
101, 97, 89, 83, 79, 73, 71, ___ a)
61
b)
65
c) 66
d) 67
8)
4, 7, 11, 18, 29, 47, ___
a)
67
b)
76
c) 84
d) 92
9)
76, 187, 298, 409, 520, ___
a)
631
b)
656
c) 701
d) 724
10) 0, 2, 5, 10, 17, 28, 41, ____
a)
50
b)
53
c) 57
d) 58
11) 36, 45, 53, 60, 66, 71, ___
a)
84
b)
78
c) 75
d) 73
12) 0, 15, 45, 90, 150, 225, ___
a)
295
b)
300
c) 315
d) 360
13) 18, 23, 25, 30, 32, 37, ___
a)
43
b)
41
c) 39
d) 38
14) 4, 7, 11, 18, 29, 47,__
a)
71
b)
76
c) 77
d) 82
15) 12, 18, 21, 27, 30, 36, 39,___
a)
43
b)
45
c) 49
d) 52
VII Class Mathematics
39
Integers
2. dŸ+¿£\q çXâDì : ‡ çXâDýì Ë ç|ŸÜ dŸ+K« <‘“ eTT+<ŠTq• dŸ+K«Å£” ÿ¿£ ç|ŸÔ«û ¿£ dŸ+K« ýñ<‘ çXâD“ì ¿£\T|Ÿ>± @sÁÎ&ƒTqT.
–<‘ (i): 7,10,13,16,19,22,,....
¿±eÚq, dŸeÖ<ó‘qeTT =22+3 = 25.
–<‘ (ii): 10, 14, 19, 25, 32 ,,....
¿±eÚq, dŸeÖ<ó‘qeTT = 32+8 = 40
–<‘ (iii): 5,7,10,15,22 ,....
ç|ŸÜ dŸ+K« esÁTdŸ ç|Ÿ<ó‘q dŸ+K«\qT ¿£\T|Ÿ>± @sÁÎ&ƒT#áTq•~.¿±eÚq, dŸeÖ<ó‘qeTT =22 + 11 = 33.
3. |˜u¾ ËH¿ì çXâD:ì ‚+<ŠTýË eTÖ&ƒedŸ+K« qT+& ç|Ÿr dŸ+K« <‘“ eTT+<ŠTq• s +&ƒT dŸ+K«\ yîTTÔáeï TT>± –+&ƒTqT.
–<‘ (i): 3,5, 8, 13, 21 ,....
‚¿£Ø&ƒ eTÖ&ƒe dŸ+K« qT+&,
3+5 = 8,
–<‘ (ii):
5 + 8 = 13,
8 + 13 = 21,
6,10,16,26,42 ,....
‚¿£Ø&ƒ eTÖ&ƒe dŸ+K« qT+&,
6+10 = 16,
ç|ŸXø•\qT çbÍ¿¡¼dt #ûjáT+& :
10 + 16 = 26,
¿±eÚq, dŸeÖ<ó‘qeTT =13 +21 =34
16 + 26 = 42, ¿±eÚq,
dŸeÖ<ó‘qeTT = 26 +42 = 68
Practice problems.
1)
12, 19, 26, 33, 40, 47, __
a)
57
b)
54
c) 52
d) 50
2)
2, 13, 24, 35, 46, 57, __
a)
65
b)
67
c)
68
d) 72
3)
61, 67, 71, 73, 79, ___
a)
89
b)
87
c) 85
d) 83
4)
3, 7, 13, 21, 31, ___
a)
43
b)
48
c) 51
d) 53
5)
8, 12, 20, 32, 52, 84, ___
a)
111
b)
126
c) 136
d) 174
6)
23, 28, 38, 53, 73, 98, ___
a)
121
b)
128
c) 135
d) 146
7)
101, 97, 89, 83, 79, 73, 71, ___ a)
61
b)
65
c) 66
d) 67
8)
4, 7, 11, 18, 29, 47, ___
a)
67
b)
76
c) 84
d) 92
9)
76, 187, 298, 409, 520, ___
a)
631
b)
656
c) 701
d) 724
10) 0, 2, 5, 10, 17, 28, 41, ____
a)
50
b)
53
c) 57
d) 58
11) 36, 45, 53, 60, 66, 71, ___
a)
84
b)
78
c) 75
d) 73
12) 0, 15, 45, 90, 150, 225, ___
a)
295
b)
300
c) 315
d) 360
13) 18, 23, 25, 30, 32, 37, ___
a)
43
b)
41
c) 39
d) 38
14) 4, 7, 11, 18, 29, 47,__
a)
71
b)
76
c) 77
d) 82
15) 12, 18, 21, 27, 30, 36, 39,___
a)
43
b)
45
c) 49
d) 52
7e ÔásÁ>·Ü >·DìÔá+
40
|ŸPsÁ’ dŸ+K«\T
2
Learning Outcomes
Content Items
The learner is able to
l
solve the multiplication and division of fractions in word
problems.
l
use algorithms to multiply and divide decimals.
l
understand the rational numbers.
l
represent the rational numbers on number line.
l
differentiate fractions and rational numbers.
l
2.0 Introduction
2.1 Multiplication and division of
fractions in word problems
2.2 Multiplication of decimals
2.3 Division of decimals
2.4 Rational numbers
2.5 Word problems on rational
numbers
solve problems related to daily life situations
involving rational numbers.
2.0 Introduction :
We know that a part of the whole is a fraction. We have learnt different types of fractions, four
fundamental operations on fractions, decimal numbers, addition and subtraction of decimal numbers
in previous classes. There are several situations in our daily life, where we use fractions, decimals
and rational numbers. Some of the situations are shown in the following pictures :
Piece of pizza
Digital weighing Machine
Petrol Bunk Readings
Let us recall what you have
learnt previous classes. Here is a
picture containing fractions.
Identify and put a tick mark
on like fractions in the picture.
VII Class Mathematics
41
Fractions, Decimals and Rational Numbers
n<ó‘«jáT+
_óH•\T, <ŠXæ+Xæ\T
eT]jáTT n¿£sÁD¡jáT dŸ+K«\T
nuó«„ dŸq |˜*Ÿ Ԑ\T
$wŸjTá n+Xæ\T
nuó²«dŸÅ£”&ƒT :
l
l
l
l
l
l
_óH•\ >·TD¿±sÁ, uó²>·VŸäsÁ |Ÿ<Š dŸeTdŸ«\qT kÍ~ó+#á>·\&ƒT.
<ŠXæ+Xæ\ >·TD¿±sÁ, uó²>·VŸäsÁ+ dŸeTdŸ«\ kÍ<óŠqýË
€ý²Z]<óyŠ TŽ qT –|ŸjÖî ÐkÍï&Tƒ .
n¿£sD
Á j
¡ Tá dŸ+K«\qT ne>±VŸ²q #ûdTŸ Å£”+{²&ƒT.
n¿£sÁD¡jáT dŸ+K«\qT dŸ+U²« ¹sK™|Õ >·T]ï+#á>·\&ƒT.
_óH•\T eT]jáTT n¿£sÁD¡jáTdŸ+K«\ uóñ<Š+ >·eT“+#áTqT
n¿£sD
Á j
¡ Tá dŸ+K«\Ôà ţL&q “Ôá« J$Ôá dŸeTdŸ«\qT
|Ÿ]wŸØ]+#á>\· &ƒT.
2.0 |Ÿ]#ájTá +
2.1 _óH•\ >·TD¿±sÁ, uó²>·VŸäsÁ |Ÿ<Š
dŸeTdŸ«\T
2.2 <ŠXæ+Xø _óH•\ >·TD¿±sÁ+
2.3 <ŠXæ+Xø _óH•\ uó²>·VäŸ sÁ+
2.4 n¿£sÁD¡jáTdŸ+K«\T
2.5 n¿£sD
Á j
¡ Tá dŸ+K«\™|Õ |Ÿ<Š dŸeTdŸ«\T
2.0 |Ÿ]#ájáT+ :
_óq•+ nq>± yîTTÔá+ï ýË ¿=+Ôá uó²>·+ n“ eTqÅ£” Ôî\TdŸT. eTq+ ‚+ÔáŔ£ eTT+<ŠT
ÔásÁ>·ÔáTýË¢ $_óq• sÁ¿±\ _óH•\T, _óH•\ýË #áÔáT]Ç<óŠ |Ÿ]ç¿ìjáT\T, <ŠXæ+Xø
_óH•\T, <ŠXæ+Xø _óH•\ jîTT¿£Ø Å£L&¿£ eT]jáTT rd¾yÔû \á >·T]+º HûsTÁ ÌÅ£”H•+. eTq<îqÕ +~q J$Ôá+ýË _óH•\T,
<ŠXæ+Xæ\T –|ŸjÖî Ð+#û dŸ+<ŠsÒÛ\T nHû¿+£ –H•sTT. ¿=“• dŸ+<ŠsÒÛ\T ç¿ì+~ ºçԐ\ýË #áÖ|¾+#á‹&†¦sTT.
|¾C²¨ýË uó²>·+
‹sÁTeÚqT ÔáÖ#û jáT+çÔá+
™|ç{ËýÙ ‹+Å£”ýË ¯&+>´
eTT+<ŠT Ôás>Á Ü· ýË HûsTÁ ÌÅ£”q•
n+Xæ\qT Èã|¿ï¾ ì Ôî#Tá ÌÅ£”+<‘+.$$<óŠ
sÁ¿±\ _óH•\T ¿£*Ð –q• ÿ¿£ ºçÔá+
‚¿£Ø&ƒ ‚eNj&+~.dŸC²Ü _óH•\qT
>·T]ï+º {ì¿ù #ûjáT+&.
7e ÔásÁ>·Ü >·DìÔáeTT
42
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
1 2 6 13
, , , have the same denominator. So, they are like fractions.
5 5 5 5
7 5 2 7
, , , have different denominators. So, they are unlike fractions.
4 3 5 3
Look at the following examples .
Example 1 : Raju used 3
1
1
litres of milk in the morning and 2 litres of milk in the evening to
2
2
prepare tea in his hotel. What was the total quanitity of milk he used?
Solution :
1 7
Milk used in the morning = 3  l
2 2
2
Milk used in the evening =
1 5
 l
2 2
Total quantity of milk he used in the hotel

Example 2 : Subtract
Solution :
7 5 7  5 12
 

 6 litres
2 2
2
2
3
7
from .
5
2
7 3
 (Are they like fractions? Why?)
2 5
7 3
, have different denominators. So, they are unlike fractions.
2 5
we have to change them into like fractions.
LCM of 2 and 5 is 10.
3 3 2 6


5 5  2 10
7 7  5 35


2 2  5 10
7 3 35 6 35  6 29
 
 

2 5 10 10
10
10

7 3 29
 
2 5 10
Look at the following example.
Example 3 : Solve the following :
VII Class Mathematics
i)
3 2
1
4 5
43
ii)
2 1
2 1
3 5
Fractions, Decimals and Rational Numbers
1 2 6 13
, , ,
5 5 5 5
7 5 2 7
, , ,
4 3 5 3
\T ÿ¹¿VŸäsÁ+ ¿£*ÐeÚH•sTT. ¿±‹{ì,¼ n$ dŸC²Ü _óH•\T.
\T $_óq• VŸäs\qT ¿£*ÐeÚH•sTT. ¿±‹{ì,¼ n$ $C²Ü _óH•\T.
ç¿ì+~ –<‘VŸ²sÁD\qT >·eT“+#á+&.
1
–<‘VŸ²sÁD 1: sE Ôáq VŸ²Ã³ýË¢ {¡ ÔájÖá sÁT #ûjTá &†“¿ì –<ŠjTá + 3 1 ©³sÁ¢ bÍ\T eT]jáTT kÍjáT+çÔá+ 2 2
2
©³sÁ¢ bÍ\T –|ŸjÖî Ð+#&ƒT. nÔá&Tƒ –|ŸjÖî Ð+ºq yîTTÔá+ï bÍ\ |Ÿ]eÖD+ m+Ôá ?
kÍ<óŠq :
–<ŠjTá + –|ŸjÖî Ð+ºq bÍ\T = 3 1  7 ©
2
2
kÍjáT+çÔá+ |ŸP³ –|ŸjÖî Ð+ºq bÍ\T = 2 12  52 ©
VŸ²Ã³ýË¢ –|ŸjÖî Ð+ºq yîTTÔá+ï bÍ\T
–<‘VŸ²sÁD 2:
kÍ<óŠq :
3
7
qT+& qT rd¾yûjáT+&.
5
2
7 3
 (n$ dŸC²Ü _óH•ý²?
2 5
7 3
,
2 5

7 5 7  5 12
 


2 2
2
2 6 ©³sÁT¢
m+<ŠTÅ£”?)
\T $_óq• VŸäs\qT ¿£*ÐeÚH•sTT. ¿±eÚq, n$ $C²Ü _óH•\T.
eTq+ y{ì“ dŸC²Ü _óH•\T>± eÖsÌ*à –+³T+~.
2 eT]jáTT 5 \ ¿£.kÍ.>·T.10
7 7  5 35


2 2  5 10
3 3 2 6


5 5  2 10
7 3 35 6 35  6 29
 
 

2 5 10 10
10
10

7 3 29
 
2 5 10
ç¿ì+~ –<‘VŸ²sÁDqT >·eT“+#á+&.
–<‘VŸ²sÁD 3: ‡ ç¿ì+~ y{ì“ kÍ~ó+#á+&.
i)
7e ÔásÁ>·Ü >·DìÔáeTT
44
3 2
1
4 5
ii)
2 1
2 1
3 5
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
Solution :
i)
3 2 3 7 3  7 21
1   

4 5 4 5 4  5 20
ii)
2 1 8 6
2 1  
3 5 3 5
=
8 5
6 5
 (reciprocal of is )
3 6
5 6
8  5 40 20
2


2
3  6 18 9
9
For multiplication of fractions, we simply multiply numerators and multiply denominators.
To divide one fraction by another fraction, we have to multiply first fraction with the reciprocal
of the second fraction.
=
1.
Look at the fractions and fill them in the table
Improper
fraction
__
Proper
fraction
1
2
__
2.
5
3
1
2
3
3 5 1 17 9
, , , ,
2 2 2 2 2
ii)
6 11 19 7 5
, , , ,
5 10 5 10 10
iii)
8 7 1 5 11
, ,3 , ,
3 6 4 3 4
Solve the following .
i)
4.
Convert improper fraction
into Mixed fraction
__
Write the following fractions in ascending order .
i)
3.
1 5 11 23 19 99
, , ,
,
,
.
2 3 9 25 100 70
3 7

4 4
ii)
5 7

6 12
iii)
7 1
1 
8 5
iv)
1
1
4 3
2
3
5
2
of
8
3
iii)
15
1
2
4
7
iv)
1
2
3 2
3
5
Simplify the following .
i)
1
of 3
4
ii)
VII Class Mathematics
45
Fractions, Decimals and Rational Numbers
kÍ<óŠq :
i)
3 2 3 7 3  7 21
1   

4 5 4 5 4  5 20
ii)
2 1 8 6
2 1  
3 5 3 5
=
8 5 6
 (
3 6 5
=
8  5 40 20
2


2
3  6 18 9
9
jîTT¿£Ø >·TD¿±sÁ $ýËeT+
5
)
6
_óH•\qT >·TDì+#\+fñ, eTq+ y{ì \y\qT >·TDì+#* eT]jáTT y{ì VŸäs\qT >·TDì+#*. ÿ¿£ _óH•“•
eTsà _óq•+Ôà uó²Ð+#\+fñ yîTT<Š{ì _óH•“•, s +&Ã_óq•+ jîTT¿£Ø >·TD¿±sÁ $ýËeT+Ôà >·TDì+#*à –+³T+~.
|ŸÚq]ÇeTsÁô nuó²«dŸ+
1.
ç¿ì+~ _óH•\qT >·eT“+#á+& eT]jáTT y{ì“ |Ÿ{¿¼ì ý£ ˓+|Ÿ+&.
dŸC²Ü _óH•\T
__
__
2.
5
3
1
2
3
3 5 1 17 9
, , , ,
2 2 2 2 2
6 11 19 7 5
, , , ,
5 10 5 10 10
ii)
iii)
8 7 1 5 11
, ,3 , ,
3 6 4 3 4
ç¿ì+~ y{ì“ kÍ~ó+#á+&.
i)
4.
__
ç¿ì+~ _óH•\qT nsÃVŸ²D ç¿£eT+ýË sjáT+&.
i)
3.
$C²Ü _óH•\T $TçXøeT
_óH•\T>± eÖsÁÌ&ƒ+
$C²Ü _óH•\T
1
2
1 5 11 23 19 99
, , ,
,
,
.
2 3 9 25 100 70
3 7

4 4
ii)
5 7

6 12
iii)
7 1
1 
8 5
iv)
1
1
4 3
2
3
ç¿ì+~ y{ì“ dŸÖ¿¡ˆŒ ¿£]+#á+&.
i)
3 ýË
1
4
e e+ÔáT
7e ÔásÁ>·Ü >·DìÔáeTT
ii)
5
8
ýË
2
3
e e+ÔáT
46
iii)
15
1
2
4
7
iv)
1
2
3 2
3
5
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
5.
Solve the following :
i)
3
3
4
ii)
82
1
7
12 2

7 7
iii)
iv)
1
9
5 2
2
11
2.1 Word problems of multiplication and division on fractions:
We come across in many daily life situations where we need to use fractions. Let us consider
some situations.
One day Chandu asked his mother for chapaties for dinner. She prepared 15 chapaties and
3
1
of chapaties to Chandu and his sister, of chapaties to Chandu’s father. Can you
5
5
guess how many chapaties remained to Chandu’s mother?
served
Let us find.
Total number of chapaties = 15
3
3
No. of chapaties served to Chandu and his sister  of 15   15  9
5
5
1
1
of 15 =  15 = 3
5
5
 No. of chapaties remained to Chandu’s mother = 15 – (9 + 3) = 3
No. of chapaties served to Chandu’s father =
Example 4 : In the school out of 180 students,
Solution :
4
of the students are boys. Find the number of girls
9
in the school?
Number of students in the school = 180
Part of the boys in the school
=
4
9
4
4
 180 = 80
of 180 =
9
9
 Number of girls = 180 – 80 = 100
Number of boys =
VII Class Mathematics
47
Fractions, Decimals and Rational Numbers
5. ç¿ì+~ y{ì“ kÍ~ó+#á+&.
i)
3
3
4
ii)
82
1
7
12 2

7 7
iii)
1
9
5 2
2
11
iv)
2.1 _óH•\ >·TD¿±sÁ, uó²>·VŸäsÁ |Ÿ<Š dŸeTdŸ«\T:
eTq “Ôá« J$Ôá+ýË _óH•\qT –|ŸjîÖÐ+#áe\d¾q ¿=“• dŸ+<ŠsÒÛ\qT rdŸTÅ£”+<‘+.
3
#á+<ŠT ÿ¿£sÃE, sçÜ uóËÈH“¿ì Ôáq Ôá*“¢ #ábÍr\T n&>±&ƒT. €yîT 15 #ábÍr\T ÔájÖá sÁT #ûdq¾ y{ìýË 5
e e+ÔáT #ábÍr\T #á+<ŠT eT]jáTT nÔᓠkþ<Š]¿ì,
Ôá*¿¢ ì m“• #ábÍr\T $TÐ*qyà }V¾²+#á>\· s?
eTq+ ‚|ŸÚÎ&ƒT Ôî\TdŸTÅ£”+<‘+.
yîTTÔá+ï #ábÍr\T R 15
1
5
e e+ÔáT #ábÍr\T #á+<ŠT Ôá+ç&¿ì e&+¦ º+~. #á+<ŠT
3
ýË
5
15 e e+ÔáT
#á+<ŠT Ôá+ç&¿ì e&+¦ ºq #ábÍr\T R 5 ýË 15 e e+ÔáT =
1
 15 = 3
5
#á+<ŠT eT]jáTT nÔᓠkþ<Š]¿ì e&+¦ ºq #ábÍr\T

1
 #á+<ŠT
Ôá*¿¢ ì $TÐ*q #ábÍr\T R 15 - (9 G 3) R 3
–<‘VŸ²sÁD 4: ÿ¿£ bÍsÄXÁ æ\ýË 180 eT+~ $<‘«sÁTýœ Ë¢
kÍ<óŠq :
u²*¿£\ dŸ+K«qT ¿£qT>=q+&.
bÍsÄXÁ æ\ýË $<‘«sÁT\œ dŸ+K« R 180
u²\TsÁ uó²>·+ R
4
9
e e+ÔáT $<‘«sÁT\œ T u²\TsÁT nsTTq € bÍsÄXÁ æ\ý˓
4
9
yîTTÔá+ï $<‘«sÁT\Æ ýË u²\TsÁ dŸ+K« R 180 ýË

u²*¿£\ dŸ+K« R 180 - 80 R 100
7e ÔásÁ>·Ü >·DìÔáeTT
3
 15  9
5
48
4
e
9
e+ÔáT =
4
 180 = 80
9
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
1
Example 5 : If the cost of 22 kg of apples (a box) in a whole sale market is `1170, then find the
2
cost of 5kg of apples.
Solution :
Cost of 22
1
kg of apples = `1170
2
= `1170  22
Cost of 1kg of apple
= `1170 
1
2
45
2
2
= `52
45
 Cost of 5kg of apples = 5  `52 = `260
= `1170 
Exercise - 2.1
1.
In Jagananna Gorumudda (MDM) scheme each student got
3
kg rice per day, find the weight of
20
the rice required for 60 students in a class per day.
2.
What is a perimeter of a equilateral triangle, if each side of a triangle is 5
3.
Surya can walk
4
If the length and breadth of a rectangular garden are
and
5
6
3
cm?
10
1
18
km in an hour. How much distance will be walk in 2 hours?
2
5
27
m
2
15
m respectively, then find the area of the garden.
2
1
Gopal bought 3 kg of potatoes in the market. If he paid `84, then find the cost of 1kg of
2
potatoes.
A car travelled 225 km in 4
1
hours with uniform speed.
2
Find the distance travelled in 1 hour.
7.
If 24 students share 4
4
kg of cake, then how much
5
cake does each one get?
VII Class Mathematics
49
Fractions, Decimals and Rational Numbers
–<‘VŸ²sÁD 5 : ÿ¿£ {ËÅ£” <ósŠ \Á <ŠT¿±D+ýË, ™|fɼ ý˓
22
€|¾ýÙ |Ÿ+&ƒ¢ yî\ ¿£qT¿ÃØ+&.
kÍ<óŠq :
22
1
¿ì.ç>±.
2
¿ì.ç>± €|¾ýÙ |Ÿ+&ƒ¢ yî\ sÁÖ.1170 nsTTq 5 ¿ì.ç>±.
€|¾ýÙ |Ÿ+&ƒ¢ yî\ R `1170.
1¿ì.ç>±. €|¾ýÙ |Ÿ+&ƒ¢ yî\ R `1170 ª
R `1170
€|¾ýÙ |Ÿ+&ƒ¢ yî\
1
2
22
45
2

2
= `52
45
= 5  `52 = `260
R ` 1170
 `5¿ì.ç>±.
1
2

nuó²«dŸ+ - 2.1
3
1. È>·qq• >ÃsÁT eTT<ŠÝ (MDM) |Ÿ<¿Š¸ +£ ýË ÿ¿ÃØ¿£Ø $<‘«]œ sÃEÅ£” 20 ¿ìýË\ _jáT«+ bõ+~q, Ôás>Á Ü· ýË >·\
yîTTÔá+ï 60 eT+~ $<‘«sÁT\œ Å£” ÿ¿£ sÃEÅ£” ¿±e*d¾q _jáT«+ ‹sÁTeÚ ¿£qT>=q+&.
2. ÿ¿£ dŸeTu²VŸQ çÜuó„TÈ+ jîTT¿£Ø ç|ŸÜ uó„TÈ+
3. dŸÖsÁ« ÿ¿£ >·+³ýË
18
5
5
3
10
¿ìýË MT³sÁT¢ q&ƒe>·\&ƒT.
™d+.MT. nsTTÔû çÜuóT„ È+ jîTT¿£Ø #áT³T¼¿=\Ôá m+Ôá?
2
1
2
>·+³ýË¢ m+Ôá <ŠÖsÁ+ q&ƒe>·\&ƒT?
4. ÿ¿£ BsÁ#é Ôá Tá sÁçkÍ¿±sÁ Ôó bõ&ƒeÚ eT]jáTT yî&\ƒ TÎ\T esÁTdŸ>±
eT]jáTT
15
2
27
MT.
2
MT nsTTq n|ŸÚÎ&ƒT € Ôó jîTT¿£Ø yîXÕ æ\«+ ¿£qT>=q+&.
5. eÖÂsØ{Ë¢ >ÃbÍýÙ
3
1
2
¿ìýËç>±eTT\ ‹+>±Þø<TŠ +|Ÿ\T ¿=qT>Ã\T #ûXæ&ƒT. y{ì¿ì nÔá&Tƒ `84 #î*+¢ ºq#Ã,
1¿ìýËç>±eTT\ ‹+>±Þø<TŠ +|Ÿ\ yî\ ¿£qT>=q+&.
1
6. ÿ¿£ ¿±sÁ T dŸ e Tyû > · + Ôà 4 2 >· + ³ýË¢ 225 ¿ì ý ËMT³sÁ T ¢
ç|ŸjáÖDì+º+~.n~ ÿ¿£ >·+³ýË ç|ŸjáÖDì+ºq <ŠÖs“•
¿£qT>=q+&.
4
7. 24 eT+~ $<‘«sÁT\œ T 4 5 ¿ìýËç>±eTT\ ¿¹ ¿ù qT dŸeÖq+>± |Ÿ+#áTÅ£”+fñ,
n|ŸÚÎ&ƒT ç|ŸÜ ÿ¿£ØsÁÖ m+Ôá uó²>·+ ¿¹ ¿ù qT bõ+<ŠTԐsÁT?
7e ÔásÁ>·Ü >·DìÔáeTT
50
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
8.
A drum contains 210 litres of water. How many
times does the boy get the water for watering
the plants with 3
1
litres of full bucket from the
2
drum?
2.2 Multiplication Decimals :
We have learnt about decimals, addition and subtraction of decimals in the previous class. Let us
review on these concepts and then learn multiplication and division of decimals in this class.
The fraction with denominators 10, 100, 1000, 10000,
etc. are decimal fractions or decimal
1
numbers. The fraction
can be written as 0.1 in decimal form (read as zero point one). So, decimal
10
number is another way of expression of a fraction.
The place value chart can also be used to understand decimals.
Look at the table below and fill up the blank places.
Decimal
Hundreds
Tens
Ones
Tenths
Hundredths
Thousandths
Numbers
(100)
(10)
(1)
1
 
 10 
 1 


 100 
 1 


 1000 
31.402
0
3
1
4
0
2
702.136
7
0
1
3
6
495.834
4
520.301
5
2
0
0
7
2
5
8
4
1
1
8
8
We can also write these decimal numbers in expanded form. Look at the following example from the
table
31. 402  30  1 
4
0
2
.


10 100 1000
Place value of 4 in 31.402 is
Similarly, we can write expand form to the remaining decimals.
4
10
Let us see how to add or subtract decimals. We have learnt that, the decimal numbers are written one
under the other with the decimal points lined up. Add or subtract as you would do to whole numbers,
then, place the decimal place in the result in line with the other decimal points.
i)
356.24 + 65.08
+
ii)
67.29 – 35.91
356.24
67.29
65.08
421.32
– 35.91
31.38
VII Class Mathematics
51
Fractions, Decimals and Rational Numbers
8. ÿ¿£ ç&ƒyTŽ ýË 210 ©³sÁ¢ úsÁT ¿£\<ŠT. yîTT¿£Ø\Å£”
1
úsÁT bþjáTT³Å£” u²\T&ƒT 3 2 ©³sÁ¢ kÍeTsÁ«Æ + >·\
“+&ƒT ‹Â¿Ø³T¼Ôà € ç&ƒyTŽ qT+º m“• kÍsÁT¢ ú{ì“
bõ+<Š>·\&ƒT?
2.2 <ŠXæ+Xø _óH•\ >·TD¿±sÁ+
eTq+ ‚+ÔáŔ£ eTT+<ŠT Ôás>Á Ü· ýË <ŠXæ+Xø _óH•\T, <ŠXæ+Xæ\ Å£L&¿£ eT]jáTT rd¾yÔû \á >·T]+º Ôî\TdŸTÅ£”H•+.
‡uó²eq\qT eTq+ dŸMT¿ì+Œ ºq ÔásTÁ yÔá <ŠXæ+Xæ\ >·TD¿±sÁ+ eT]jáTT uó²>·VäŸ sÁ+ >·T]+º ‡ Ôás>Á Ü· ýË
Ôî\TdŸTÅ£”+<‘+.
10, 100, 1000, 10000,... yîTT<ŠýqÕÉ $ VŸäsÁ+>± >·\ _óH•\qT <ŠXæ+Xø _óH•\T ýñ<‘ <ŠXæ+Xø dŸ+K«\T
1
n+{²sÁT. 10 nHû _óH•“• 0.1>± <ŠXæ+Xø sÁÖ|Ÿ+ýË sjáTe#áTÌ (dŸTH• bÍsTT+{Ù ÿ¿£{>ì ± #á<eŠ ÚԐ+).
¿£qT¿£ <ŠXæ+Xø+qT, _óq•+ jîTT¿£Ø eTsà sÁÖ|Ÿ+>± e«¿£ï |ŸsTÁ kÍïeTT. <ŠXæ+Xø dŸ+K«\qT nsÁ+œ #ûdTŸ ¿Ãe&ƒ+ ¿=sÁŔ£
k͜q $\Te |Ÿ{¿¼ì q£ T –|ŸjÖî Ð+#áe#áTÌ.
~>·Te |Ÿ{¿¼ì q£ T >·eT“+º U²°\qT |ŸP]+#á+&.
<ŠXæ+Xø
e+<Š\T
|Ÿ<TŠ \T
ÿ¿£³T¢
<ŠXæ+Xæ\T
XøԐ+Xæ\T
dŸV²Ÿ çkÍ+Xæ\T
dŸ+K«\T
(100)
(10)
(1)
1
 
 10 
 1 


 100 
 1 


 1000 
31.402
702.136
495.834
520.301
0
3
1
4
0
2
7
0
1
3
6
4
5
8
4
5
2
0
1
0
7
2
1
8
8
‡ <ŠXæ+Xø dŸ+K«\qT eTq+ $dŸsï DÁ sÁÖ|Ÿ+ýË Å£L&† sjáTe#áTÌ. |Ÿ{¿¼ì £ qT+º ~>·Te –<‘VŸ²sÁDqT >·eT“+#á+&.
31. 402  30  1 
4
0
2


.
10 100 1000
4
31.402 ýË 4 jîTT¿£Ø k͜q $\Te 10
n<û $<óŠ+>±, eTq+ $TÐ*q <ŠXæ+Xø dŸ+K«\qT $dŸsï D
Á sÁÖ|Ÿ+ýË sjáTe#áTÌ.
<ŠXæ+Xæ\qT eTq+ dŸ+¿£\q+ eT]jáTT e«e¿£\q+ #ûjTá &ƒ+ mý²Hà #áÖ<‘Ý+. |ŸPs’+¿±\ ýË dŸ+¿£\q+,
e«e¿£\q+ #ûdq¾ fñ,¢ <ŠXæ+Xø dŸ+K«\qT Å£L&† <ŠXæ+Xø _+<ŠTeÚ\ esÁTdŸýË ÿ¿£<‘“ ¿ì+<Š ÿ¿£{ì eÚ+&û $<ó+Š >±
çyd¾, <ŠXæ+Xø _+<ŠTeÚqT n<û esÁTdŸýË dŸeÖ<ó‘q+ ýË Å£L&† –+#*.
i)
356.24 + 65.08
356.24
+ 65.08
421.32
7e ÔásÁ>·Ü >·DìÔáeTT
ii)
67.29 – 35.91
67.29
– 35.91
31.38
52
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
Let us consider a daily life situation
Dheeraj of 7th class went to the market with his father to
purchase a cloth for stitching of shirts. If 1.5 metre cloth is needed
to each shirt, then how much cloth is required for 3 shirts?
To find this, we need to understand multiplication of decimals.
There are many situations in real life where we need to multiply
decimal numbers.
Let us learn multiplication of decimals in 3 methods.
2.2.1 Multiplication of Decimal number with Whole number :
Method - 1: (Converting decimal in to a fraction) Method – 2 : (Multiply and put decimals)
3  1.5
3  1.5
15
Step 1 : Multiply whole numbers ignoring
 3
decimals. 3  15 = 45
10
Step 2 : Then mark the decimal point to the
product from right most to left
according to the number of decimal
places in the given decimal. Here
only one decimal place is there.
 3  1.5 = 4.5
45
10
= 4.5

Method - 3 : (Pictorial representation)
3 × 1.5
1.5
1.5
4.5
Fig. 2.2
1.5
Fig. 2.1
 5
Each 1.5 decimal represents 1 whole and 5tenths   parts. So, the shaded region shows 3  1.5 in
 10 
Fig. 2.1. We know that repeated addition is multiplication. Hence in adding of three 1.5 decimals, we
get 4 whole and one 0.5 decimal part (Fig.2.2)
3  1.5 = 4.5
Example 6 : If one side of a square is 3.8 cm, then find its perimeter?
Solution :
A
Number of sides to a square = 4
3.8 cm
B
Each side of a square is equal.
Side of a square = 3.8 cm
 Perimeter of a square = 4  side
= 4  3.8 = 15.2 cm
VII Class Mathematics
53
C
D
Fractions, Decimals and Rational Numbers
eTq “Ôá« J$Ôá+ýË ÿ¿£ dŸ+<ŠsÒۓ• rdŸTÅ£”+<‘eTT.
7eÔás>Á Ü· #á<TŠ eÚÔáTq• BósCÁ Ù Ôáq Ôá+ç&Ôà ¿£*d¾ #=¿±Ø\T ¿=qT>Ã\T
#ûd+<ŠTÅ£” eÖÂsسTÅ£” yîÞ²¢&Tƒ . ç|ŸÜ #=¿±ØÅ£” 1.5 MT³sÁ¢ ‹³¼ nedŸs+Á
nsTTÔû, 3 #=¿±ØnÅ£” m+Ôá ‹³¼ nedŸs+Á neÚÔáT+~?
B““ ¿£qT>=q&ƒ+ ¿=sÁŔ£ , <ŠXæ+Xæ\ >·TD¿±s“• nsÁ+œ #ûdTŸ ¿Ãy*.
“Ôá«J$Ôá+ýË <ŠXæ+Xø _óH•\T >·TD¿±sÁ+ #ûjTá e\d¾q dŸ+<ŠsÒÛ\T nHû¿+£
–+{²sTT.<ŠXæ+Xæ\qT 3 |Ÿ<ÔÝŠÆ Tá \ýË >·TDì+#á&†“• eTq+ HûsTÁ ÌÅ£”+<‘+.
2.2.1 |ŸPs’+¿£+ Ôà <ŠXæ+Xø _óH•\ >·TD¿±sÁ+:
|Ÿ<ŠÆÜ 1: (<ŠXæ+X擕 _óq•+>± eÖsÁÌ&ƒ+)
|Ÿ<ÜƊ 2: (>·TDì+º, <ŠXæ+Xø _+<ŠTeÚqT ™|³T¼³)
3 I 1.5
kþbÍq+ 1: <ŠXæ+Xø _+<ŠTeÚÔà dŸ+‹+<ó+Š ýñŔ£ +&†
|ŸPsÁd’ +Ÿ K«\qT >·TDì+#á+&.
3 I 15 R 45
kþbÍq+ 2: ‚eNj&ƒ¦ <ŠXæ+Xæ\ýË, <ŠXæ+Xø k͜H\
dŸ+K«Å£” nqT>·TD+>±, Å£”&yîÕ|ŸÚ qT+º
m&ƒeTÅ£” <ŠXæ+Xø _+<ŠTeÚqT >·T]ï+#á+&.
‚¿£Ø&ƒ ¿¹ e\+ ÿ¿£ <ŠXæ+Xø k͜q+ eÖçÔáyTû
–+~.
 3  1.5 = 4.5
3  1.5
 3
15
10
45
10
= 4.5

|Ÿ<ŠÆÜ 3: (|Ÿ³sÁÖ|Ÿ |Ÿ<ŠÆÜ)
3 I 1.5
1.5
1.5
4.5
Fig. 2.2
1.5
|ŸFig.
³+ 2.1
2.1
|Ÿ³+ 2.2
 5
ç|ŸÜ 1.5 <ŠXæ+Xø+ 1 dŸ+|ŸPsÁ’ eT]jáTT 5 <ŠXæ+Xø  10  uó²>±\qT ¿£*Ð eÚ+~. |Ÿ³+ 2.1 ýË 3 I 1.5 “ w&Ž
#ûjTá ‹&q çbÍ+Ôá+ #áÖ|ŸÚÔáT+~. |ŸÚHsÁeÔá dŸ+¿£\q+ >·TD¿±sÁ+ n“ eTqÅ£” Ôî\TdŸT. n+<ŠTe\¢ eTÖ&ƒT 1.5
<ŠXæ+Xæ\qT ¿£\|Ÿ>± eTqÅ£” 4 |ŸP]ï uó²>±\T eT]jáTT ÿ¿£ 0.5 <ŠXæ+Xø uó²>·+ \_ódTŸ +ï ~. (|Ÿ³+ 2.2)
3  1.5 = 4.5
–<‘VŸ²sÁD 6 : ÿ¿£ #áÔTá sÁçdŸ+ jîTT¿£Ø uóT„ È+ 3.8™d+.MT. nsTTÔû, <‘“ #áT³T¼¿=\ÔáqT
kÍ<óŠq :
¿£qT>=q+&.
#áÔTá sÁçk͓¿ì uóT„ C²\ dŸ+K« R 4
#áÔTá sÁçdŸ+ jîTT¿£Ø ç|ŸÜ uóT„ È+ dŸeÖq+.
#áÔTá sÁçdŸ+ jîTT¿£Ø uóT„ È+ R 3.8 ™d+.MT.
#áÔTá sÁçdŸ+ jîTT¿£Ø #áT³T¼¿=\Ôá R 4 I uóT„ È+
R 4 I 3.8 R 15.2 ™d+.MT.
7e ÔásÁ>·Ü >·DìÔáeTT
54
A 3.8 ™d+.MT. B
C
D
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
Find the product:
1)
32.5 × 8
2) 94.62 × 7
3) 109.761 × 31
4)
61 × 2.39
2.2.2. Multiplication of decimal with 10, 100, 1000
You may wonder, if you know what would happen, if a decimal number is multiplied by 10, 100,
1000 etc.
Let us consider the following multiplications. we can observe some patterns here.
24183
10
1000
=
24183
= 241.83
100
24.183 x 100 =
24183
100
1000
=
24183
= 2418.3
10
24.183 x 1000 =
24183
24183
= 1000 =
= 24183
1000
1
Ø
24.183 x 10
Ø
Ø
=
If we multiply a decimal number by 10 then we shift the decimal point 1 place to the right in the
product. Eg : 12.56  10 = 125.6
If we multiply a decimal number by 100 then we shift the decimal point 2 places to the right in the
product Eg : 12.56  100 = 1256.0 = 1256
If we multiply a decimal number by 1000 then we shift the decimal point 3 places to the right in the
product Eg : 12.56  1000 = 12560.0 = 12560
To multiply decimal numbers by 10, 100, 1000... just shift the decimal point
in the product as many places to the right, as number of zeroes after 1.
i) 239.27  10
Example 7 : Find
Solution :
i)
ii) 5.305  100
iii)
23.1  1000
239.27  10 (number of zeroes in 10 is one. So, decimal point has to shift to one
place right in the product).
 239.27  10 = 2392.7
ii)
5.305  100 = 530.5
iii)
23.1  1000 = 23100.0
= 23100
1) Find 27.35 × 10
27.35 × 100
27.35 × 1000
2) Find the values of the following .
i)
26.59 × 10
ii)
VII Class Mathematics
206.5 ×100
iii)
206.5 × 1000
iv)
10.001 × 1000
55
Fractions, Decimals and Rational Numbers
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
\u²Æ“• ¿£qT>=q+&:
1) 32.5 I 8
2) 94.62 I 7
3) 109.761 I 31
4) 61 I 2.39
2.2.2. 10, 100, 1000 yîTT<Š\>·T y{ìÔà <ŠXæ+Xæ\ >·TD¿±sÁ+:
ÿ¿£ <ŠXæ+Xø dŸ+K«qT 10, 100, 1000, ... yîTT<ŠýqÕÉ y{ìÔà >·TDì+#á‹&q|ŸÚ&ƒT @$T ÈsÁT>·TÔáT+<à Ôî*dï
€XøÌsÁ«+>± n“|¾dTŸ +ï ~.
‡ ç¿ì+~ >·TD¿±s“• |Ÿ]o*<‘Ý+. ‚¿£Ø&ƒ eTq+ ¿=“• neT]¿£\qT |Ÿ]o*+#áe#áTÌqT.
24183
 10
1000
=
24183
= 241.83
100
24.183 x 100 =
24183
100
1000
=
24183
= 2418.3
10
24.183 x 1000 =
24183
24183
= 1000 =
= 24183
1000
1
Ø
24.183 x 10
Ø
Ø
=
<ŠXæ+Xø dŸ+K«qT 10Ôà >·TDì+ºq|ŸÚÎ&ƒT, eTq+ <ŠXæ+Xø _+<ŠTeÚ ÿ¿£ k͜H“• Å£”&y|Õî ڟ Å£” eÖsÁTkÍï+.
–<‘ : 12.56 I 10 R 125.6
<ŠXæ+Xø dŸ+K«qT 100Ôà >·TDì+ºq|ŸÚÎ&ƒT, <ŠXæ+Xø _+<ŠTeÚ 2 k͜H\qT Å£”& yî|Õ ÚŸ Å£” eÖsÁTkÍï+.
–<‘ : 12.56 I 100 R 1256.0 R 1256
ÿ¿£ <ŠXæ+Xø dŸ+K«qT 1000Ôà >·TDì+ºq|ŸÚÎ&ƒT, eTq+ <ŠXæ+Xø _+<ŠTeÚ eTÖ&ƒT k͜H\qT Å£”& yî|Õ ÚŸ Å£”
eÖsÁTkÍï+. –<‘: 12.56 I 1000 R 12560.0 R 12560
<ŠXæ+Xød+Ÿ K«\qT10, 100, 1000,.... yîTT<ŠýqÕÉ y{ìÔà >·TDì+#\+fñ 1 ÔásTÁ yÔá dŸTH•\
dŸ+K«Å£” dŸeÖq dŸ+K«ýË <ŠXæ+Xø _+<ŠTeÚqT Å£”&y|Õî ڟ Å£” eÖsÁTkÍï+.
–<‘VŸ²sÁD 7: i) 239.27 I 10 ii) 5.305 I 100 iii) 23.1 I 1000 \qT ¿£qT>=q+&.
kÍ<óŠq :
i) 239.27 I 10 (10 ýË dŸTH•\ dŸ+K« R 1. n+<ŠTe\¢, <ŠXæ+Xø _+<ŠTeÚqT Å£”&y|
Õî ڟ Å£” ÿ¿£
k͜H“¿ì eÖsÁTÎ #ûjTá ‹&ƒTÔáT+~)
 239.27  10 = 2392.7
ii)
iii)
5.305 I 100
 5.305 I 100 = 530.5
23.1 I 1000
 23.1 I 1000 = 23100.0
=23100
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
1)
27.35 I 10
27.35 I 100
27.35 I 1000 \qT
2. ç¿ì+~ $\Te\qT ¿£qT>=q+&.
i)
26.59 × 10
7e ÔásÁ>·Ü >·DìÔáeTT
ii)
206.5 ×100
¿£qT>=q+&.
iii)
56
206.5 × 1000
iv)
10.001 × 1000
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
2.2.3. Multiplication of two decimal numbers:
Let us find multiplication of any to decimals. Eg. 0.4 × 0.3
Method 1: (Converting decimal into a fraction)
0.4 × 0.3 =
Method 2 :
Step 1 :
Step 2 :
Method 3:
4 3
43
12
 

 0.12
10 10 10  10 100
 0.4 × 0.3 = 0.12
(Multiply and put decimals)
0.4 × 0.3
0.4 (  1decimal place)
× 0.3 (  1decimal place)
Multiply whole numbers ignoring decimal points. 4 × 3 = 12
Mark the decimal point in the product from right most to left according to the total
number of decimal places of the given numbers. Here it is two.
Hence, 0.4 × 0.3 = 0.12 (  2 decimal places)
(Pictorial representation)
0.4 × 0.3
Let us find the product pictorially in Fig 2.3 and 2.4.
0.3
Fig. 2.3
We know that 0.4 × 0.3 =
Pictorially
0.12
Fig. 2.4
4 3
4
3

(or)
of
10 10
10
10
4
3
12

of
10
10 100
3
(or) 0.3
10
(Fig.2.3). Again divide each of these three equal parts into ten equal parts horizontally and take four
from each of it.
We divide rectangle into ten equal parts and shade three parts out of it, to get
4
3
of
(or) 0.3 × 0.4.
10
10
Since, there are 12 double shaded parts out of 100, they represent 0.12 (Fig.2.4).
 0.4 × 0.3 = 0.12
Now, we get
VII Class Mathematics
57
Fractions, Decimals and Rational Numbers
2.2.3 s +&ƒT <ŠXæ+Xæ\ >·TD¿±sÁ+:
@yîHÕ  s +&ƒT <ŠXæ+Xæ\ jîTT¿£Ø >·TD¿±sÁ+ >·T]+º eTq+ Ôî\TdŸTÅ£”+<‘+.
|Ÿ<ÜƊ 1 : (<ŠXæ+X擕 _óq•+>± eÖsÁÌ&ƒ+)
0.4 I 0.3
4
43
3
–<‘: 0.4 I 0.3
12
0.4 I 0.3 R 10  10  10 10  100  0.12
 0.4 I 0.3 R 0.12
|Ÿ<ÜƊ 2 : (>·TDì+º, <ŠXæ+Xø _+<ŠTeÚqT ™|fñɼ |Ÿ<ŠÆÜ)
0.4 I 0.3
0.4 (  1<ŠXæ+Xø k͜q+)
× 0.3 (  1<ŠXæ+Xø k͜q+)
kþbÍq+ 1: <ŠXæ+Xø _+<ŠTeÚ\qT |Ÿ{+¼ì #áT¿ÃÅ£”+&† |ŸPsÁd’ +Ÿ K«\qT >·TDì+#á&+ƒ . 4 I 3 R 12
kþbÍq+ 2: ‚eNj&ƒ¦ dŸ+K«\ jîTT¿£Ø <ŠXæ+Xøk͜H\ dŸ+K«Å£” nqT>·TD+>± Å£”&y|Õî ڟ qT+º m&ƒeTyî|Õ ÚŸ Å£” <ŠXæ+Xø
_+<ŠTeÚqT >·T]ï+#á+&. ‚¿£Ø&ƒ s +&ƒT <ŠXæ+Xæ\T (1G1) –H•sTT.
n+<ŠTe\¢ 0.4 I 0.3 R 0.12 (  2 <ŠXæ+Xøk͜H\T)
|Ÿ<ÜƊ 3:
(|Ÿ³sÁÖ|Ÿ |Ÿ<ÜƊ )
0.4 I 0.3
|Ÿ³+ 2.3 eT]jáTT 2.4 \ýË 0.4, 0.3 \ \u²Æ“• ºçÔá|³
Ÿ +ýË #áÖ|¾q³T¢ ¿£qT>=+<‘+.
0.3
0.12
|Ÿ³+ 2.3
0.4 × 0.3 =
4 3

10 10
3
|Ÿ³+ 2.4
(ýñ<‘)
4
3
10
ýË
4
10
n“ eTqÅ£” Ôî\TdŸT.
12
ºçÔá|sŸ +Á >± 10 ýË 10  100
3
eTq+ BsÁ#é Ôá Tá sÁçk͓• |Ÿ~ dŸeÖq uó²>±\T>± $uó›„ +#+ eT]jáTT 10 (ýñ<‘) 0.3 bõ+<Š&+ƒ ¿=sÁŔ£ eTÖ&ƒT
uó²>±\qT w&Ž#kû Í+(|Ÿ³+2.3). ‡ eTÖ&ƒT dŸeÖq uó²>±\qT ¿ì܌ È dŸeÖ+Ôás+Á >± |Ÿ~ dŸeÖq uó²>±\T>±
$uó›„ +º, ç|ŸÜ <‘“ qT+º H\T>·T uó²>±\T
3
4
10
rdŸT¿=+<‘+.
4
‚|ŸÚÎ&ƒT, eTqÅ£” 0.3 I 0.4 (ýñ<‘) 10 ýË 10 \_ódŸTï+~.
100 uó²>±\TýË 12 –eTˆ&>± w&Ž #ûd¾q uó²>±\TH•sTT ¿£qT¿£, n$ 0.12 Å£” çbÍܓ<óŠ«+ eV¾²kÍïsTT (|Ÿ³+
2.4).
 0.4 × 0.3 = 0.12
7e ÔásÁ>·Ü >·DìÔáeTT
58
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
Example 8 : Bindu went to vegetable market with her mother to buy onions. If the cost of Onions is
`18.50 per kg, then find the cost of 3.5 kg of Onions.
Step-1:Multiply whole numbers ignoring
Solution : Cost of 1 kg Onions = `18.50
decimals 35  1850 = 64750
Cost of 3.5 kg of Onions = 3.5 × `18.50
Step-2: As there are total (1 + 2)
(1decimal place) (2 decimal places)
3 decimals, put decimal point after three
= 64.750
digits from the right most to the product.
 Cost of 3.5 kg of Onions = `64.75
So, 3.5  18.50 = 64.750
Example 9 : One day Karthik went to petrol bunk to fill
petrol in his car. If the cost of petrol is `83.21
per litre, then how much he has to pay to fill
the full tank with 36.2 litres?
Solution :
Cost of 1 litre petrol = `83.21
Cost of 36.2 litre petrol = 36.2 × `83.21
= `3012.202
 Cost of 36.2 litres of petrol = `3012.20 (rounded to two decimals). (Why?)
The number of decimal digits in the product of any two decimal numbers
is equal to the sum of decimal digits that are multiplied.
Find the product of the following.
i) 69.2 × 2.5 ii) 20.61 × 3.09
iii) 658.321 × 43.2
iv) 206.005 × 0.07
Here is an activity to multiply two decimals using grid. Cardboard,
white chart paper, ruler, pencil, fevicol, sketch pens of different
colours are needed.
D
A
Fig.1
C
D
B
A
Fig.2
C
D
B
A
C
Fig.3
B
1.
Take a cardboard of convenient size and paste a white chart on it.
2.
Make 10 x 10 grids on it and label the corners of the grid as A, B, C and D as shown in Fig.1
VII Class Mathematics
59
Fractions, Decimals and Rational Numbers
–<‘VŸ²sÁD 8: –*¢bÍjáT\T ¿=qT>Ã\T #ûjTá &†“¿ì _+<ŠT Ôáq Ôá*Ô¢ à ¿£*d¾ Å£LsÁ>±jáT\ eÖÂsØ{ÙŔ£ yî[+¢ ~.
–*¢bÍjáT\ <ósŠ Á ¿ìýËÅ£” `18.50 nsTTÔû, 3.5 ¿ìýËç>±eTT\ kþbÍq+`1: <ŠXæ+Xø _+<ŠTeÚÔà dŸ+‹+<óŠ+ ýñÅ£”+&†
–*¢bÍjáT\ jîTT¿£Ø <ósŠ Á ¿£qT>=q+&.
|ŸPsÁ’dŸ+K«\qT >·TDì+#á+& .
kÍ<óŠq :
1 ¿ìýËç>±eTT\ –*¢bÍjáT\ <ósŠ Á R `18.50
35 I 1850 R 64750
kþbÍq+`2: ‚eNj&q dŸ+K«\ <ŠXæ+Xø k͜H\T 1G2 R 3
3.5 ¿ìýËç>±eTT\ –*¢bÍjáT\ <ósŠ Á R 3.5 I ` 18.50
¿£qT¿£ \‹Ý+qÅ£” Å£”& yîÕ|ŸÚ qT+& m&ƒeT yîÕ|ŸÚÅ£” eTÖ&ƒT
(1<ŠXæ+Xø k͜q+) (2 <ŠXæ+Xø k͜H\T) k͜H\ ÔásÇÔá <ŠXæ+Xø _+<ŠTeÚqT >·T]ï+#á+&.
3.5 I 18.50 R 64.750
R 64.750
 3.5 ¿ìýËç>±eTT\ –*¢bÍjáT\ <ós
Š Á R ` 64.75
–<‘VŸ²sÁD 9: z sÃE ¿±¯ï¿ù Ôáq ¿±sÁTýË ™|ç{ËýÙ “+|Ÿ&†“¿ì ™|ç{ËýÙ
‹+¿ùŔ£ yîÞ²¢&Tƒ . ™|ç{ËýÙ <ósŠ Á ç|Ÿr ©³sÁTÅ£” ` 83.21
nsTTÔû, 36.2 ©³sÁÔ¢ à |˜ÚŸ ýÙ {²«+¿ù “+|Ÿ&+ƒ ¿=sÁŔ£
njûT« KsÁTÌqT ¿£qT>=q+&?
kÍ<óŠq :
1 ©³sY ™|ç{ËýÙ <ósŠ Á R `83.21
36.2 ©³sÁ¢ ™|ç{ËýÙ <ósŠ Á R 36.2 I ` 83.21
R ` 3012.202
 36.2 ©³sÁ¢ ™|ç{ËýÙ <ós
Š Á R ` 3012.20 (Âs+&ƒT <ŠXæ+Xæ\TÅ£” dŸesÁD #ûjTá ‹&+~). m+<ŠTÅ£”?
s +&ƒT <ŠXæ+Xø dŸ+K«\qT >·TDì+ºq|ŸÚ&ƒT \‹Æ+ýË <ŠXæ+Xø k͜H\ dŸ+K«,
>·TDì+#á‹&q dŸ+K«\ <ŠXæ+Xø k͜H\ dŸ+K«\ yîTTԐì dŸeÖq+.
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
ç¿ì+<Š \u²Æ\qT ¿£qT>=q+&.
i) 69.2 × 2.5 ii) 20.61 × 3.09
‚~ #û<‘Ý+!
iv) 206.005 × 0.07
çÐ&Ž ||ŸsY qT –|ŸjÖî Ð+º s +&ƒT <ŠXæ+Xæ\qT >·TD¿±sÁ+ #ûjTá &†“¿ì ¿£ Ôá«eTT.‡
¿£Ôá«+ #ûjTá &†“¿ì ¿±sY¦ uËsY,¦ Ôî\T|ŸÚ sÁ+>·T #sY¼ ||Ÿs,Y sÁÖ\sY, ™|“àýÙ, ™|$˜ ¿±ýÙ,
$_óq• sÁ+>·TýË¢ ™dØ#Y ™|qT•\T nedŸs+Á neÚԐsTT.
¿£Ôá«+
D
A
iii) 658.321 × 43.2
Fig.1
C
D
B
A
Fig.2
C
D
B
A
C
Fig.3
B
1. ÔáÐq ¿=\Ôá\Ôà ÿ¿£ n³¼eTT¿£ØqT rdŸT¿=“ <‘“™|Õ Ôî\“¢ #sÁTq¼ T nÜ¿ì+#á+&.
2. 10 I 10 çÐ&ƒq¢ T <‘“™|Õ ^jáT+& eT]jáTT |Ÿ³+-1ýË #áÖ|¾+ºq $<ó+Š >± çÐ&Ž jîTT¿£Ø osü\qT A, B, C
eT]jáTT D n“ >·T]ï+#á+&.
7e ÔásÁ>·Ü >·DìÔáeTT
60
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
3.
Shade three horizontal strips using a sketch pen say, red colour starting from the bottom as shown
in Fig.2. The portion shaded in red colour (horizontal strips) represents
4.
3
= 0.3
10
Shade four vertical strips using a sketch pen say, blue colour starting from right most corner as
shown in Fig.3. The portion shaded in blue colour (vertical strips) represents
5.
In Fig.3 portion shaded both in red and blue colours represents
4
= 0.4
10
12
= 0.12,
100
thus, 0.3 × 0.4 = 0.12
6.
Repeat this activity by taking different numbers of horizontal and vertical strips to represent the
product of the pairs of decimals such as 0.5 × 0.6 and 0.2 × 0.8 etc.
I am a decimal number, who is half of one fourth of 100. Who am I?
7532.4
Observe the figure. Fill the blue boxes
with suitable decimal numbers.
10
x
00
x1
75.324
0x
00
10
x1
00
0
Exercise - 2.2
1.
Find the product of the following.
i)
2.
3.
23.4 × 6
ii)
681.25 × 9
iii) 53.29 × 14
iv) 8 × 2.52
v) 25 × 2.013
Fill the blanks in the table.
Multiplications
Product
36.21 × 10
362.1
23.104 × 100
______
6.24 × ______
6240.0
______ × 1000
21.05
9.234 × 100
______
1.3004 × ______
1300.4
______ × 10
59.001
Find the product.
i)
5.1 × 8.1
ii) 63.205 × 0.27
VII Class Mathematics
iii) 1.321 × 0.9 iv) 6.51 × 0.99 v) 837.6 × 0.006
61
Fractions, Decimals and Rational Numbers
3. eTÖ&ƒT ¿ì܌ È dŸeÖ+ÔásÁ |Ÿ{\¼¡ qT, |Ÿ³+-2 ýË #áÖ|¾q $<ó+Š >± ^º ~>·Te qT+º msÁT|ŸÚ sÁ+>·TýË w&Ž
#ûjáT+&.w&Ž #ûjáT‹&q uó²>·+ (¿ìŒÜÈ dŸeÖ+ÔásÁ |Ÿ{¡¼\T)
3
= 0.3 Å£”
10
çbÍܓ<óŠ«+ eV¾²dŸTï+~.
4. |Ÿ³+-3 ýË #áÖ|¾q $<ó+Š >± Å£”& yî|Õ ÚŸ qT+& ú\+ sÁ+>·T ™dØ#Y ™|HŽ –|ŸjÖî Ð+º H\T>·T “\TeÚ |Ÿ{\¼¡ qT
w&Ž #ûjTá +&.ú\+ sÁ+>·T ýË w&Ž #ûjTá ‹&q uó²>·+ (“\TeÚ |Ÿ{\¼¡ T)
5. |Ÿ³+-3 ýË msÁT|ŸÚ eT]jáTT ú\+ sÁ+>·T\ýË
12
= 0.12 qÅ£”
100
4
= 0.4 Å£” çbÍܓ<óŠ«+ eV¾²dŸTï+~.
10
çbÍܓ<ó«Š + eV¾²dŸT+ï ~.
€ $<ó+Š >± 0.3 × 0.4 = 0.12
6. 0.5 I 0.6 eT]jáTT 0.2 I 0.8 e+{ì <ŠXæ+Xø ÈÔá\ \‹Æ+ Ôî\TdŸT¿ÃqT³Å£” ¿ì܌ È dŸeÖ+ÔásÁ eT]jáTT
“\TeÚ |Ÿ{\¼¡ jîTT¿£Ø $_óq• dŸ+K«\qT rdŸT¿=“ ‡ ¿£Ԑ«“• |ŸÚqseÔá+ #ûjTá +&.
HûH=¿£ <ŠXæ+Xø dŸ+K«qT. 100ýË H\T>à e+ÔáTýË dŸ>+· >± eÚ+{²qT. HûqT mesÁT?
7532.4
|Ÿ³+ >·eT“+#á+&. ÔáÐq <ŠXæ+Xø dŸ+K«\ÔÃ
ú\+ >·&Tƒ \qT “+|Ÿ+&.
nHûÇw¾<‘Ý+
10
x
00
x1
75.324
0x
00
10
x1
00
0
nuó²«dŸ+ ` 2.2
1. ç¿ì+~ y{ì \u²Æ“• ¿£qT>=q+&.
i)
23.4 × 6
ii)
681.25 × 9
2. |Ÿ{¿¼ì ý£ Ë U²°\qT “+|Ÿ+&.
iii) 53.29 × 14
iv) 8 × 2.52
>·TD¿±sÁ+
\‹Ý+
36.21 × 10
362.1
23.104 × 100
______
6.24 × ______
6240.0
______ × 1000
21.05
9.234 × 100
______
1.3004 × ______
1300.4
______ × 10
59.001
v) 25 × 2.013
3. \u²Æ“• ¿£qT>=q+&.
i)
5.1 × 8.1
7e ÔásÁ>·Ü >·DìÔáeTT
ii) 63.205 × 0.27
iii) 1.321 × 0.9 iv) 6.51 × 0.99 v) 837.6 × 0.006
62
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
4.
5.
6.
7.
8.
Rithesh reads a book for 2.5 hours everyday. If he reads the entire
book in a Week, then how many hours all together are required for
him read to the book?
Find the area of the rectangle whose length and breadth are 5.3cm,
2.7cm respectively.
If the cost of each cement bag is `326.50, then find the cost of 24 bags
of cement.
Dharmika purchased chudidhar material of 1.40m at the rate
of `152.5per metre. Find the amount to be paid.
If a picture chart costs `4.25. Amrutha wants to buy 16 charts to
make an album. How much money does she has to pay?
2.3 Division of Decimals :
Sujatha wants to decorate her class room with coloured paper strips
each of 2.25m length. If she has a coloured paper strip roll of 13.5m
length, how many pieces of paper strips of required length will she
get from the strip roll? Sujatha expressed it as 13.5 ÷ 2.25.
Hence, we have to understand how to divide decimal numbers. There are many situations in
real life where we need to divide decimal numbers.
2.3.1 Division of Decimal numbers by 10, 100, 1000,
etc.
Let us observe the following patterns .
1) 23.5 ÷ 10 =
235 10 235 1 235  1 235
 
 

 2.35
10
1
10 10 10 10 100
2) 23.5 ÷ 100 =
3)
235 100 235 1
235  1
235





 0.235
10
1
10 100 10 100 1000
23.5 ÷ 1000 =
235 1000 235
1
235 1
235





 0.0235
10
1
10 1000 10  1000 10000
Like this, you may fill the following blanks .
169.28 ÷ 10 = 16.928
525.9 ÷ 10 =_____
169.28 ÷ 100 = 1.6928
525.9 ÷ 100 = _____
169.28 ÷ 1000 = _____
525.9 ÷ 1000 = _____
While dividing a decimal number by 10 or 100 or 1000 or the digits of
the number and the digits of quotient are same, but the decimal point in the
quotient shifts to the left by as many places as there are zeros after 1.
Find the following.
i) 81.5 ÷ 10 ii) 4901.2 ÷ 100
VII Class Mathematics
63
iii) 7301.3 ÷ 1000 iv) 1.2 ÷ 100
Fractions, Decimals and Rational Numbers
4. ]Ôûwt ç|ŸÜsÃp s +&ƒTq•sÁ >·+³\bͳT ÿ¿£ |ŸÚdŸ¿ï ±“• #á<TŠ eÚԐ&ƒT. ÿ¿£ ysÁ+
ýË € |ŸÚdŸ¿ï +£ qT nÔáqT |ŸP]ï>± #á~$Ôû, yîTTÔá+ï m“• >·+³\T #á~y&ƒT?
5. bõ&ƒeÚ eT]jáTT yî&\ƒ TÎ\T esÁTdŸ>± 5.3 ™d+.MT. eT]jáTT 2.7 ™d+.MT >± –
q• BsÁ#é Ôá áTsÁçdŸ+ jîTT¿£Ø yîXÕ æ\«+ ¿£qT>=q+&.
6. ÿ¿£ d¾yTî +{Ù ‹kÍï <ósŠ Á `326.50 nsTTq#à 24 u²«>·T\ d¾yTî +{Ù ‹kÍï\ <ósŠ Á qT
¿£qT>=q+&.
7. <ó‘]ˆ¿£ #áT&<‘ó sY yîT{¡]jáTýÙ qT, ÿ¿£ MT³sÁTÅ£” `152.5 #=|ŸÚÎq 1.40MT.
¿=qT>Ã\T #ûd+¾ ~. <ó‘]ˆ¿£ #î*+¢ #*àq yîTTԐ ¿£qT>=q+&.
8. neTÔá ÿ¿£ €\ÒyŽT ÔájÖá sÁT #ûjTá &†“¿ì 16 #ósÁT\¼ qT ¿=qT>Ã\T #ûjÖá \“
nqTÅ£”+³T+~. ÿ¿£ |¾¿Ì£ sY #ósÁT¼ <ósŠ Á `4.25 nsTTÔû €yîT m+Ôá &ƒ‹TÒ
#î*+¢ #*à –+³T+~?
2.3 <ŠXæ+Xø dŸ+K«\ uó²>·VŸäsÁ+:
dŸTC²Ôá Ôáq Ôás>Á Ü· >·~“ 2.25 MT. bõ&ƒeÚq• sÁ+>·T ¿±ÐÔá|ڟ |Ÿ{\¼ì ÔÃ
n\+¿£]+#\“ nqTÅ£”+{Ë+~. 13.5 MT. bõ&ƒeÚ >·\ sÁ+>·T ¿±ÐԐ\
çd¾|¼ t sÃýÙ qT €yîT ¿£*Ð –q•³¢sTTÔû, çd¾|¼ t sÃýÙ qT+º €yîT m“• eTT¿£Ø\ sÁ+>·T ¿±ÐÔá|ڟ |Ÿ{\¼ì qT ¿±e*d¾q
¿Ã\Ôá\Ôà #ûjTá >·\<ŠT? B““ dŸTC²Ôá 13.5 ª 2.25 >± e«¿£+ï #ûd+¾ ~.
‡ $<ó+Š >± “È J$Ôá+ýË eTq+ <ŠXæ+Xø dŸ+K«\qT uó²>·VäŸ sÁ+ #ûjÖá *àq |Ÿ]d¾Ôœ Tá \T #ý² eÚ+{²sTT.
¿±eÚq, ‚|ŸÚÎ&ƒT eTq+ <ŠXæ+Xø dŸ+K«\qT @$<ó+Š >± uó²>·VäŸ sÁ+ #ûjÖá ýË Ôî\TdŸTÅ£”+<‘+.
2.3.1 10, 100, 1000, .... yîTT<ŠýqÕÉ y{ìÔà <ŠXæ+Xø _óH•\ uó²>·VäŸ sÁ+.
¿ì+~ neT]Å£\qT eTq+ |Ÿ]o*<‘Ý+.
1) 23.5 ÷ 10 =
235 10 235 1 235  1 235
 
 

 2.35
10
1
10 10 10  10 100
2) 23.5 ÷ 100 =
3)
235 100 235 1
235  1
235





 0.235
10
1
10 100 10 100 1000
23.5 ÷ 1000 =
235 1000 235
1
235  1
235





 0.0235
10
1
10 1000 10 1000 10000
‚<û $<ó+Š >±Hû MTsÁT ç¿ì+~ U²°\qT |ŸP]+#áe#áTÌqT.
169.28 ÷ 10 = 16.928
169.28 ÷ 100 = 1.6928
169.28 ÷ 1000 = _____
525.9 ÷ 10 = _____
525.9 ÷ 100 = _____
525.9 ÷ 1000 = _____
ÿ¿£ <ŠXæ+Xø dŸ+K«qT 10 ýñ<‘ 100 ýñ<‘ 1000 ýñ<‘
\Ôà uó²Ð+ºq|ŸÚÎ&ƒT € dŸ+K«
jîTT¿£Ø n+¿\T eT]jáTT uó²>·|\Ÿ˜ + jîTT¿£Ø n+¿\T ÿ¹¿ $<ó+Š >± –+{²sTT. ¿±“ 1 ÔásTÁ yÔá
m“• dŸTH•\T eÚq•jîÖ n“• k͜H\T m&ƒeT yî|Õ ÚŸ Å£” <ŠXæ+Xø _+<ŠTeÚ ÈsÁT>·TÔáT+~.
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
ç¿ì+~ y{ì“ ¿£qT>=q+&.
i) 81.5 ÷ 10
7e ÔásÁ>·Ü >·DìÔáeTT
ii) 4901.2 ÷ 100
64
iii) 7301.3 ÷ 1000 iv) 1.2 ÷ 100
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
2.3.2 Division of Decimal numbers by whole numbers:
John dig a well of 4.8m in 3 hours. How much depth he digs in 1 hour on an average?
For this, we have to find 4.8 ÷ 3.
Let us find 4.8 ÷ 3 in 3 methods.
Method 2 : (Divide and put decimal)
Method 1 : (Change decimal into fraction)
(1 decimal place) 4.8 ÷ 3
48 3

4.8 ÷ 3 =
10 1
=
48 1

10 3
=
16
 1.6
10
Step 1 : Divide whole number by ignoring
decimals. 48 ÷ 3 = 16
Step 2 : Then mark the decimal point to the
quotient from right most to left
according to the number of decimal
places in the given decimal.
(why?)
 4.8 ÷ 3 = 1.6
 4.8 ÷ 3 = 1.6
So, John digs in 1hour = 1.6m
Method 3 : (Pictorial representation)
4.8 ÷ 3
1.6
Fig.2.7
4.8
Fig.2.5
1.6
1.6
Fig.2.6
Take four shaded units and eight parts from ten equal parts of 5th unit which represents 4.8 in
Fig.2.5. We have to divide these 4.8 into three equal groups. So, divide three units as one to each
group (Fig.2.6). Remaining fourth unit can’t be shared directly. Hence divide it into ten equal parts.
Now a total of 18 tenth parts are available to share among three equal groups. Divide and share six
tenth (0.6) parts to each group. Therefore 4.8 divides into 3 equal groups as each group has one unit
and six tenth parts which totally represents 1.6.
Hence 4.8 ÷ 3 = 1.6
Find the following :
i) 69.4 ÷ 2 ii) 56.32 ÷ 8 iii)
6.5 ÷ 4
iv) 108.7 ÷ 5
2.3.3 Division of Decimal numbers by decimal numbers:
Srivalli is doing a part-time job at a milk dairy. One day she sold 32.5 litres of milk to the
customers. If she received `1641.25 from them, find the cost of 1 litre milk?
To find this, we need to find `1641.25 ÷ 32.5
VII Class Mathematics
65
Fractions, Decimals and Rational Numbers
2.3.2 |ŸPs’+¿±\Ôà <ŠXæ+Xø _óH•\ uó²>±VŸäsÁ+ :
C²HŽ 3 >·+³ýË¢ 4.8 MT³sÁ¢ u²$“ ÔáyÇ&ƒT. dŸ>³
· Tq 1 >·+³ýË nÔá&Tƒ m+Ôá ýËÔáT ÔáeÇ>·\&ƒT?
‚+<ŠT¿ÃdŸ+ 4.8 ª 3 qT ¿£qT>=H*.
eTq+ 4.8 ª 3 qT 3 |Ÿ<ÔƊ Tá ýË¢ ¿£qT>=Háe#áTÌ.
|Ÿ<Ü݊ 1: (<ŠXæ+X擕 _óq•+>± eÖsÁÌ&ƒ+)
4.8 ª 3 R
48 3

10 1
=
48 1

10 3
=
16
 1.6
10
(m+<ŠTÅ£”?)
|Ÿ<ÜƊ 2: (uó²Ð+º, <ŠXæ+Xø _+<ŠTeÚqT ™|³T¼³)
4.8 ª 3 (1 <ŠXæ+Xø k͜q+)
kþbÍq+ 1: <ŠXæ+Xø _+<ŠTeÚÔà dŸ+‹+<ó+Š ýñŔ£ +&†
|ŸPsÁd’ +Ÿ K«\qT uó²Ð+#á+&. 48 ª 3 R 16
kþbÍq+ 2: ‚eNj&ƒ¦ <ŠXæ+Xø+ýË <ŠXæ+Xø k͜H\ dŸ+K«Å£”
nqT>·TD+>± Å£”&y|Õî ڟ qT+º m&ƒeTyî|Õ ÚŸ Å£”
<ŠXæ+Xø _+<ŠTeÚqT >·T]ï+#á+&?
 4.8 ª 3 R 1.6
 4.8 ÷ 3 = 1.6
¿±eÚq, C²HŽ 1>·+³ýË çÔá$Çq~ R 1.6MT.
|Ÿ<ÜƊ 3:
(|Ÿ³sÁÖ|Ÿ |Ÿ<ÜƊ )
4.8 ª 3
1.6
4.8
|Ÿ³+ 2.7
|Ÿ³+ 2.5
1.6
1.6
|Ÿ³+ 2.6
|Ÿ³+ 2.5ýË #áÖ|¾q³T¢ w&Ž #ûjTá ‹& –q• H\T>·T jáT֓{Ù\qT eT]jáTT 5e jáT֓{Ù ý˓ |Ÿ~ dŸeÖq
uó²>±\ýË m“$T~ uó²>±\qT rdŸT¿Ã+&. ÿ¿£ jáT֓{Ù 10 dŸeÖq uó²>±\Å£” çbÍܓ<ó«Š + eV¾²dŸT+ï ~. ‡ 4.8 “
eTÖ&ƒT dŸeÖq uó²>±\T>± $uó›„ +#*à –+³T+~. n+<ŠTe\¢, ç|ŸÜ ç>·Ö|ŸÚÅ£” eTÖ&ƒT jáT֓{Ù\qT ÿ¿=¿£Ø{ì #=|ŸÚÎq
|Ÿ+#á+& (|Ÿ³+ 2.6). $TÐ*q ÿ¿£ jáT֓{ٓ eTq+ HûsTÁ >± |Ÿ+#áýeñ TT. n+<ŠTe\¢ <‘““ |Ÿ~ dŸeÖq uó²>±\T>±
$uó›„ +#á+&. ‚|ŸÚÎ&ƒT yîTTÔá+ï 18 uó²>±\T eTÖ&ƒT ç>·Ö|ŸÚ\Å£” dŸeÖq |Ÿ+#áT¿Ãe&†“¿ì n+<ŠTu²³TýË –H•sTT. ç|ŸÜ
ç>·Ö|ŸÚÅ£” €sÁT |Ÿ<eŠ e+ÔáT uó²>±\(0.6) #=|ŸÚÎq |Ÿ+#á+&. n+<ŠTe\¢ 4.8 qT 3 dŸeÖq ç>·Ö|ŸÚ\T>± $uóȄ q
#ûdq¾ |ŸÚ&ƒT, ç|ŸÜ ç>·Ö|ŸÚÅ£” ÿ¿£ jáT֓{Ù eT]jáTT €sÁT |Ÿ<Ãe+ÔáT uó²>±\T eÚ+&ƒTqT.B““ 1.6 >± dŸÖº+#áe#áTÌqT.
n+<ŠTe\¢ 4.8ª 3 R 1.6
ú |ç >Ÿ Ü
· “
ç¿ì+~ y{ì“ ¿£qT>=q+&.
Ôî\TdŸT¿Ã
i) 69.4 ÷ 2 ii) 56.32 ÷ 8 iii) 6.5 ÷ 4 iv) 108.7 ÷ 5
2.3.3 <ŠXæ+Xø dŸ+K«qT eTsà <ŠXæ+Xø dŸ+K«Ôà uó²>·VäŸ sÁ+:
çoe*¢ bÍ\ &îsTT¯ýË bÍsY¼ fÉ+® –<ë>·+ #ûjTá T#áTq•~. ÿ¿£sÃE €yîT 32.5 ©³sÁ¢ bÍ\qT $“jîÖ>·<‘sÁT\Å£”
n$Tˆq~.
n$Tˆq bÍ\Å£” y] qT+º `1641.25 qT €yîT rdŸTÅ£”q•³¢sTTÔû, 1 ©³sÁT bÍ\ <ósŠ qÁ T ¿£qT>=q+&?
1 ©³sÁT bÍ\ <ósŠ Á ¿£qT>=q&ƒ+ ¿=sÁŔ£ , eTq+ `1641.25 ª 32.5 qT ¿£qT>=H*àq nedŸs+Á eÚ+~.
7e ÔásÁ>·Ü >·DìÔáeTT
66
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
This is division of decimal number by decimal number. First let us learn through simple
example. 0.8ª 0.2 We can do it in three methods:
Method - 2 : (Divide and put decimals)
Method - 1 : (Change decimal into fraction)
0.8 ÷ 0.2
0.8 ÷ 0.2
Step1:
Divisor (0.2) has one decimal point.so,
8 2
8 10
multiply both with 10, then, the divisor
   
becomes a whole number.
10 10 10 2
= (0.8 × 10) ÷ (0.2 × 10)
8
=8÷2
 4
2
Step 2 : Now we can divide by 2.
8÷2=4
 0.8 ÷ 0.2 = 4
Method - 3 : (Pictorial representation)
0.8 ÷ 0.2
Fig. 2.9
Fig. 2.8
Fig. 2.10
Eight from ten equal parts of a unit represents 0.8 in Fig.2.8. We have to divide 0.8 into
0.2 to each group (Fig.2.9). How many such equal groups can formed? Yes, 4 groups (Fig.2.10).
Therefore 0.8 divides in to 4 equal groups as each group having 0.2.
Hence 0.8 ÷ 0.2 = 4.
Let us observe one more example 0.341 ÷ 1.1
Step1: Divisor 1.1 has one decimal point. So, multiply both with 10, so the divisor is a whole number.
= (0.341 × 10) ÷ (1.1 × 10)
= 3.41 ÷ 11 (2 decimal places to dividend)
Step 2 : We have to ignore the decimal places in the dividend so long as we remember to put it back
later. So, do the calculation without decimal point.
341 ÷ 11 = 31
Step 3: Now, mark the decimal point in the quotient from right most to left, according to the total
number of decimal places of the dividend.
 0.341 ÷ 1.1 = 0.31
Now we calculate `1641.25 ÷ 32.5 for 1 litre of milk.
Step 1: Divisor (32.5) has one decimal place. Multiply both numerator and denomirator by 10, so the
divisor is a whole number.
(1641.25 × 10) ÷ (32.5 × 10)
= 16412.5 ÷ 325 (1 decimal place in the dividend)
Step 2 :
164125 ÷ 325 = 505
Step 3 :
`1641.25 ÷ 32.5 = `50.5
Cost of 1litre milk = `50.50
VII Class Mathematics
67
Fractions, Decimals and Rational Numbers
n+fñ <ŠXæ+Xø dŸ+K«qT eTsà <ŠXæ+Xø dŸ+K«Ôà uó²>·VäŸ sÁ+ #ûjÖá *. B““ yîTT<Š³ dŸsÞÁ yø Tî q® –<‘VŸ²sÁD <‘Çs
Ôî\TdŸTÅ£”+<‘+. 0.8ª 0.2 B““ eTq+ eTÖ&ƒT |Ÿ<ÔƊ Tá ýË¢ #ûjTá e#áTÌ.
|Ÿ<ÜƊ 2 : (uó²Ð+º, <ŠXæ+Xø _+<ŠTeÚqT ™|³T¼³)
|Ÿ<ÜƊ 1 : (<ŠXæ+X擕 _óq•+>± eÖsÁÌ&ƒ+)
0.8ª0.2
0.8 ª 0.2
kþbÍq+1:$uó²È«+ 0.2 ÿ¿£ <ŠXæ+Xø+ ¿£*Ð eÚ+~.
8 2
8 10
¿±eÚq, s +&+{ìú 10Ôà >·TDì+#áe#áTÌ,
   
10 10 10 2
n|ŸÚÎ&ƒT $uó²È«+, |ŸPsÁ’ dŸ+K« neÚÔáT+~.
R
(0.8 I 10) ª (0.2 I 10)
8
 4
R8ª2
2
kþbÍq+ 2: ‚|ŸÚÎ&ƒT eTq+ 2 Ôà uó²Ð+#áe#áTÌqT.
8 ª 2 R 4  0.8 ª 0.2 R 4
|Ÿ<ÜƊ 3 : (|Ÿ³sÁÖ|Ÿ |Ÿ<ÜƊ )
0.8 ª 0.2
|Ÿ³+ 2.8
|Ÿ³+ 2.9
|Ÿ³+ 2.10
|Ÿ³+ 2.8ýË #áÖ|¾q $<ó+Š >± ÿ¿£ jáT֓{Ù jîTT¿£Ø |Ÿ~ dŸeÖq uó²>±\ qT+º m“$T~ uó²>±\T (€Å£”|Ÿ#Ìá sÁ+>·T
uó²>±\T) 0.8Å£” çbÍܓ<ó«Š + eV¾²+#áTqT. eTq+ 0.8 qT ç|ŸÜ dŸeTÖVŸ²eTTqÅ£” 0.2 e#ûÌ$<ó+Š >± $uó›„ +#*à –
+³T+~ (|Ÿ³+ 2.9). dŸeÖq dŸeTÖVŸ²\T n³Te+{ì$ m“• @sÁÎ&ƒ>\· eÚ? 4 dŸeÖq dŸeTÖVŸ²\T (|Ÿ³+ 2.10)
@sÁÎ&ƒTqT. n+<ŠTe\¢ 0.8 qT, ç|ŸÜ dŸeTÖVŸ²eTT ýË 0.2 eÚ+&û $<ó+Š >± 4 dŸeÖq dŸeTÖVŸä\T>± $uó›„ +#áe#áTÌqT.
 0.8 ª 0.2 R 4
eTq+ ‚+¿=¿£ –<‘VŸ²sÁD |Ÿ]o*<‘Ý+ 341 ª 1.1
kþbÍq+ 1: $uó²È¿£+ 1.1 ÿ¿£ <ŠXæ+Xø _+<ŠTeÚqT ¿£*Ð –+~.
¿±eÚH, s +&+{ìú 10Ôà >·TDì+ºÔû, $uó²È¿£+ |ŸPsÁd’ +Ÿ K« >± eÖsÁTqT.
R (0.341 I 10) ª (1.1 I 10)
R 3.41 ª 11 ($uó²È«+Å£” 2 <ŠXæ+Xæ\T eÚH•sTT)
kþbÍq+ 2: $uó²È«+Å£” >·\ <ŠXæ+Xø k͜H\qT eTq+ $dŸˆ]+#*. <ŠXæ+Xø _+<ŠTeÚ ýñŔ£ +&† uó²Ð+#á+&.
341 ª 11 R 31
kþbÍq+ 3: ‚|ŸÚÎ&ƒT, uó²>·|\Ÿ˜ +Å£” Å£”&y|Õî ڟ qT+º m&ƒeTyî|Õ ÚŸ Å£” <ŠXæ+Xø _+<ŠTeÚqT >·T]ï+#á+&. ($uó²È«+Å£” >·\
<ŠXæ+Xø k͜H\ dŸ+K« Å£” nqT>·TD+>±)
 0.341 ª 1.1 R 0.31
‚|ŸÚ&ƒT eTq+ 1© bÍ\ <ósŠ Á ¿£qT>=qT³Å£” ` 1641.25 ª 32.5 qT >·D<ì ‘Ý+.
kþbÍq+ 1: $uó²È¿£+ (32.5) ÿ¿£ <ŠXæ+Xø k͜H“• ¿£*Ð –+~. n+<ŠTe\¢, s +&+{ì“ 10Ôà >·TDì+#á+&.
R (1641.25 I 10) ª (32.5 I 10)
R 16412.5 ª 325 ($uó²È«+ýË 1 <ŠXæ+Xø k͜q+ )
kþbÍq+ 2: 164125 ª 325 R 505
 1© bÍ\ <ós
Š Á R ` 50.50
kþbÍq+ 3: `1641.25 ª 32.5 R ` 50.5
7e ÔásÁ>·Ü >·DìÔáeTT
68
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
Example 10: Madhuri is studying 7th class in Visakhapatnam. Her school teachers organised a tour
to Araku valley by bus. Bus covered a distance of 98.5 km in 2.5 hours. If the bus is
travelled at the same speed in the journey then find the distance travelled in 1hour.
Solution : Distance travelled by bus = 98.5 km
Time taken to travel this distance = 2.5 hours
 Bus travelled in 1 hour = 98.5 ÷ 2.5
=
985
 39.4 km
25
 Bus travelled in 1 hour = 39.4 km
Solve the following :
i) 0.45 ÷ 0.9 ii) 2.125 ÷ 0.05
iii) 94.3 ÷ 0.004 iv) 10.25 ÷ 0.2
Exercise - 2.3
1)
Fill in the blanks in the table.
Example :
2)
3)
4)
5)
6)
7)
Division
Quotient
362.21 ÷ 10
36.221
5636.1 ÷ 100
_____
374.9 ÷ _____
0.3749
_____ ÷ 1000
2.0164
123.0 ÷ 100
_____
1300.7 ÷ _____
1.3007
_____ ÷ 10
59.001
Solve the following.
i) 5.51 ÷ 2
ii) 38.4 ÷ 3
iii) 57.39 ÷ 6
iv) 562.1 ÷ 11 v) 0.7005 ÷ 5
vi) 9.99 ÷ 3
vii) 13 ÷ 6.5
viii) 10.01 ÷ 11 ix) 8 ÷ 0.32
x) 320.1 ÷ 33
Solve the following divisions.
i) 78.24 ÷ 0.2
ii) 4.845 ÷ 1.5
iii) 0.246 ÷ 0.6
iv) 563.2 ÷ 2.2
v) 0.026 ÷0.13
vi) 4.347 ÷ 0.09
vii) 3.9 ÷ 0.13
viii) 20.32 ÷ 0.8
ix) 24.4÷ 6.1
x) 2.164 ÷ 0.008
Solve the following.
i) Divide 39.54 by 6 ii) Divide 7.2 by 10 iii) Divide 5.2 by 1.3
Sekhar travelled 154.5 km in 5 hours with uniform speed on his bike.
How much distance does he travel in one hour?
If a mason worked 100 hours in 12.5 days to construct a wall, then
how many hours he totally worked in a day?
If the cost of dozen eggs is `61.80, then find the cost of an egg.
VII Class Mathematics
69
Fractions, Decimals and Rational Numbers
–<‘VŸ²sÁD 10 :eÖ<óTŠ ] $XæK|Ÿ³D¼ +ýË 7e Ôás>Á Ü· #á<TŠ eÚÔÃ+~. €yîT bÍsÄXÁ æ\ –bÍ<ó‘«jáTT\T ‹dŸTàýË nsÁŔ£ ýËjáTÅ£” $VŸäsÁjÖá çÔÅá ”£
kÍ<óŠq:
@sÎ³T¢ #ûXæsÁT. ‹dŸTà 2.5 >·+³ýË¢ 98.5 ¿ìýËMT³sÁ¢ <ŠÖs“• ç|j
Ÿ Öá Dì+º+~. ‹dŸTà n<û yû>+· Ôà ç|j
Ÿ Öá Dì+ºq³¢sTTÔû,
1>·+³ýË ç|j
Ÿ Öá Dì+ºq <ŠÖs“• ¿£qT>=q+&?
‹dŸTà ç|ŸjÖá Dì+ºq <ŠÖsÁ+ R 98.5 ¿ì.MT.
‡ <ŠÖsÁ+ ç|ŸjÖá Dì+#á&†“¿ì |Ÿ{q¼ì
dŸeTjáT+ R 2.5 >·+³\T
 1 >·+³ýË ‹dŸTà ç|ŸjÖ
á Dì+ºq <ŠÖsÁ+ R 98.5 ª 2.5
=
985
 39.4 ¿ì.MT.
25
 ‹dŸTà
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
1 >·+³ýË ç|ŸjÖá Dì+ºq <ŠÖsÁ+ R 39.4 ¿ì.MT.
ç¿ì+~ y{ì“ kÍ~ó+#á+&:
i) 0.45 ÷ 0.9
ii) 2.125 ÷ 0.05
iii) 94.3 ÷ 0.004 iv) 10.25 ÷ 0.2
nuó²«dŸ+ ` 2.3
1) |Ÿ{¿¼ì £ ý˓ U²°\qT “+|Ÿ+&.
–<‘VŸ²sÁD :
2) ç¿ì+~ y{ì“ kÍ~ó+#á+&.
i) 5.51 ÷ 2
vi) 9.99 ÷ 3
uó²>·VäŸ sÁ+
uó²>·|\Ÿ˜ +
362.21 ÷ 10
36.221
5636.1 ÷ 100
______
374.9 ÷ ______
0.3749
______ ÷ 1000
2.0164
123.0 ÷ 100
______
1300.7 ÷ ______
1.3007
______ ÷ 10
59.001
ii) 38.4 ÷ 3
vii) 13 ÷ 6.5
3) ç¿ì+~ |s=Øq• uó²>±VŸäs\qT #ûjTá +&.
i) 78.24 ÷ 0.2
v) 0.026 ÷0.13
ix) 24.4÷ 6.1
ii)
vi)
x)
iii) 57.39 ÷ 6
viii) 10.01 ÷ 11
4.845 ÷ 1.5
4.347 ÷ 0.09
2.164 ÷ 0.008
iii)
vii)
iv) 562.1 ÷ 11
ix) 8 ÷ 0.32
0.246 ÷ 0.6
3.9 ÷ 0.13
v)
x)
0.7005 ÷ 5
320.1 ÷ 33
iv) 563.2 ÷ 2.2
viii) 20.32 ÷ 0.8
4) ç¿ì+~ y{ì“ kÍ~ó+#á+&.
i) 39.54qT 6Ôà uó²Ð+#á+&.
ii) 7.2“ 10Ôà uó²Ð+#á+&.
iii) 5.2“ 1.3Ôà uó²Ð+#á+&.
5) XâKsY Ôáq uÉ¿Õ ™ù |Õ dŸeTyû>+· Ôà 5 >·+³ýË¢ 154.5 ¿ìýËMT³sÁT¢ ç|ŸjÖá Dì+#&ƒT. ÿ¿£
>·+³ýË m+Ôá <ŠÖsÁ+ ç|ŸjÖá Dì+#á>\· &ƒT?
6) ÿ¿£ Ԑ|¾ yûTçd¾ï >Ã&ƒqT “]ˆ+#á&†“¿ì 12.5 sÃEýË¢ 100 >·+³\T |Ÿ“#ûd,ï nÔáqT
sÃEÅ£” m“• >·+³\T |Ÿ“#ûXæ&ƒT?
7) &ƒÈHŽ >·T&ƒT¢ K¯<ŠT ` 61.80 nsTTÔû, ÿ¿£ >·T&ƒT¦ jîTT¿£Ø <ósŠ Á ¿£qT>=q+&.
7e ÔásÁ>·Ü >·DìÔáeTT
70
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
8)
If the price of a tablet strip containing 10 tablets is `26.5, then find the price of
each tablet.
2.4.Rational numbers:
Earlier, we have learnt that some situations like height and depth which are
represented by positive and negative integers respctively. We can represent height of 750 m
3
3
km. and depth of 750m below sea level can be represented as  km.
4
4
Which is neither an integer nor a fraction.There are many situations similar to the above
situation neither integers nor fractions. So, we need to extend our number system to include
such numbers.
above sea level as
p
A number which can be written in the form of q , where p and q
are integers and q  0 is a rational number. The set of rational
numbers is denoted by ‘Q’. A rational number may be positive,
zero or negative.
1/3
Q
Examples : –
0.3
3
7
4.5151...
..., 4, –3, –2, –1 0
9
4
1, 2, 3, 4, 5, ...
N
W
Z
5
2
1
9 5
1
, 1 , –2, 100, ,
, 0, 2 , 0.35, –1.92 etc. are rational numbers.
4
2 9
5
1 5
9
1
,
are negative rational numbers and , 2 are positve rational numbers.
4 9
2
5
p
Any number is in the form of q where both p, q are integers and q  0 is called a rational
number.
3
3
3
or
as 
5
5
5
p
2.4.1 Standard form of rational number: A rational number is said to be in standard form,
q
if p, q are integers having no common divisor (except 1) and q  0.
We may write negative rational numbers
Example :
1 21 6
,
,
etc.
4 8 13
42
Example 11 : Find the standard form of rational number
.
18
Solution : 
42 42  6
7


18 18  6
3
Now 7 and 3 have no common divisors. So, 
VII Class Mathematics
71
If HCF of p and q is 1, then
p
is said to
q
be in the standard form.
rational number
7
42
is standard form to
.
3
18
Fractions, Decimals and Rational Numbers
8) 10 {²uÉ¢{Ù(eÖçÔá)\qT ¿£*Ð –q• {²uÉ¢{Ù çd¾¼|t <óŠsÁ `26.5 nsTTÔû, ÿ¿£ {²uÉ{¢ Ù <ósŠ qÁ T
¿£qT>=q+&.
2.4. n¿£sÁD¡jáT dŸ+K«\T :
mÔáTï eT]jáTT ýËÔáT e+{ì$ esÁTdŸ>± <óqŠ eT]jáTT sÁTD |ŸPs’+¿±\#û dŸÖº+#á‹&ƒÔjáT“
3
eTT+<û Ôî\TdŸTÅ£”H•eTT. dŸeTTç<Š eT{²¼“¿ì ™|qÕ 750 MT. mÔáTqï T _óH•\ýË 4 ¿ì.MT.>±qT,dŸeTTç<Š eT{²¼“¿ì
3
~>·Teq 750MT. ýËÔáTqT  4 ¿ì.MT.>± dŸÖº+#áe#áTÌ. ‚~ |ŸPs’+¿£+ ¿±<ŠT eT]jáTT _óq•+ Å£L&† ¿±<ŠT.
|ŸPs’+¿±\T eT]jáTT _óH•\qT ¿£*Ð –+&ƒ“ “Ôá« J$Ôá |Ÿ]d¾Ôœ Tá \T nHû¿+£ –H•sTT. ¿±‹{ì,¼ n³Te+{ì
dŸ+K«\qT #ûsÌÁ &†“¿ì eTq dŸ+U²« e«edŸœ jîTT¿£Ø $dŸsï D
Á nedŸs+Á . p eT]jáTT q \T |ŸPsÁ’dŸ+K«\T eT]jáTT
q  0 \T
nsTTÔû
p
q
sÁÖ|Ÿ+ýË sjáT>·\ dŸ+K«\qT n¿£sD
Á j
¡ Tá dŸ+K«\T
n+{²sÁT. n¿£sDÁ j
¡ Tá dŸ+K«\ dŸ$Tܓ »Qµ Ôà dŸÖºkÍï+. ÿ¿£ n¿£sDÁ j
¡ Tá dŸ+K«
<óHŠ Ôሿ£+ ýñ<‘ ‹TTD²Ôሿ£+ ýñ<‘ XøSq«+ ¿±y=#áTÌqT.
–<‘VŸ²sÁD\T: – 1 , 1 , –2, 100, 9 , 5 , 0, 2 1 , 0.35, –1.92 yîTT<ŠýqÕÉ $
4
2 9
5
0.3
3
7
4.5151...
..., 4, –3, –2, –1 0
1/3
Q
9
4
Z
1, 2, 3, 4, 5, ...
N
W
5
2
n¿£sD
Á j
¡ Tá dŸ+K«\T.
1 5
,
4 9
@<îÕq dŸ+K«
n+{²sÁT.
p
q
\T ‹TTD n¿£sD
Á j
¡ Tá dŸ+K«\T eT]jáTTT
9
1
,2
2
5
\T <óŠq n¿£sÁD¡jáT dŸ+K«\T.
sÁÖ|Ÿ+ýË eÚ+³Ö p eT]jáTT q \T |ŸPsÁ’dŸ+K«\T nsTT, q  0 nsTTÔû
‹TTD n¿£sD
Á j
¡ Tá dŸ+K«
3
5
ýñ<‘
3
\qT
5
eTq+
3
>±
5
p
q
qT n¿£sD
Á j
¡ Tá dŸ+K«
sjáTe#áTÌ.
2.4.1.n¿£sD
Á j
¡ Tá dŸ+K«\ bç ÍeÖDì¿£ sÁÖ|Ÿ+ : p, q \T –eTˆ& ¿±sÁD²+¿£+ ýñ“ (1$TqVŸä) |ŸPsÁd’ +Ÿ K«\T nsTTÔû
p
p eT]jáTT q jîTT¿£Ø >·.kÍ.uó². 1
n¿£sÁD¡jáTdŸ+K« q (q  0) qT çbÍeÖDì¿£ sÁÖ|Ÿ+ýË –+<Š“
p
nsTTÔû n¿£sD
Á j
¡ Tá dŸ+K« q (q  0)qT
1 21 6
#î|Ο e#áTÌqT. –<‘VŸ²sÁD : 4 , 8 , 13 yîTT<ŠýÉÕq$.
çbÍeÖDì¿£
sÁÖ|Ÿ+ýË –+<Š“ #î|Ο e#áTÌqT.
42
–<‘VŸ²sÁD 11: n¿£sÁD¡jáTdŸ+K« 18 Å£” çbÍeÖDì¿£ sÁÖb͓• ¿£q>=q+&.
kÍ<óŠq :

42 42  6
7


18 18  6
3
‚|Ÿð&ƒT 7, 3\Å£” –eTˆ& ¿±sÁD²+¿±\T ýñeÚ. n+<ŠTe\¢
7e ÔásÁ>·Ü >·DìÔáeTT
72

7
3
nHû~
42
Å£” çbÍeÖDì¿£ sÁÖ|Ÿ+.
18
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
2.4.2 Equivalent rational numbers: We get equivalent rational numbers if we multiply or
divide both numerator and denominator with same number.
1 1 3 3

 ,
2 23 6
1 1 2 2

 ,
2 22 4
So, equivalent rational numbers of
45 45  3 15


,
18
18  3
6
1 1 4 4


2 24 8
1
2 3 4
are , , ,...... etc.
2
4 6 8
45 45  9 5


18
18  9
2
So, equivalent rational numbers of
45
15 5
are
,
,
18
2
6
Example 12 : What are the equivalent rational numbers of
(i )
Solution :
numerator as –20
Rational number =
(i)
(ii) numerator
etc.
5
with
2
as –35?
5
2
To get numerator as –20, we have to multiply numerator and denominators with 4
5 5  4 20


2
2 4
8
(ii)
To get numerator as -35, we have to multiply numerator and denominators
with 7
5 5  7 35


2
2 7
14
Example 13: What are the equivalent rational numbers to
(i)
Solution :
(i )
Denominator as 40
Rational number =
(ii)
3
with
8
Denominator as 800?
3
8
To get denominator as 40, we have to multiply numerator and denominators with 5.
3 3  5 15


8 8  5 40
(ii)
To get denominator as 800, we have to multiply numerator and denominators with 100.
3 3 100 300


8 8 100 800
VII Class Mathematics
73
Fractions, Decimals and Rational Numbers
2.4.2.dŸeÖq n¿£sD
Á j
¡ Tá dŸ+K«\T : n¿£sD
Á j
¡ Tá dŸ+K« jîTT¿£Ø \e+ eT]jáTT VŸäsÁ+ s +&+{ìú ÿ¹¿ dŸ+K«ÔÃ
>·TDì+ºH ýñ<‘ uó²Ð+ºH eTq+ dŸeÖq n¿£sD
Á j
¡ Tá dŸ+K«\qT bõ+<ŠTԐ+.
1 1 3 3

 ,
2 23 6
1 1 2 2

 ,
2 22 4
¿±eÚq,
1
2
Å£” dŸeÖqyîTq® n¿£sD
Á j
¡ Tá dŸ+K«\T
45 45  3 15


,
18
18  3
6
2 3 4
, , ,...... yîTT<ŠýÉÕq$.
4 6 8
45 45  9 5


18
18  9
2
45
Å£” dŸeÖqyîTq® n¿£sD
Á j
¡ Tá dŸ+K«\T 15 , 5 ,
18
2
6
–<‘VŸ²sÁD 12 : (i) \e+ -20 (ii) \e+ -35 >± –+&ƒTq³T¢
¿±eÚq,
5
2
1 1 4 4


2 24 8
yîTT<ŠýÉÕq$.
Å£” dŸeÖq n¿£sD
Á j
¡ Tá dŸ+K«\T ¿£qT>=q+&?
kÍ<óŠq: n¿£sDÁ j
¡ Tá dŸ+K« R
5
2
(i) \e+ -20>± bõ+<Š&+ƒ ¿=sÁŔ£ , eTq+ \e+ eT]jáTT VŸäsÁ+ s +&+{ìú 4Ôà >·TDì+#*à
–+³T+~.
5 5  4 20


2
2 4
8
(ii) \e+ -35>± bõ+<Š&+ƒ ¿=sÁŔ£ , eTq+ \e+ eT]jáTT VŸäsÁ+ s +&+{ìú 7Ôà >·TDì+#*à
5 5  7 35


2
2 7
14
(i) VŸäsÁ+ 40 >±, (ii) VŸäsÁ+
–+³T+~.
–<‘VŸ²sÁD 13 :
3
Å£”
8
800 >± –+&ƒTq³T¢
dŸeÖqyîTq® n¿£sD
Á j
¡ Tá dŸ+K«\T ¿£qT>=q+&?
3
kÍ<óŠq: n¿£sDÁ j
¡ Tá dŸ+K« R 8
(i)
(ii)
VŸäsÁ+ 40>± bõ+<Š&+ƒ ¿=sÁŔ£ , eTq+ 5Ôà \e+ eT]jáTT VŸäsÁ+ s +&+{ìú >·TDì+#*à –+³T+~.
3 3  5 15


8 8  5 40
VŸäsÁ+ 800>± bõ+<‘\+fñ, eTq+ 100Ôà \e+ eT]jáTT VŸäsÁ+ Âs+&+{ì“ >·TDì+#*à
–+³T+~.
3 3 100 300


8 8 100 800
7e ÔásÁ>·Ü >·DìÔáeTT
74
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
1)
Write equivalent rational numbers of
i)
2)
2
5
ii)
5
7
11
and iv)
2
iii)
Find the stanard form of
20
9
40
.
48
Here are equivalent rational numbers containing 1 to 9 digits only once.
One such example is :
another example :
2 3 58
 
6 9 174
2 3 79
 
4 6 158
can you write some more?
2.4.3 Comparing rational numbers: We knew that to compare fractions, they should be like
fractions. If they are unlike, convert them into like fractions by using their equivalent fractions.
Like that we can compare rational numbers too.
For example to compare
3
5
and
, we have to write equivalent rational numbers of both.
4
7
3 6 9 12 15 18 21






=
4
8 12 16
20
24
28
5 10 15 20



=
6
14
21
28
Now, we can compare
.
21
20
and
as they have same denominators.
28
28
20
21
>
28
28

5 3

7
4
7
9
or
?
3
2
To find smaller number, we have to write equivalent rational numbers of both
Example14 : Which is smaller
Solution :
7 14

3
6
9 18 27


2
4
6
Now, we can say that
VII Class Mathematics
27 14

(' –27 is smaller than –14)
6
6
75

9 7

2
3
Fractions, Decimals and Rational Numbers
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
1) ¿ì+~ y{ì¿ì dŸeÖq n¿£sD
Á j
¡ Tá dŸ+K«\qT sjáT+&.
2
5
i)
2)
nHûÇw¾<‘Ý+
40
Å£”
48
ii)
5
7
iii)
11
2
eT]jáTT
iv)
20
9
çbÍeÖDì¿£ n¿£sD
Á j
¡ Tá dŸ+K«qT ¿£qT>=q+&.
1 qT+& 9 n+¿\T ÿ¹¿kÍ] –|ŸjÖî Ð+º dŸeÖq n¿£sDÁ j
¡ Tá dŸ+K«\T ¿ì+<Š ‚eNj&†¦sTT.
n³Te+{ì ÿ¿£ –<‘VŸ²sÁD:
2 3 58
 
6 9 174
eTsÿ£ –<‘VŸ²sÁD
2 3 79
 
4 6 158
MTsÁT eT]¿=“• sjáT>·\s?
2.4.3 n¿£sÁD¡jáT dŸ+K«\qT bþ\Ì&ƒ+ : _óH•\qT bþ\Ì&†“¿ì n$ dŸC²Ü _óH•\T>± –+&†\“ eTqÅ£”
Ôî\TdŸT. ÿ¿£yÞû ø n$ $C²Ü _óH•\T>± –+fñ y{ì dŸeÖq _óH•\qT >·T]ï+º y{ì <‘Çs dŸC²Ü _óH•\T>±
eÖsÁTÌÅ£”“, ÔásTÁ yÔá bþ\ÌeýÉq“ Å£L&† eTqÅ£” Ôî\TdŸT. n¿£sD
Á j
¡ Tá dŸ+K«\qT Å£L&† n<û $<ó+Š >± eTq+
bþ\Ì>·\+.
–<‘VŸ²sÁDÅ£”
3
4
eT]jáTT
5
\qT bþ\Ì&†“¿ì, eTq+ ‡ s +&+{ì¿ì dŸeÖq n¿£sDÁ j
¡ Tá dŸ+K«\T çyjáÖ*.
7
3 6 9 12 15 18 21






=
4
8 12 16
20
24
28
5 10 15 20



=
6
14
21
28
‚|Ÿð&ƒT eTq+
21
28
eT]jáTT
.
20
\Å£”
28
ÿ¹¿ VŸäsÁ+ –q•+<ŠTq bþ\Ìe#áTÌ.
20
21
>
28
28

7
5 3

7
4
9
–<‘VŸ²sÁD 14: 3 , 2 \ ýË @~ ºq•~?
kÍ<óŠq:
ºq• dŸ+K«qT ¿£qT>=q&ƒ+ ¿=sÁŔ£ , eTq+ s +&+{ì¿¡ dŸeÖq n¿£sD
Á j
¡ Tá dŸ+K«\qT çyjáÖ*.
7 14

3
6
9 18 27


2
4
6
‚|Ÿð&ƒT, eTq+ ¿ì+<Š $<ó+Š >± #î|Ο e#áTÌqT.
7e ÔásÁ>·Ü >·DìÔáeTT
27 14

(' –27, –14 ¿£+fñÉ
6
6
76
ºq•~)

9 7

2
3
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
1)
Find which is bigger?
i)
2)
2
or 0
5
2
1
or
3
3
ii)
Find which rational number is smaller
iii)
5
7
or
2
8
13
17
or
?
4
6
Any negative rational number is smaller than zero and all positive rational numbers
2.4.4 Rational numbers on number line :
Abhiram drew a number line and represented integers on it. His classmate Akhila said that
the numbers which are on that number line are rational numbers too. Do you agree with her?
–2
–3
–1
0
1
2
3
All integers are rational numbers but all rational numbers need not integers
Rational numbers can be represented on number line by just following simple steps:
Ø
Before going to the number line don’t forget to check the negative or positive sign of the
rational number.
Ø
Positive rational numbers are always represented on the right side of the zero on the
number line.
Ø
Negative rational numbers are always represented on the left side of the zero on the
number line.
Ø
Rational numbers which are proper fractions always lie in between 0 to –1 or 0 to 1.
Ø
Rational numbers which are improper fractions always lie less than –1 (if they have negative
sign) or lie greater than 1 (if they have positive sign).
Ø
If the rational number which are improper fractions,we have to change them into mixed
fractions, to find between which whole numbers, the rational number exists.
Example15 : Represent
3
on number line
4
Solution :
Step 1:
3
is a positive number so, it lies on the right side of the zero on number line.
4
Step 2:
Since,
3
is a proper fraction (numerator lesser than denominator), it lies in between 0 and 1.
4
VII Class Mathematics
77
Fractions, Decimals and Rational Numbers
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
1) @~ ™|<ŠÝ<à ¿£qT>=q+&.
2)
i)
2
5
13
4
ýñ<‘
ýñ<‘
0
2
3
ii)
17
6
ýñ<‘
1
3
iii)
5
2
ýñ<‘
7
8
\ýË @ n¿£sD
Á j
¡ Tá dŸ+K« ºq•<à ¿£qT>=q+&.
@<îÕH ‹TTD n¿£sD
Á j
¡ Tá dŸ+K«, dŸTH• ¿£+fñ eT]jáTT n“• <óqŠ n¿£sD
Á j
¡ Tá dŸ+K«\ ¿£+fñ ºq•~.
2.4.4 dŸ+U²« ¹sK™|Õ n¿£sÁD¡jáT dŸ+K«\T :
n_ósyŽT dŸ+U²« s¹ K ^d¾ <‘“™|Õ |ŸPsÁ’ dŸ+K«\qT >·T]ï+#&ƒT. € dŸ+U²« s¹ K™|Õ –q• dŸ+K«\T n¿£sD
Á j
¡ Tá
dŸ+K«\T Å£L&ƒ neÚԐjáT“ n“ Ôáq ¿±¢dyt Tû {Ù nÏ\ #î|ξ +~. MTsÁT €yîTÔà @¿¡u$„ó kÍïs?
–3
–2
–1
0
1
2
3
n“• |ŸPsÁ’ dŸ+K«\T n¿£sD
Á j
¡ Tá dŸ+K«\T ¿±ú n“• n¿£sD
Á j
¡ Tá dŸ+K«\T |ŸPsÁ’ dŸ+K«\T ¿±qedŸs+Á ýñ<TŠ .
¿=“• dŸsÞÁ yø Tî q® kþbÍH\qT bÍ{ì+#á&+ƒ <‘Çs n¿£sD
Á j
¡ Tá dŸ+K«\qT dŸ+U²« s¹ K™|Õ Ôû*¿£>± #áÖ|¾+#áe#áTÌ.
Ø dŸ+K«¹sKÅ£” yîÞ&
¢ø †“¿ì eTT+<ŠT n¿£sD
Á j
¡ Tá dŸ+K« jîTT¿£Ø ‹TTD ýñ<‘ <óŠq >·TsÁTïqT >·T]ï+#á&ƒ+ eT]Ìbþe<ŠTÝ.
Ø <óH
Š Ôሿ£ n¿£sD
Á j
¡ Tá dŸ+K«\T m\¢|ڟ Î&ƒÖ dŸ+U²« s¹ K™|Õ dŸTH•Å£” Å£”&y|Õî ڟ q eÚ+{²sTT.
Ø ‹TTD n¿£sD
Á j
¡ Tá dŸ+K«\T m\¢|ڟ Î&ƒÖ dŸ+U²« s¹ K™|Õ dŸTH• Å£” m&ƒeTyî|Õ ÚŸ q eÚ+{²sTT.
Ø ç¿£eT _óH•\T>± –+&û n¿£sD
Á j
¡ Tá dŸ+K«\T m\¢|ڟ Î&ƒÖ 0 qT+& -1 ýñ<‘ 0 qT+& 1 \ eT<ó«Š ýË –+{²sTT.
Ø n|Ÿç¿£eT _óH•\T >± n¿£sÁD¡jáT dŸ+K«\T ‹TTD >·TsÁTï ¿£*Ð –q•³¢sTTÔû -1 ¿£+fñ ÔáŔ£ Øe>± –+{²sTT ýñ<‘
ÿ¿£yÞû ø n$ <óHŠ Ôሿ£ >·TsÁTï ¿£*Ð –q•³¢sTTÔû 1 ¿£+fñ mÅ£”Øe>± –+{²sTT.
Ø n|Ÿç¿£eT _óH•\T>± eÚ+&û n¿£sD
Á j
¡ Tá dŸ+K«\T –q•³¢sTTÔû, y{ì“ eTq+ $TçXøeT _óH•\T>± eÖ]Ìq#Ã, n$
@ |ŸPsÁ’ dŸ+K«\ eT<ó«Š –+&ƒTHà dŸT\uó+„ >± ¿£qT>=qe#áTÌqT.
–<‘VŸ²sÁD 15 :
3
qT
4
dŸ+U²« s¹ K™|Õ >·T]ï+#á+&.
kÍ<óŠq :
kþbÍq+ 1:
3
4
nHû~ <óŠq dŸ+K«, dŸ+U²« ¹sK™|Õ dŸTH• Å£” Å£”&yîÕ|ŸÚq –+³T+~.
kþbÍq+ 2:
3
4
nHû~ ç¿£eT _óq•+ ¿£qT¿£ (VŸäsÁ+ ¿£+fñ \e+ ÔáŔ£ Øe), ‚~ 0 eT]jáTT 1 \ eT<ó«Š –+³T+~.
7e ÔásÁ>·Ü >·DìÔáeTT
78
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
Step 3:
we have to divide number line in to four equal parts in between 0 and 1 and third part of the
four parts is represented as
Example16: Represent
3
.
4
5
on number line
2
Solution :
Step 1:
Step 2:
Step 3:
5
is a negative number. So, it lies on the left side of the zero on number line.
2
5
5
1
is an improper fraction, it should be changed into mixed fraction
= 2 so,
2
2
2
it lies between –2 and –3.
Since
We have to divide number line in between –2 to –3 into two equal parts and first part of the
1
two parts is represented as 2 .
2
–5
2
–3
–2
1
2
–2
1)
–1
0
1
Represent these rational numbers on number line.
5
8
i)
ii)
9
4
iii)
11
2
iv)
21
8
Exercise - 2.4
1)
Write the equivalent rational numbers of 
i)
2)
ii)
–20
Write the equivalent rational numbers of
i)
3)
–15
36
ii)
5
with numerator as given below ,
3
4
with denominator as given below,
9
90
Identify equivalent rational numbers in the following.
2 5 0 4 10 8 2 6 15 0
, , ,
, , ,
, , ,
3 4 3 6 8 12 3 9 12 5
VII Class Mathematics
79
Fractions, Decimals and Rational Numbers
kþbÍq+ 3: 0 eT]jáTT 1 eT<ó«Š dŸ+U²«¹sKqT H\T>·T dŸeÖq uó²>±\T>± #ûd¾, H\T>·T uó²>±\ jîTT¿£Ø eTÖ&Ã
uó²>·+qT
3
>±
4
>·T]ïkÍïeTT.
–<‘VŸ²sÁD 16 : dŸ+U²«¹sK™|Õ
kÍ<óŠq :
5
qT
2
>·T]ï+#á+&?
kþbÍq+ 1:
5
nHû~
2
kþbÍq+ 2:
5
5
1
nHû
~
n|Ÿ
ç
¿£
e
T
_ó
q
•+
¿£
q
T¿£
,
B““
$TçXø
e
T
_ó
q
•+>±
eÖsÌ*.
= 2
2
2
2
kþbÍq+ 3:
‹TTD dŸ+K«. ¿±eÚq, dŸ+U²«¹sK™|Õ dŸTH• Å£” m&ƒeTyî|Õ ÚŸ q –+³T+~.
n+<ŠTe\¢, ‚~
-2 eT]jáTT -3 eT<ó«Š –+³T+~.
eTq+ -2 qT+º -3 eT<ó«Š dŸ+U²«¹sKqT s +&ƒT dŸeÖq uó²>±\T>± $uó›„ kÍï+ eT]jáTT s +&ƒT
uó²>±\ ýË yîTT<Š{ì uó²>±“•
2
–5
2
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
–3
–2
1
2
–2
1
>±
2
>·T]ïkÍï+.
–1
0
1
1) ‡ ¿ì+~ n¿£sD
Á j
¡ Tá dŸ+K«\qT dŸ+U²« s¹ K™|Õ >·T]ï+#á+&?
i)
5
8
ii)
9
4
iii)
11
2
iv)
21
8
nuó²«dŸ+ ` 2.4
1) ¿ì+<Š ‚ºÌq \e+ e#áTÌq³T¢
i)
-15

5
Å£”
3
ii)
dŸeÖq n¿£sD
Á j
¡ Tá dŸ+K«\qT sjáT+&.
-20¢
4
2) ¿ì+<Š ‚ºÌq VŸäsÁ+ e#áTÌq³T¢ 9 Å£” dŸeÖq n¿£sD
Á j
¡ Tá dŸ+K«\qT sjáT+&.
i)
36
ii) 90
3) ç¿ì+~ y{ìýË dŸeÖq n¿£sD
Á j
¡ Tá dŸ+K«\qT >·T]ï+#á+&.
2 5 0 4 10 8 2 6 15 0
, , ,
, , ,
, , ,
3 4 3 6 8 12 3 9 12 5
7e ÔásÁ>·Ü >·DìÔáeTT
80
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
4)
Write the following rational numbers in ascending order.
4 2 1
2
, , , 0,
9 9 3
3
3 1 9 11
, , ,
on the same number line.
5 5 5 5
5)
Represent
6)
Compare the following pair of rational numbers.
i)
7)
8)
2
3
 ,
7
7
5 5
,
9 8
ii)
iii)

13 7
,
12
6
Which of the following is False?
1.
Every natural number is a rational number but every rational number need not be a natural
number.
2.
Every rational number is an integer but every integer need not be a rational number.
3.
Every integer is a rational number but every rational number need not be an integer.
4.
Every fraction is a rational number but every rational number need not be a fraction.
Ramya said 
3
is in between –1 and –2 on number line. Do you agree with Ramya? Why?
2
You know that different operations with the same pair of rational
numbers usually give different answers. Observe the following
calculations which are some interesting exceptions in rational numbers.
1)
11 11 11 11
  
6 5 6 5
2)
169 13 169 13
 

30 15 30 15
Can you tell some more like these?
2.5 Word problems on rational numbers:
Pranavi’s father brought a fruit box from market, which contains three types of fruits weighing
73
58
19
kg in all. If
kg of these are apples,
kg are oranges and the rest are grapes. What is the
9
3
6
weight of the grapes in the box?
Let us learn, how to find the weight of the grapes
Weight of a box =
Weight of apples =
VII Class Mathematics
58
kg
3
73
kg
9
81
Fractions, Decimals and Rational Numbers
4) ç¿ì+~ n¿£sD
Á j
¡ Tá dŸ+K«\qT €sÃVŸ²D ç¿£eT+ýË sjáT+&.
4 2 1
2
, , , 0,
9 9 3
3
5) ÿ¹¿ dŸ+U²« s¹ K™|Õ
3 1 9 11
, , ,
5 5 5 5
\qT >·T]ï+#á+&.
6) ç¿ì+~ n¿£sD
Á j
¡ Tá »dŸ+K«\ ÈÔáµ \qT bþ\Ì+&.
2
3
 ,
7
7
i)
ii)
5 5
,
9 8
iii)

13 7
,
12
6
7) ‡ ¿ì+~y{ìýË @~ ndŸÔ«á +?
1.ç|ŸÜ dŸV²Ÿ È dŸ+K« n¿£sDÁ j
¡ Tá dŸ+K« neÚÔáT+~, ¿±“ ç|ŸÜ n¿£sDÁ j
¡ Tá dŸ+K« dŸV²Ÿ È dŸ+K« ¿±qedŸs+Á ýñ<TŠ .
2.ç|ŸÜ n¿£sDÁ j
¡ Tá dŸ+K« ÿ¿£ |ŸPsÁd’ +Ÿ K« neÚÔáT+~, ¿±“ ç|ŸÜ |ŸPsÁd’ +Ÿ K« n¿£sDÁ j
¡ Tá dŸ+K« ¿±qedŸs+Á ýñ<TŠ .
3.ç|ŸÜ |ŸPsÁd’ +Ÿ K« ÿ¿£ n¿£sD
Á j
¡ Tá dŸ+K« neÚÔáT+~, ¿±“ ç|ŸÜ n¿£sD
Á j
¡ Tá |ŸPsÁd’ +Ÿ K« ¿±qedŸs+Á ýñ<TŠ .
4.ç|ŸÜ _óq•+ ÿ¿£ n¿£sD
Á j
¡ Tá dŸ+K« neÚÔáT+~, ¿±“ ç|ŸÜ n¿£sD
Á j
¡ Tá dŸ+K« _óq•+ ¿±qedŸs+Á ýñ<TŠ .
8)

3
dŸ+U²«
2
s¹ K™|Õ -1 eT]jáTT -2 \ eT<ó«Š –+³T+~ n“ sÁeT« nq•~. sÁeT«Ôà MTsÁT @¿¡u$„ó dŸTqï •s?
m+<ŠTÅ£”?
ÿ¹¿ ÈÔá n¿£sD
Á j
¡ Tá dŸ+K«\Ôà yûsTÁ yûsTÁ . ç|Ÿç¿ìjTá \Å£” yûsTÁ yûsTÁ dŸeÖ<ó‘H\T
–+{²jáT“ MTÅ£” Ôî\TdŸT. ¿=“• n¿£sD
Á j
¡ Tá dŸ+K«ýË¢ €dŸ¿¿ïì s£ yÁ Tî q® $TqVŸäsTT+|ŸÚ\T
eÚ+{²sTT. ‡ ~>·Te ç|Ÿç¿ìjTá \qT >·eT“+#á+&.
€ý˺<‘Ý+!
1)
11 11 11 11
  
6 5 6 5
2)
169 13 169 13
 

30 15 30 15
‚³Te+{ì$ eT]¿=“• MTsÁT #î|Ο >·\s?
2.6n¿£sÁD¡jáT dŸ+K«\™|Õ |Ÿ<Š dŸeTdŸ«\T:
ç|ŸD$ Ôá+ç&
58
3
¿ìýËç>±eTT\ ‹sÁTeÚq• eTÖ&ƒT sÁ¿±\ |Ÿ+&ƒq¢ T ¿£*Ð –q• |Ÿ+&ƒ¢ ™|fÉq¼ T eÖÂsØ{Ù qT+º
rdŸT¿=#ÌsÁT. M{ìýË (
73
9
¿ì.ç>±\T €|¾ýÙà nsTTÔû,
19
¿ì.ç>±\T
6
H]+È, $TÐ*q$ ç<‘¿£|Œ +Ÿ &ƒT¢ nsTTÔû
™|fÉý¼ Ë eÚq• ç<‘¿£|Œ +Ÿ &ƒT¢ jîTT¿£Ø ‹sÁTeÚ m+Ôá?
ç<‘¿£Œ |Ÿ+&ƒ¢ ‹sÁTeÚqT mý² ¿£qT>=HýË Ôî\TdŸTÅ£”+<‘+.
™|fɼ ‹sÁTeÚ =
58
3
€|¾ýÙà ‹sÁTeÚ =
7e ÔásÁ>·Ü >·DìÔáeTT
¿ì.ç>±.
73
9
¿ì.ç>±.
82
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
Weight of oranges =
19
kg
6
Total weight of apples and oranges =
73 19

9
6

146 57 203


kg
18
18
18
 Weight of grapes = Weight of box – Weight of apples and oranges
=
58 203 348 203 348  203 145





kg.
3 18
18
18
18
18
Example 17 : An examination has 10 questions. Each correct answer gets 1mark and for each wrong
1
negative mark. If Kalisha done 5 questions right and 5 questions wrong,
2
what would be her total marks?
answer
Solution :
For the 5 correct answers, she got marks
=5  1=5
5
 1
For remaining 5 wrong answers, she got marks = 5      
2
 2
10  5 5
1
 5
 2
So, total marks Kalisha got = 5      
2
2
2
 2
1.
How much should be subtracted from
2.
Divide sum of
5
7
to get  ?
6
2
5
 3
9
and    by .
2
2
 2
Exercise - 2.5
1.
The temperature is raised by
2.
If the cost of
45 0
 4
C from    0C. What would be the new temperature?
14
 7
5
metres of cloth is `235. What is the cost of 1 metre of cloth?
2
VII Class Mathematics
83
Fractions, Decimals and Rational Numbers
H]+È\ ‹sÁTeÚ =
19
6
¿ì.ç>±.
73 19

9
6
n|¾ýÙà eT]jáTT H]+È |Ÿ+&ƒ¢ jîTT¿£Ø yîTTÔá+ï ‹sÁTeÚ =

 ç<‘¿£Œ|Ÿ+&ƒ¢
‹sÁTeÚ
= ™|fɼ
=
146 57 203


18
18
18
¿ì.ç>±.
‹sÁTeÚ – n|¾ýÙà eT]jáTT H]+È |Ÿ+&ƒ¢ jîTT¿£Ø yîTTÔá+ï ‹sÁTeÚ
58 203 348 203 348  203 145





3 18
18
18
18
18
¿ì.ç>±.
–<‘VŸ²sÁD 17: ÿ¿£ |Ÿ¯¿£ýŒ Ë 10 ç|ŸX•ø \T –H•sTT. ç|ŸÜ dŸs qÕ dŸeÖ<ó‘H“¿ì 1eÖsÁTØ eT]jáTT ç|ŸÜ Ôá|ڟ Î
1
dŸeÖ<ó‘H“¿ì 2 ‹TTDeÖsÁTØ –+³T+~. ÿ¿£yÞû ø K©cÍ 5 ç|ŸX•ø \Å£” dŸs qÕ dŸeÖ<ó‘H\T eT]jáTT
5 ç|ŸX•ø \Å£” Ôá|ڟ Î>± dŸeÖ<ó‘H\T sd¾ –q•³¢sTTÔû, €yîT bõ+~q yîTTÔá+ï eÖsÁTØ\T m“•?
5 dŸs qÕ dŸeÖ<ó‘H\Å£” €yîT bõ+~q eÖsÁTØ\T
= 5  1=5
kÍ<óŠq:
$TÐ*q 5 Ôá|ڟ Î dŸeÖ<ó‘H\Å£” €yîT bõ+~q eÖsÁTØ\T = 5    1    5
 2
 K©cÍ
nHûÇw¾<‘Ý+
2
5
10  5 5
1
 2
 
2
2
2
 2
bõ+~q yîTTÔá+ï eÖsÁTØ\T = 5   
1.
7
2
qT+& m+Ôá rd¾ydû q¾
2.
5
2
eT]jáTT
 3
 
 2

5
6
edŸTï+~?
jîTT¿£Ø yîTTԐ
9
2
Ôà uó²Ð+#á+&.
nuó²«dŸ+ ` 2.5
1. –cþ’ç>·Ôá
2.
5
MT³sÁ¢
2
 40
  C
 7
qT+&
45 0
C ™|]Ð+~.
14
ç|ŸdŸTïÔá –cþ’ç>·Ôá m+Ôá?
‹³¼ K¯<ŠT `235 nsTTq ÿ¿£ MT³sÁT ‹³¼ K¯<ŠT m+Ôá?
7e ÔásÁ>·Ü >·DìÔáeTT
84
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
3.
4.
161
1
. What is the cost of 5 litres of diesel?
2
2
An examination has 20 questions. Each correct answer gets 1 mark and each wrong answer
One litre of diesel costs `
1
negative mark. If Kundan writes 13 right answers and 7 wrong answers, what would
2
be his total marks?
gets
5.
 3
One day in the winter season, the evening temperature in New Yark city was    0C. If
 2
the temperature further fell down 4 more degrees at mid-night. Then, find the temperature at
mid-night.
6.
Simplify the following using BODMAS rule :
7.
Divide the sum of
4 2 5 4 4
    .
3 5 3 9 3
4
2
3
8
and by the product of
and .
5
5
10
5
Make two dice with cartboard or wood. Paste colour chart paper to the all faces to the each
dice. Write any three positive and three negative rational numbers on faces of each dice. Now
in group each time two students throw the dice who are sitting oppostie side to each other. Write
the come up number on the dice in the table given and do the four fundamental operations with
those two rational numbers. Submit the filled table to the teacher.
Rational number-1 (come Rational number-2
up on the dice 1)
(come up on the dice 2)
Eg : 2/3
5/3
Addition
Subtraction Multiplication Division
2/3 + 5/3 =
2/3–5/3
7/3
= –3/3 = –1
2/3 x 5/3
= 10/9 = 5/3
John Napier of Merchiston (UK) was born in the year 1550. He started his formal education
at the age of 13 as was the common tradition of that time. However he soon dropped out
of school and travelled to Europe. He returned to Scotland at the age of 21.
John Napier is founder of logarithms and he became so after spending long hours, doing
lengthy calculations of astronomy.
He also invented the so-called ‘Napier’s strips’ which were devices that could be used as
calculators.
Napier also made improvements to the idea of the decimal fraction by starting the use of
decimal point, a practice that very soon became common throughout Britain.
Napier was widely recognised for his work in mathematics and astronomy. He died in the
year 1617.
VII Class Mathematics
85
2/3  5/3
= 2/5
John Napier
1550 - 1617
Fractions, Decimals and Rational Numbers
3. ÿ¿£ ©³sÁT &ž›ýÙ <ósŠ Á `
161
2
nsTTÔû
5
1
2
©³sÁ¢ &ž›ýÙ <ósŠ Á m+Ôá?
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1
2
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bõ+~q yîTTÔá+ï eÖsÁTØ\T m+Ôá?
5. oԐ¿±\+ýË ÿ¿£ sÃE, qÖ«jáÖsYØ q>·s+Á ýË kÍjáT+çÔá+ –cþ’ç>·Ôá
cþ’ç>·Ôá 40C ÔáÐ+Z ~. nsÁsÆ çÜ dŸeTjáT+ýË –cþ’ç>·Ôá qT ¿£qT>=q+&.
6.
BODMAS
7.
4
5
“jáTeT+ –|ŸjÖî Ð+º ~>·Te y{ì“ dŸÖ¿¡ˆŒ ¿£]+#á+&:
eT]jáTT
2
\
5
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yîTTÔá+ï qT,
3
10
eT]jáTT
8
5
 3 0
   C.
 2
nsÁsÆ çÜ njûT«dŸ]¿ì –
4 2 5 4 4
    .
3 5 3 9 3
\ \‹Æ+Ôà uó²Ð+#á+&.
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bͺ¿£\ ™|uÕ ²ó >·+ –q• dŸ+K«\qT çyd¾, € s +&ƒT n¿£sDÁ j
¡ Tá dŸ+K«\Ôà #áÔTá ]Ç<óŠ |Ÿ]ç¿ìjTá \qT #ûjTá +&. Èy‹T\qT
|Ÿ{켿£ýË “+|Ÿ+&. “+|¾q |Ÿ{쿼 £qT MT {¡#ásYÅ£” dŸeT]Î+#á+&.
n¿£sDÁ j¡ Tá dŸ+K« - 2
n¿£sDÁ j
¡ Tá dŸ+K« -1
dŸ+¿£\q+ e«e¿£\q+ >·TD¿±sÁ+ uó²>·V²Ÿ sÁ+
(bͺ¿£ -1 ™|Õ eÚq• dŸ+K«) (bͺ¿£ -2 ™|Õ eÚq• dŸ+K«)
Eg : 2/3
5/3
2/3 + 5/3 =
2/3–5/3
7/3
= –3/3 = –1
2/3 x 5/3
= 10/9 = 5/3
2/3  5/3
= 2/5
#]çÔá¿£ n+XøeTT
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Ü]Ð e#Ì&ƒT. »»C²HŽ Hû|j
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eÚq• dŸeTdŸ«\qT n~ó>$· T+#û ç¿£eT+ýË »»ý²>·]<¸+Š µµ\qT ¿£qT>=H•&ƒT. nÔáqT »»Hû|j
¾ Tá sY |Ÿ{\¼¡ Tµµ n“
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¾ Tá sYµµ <ŠXæ+Xø
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ç_³HŽ n+Ôá{² #ý² kÍ<ó‘sÁDyîT+® ~. »»Hû|j
¾ Tá sYµµ >·DÔì +á eT]jáTT K>ÃÞø Xæçd+ïŸ ýË #ûdq¾ ¿£w¾¿ì ç|ԟ «û ¿£+>±
>·T]ï+|ŸÚ bõ+<‘&ƒT. nÔáqT 1617 dŸ+eÔáàsÁ+ýË eTsÁD+ì #&ƒT.
7e ÔásÁ>·Ü >·DìÔáeTT
86
John Napier
1550 - 1617
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
Fill the boxes with suitable words. Clues are given in the below :
CROSS WORD PUZZLE
DOWN:
ACROSS :
p
form where p, q are integers
q
1.
a number in
3.
and q  0.
type of rational numbers which are left side of
zero on the number line.
5.
place of 4 in 39.146
6.
another form of fractions with denominators
10, 100, 1000, .
1.
Choose the correct answer?
i)
type of rational numbers
4.
type of fractions which are same
denominator.
7.
the dot or point which separates whole
number part and fractional part
8.
moving side of a decimal point when a
decimal multiply with 10.
The set of rational numbers are denoted by
a)
ii)
1 2 3
,
,
.
3 6 12
2.
N
b)
W
c)
Z
d)
Q
Number of decimal places in the product of 48.23 x 0.2
a)
2
b)
3
c)
1
d)
5
iii) Number of decimal places to the quotient of 537.1 ÷10
a)
1
b)
2
c)
4
d)
3
iv) A rational number can be
a)
negative
VII Class Mathematics
b)
positive
c) zero
87
d)
all the above
Fractions, Decimals and Rational Numbers
¿ì+<Š ‚eNj&q €<ó‘s\Ôà >·&Tƒ \qT dŸ]jîT® q |Ÿ<‘\Ôà “+|Ÿ+&.
|Ÿ›ýÙ fÉ®yŽT
>·DìÔá |Ÿ<Š ¹¿[
¿ì+<Š ‚eNj&q €<ó‘s\Ôà >·&Tƒ \qT dŸ]jîT® q |Ÿ<‘\Ôà “+|Ÿ+&.
“\TeÚ :
n&ƒ+¦ :
1) ÿ¹¿ VŸäsÁ+ ¿£*Ð eÚq• n¿£sD
Á j
¡ Tá dŸ+K«\T
1 2 3
,
,
1)
e+{ì _óH•\T
3
6
12
2) 10, 100, 1000, ... e+{ì$ VŸäsÁ+ >± >·\ _óH•\T
2) <ŠXæ+Xø _óq•+ýË |ŸPs’+¿£ uó²>±“•, <ŠXæ+Xø uó²>±“• yûsTÁ #ûd #áT¿£Ø
3) @<îHÕ  <ŠXæ+Xø dŸ+K« qT10 #û uó²Ð+ºq|ŸÚ&ƒT
5) 39.149 nHû <ŠXæ+Xø _óq•+ ýË 4 eÚq• k͜q+
4) dŸ+U²« s¹ K ™|Õ »»0µµ Å£” m&ƒeT yî|Õ ÚŸ q >·\$
6) p, q \T |ŸPsÁ’ dŸ+K«\T eT]jáTT q  0 nsTTÔû,
<‘“ <ŠXæ+Xø _+<ŠTeÚ m³T yîÕ|ŸÚ È]>·TqT?
p
sÁÖ|Ÿ+ ýË >·\ dŸ+K«\T.
q
jáT֓{Ù nuó²«d+Ÿ
1. dŸs qÕ dŸeÖ<ó‘q+ m+#áT¿Ã+&?
i) n¿£sD
Á j
¡ Tá dŸ+K«\ dŸ$Tܓ @ n¿£sŒ +Á Ôà dŸÖºkÍïsTÁ ?
ii)
iii)
iv)
a)
N
b)
W
c)
Z
d)
Q
a)
2
b)
3
c)
1
d)
5
a)
1
b)
2
c)
4
d)
3
48.23 I 0.2 jîTT¿£Ø \‹Æ+ ýË <ŠXæ+Xø uó²>·+ ý˓ n+¿\ dŸ+K«
537.1 ª 10 jîTT¿£Ø uó²>·|ýŸ˜ ²“¿ì <ŠXæ+Xø uó²>·+ ý˓ n+¿\ dŸ+K«
m<îHÕ  ÿ¿£ n¿£sD
Á j
¡ Tá dŸ+K«.............>± eÚ+&ƒe#áTÌ.
a) sÁTD²Ôሿ£+ b)
<óHŠ Ôሿ£+
c)
dŸTq•
7e ÔásÁ>·Ü >·DìÔáeTT
88
d)
™|Õ eú•
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
2.
Fill in the blanks:
i)
The numbers written in the form of
p
, where p, q are integers and q  0 are
q
..numbers
ii)
3)
4)
0.11  0.11 =
..
iii)
Standard form of 
15
=
6
.
2
iv) Equivalent fraction to  =
3
Find the product:
i) 2.1 × 6.3
ii) 43.205 × 1.27 iii) 7.641 × 3.5
iv) 5.24 × 0.99
Solve the following:
i) 61.24 ÷ 0.4 ii) 23.45 ÷ 1.5
iii)
0.312 ÷ –0.6 iv) 32.2 ÷ 2.2
5)
1
Mutiply 0.04 by  .
2
6)
Standard form of 
7)
A Bus travelled 300 km in 7
15
.
35
1
hours with uniform speed. Find how many kilometres it travelled in
2
1 hour?
8)
9)
1
1
of her money on notebooks and of the remaining on statio3
4
nery items. How much money is left with her?
Suvarna had `300. She spent
One litre of diesel costs `84.65. What is the cost of 12.5 liters of diesel?
10) Write any five equivalent rational numbers of
11) Represent
9
.
5
2 3 1 3
, , , on same number line.
5 5 5 5
13
1
12) Find which rational number is smaller 1 or  .
6
4
1.
To add or subtract fractions they must have same denominator (like fractions).
2.
For multiplication of fractions, we simply multiply numerators and multiply denominators.
3.
To divide one fraction by another fraction, we have to multiply one fraction with the reciprocal
of the another fraction.
VII Class Mathematics
89
Fractions, Decimals and Rational Numbers
2. U²°\qT “+|Ÿ+&:
p
, p,q \T
q
i)
ii)
|ŸPsÁ’dŸ+K«\T eT]jáTT q
0.11 I 0.11
iv)
 0 sÁÖ|Ÿ+ýË
iii)

sjáT‹&ƒ¦ dŸ+K«\T...............dŸ+K«\T
15
6
Å£” çbÍeÖDì¿£ sÁÖ|Ÿ+ R
2
Å£” dŸeÖqyîTq® _óq•+ R
3

3) \u²Æ“• ¿£qT>=q+&:
i)
2.1 × 6.3
ii)
43.205 × 1.27 iii) 7.641 × 3.5
iv) 5.24 × 0.99
4) ç¿ì+~ y{ì“ kÍ~ó+#á+&.
i)
61.24 ÷ 0.4
5) 0.04qT
6)

23.45 ÷ 1.5
iii)
0.312 ÷ –0.6 iv)
32.2 ÷ 2.2
1
Ôà >·TDì+#á+&.
2

15
jîTT¿£Ø
35
7) ÿ¿£ ‹dŸTà
ii)
7
çbÍeÖDì¿£ sÁÖ|Ÿ+.
1
2
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ç|ŸjÖá Dì+ºq<à ¿£qT>=q+&.
8) dŸTesÁ’ <Š>·ZsÁ `300 eÚH•sTT. €yîT Ôáq <Š>sZ· Á eÚq• &ƒ‹TÒýË
1
3
e+ÔáT “ HÃ{Ù |ŸÚdŸ¿ï ±\ ¿=sÁŔ£ eT]jáTT
1
$TÐ*q &ƒ‹TÒýË 4 e+ÔáT dw¼ qŸ ¯ edŸTeï Ú\ ¿=sÁŔ£ KsÁTÌ ™|{ì+¼ ~. €yîT e<ŠÝ m+Ôá &ƒ‹TÒ $TÐ* –+~?
9) ÿ¿£ ©³sÁT &ž›ýÙ <ósŠ Á `84.65 nsTTq 12.5 ©³sÁ¢ &ž›ýÙ K¯<ŠT m+Ôá?
10)
9
Å£”
5
@yîHÕ  ×<ŠT dŸeÖq n¿£sD
Á j
¡ Tá dŸ+K«\qT çyjáT+&.
11) ÿ¹¿ dŸ+U²« s¹ K MT<Š
12)
1
1
4
ýñ<‘

13
6
2 3 1 3
, , ,
5 5 5 5
\qT >·T]ï+#á+&.
\ýË @ n¿£sD
Á j
¡ Tá dŸ+K« ºq•~.
>·TsÁTï+#áT¿Ãy*àq
n+Xæ\T
1. _óH•\qT ¿£\|Ÿ&†“¿ì ýñ<‘ rd¾yÔû á #ûjTá T³Å£” n$ ÿ¹¿ VŸäsÁ+ ¿£*Ð –+&†*.(dŸC²Ü _óH•\T)
2. _óH•\ >·TD¿±sÁ+ #ûjTá T³¿=sÁŔ£ , eTq+ y{ì \y\qT eT]jáTT VŸäs\qT >·TDìkÍï+.
3. ÿ¿£ _óH•“• eTsà _óq•+Ôà uó²Ð+#\+fñ, ÿ¿£ _óH•“• eTsà _óq•+ jîTT¿£Ø >·TD¿±sÁ $ýËeT+ Ôà >·TDì+#*à
–+³T+~.
7e ÔásÁ>·Ü >·DìÔáeTT
90
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
4.
To multiply decimal numbers by 10, 100, 1000, shift the decimal point in the product as
many places to the right as number of zeroes after 1.
5.
The number of decimal digits in the product of any two decimal numbers is equal to the sum
of decimal digits that are multiplied.
6.
While dividing a decimal number by 10, 100 or 1000, the digits of the number and the
quotient are same but the decimal point in the quotient shifts to the left by as many places as
there are zeros 1 after.
7.
A rational number
8.
divisor and q  0.
Before going to the number line don't forget to check the negative or positive sign of the
rational number.
9.
Rational numbers which are proper fractions are always lie in between 0 to 1 or 0 to –1.
p
is said to be standard form, if p, q are Integers having no common
q
10. Rational numbers which are improper fractions are always lie greater than 1(if they have
positive sign) or lie less than –1 (if they have negative sign).
11. If the rational numbers which are improper fractions,we have to change them in to mixed
fractions, to find between which whole numbers the rational number exists.
Number Series - 2
1.
2.
3.
4.
Adding or subtract of natural numbers:
Eg: 6, 7, 9, 12, 16, 21,
Explanation: (6+1), (7+2), (9+3), (12+4), (16+5)
a) 21 b) 25 c) 27 d) 28
so, next number is (21 + 6) = 27
Add the pattern:
Eg: 10, 20, 40, 70, 110,
Explanation: (10+10), (20+20), (40+30), (70+40)
a) 160 b) 180 c) 150 d) 210
so, next number is (110 + 50) = 160
Subtracting or adding odd numbers:
Eg: 27, 26, 23, 18, 11,
Explanation: (27 – 1), (26 – 3), (23 – 5), (18 – 7)
a) 4 b) 2 c) 9 d) 5
so, next number is (11 – 9) = 2
Multiply with a fixed number
Eg: 5, 15, 45, 135, 405,
a) 1200 b) 1215 c) 850 d) 925
5.
Explanation: (5  3), (15  3), (45  3),(135  3),
so, next number is (405  3) = 1215
Multiply and add with same natural numbers:
Eg: 5, 6, 14, 45
(2016.NMMS)
a) 184 b) 180 c) 176 d) 225
VII Class Mathematics
Explanation: (5  1) + 1, (6  2) +2, (14  3) +3,
so, next number is (45  4) + 4 = 184
91
Fractions, Decimals and Rational Numbers
4. <ŠXæ+Xø dŸ+K«\qT 10, 100, 1000Ôà >·TDì+#\+fñ, dŸ+K« eT]jáTT uó²>·|\Ÿ˜ + jîTT¿£Ø n+¿\T ÿ¹¿$<ó+Š >±
–+{²sTT nsTTÔû, 1 ÔásTÁ yÔá dŸTH•\ dŸ+K«Å£” dŸeÖq+>± <ŠXæ+Xø _+<ŠTeÚqT Å£”&y|Õî ڟ Å£” eÖsÁÌ+&.
5. s +&ƒT <ŠXæ+Xø dŸ+K«\qT >·TDì+ºq|ŸÚ&ƒT \‹Æ+ýË <ŠXæ+Xø k͜H\ dŸ+K«, >·TDì+#á‹&q dŸ+K«\ <ŠXæ+Xø
k͜H\ dŸ+K«\ yîTTԐì dŸeÖq+.
6. ÿ¿£ <ŠXæ+Xø dŸ+K«qT 10, 100 ýñ<‘ 1000Ôà uó²Ð+#û³|ŸÚÎ&ƒT, dŸ+K« eT]jáTT uó²>·|\Ÿ˜ + jîTT¿£Ø n+¿\T
ÿ¹¿$<ó+Š >± –+{²sTT nsTTÔû, 1 ÔásTÁ yÔá dŸTH•\ dŸ+K«Å£” dŸeÖq+>± <ŠXæ+Xø _+<ŠTeÚqT m&ƒeT yî|Õ ÚŸ Å£”
eÖsÁÌ+&.
7.
8.
9.
10.
11.
p
\T
q
–eTˆ& ¿±sÁD²+¿£+ ýñ“ |ŸPs’+¿±\T eT]jáTT q
 0.
nsTTÔû @<îHÕ  n¿£sD
Á j
¡ Tá dŸ+K« p, q qT
çbÍeÖDì¿£ sÁÖ|Ÿ+ýË eÚ+~ n“ n+{²+.
dŸ+K«¹sKÅ£” yîÞ&¢ø †“¿ì eTT+<ŠT n¿£sD
Á j
¡ Tá dŸ+K« jîTT¿£Ø sÁTD ýñ<‘ <óqŠ >·TsÁTqï T Ôá“F #ûjTá &ƒ+ eT]Ìbþe<ŠT.Ý
ç¿£eT _óH•\T e+{ì n¿£sD
Á j
¡ Tá dŸ+K«\T m\¢|ڟ Î&ƒÖ 0 qT+º 1 ýñ<‘ 0 qT+º -1 eT<ó«Š –+{²sTT.
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n|Ÿç¿£eT _óH•\T e+{ì n¿£sDÁ j
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¡ Tá dŸ+K« –q•<à ¿£qT>=H*.
dŸ+U²« çXâDT\T ` 2
1. dŸVŸ²È dŸ+K«\qT ¿£\|Ÿ&ƒ+ ýñ<‘ rd¾yûjáT&ƒ+:
–<‘: 6, 7, 9, 12, 16, 21,
$esÁD: (6G1), (7G2), (9G3), (12G4), (16G5)
a) 21 b) 25 c) 27 d) 28
¿±eÚq ÔásTÁ yÔá e#ûÌ dŸ+K« (21 G 6) R 27
2. ÿ¿£ ç¿£eT+ýË dŸ+K«\qT ¿£\|Ÿ&+ƒ :
$esÁD: (10G10), (20G20), (40G30), (70G40)
–<‘: 10, 20, 40, 70, 110,
a) 160 b) 180 c) 150 d) 210
¿±eÚq ÔásTÁ yÔá e#ûÌ dŸ+K« (110 + 50) = 160
3. uñd¾ dŸ+K«\qT ¿£\|Ÿ&ƒ+ ýñ<‘ rd¾yûjáT&ƒ+
–<‘: 27, 26, 23, 18, 11,
$esÁD: (27-1), (26-3), (23-5), (18-7)
a) 4 b) 2 c) 9 d) 5
ÔásTÁ yÔá e#ûÌ dŸ+K« (11-9) R 2
4. ÿ¿£ d¾sœ Á dŸ+K« Ôà >·TDì+#á&+ƒ
–<‘: 5, 15, 45, 135, 405,
$esÁD: (5 I 3), (15 I 3), (45 I 3), (135 I 3),
a) 1200 b) 1215 c) 850 d) 925
¿±eÚq ÔásTÁ yÔá e#ûÌ dŸ+K« R 1215
5. ÿ¿£ dŸ+K«Ôà >·TDì+º n<û dŸ+K«qT ¿£\|Ÿ&+ƒ
–<‘: 5, 6, 14, 45 (2016.NMMS) $esÁD: (5I1) G 1, (6I2) G 2, (14 I 3) G3,
a) 184 b) 180 c) 176 d) 225
¿±eÚq ÔásTÁ yÔá e#ûÌ dŸ+K« R 184
7e ÔásÁ>·Ü >·DìÔáeTT
92
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¡ Tá dŸ+K«\T
6.
Multiply and add with a different fixed numbers
Eg: 3, 9, 21, 45, 93
a) 184 b) 187 c) 186 d) 189
7.
Multiply with fixed number and add different numbers:
Explanation: (12  2) +1, (25  2) +2, (52  2) +3,
so, next number is (107  2) + 4 = 218
Eg: 12, 25, 52, 107
a) 196 b) 207 c) 214 d) 218
8.
Multiply the sequence number (2016 NMMS) :
Eg: 7, 14, 42, 168, 840
a) 1680 b) 5040 c) 760 d) 4200
9.
Explanation: (3  2) +3, (9  2) +3, (21  2) + 3,
(45  2) + 3, so, next number is (92  2) +3 = 187
Explanation: (7  2), (14  3), (42  4), (168  5)
so, next number is (840  6) = 5040
Dividing with a fixed number
Eg: 256, 128, 64, 32, 16,
Explanation: (256/2), (128/2), (64/2),(32/2),
a) 8
so, next number is (16/2) = 8
b) 4
c) 16 d) 10
10. Multiply and divide with a different fixed numbers.
Eg: 12, 60, 30, 150, 75,
Explanation: (12  5), (60/2), (30  5),(150/2),
..
so, next number is (75  5) = 375
a) 325 b) 150 c) 375 d) 300
Practice problems:
1)
15, 27, 39, 51, 63,
a) 85
b)
75
c) 65
d) 73
2)
2, 5, 10, 17, 26, 37,
a) 48
b)
75
c) 50
d) 73
3)
1, 6, 16, 31, 51, 76,
a) 95
b)
86
c) 91
d) 96
4)
13, 14, 16, 20, 28, 44,
a) 76
b)
75
c) 87
d) 73
5)
28, 25, 30, 27, 32, 29,
a) 26
b)
24
c) 34
d) 32
6)
3, –6, 12, -24, 48, -96,
a) 192
b)
–102
c) –192
d) 106
7)
1, 2, 6, 24, 120, 720,
a) 920
b)
5040
c) 1040
d) 4320
8)
63, 64, 67, 72, 79,
a) 88
b)
86
c) 87
d) 98
9)
9, 10, 22, 69, 280,
a) 1205
b)
1425
c) 1400
d) 1405
10) 729, 243, 81, 27,
a) 65
b)
18
c) 9
d) 73
11) 5, 15, 35, 75, 155,
a) 215
b)
305
c) 315
d) 265
a) 10
b)
20
c) 18
d) 35
13) 20, 10, 10, 20, 80,
a) 320
b)
640
c) 400
d) 80
14) 7, 10, 8, 11, 9, 12,
a) 8
b)
14
c) 15
d) 10
15) 34, 30, 28, 24, 22, 18,
a) 16
b)
14
c) 20
d) 15
16) 13, 26, 28, 56,
a) 60
b)
112
c) 58
d) 15
17) 2020, 2021, 2023, 2026, 2030,
a) 2040
b)
2035
c) 2034
d) 2032
18) 6, 12, 48,
a) 288
b)
75
c) 50
d) 480
19) 2, 2, 6, 30, 210,
a) 150
b)
96
c) 192
d) 1890
20) 9, 3, 18, 6, 36,
a) 12
b)
24
c) 72
d) 1
12) 240,240, 120, 40,
,2
,116
, 2304
VII Class Mathematics
93
Fractions, Decimals and Rational Numbers
6. $$<óŠ dŸ+K«\Ôà >·TDì+º eT]jáTT d¾œsÁ dŸ+K«qT ¿£\|Ÿ&ƒ+.
–<‘: 3, 9, 21, 45, 93
$esÁD: (3I2)G3, (9 I 2)G3, (21I 2)G3,(45I2)G3
a) 184 b) 187 c) 186 d) 189
¿±eÚq ÔásTÁ yÔá e#ûÌ dŸ+K« R 187
7. d¾sœ Á dŸ+K«Ôà >·TDì+º, esÁTdŸ dŸ+K«\qT ¿£\|Ÿ&+ƒ .
–<‘: 12, 25, 52, 107
$esÁD: (12 I 2) G 1, (25 I 2) G 2, (52 I 2) G 3,
a) 196 b) 207 c) 214 d) 218
¿±eÚq ÔásTÁ yÔá e#ûÌ dŸ+K« (107 I 2) G 4 R 218
8. esÁTdŸ dŸ+K«\Ôà >·TDì+#á&+ƒ (2016 NMMS) :
–<‘: 7, 14, 42, 168, 840
$esÁD: (7I2), (14I3), (42I4), (168I5)
a) 1680 b) 5040 c) 760 d) 4200 ¿±eÚq ÔásTÁ yÔá e#ûÌ dŸ+K« (840I6) R 5040
9. d¾sœ Á dŸ+K«Ôà uó²Ð+#á&+ƒ .
–<‘: 256, 128, 64, 32, 16,
$esÁD: (256/2), (128/2), (64/2),(32/2), .....
a) 8 b) 4 c) 16 d) 10
¿±eÚq ÔásTÁ yÔá e#ûÌ dŸ+K« (16/2) R 8
10. ÿ¿£ d¾sœ Á dŸ+K«Ôà >·TDì+#á&+ƒ ÔásTÁ yÔá eTs=¿£ d¾sœ Á dŸ+K«Ôà uó²Ð+#á&+ƒ .
–<‘: 12, 60, 30, 150, 75, ..
$esÁD: (12 I 5), (60/2), (30I5), (150/2), .....
a) 325 b) 150 c) 375 d) 300
¿±eÚq ÔásTÁ yÔá e#ûÌ dŸ+K« (75 I 5) R 375
kÍ<óHŠ  dŸeTdŸ«\T:
1)
15, 27, 39, 51, 63,
a) 85
b)
75
c) 65
d) 73
2)
2, 5, 10, 17, 26, 37,
a) 48
b)
75
c) 50
d) 73
3)
1, 6, 16, 31, 51, 76,
a) 95
b)
86
c) 91
d) 96
4)
13, 14, 16, 20, 28, 44,
a) 76
b)
75
c) 87
d) 73
5)
28, 25, 30, 27, 32, 29,
a) 26
b)
24
c) 34
d) 32
6)
3, –6, 12, -24, 48, -96,
a) 192
b)
–102
c) –192
d) 106
7)
1, 2, 6, 24, 120, 720,
a) 920
b)
5040
c) 1040
d) 4320
8)
63, 64, 67, 72, 79,
a) 88
b)
86
c) 87
d) 98
9)
9, 10, 22, 69, 280,
a) 1205
b)
1425
c) 1400
d) 1405
10) 729, 243, 81, 27,
a) 65
b)
18
c) 9
d) 73
11) 5, 15, 35, 75, 155,
a) 215
b)
305
c) 315
d) 265
a) 10
b)
20
c) 18
d) 35
13) 20, 10, 10, 20, 80,
a) 320
b)
640
c) 400
d) 80
14) 7, 10, 8, 11, 9, 12,
a) 8
b)
14
c) 15
d) 10
15) 34, 30, 28, 24, 22, 18,
a) 16
b)
14
c) 20
d) 15
16) 13, 26, 28, 56,
a) 60
b)
112
c) 58
d) 15
17) 2020, 2021, 2023, 2026, 2030,
a) 2040
b)
2035
c) 2034
d) 2032
18) 6, 12, 48,
a) 288
b)
75
c) 50
d) 480
19) 2, 2, 6, 30, 210,
a) 150
b)
96
c) 192
d) 1890
20) 9, 3, 18, 6, 36,
a) 12
b)
24
c) 72
d) 1
12) 240,240, 120, 40,
,116
, 2304
7e ÔásÁ>·Ü >·DìÔáeTT
,2
94
_óH•\T, <ŠXæ+Xæ\T eT]jáTT n¿£sDÁ j
¡ Tá dŸ+K«\T
p
n
4
m
Content Items
Learning Outcomes
The learner is able to
l
understand the pair of angles based on their properties
as complementary, supplementary, conjugate,
adjacent, linear and vertically opposite angles.
l
find the value of complementary, supplementary and
conjugate angles of the given angle.
l
give examples of various pairs of angles.
l
explain the properties of various pairs of angles
formed when a transversal cuts two lines.
l
identify the pairs of angles and parallel lines in
surroundings.
l
solve the problems involving angles made by
transversal on parallel lines.
l
visualise and establish the relation between the pairs
of angles formed by transversal with two parallel lines.
4.0 Introduction
4.1 Pair of angles
4.2 Adjacent angles
4.3 Vertically opposite angles
4.4 Angles made by a
transversal.
4.0 Introduction :
Look at the following picture, what do you observe? You will observe the concepts like point, line
segment, line, ray, angle, types of angles, parallel lines, perpendicular lines, intersecting lines that we
have learnt in the previous classes. Can you name them? In the picture, A is a point, AB is a Line
segment. If we extend the two end points of line segment in either direction endlessly, we will get a
HJJG
line AB . Angles formed at corners PQR is an angle. ‘l’ and ‘m’ are parallel lines.
You also observe that a carpenter making
chairs, tables, interiors etc. Do you know why they
look so beautiful, perfect and strong? What happens
if the marked angles are not equal? Do you observe
where these lines and angles used in the picture?
Lines and angles are observed in all aspects
of our life. Carpenters, Engineers, Architects etc.,
are using the concept of lines and angles in their
works. Let us learn more about lines and angles in
this chapter. Before going to discuss these concepts,
let us recall the concepts which we have learnt in the previous class by the following exercise:
VII Class Mathematics
137
Lines and Angles
¹sK\T
nuó²«dŸÅ£”\T :
l
l
l
l
l
l
l
eT]jáTT
¿ÃD²\T
nuó„«dŸq |˜Ÿ*Ԑ\T
$wŸjáÖ+Xæ\T
|ŸPsÁ¿£, dŸ+|ŸPsÁ¿£, dŸ+jáTT>·ˆ, €dŸq•, ¹sFjáT eT]jáTT
osü_eó TTK ¿ÃD²\T >·T]+º ne>±VŸ²q #ûdTŸ ¿=+{²sÁT.
‚ºÌq ¿ÃD²“¿ì |ŸPsÁ¿£, dŸ+|ŸPsÁ¿£, dŸ+jáTT>·ˆ¿ÃD²\qT
¿£qT>=+{²sÁT.
$$<óŠ ¿ÃD²\ ÈÔá\Å£” –<‘VŸ²sÁD\T ‚kÍïsÁT.
ÿ¿£ ÈÔá dŸsÞÁ s¹ø K\™|Õ ÜsÁ«>´s¹ K #ûjTá T ¿ÃD²\ÈÔá\ \¿£D
Œ ²\qT
$e]kÍïsÁT.
Ôáq |Ÿ]dŸsýË¢ >·\ ¿ÃDÈÔá\qT eT]jáTT dŸeÖ+ÔásÁ ¹sK\qT
>·T]ïkÍïsÁT.
dŸeÖ+ÔásÁ ¹sK\™|Õ ÜsÁ«>´¹sK #ûjáTT ¿ÃD²\Å£” dŸ+‹+~ó+ºq
dŸeTdŸ«\qT kÍ~ókÍïsÁT.
dŸ e Ö+Ôá s Á ¹ s K\™|Õ ÜsÁ « >· ¹ s K #û j á T T ¿ÃD²\ ÈÔá \ qT
ç<Š T o«¿£ ] +#á T qT eT]já T T € ¿ÃD²\ ÈÔá \ eT<ó Š « >· \
dŸ+‹+<󑓕 k͜|¾kÍïsÁT.
4.0. |Ÿ]#ájáT+
4.1 ¿ÃD²\ ÈÔá
4.2. €dŸq• ¿ÃD²\T
4.3. osü_eó TTK ¿ÃD²\T
4.4. ÜsÁ«ç¹>K#û @sÁÎ&ƒT¿ÃD²\T
4.0 |Ÿ]#ájTá +: ç¿ì+<Š
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HûsÁTÌÅ£”q• _+<ŠTeÚ, ¹sU²K+&ƒ+, ¹sK, ¿ìsÁD+, ¿ÃD+, ¿ÃD²\ sÁ¿±\T, dŸeÖ+ÔásÁ ¹sK\T, \+‹¹sK\T, K+&ƒq
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dŸsÁÞø¹sK AB e#áTÌqT. eTÖ\\ jáT+<ŠT ¿ÃD²\T @sÁÎ&q$. PQR ÿ¿£ ¿ÃD+. ‘l’ eT]jáTT ‘m’ \T dŸeÖ+ÔásÁ
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>·eT“+º –+{²sÁT.
n$ n+Ôá n+<Š+>±, ‹\+>± m+<ŠTÅ£” ¿£“|¾kÍïjÖî MTÅ£”
Ôî\TkÍ? ÿ¿£yÞû ø eÖsYØ #ûjTá ‹&ƒ¦ ¿ÃD²\T dŸeÖq+>± ýñ“#Ã
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m¿£Ø&ƒ –|ŸjîÖÐ+#á‹&†¦jîÖ MTsÁT >·eT“+#s? eTq
“Ôá«J$Ôá+ýË n“• dŸ+<ŠsÒÛ\ýË dŸsÁÞø¹sK\T eT]jáTT
¿ÃD²\TqT eTq+ >·eT“kÍïeTT. eç&ƒ+>·T\T, ‚+ÈúsÁT¢,
€]ØfÉ¿\¼ù T yîTT<ŠýqÕÉ ysÁT n+<ŠsÖÁ y] jîTT¿£Ø |ŸqT\ jáT+<ŠT
dŸsÞÁ s¹ø K\T eT]jáTT ¿ÃD²\ uó²eqqT –|ŸjÖî ÐdŸÖï –+{²sÁT.
‡ n<ó‘«jáT+ýË dŸsÞÁ s¹ø K\T eT]jáTT ¿ÃD²\ >·T]+º eTq+
eT]+Ôá HûsÁTÌÅ£”+<‘+. ‡ uó²eq\qT #á]Ì+#á&†“¿ì eTT+<ŠT, ~>·Te nuó²«dŸ+ #ûjáT&ƒ+ <‘Çs >·Ôá ÔásÁ>·ÜýË
HûsÁTÌÅ£”q• uó²eq\qT eTq+ >·TsÁTï #ûdŸTÅ£”+<‘+.
7e ÔásÁ>·Ü >·DìÔáeTT
138
s¹ K\T eT]jáTT ¿ÃD²\T
1.
Observe the figure and name the points, line
segments, rays and lines from the figure.
2.
l
Observe the figure and write intersecting lines
and concurrent lines.
3.
Draw a line segment PQ = 6.3 cm.
4.
Name any three possible angles in the figure.
5.
Write the type of angles you observed in the given clock.
(i)
(ii)
(iii)
(iv)
(v)
6. One right angle is equal to _______ degrees.
7. Write any 2 acute angles and any 2 obtuse angles.
8. Observe the parallel and perpendicular lines
in the given figure. Write them using symbols || , .
9. Measure angle  AOB with protractor.
Note : While referring to the measure of an angle ABC, we shall write mÐABC as simply
ÐABC .The context will make it clear, whether we are referring to the angle or its measure.
VII Class Mathematics
139
Lines and Angles
|ŸÚq]ÇeTsÁô nuó²«dŸ+
1. |Ÿ{²“• |Ÿ]o*+#á+&. |Ÿ³+ýË >·\ _+<ŠTeÚ\T, ¹sU²K+&†\T,
¿ìsÁD²\T eT]jáTT dŸsÁÞø¹sK\qT çyjáT+&.
2. |Ÿ³+ |Ÿ]o*+#á+&. |Ÿ³+ýË >·\ K+&ƒqs¹ K\qT eT]jáTT
$T[Ôás¹ K\qT çyjáT+&.
l
3. PQ = 6.3 ™d+.MT. bõ&ƒeÚ >·\ s¹ U²K+&†“• ^jáT+&.
4. ç|Ÿ¿Ø£ |Ÿ³+ýË >·\ @yîHÕ  eTÖ&ƒT ¿ÃD²\qT |s=Øq+&.
5. ‚ºÌq >·&jáÖsÁ+ýË MTsÁT >·T]ï+ºq ¿ÃD+ sÁ¿±“• çyjáT+&.
(i)
(ii)
(iii)
(v)
(iv)
6. ÿ¿£ \+‹¿ÃD+ _______ &ç^\Å£” dŸeÖq+.
7. @yîHÕ  s +&ƒT n\οÃD²\T eT]jáTT s +&ƒT n~ó¿¿£ ÃD²\TqT çyjáT+&.
8. ‚eNj&ƒ¦ |Ÿ³+ýË dŸeÖ+ÔásÁ¹sK\qT eT]jáTT \+‹¹sK\qT
>·T]ï+#á+&. y{ì“ >·TsÁTï\T || , \qT –|ŸjÖî Ð+º çyjáT+&.
9. ¿ÃD+ AOB “ ¿ÃDe֓“ dŸVäŸ jáT+Ôà ¿=*º çyjáT+&.
>·eT“¿£ : ÿ¿£ ¿ÃD+ ABC jîTT¿£Ø ¿=\ÔáqT dŸÖº+#á&†“¿ì, mÐABC“ Å£”|
¢ +ïŸ >± ÐABC >± çykÍïeTT.
n~ ¿ÃDeÖ ýñ¿£ <‘“ ¿=\ÔáqT ç|ŸkÍï$dŸTïH•eÖ nHû $wŸjá֓• €dŸ+<ŠsÁÒÛ+ dŸÎwŸ¼+ #ûdŸTï+~.
7e ÔásÁ>·Ü >·DìÔáeTT
140
s¹ K\T eT]jáTT ¿ÃD²\T
4.1 Pair of angles :
A pair of angles means two angles. Let ABC, PQR is a pair of angles and ABC = 600,
PQR= 450 .What is the sum of these two angles? The sum of these two angles is 600 + 450
= 1050 . In this way we can add angles of same units.
Observe the given figure.
Measure the angles AOB, BOC and AOC.
Here, AOB = ____, BOC=____, AOC = ____
What do you observe?
The sum of pair of angles AOB and BOC is AOC.
We write, AOB + BOC = AOC
Take 15 pieces of a white paper. Write each pair of angles given below on each piece of paper.
Fold each paper separately and put them in a box. Ask each student to come and take one
folded piece of paper. Ask students to find the sum of angles written on them and write the
sum on the paper.
i) 530, 270
ii) 480, 1320
iii) 140,760
iv) 900, 2700
v) 2000,1000
vi) 260, 640
vii) 630, 1170
viii) 1110, 600
ix) 1440, 360
x) 1800,1800
xi) 1000, 2600
xii) 400, 350
xiii) 450,450
xiv) 890, 2710
xv) 900, 900.
How many of you got a sum of 900? They are the pair of Complementary angles.
How many of you got a sum of 1800? They are the pair of Supplementary angles.
How many of you got a sum of 3600? They are the pair of Conjugate angles.
Remaining pair of angles are not complementary or supplementary or conjugate pair of
angles.
The types of pair of angles based on the sum of pair of angles:
Pair of angles
Complementary angles:
Examples
i) 300,600
If the sum of two angles is 900, then the
0
0
angles are called as complementary angles ii) 51 ,39
to each other. i.e., If A +B = 900, then iii) ___,___
A is complementary angle to B and
B is complementary angle to A .
iv) ___,___
Ex: 200 + 70 0 = 900 . So, 20 0 is the v) ___,___
complementary angle of 700 and 700 is
the complementary angle of 200.
Write 3 pairs on
your own
VII Class Mathematics
141
Examples in figures
i)
ii)
Lines and Angles
4.1. ¿ÃD²\ ÈÔá:
¿ÃD²\ ÈÔá nq>± s +&ƒT ¿ÃD²\T n“ nsÁ+œ . ABC, PQR \T ¿ÃD²\ ÈÔá. y{ì ¿=\Ôá\T ABC = 600,
PQR= 450 nqT¿=“q ‡ s
 +&ƒT ¿ÃD²\ yîTTÔá+ï m+Ôá? ‡ s +&ƒT ¿ÃD²\ yîTTÔá+ï 600 + 450 = 1050
neÚÔáT+~. ‡ $<óŠ+>± eTq+ ÿ¹¿ ç|ŸeÖD+ >·\ ¿ÃD²\qT ¿£\|Ÿe#áTÌ.
‚ºÌq |Ÿ{²“• >·eT“+#á+&.
¿ÃD²\T AOB, BOC eT]jáTT AOC \qT ¿=\e+&.
‚¿£Ø&ƒ, AOB = ____, BOC=____, AOC = ____
MTsÁT @$T >·eT“+#sÁT?
AOB eT]jáTT BOC ¿ÃD²\ ÈÔá yîTTÔáï+ AOC ¿ì dŸeÖq+.
B““ AOB + BOC = AOC n“ çykÍï+.
‚~ #û<‘Ý+!
¿£Ôá«+
15 Ôî\¢ ¿±ÐÔá+ eTT¿£Ø\T rdŸT¿Ã+&. ç|ŸÜ ¿±ÐÔá+ eTT¿£Ø ™|Õ ~>·Te ‚eNj&ƒ¦ ÿ¿=Ø¿£Ø ¿ÃD²\ ÈÔáqT çyjáT+&.
ç|ŸÜ ¿±ÐԐ“• $&>± eT&º, y{ì“ ÿ¿£ u²Å£”àýË –+#á+&. ç|ŸÜ $<‘«]œ eºÌ ÿ¿£ eT&ƒÔá ¿±ÐÔá+ eTT¿£ØqT
rdŸT¿ÃeT“ #î|Ο +&. y{ì™|Õ sd¾q ¿ÃD²\ yîTTԐ ¿£qT>=qeT“ $<‘«sÁT\œ qT n&ƒ>+· & eT]jáTT € yîTTԐ
¿±ÐÔá+™|Õ sjáTeT“ #î|ŸÎ+&.
i) 530, 270
vi) 260, 640
xi) 1000, 2600
ii) 480, 1320
vii) 630, 1170
xii) 400, 350
iii) 140,760
viii) 1110, 600
xiii) 450,450
iv) 900, 2700
ix) 1440, 360
xiv) 890, 2710
v) 2000,1000
x) 1800,1800
xv) 900, 900.
MTýË m+ÔáeT+~¿ì ¿ÃD²\ yîTTÔáï+ 900 eºÌ+~? n$ ¿=“• |ŸPsÁ¿£ ¿ÃD²\ ÈÔá\T.
MTýË m+ÔáeT+~¿ì ¿ÃD²\ yîTTÔáï+ 1800 eºÌ+~? n$ ¿=“• dŸ+|ŸPsÁ¿£ ¿ÃD²\ ÈÔá\T.
MTýË m+ÔáeT+~¿ì ¿ÃD²\ yîTTÔáï+ 3600 eºÌ+~? n$ ¿=“• dŸ+jáTT>·ˆ ¿ÃD²\ ÈÔá\T.
$TÐ*q ¿ÃD²\ ÈÔá\T |ŸPsÁ¿£ ¿ÃD²\T ýñ<‘, dŸ+|ŸPsÁ¿¿£ ÃD²\T ýñ<‘, dŸ+jáTT>·ˆ ¿ÃD²\ ÈÔá\T ¿±eÚ.
Âs+&ƒT¿ÃD²\ yîTTÔáï+qT ‹{ì¼ ¿ÃD²\ ÈÔá\ sÁ¿±\T:
¿ÃD²\ ÈÔá
|ŸPsÁ¿£ ¿ÃD²\T:
Âs+&ƒT ¿ÃD²\T yîTTÔáï+ 90 nsTTÔû, € s +&ƒT
¿ÃD²\TqT ÿ¿£<‘“¿=¿£{ì |ŸPsÁ¿£ ¿ÃD²\T
n+{²sÁT. nq>± A +B = 900 nsTTq#Ã
A jîTT¿£Ø |ŸPsÁ¿£ ¿ÃD+ B n“ eT]jáTT
B jîTT¿£Ø |ŸPsÁ¿£ ¿ÃD+ A n“ n+{²sÁT.
–<‘ : 200 G 700 R 900, ¿£qT¿£ 200 jîTT¿£Ø
|ŸPsÁ¿£ ¿ÃD+ 700 eT]jáTT 700 jîTT¿£Ø |ŸPsÁ¿£
¿ÃD+ 200 neÚÔáT+~.
7e ÔásÁ>·Ü >·DìÔáeTT
0
|Ÿ³eTTýË –<‘VŸ²sÁD\T
–<‘VŸ²sÁD\T
i) 300,600
i)
ii) 510,390
iii) ___,___
iv) ___,___
v) ___,___
ii)
MTsÁT dŸÇ+Ôá+>± 3 ÈÔá\T
çyjáT+&.
142
s¹ K\T eT]jáTT ¿ÃD²\T
Pair of angles
Examples
Examples in figures
Supplementary angles: If the sum of i) 1200, 600
two angles is 1800, then the angles are
called as supplementary angles to each ii) 1620, 180
other i.e., If A +B = 1800, then A
is supplementary angle to B and iii) ___,___
B is supplementary angle to A.
iv) ___,___
Eg: 1000 + 800 = 1800. So, 1000 is the
supplementary angle of 800 and 800 is v) ___,___
the supplementary angle of 1000.
Conjugate angles:
If the sum of two angles is 360 , then
the angles are called as conjugate angles
to each other i.e., If A +B = 3600,
then A is conjugate angle to B and
B is conjugate angle to A.
0
Eg: 2000 + 1600 = 3600. So, 2000 is the
conjugate angle of 1600 and 1600 is the
conjugate angle of 2000.
i)
ii)
i) 3300, 300
ii) 1510, 2090
iii) ___,___
iv) ___,___
v) ___,___
1. Find the complementary angles of i) 270 ii) 430 iii) k0 iv) 20
2. Find the supplementary angles of
3. Find the conjugate angles of
i) 130 ii) 970 iii) a0
i) 740
ii) 1800 iii) m0
iv) 460
iv) 3000
(i) Umesh said, “Two acute angles cannot form a pair of
supplementary angles.” Do you agree? Give reason.
(ii) Lokesh said, “Each angle in any pair of complementary angles is
always acute.” Do you agree? Justify your answer.
Example 1 : In the given figure B and E are complementary angles. Find the value of x.
Solution :
From the figure, B = x + 100 and E =350
Since B and E are complementary angles,
B + E =90°
x + 100 + 350 = 900
x + 450 = 900
x = 900 – 450
 x = 450
VII Class Mathematics
143
Lines and Angles
¿ÃD²\ ÈÔá
dŸ+|ŸPsÁ¿£ ¿ÃD²\T:
Âs+&ƒT ¿ÃD²\ yîTTÔáï+ 180 nsTTÔû € Âs+&ƒT
¿ÃD²\TqT ÿ¿£<‘“¿=¿£{ì dŸ+|ŸPsÁ¿£ ¿ÃD²\T
n+{²sÁ T . nq>± A +B = 180 0
nsTTq#à A jîTT¿£Ø dŸ+|ŸPsÁ¿£ ¿ÃD+ B
n“ eT]jáTT B jîTT¿£Ø dŸ+|ŸPsÁ¿£ ¿ÃD+
A n“ n+{²sÁT.
–<‘: 1000 G 800 R 1800, ¿£qT¿£ 1000
jîTT¿£Ø dŸ+|ŸPsÁ¿£ ¿ÃD+ 800 eT]jáTT 800
jîTT¿£Ø dŸ+|ŸPsÁ¿£ ¿ÃD+ 1000 neÚÔáT+~.
0
dŸ+jáTT>·ˆ ¿ÃD²\T:
Âs+&ƒT ¿ÃD²\ yîTTÔáï+ 360 nsTTÔû € s +&ƒT
¿ÃD²\T qT ÿ¿£<‘“¿=¿£{ì dŸ+jáTT>·ˆ ¿ÃD²\T
n+{²sÁ T . nq>± A + B = 360 0
nsTTq#à A jîTT¿£Ø dŸ+jáTT>·ˆ ¿ÃD+ B
n“ eT]jáTT B jîTT¿£Ø dŸ+jáTT>·ˆ ¿ÃD+ A
n“ n+{²sÁT.
–<‘: 2000 G 1600 R 3600 ¿£qT¿£ 2000
jîTT¿£Ø dŸ+jáTT>·ˆ ¿ÃD+ 1600 eT]jáTT 1600
jîTT¿£Ø dŸ+jáTT>·ˆ ¿ÃD+ 2000 neÚÔáT+~.
ú |ç >Ÿ Ü· “
Ôî\TdŸT¿Ã
0
–<‘VŸ²sÁD\T
i) 1200, 600
|Ÿ³eTTýË –<‘VŸ²sÁD\T
i)
ii) 1620, 180
iii) ___,___
iv) ___,___
ii)
v) ___,___
i) 3300, 300
ii) 1510, 2090
iii) ___,___
iv) ___,___
v) ___,___
1. ‚ºÌq ¿ÃD²\Å£” |ŸPsÁ¿£ ¿ÃD²\qT ¿£qT>=q+&.i) 270 ii) 430 iii) k0 iv) 20
2. ‚ºÌq ¿ÃD²\Å£” dŸ+|ŸPsÁ¿£ ¿ÃD²\qT ¿£qT>=q+&. i) 130 ii) 970 iii) a0 iv) 460
3. ‚ºÌq ¿ÃD²\Å£” dŸ+jáTT>·ˆ ¿ÃD²\qT ¿£qT>=q+&. i) 740 ii) 1800 iii) m0 iv) 3000
i)
»»Âs+&ƒT n\οÃD²\T, dŸ+|ŸPsÁ¿£ ¿ÃD²\ ÈÔáqT @sÁÎsÁ#ýá eñ Úµµ n“ –yûTwt nH•&ƒT.
MTsÁT n+^¿£]kÍïs? ¿±sÁD+ Ôî\Î+&.
ii) »»|ŸPsÁ¿¿
£ ÃD²\ ÈÔáýË ç|ŸÜ¿ÃD+ m\¢|ڟ Î&ƒÖ n\οÃDyûTµµ n“ ý˹¿wt nH•&ƒT.
MTsÁT n+^¿£]kÍïs? MT dŸeÖ<ó‘H“• dŸeT]œ+#á+&.
nHûÇw¾<‘Ý+
–<‘VŸ²sÁD 1: |Ÿ³+ýË B eT]jáTT E \T |ŸPsÁ¿£ ¿ÃD²\T nsTTq x jîTT¿£Ø $\TeqT ¿£qT>=q+&.
kÍ<óŠq:
‚ºÌq |Ÿ³+ qT+&
eT]jáTT E =350
B eT]jáTT E \T |ŸPsÁ¿£ ¿ÃD²\T ¿£qT¿£, B + E =90°
B = x + 100
x + 100 + 350 = 900
x + 450 = 900
x = 900 – 450
 x = 450
7e ÔásÁ>·Ü >·DìÔáeTT
144
s¹ K\T eT]jáTT ¿ÃD²\T
Example 2:
If the ratio of supplementary angles is 4:5, then find the two angles.
Solution:
Given ratio of supplementary angles = 4 : 5
sum of the parts in the ratio = 4 + 5 = 9
sum of the supplementary angles = 1800
4
× 1800 = 800
9
first angle =
second angle =
5
× 1800 = 1000
9
Exercise - 4.1
1.
Find which of the following pair of angles are complementary and which are supplementary?
(i)
(iii)
(ii)
(vi)
(v)
(iv)
2.
If the ratio of two complementary angles is 2 : 3, then find the two angles.
3.
In the figure, A and Q are complementary
angles. Find the value of x.
360
4.
If A and B are conjugate angles and A = B. Find the two angles.
5.
Draw a pair of complementary angles and a pair of supplementary angles.
6.
Teacher asked students to draw the complementary angle to the given angle. The students
have drawn as follows. Who is correct?
30
30
0
VII Class Mathematics
0
50
0
900
145
600
30
0
Lines and Angles
–<‘VŸ²sÁD 2 : dŸ+|ŸPsÁ¿£ ¿ÃD²\ “wŸÎÜï 4 : 5 nsTTq € s +&ƒT ¿ÃD²\qT ¿£qT>=q+&.
kÍ<óŠq :
‚ºÌq dŸ+|ŸPsÁ¿£ ¿ÃD²\ “wŸÎÜï R 4 : 5
“wŸÎÜïýË uó²>±\ jîTT¿£Ø yîTTÔáï+ R 4 G 5 R 9
dŸ+|ŸPsÁ¿£ ¿ÃD²\ yîTTÔáï+ R 1800
yîTT<Š{ì ¿ÃD+ =
4
× 1800 = 800
9
s +&ƒe ¿ÃD+ =
5
× 1800 = 1000
9
nuó²«dŸ+ ` 4.1
1. ç¿ì+~ ¿ÃD²\ ÈÔá\ýË @$ |ŸPsÁ¿£ ¿ÃD²\T, @$ dŸ+|ŸPsÁ¿£ ¿ÃD²\ ÈÔáýË ¿£qT>=q+&?
(i)
(iii)
(ii)
(vi)
(v)
(iv)
2. s +&ƒT |ŸPsÁ¿£ ¿ÃD²\ “wŸÎÜï 2 : 3 nsTTÔû € s +&ƒT ¿ÃD²\qT ¿£qT>=q+&.
3. ‚ºÌq |Ÿ³+ýË A eT]jáTT Q \T |ŸPsÁ¿¿£ ÃD²\T nsTTq
x jîTT¿£Ø $\TeqT ¿£qT>=q+&.
360
4. A eT]jáTT B \T dŸ+jáTT>·ˆ ¿ÃD²\T eT]jáTT A = B nsTTq € s +&ƒT ¿ÃD²\qT ¿£qT>=q+&.
5. ÿ¿£ ÈÔá |ŸPsÁ¿£ ¿ÃD²\T, ÿ¿£ ÈÔá dŸ+|ŸPsÁ¿£ ¿ÃD²\ |Ÿ³eTT\qT ^jáT+&.
6. –bÍ<ó‘«jáTT&ƒT Ôáq $<‘«sÁT\œ Å£” ÿ¿£¿ÃD²“• ‚ºÌ € ¿ÃD²“¿ì |ŸPsÁ¿£ ¿ÃD²“• ^jáTeT“ #î™|ÎqT. $<‘«sÁT\œ T
ç¿ì+~ $<óŠ+>± ^ºÜ] nsTTq y]ýË mesÁT dŸÂsÕq $<óŠ+>± ^#îqT?
300
30
0
‚ºÌq ¿ÃDeTT
7e ÔásÁ>·Ü >·DìÔáeTT
50
0
uó²sÁÜ
90
0
600
sE
146
dŸTC²Ôá
300
‡XøÇsY
s¹ K\T eT]jáTT ¿ÃD²\T
7.
In the given figure B and E are supplementary angles. Find x.
8.
Ashritha said, “In the pair of supplementary angles, one angle must be obtuse angle.” Do
you agree? Give reason.
9.
Find the angle which is 400 more than its supplementary angle?
10. Srinu said, “Two obtuse angles cannot be supplementary.” Do you agree? Justify your answer.
4.2 Adjacent angles:
Look at the picture of lemon piece. What do you observe?
At the center, we find many angles lying next to one another.
Observe the angles AOB and BOC. What is the vertex of
AOB? What is the vertex of BOC? Observe that ‘O’ is the
JJJG
common vertex of both angles. OB is the common arm of both
JJJG
angles, and both angles lie on either side of OB . These are adjacent
angles.
‘Two angles are said to be adjacent angles, if they have a common vertex, common
arm and lie on either side of the common arm.’
Observe the adjacent angles in the given pictures:
In the figure, the common vertex of KPM and MPN is P.
JJJJG
Common arm is PM . Both angles KPM and MPN lie on
JJJJG
either side of PM . So, KPM and MPN is a pair of adjacent
angles.
Note:
VII Class Mathematics
1.
The interior of adjacent angles must have no
common points.
2.
The adjacent angles need not be complementary
or supplementary.
147
Lines and Angles
7. ‚ºÌq |Ÿ³+ýË B eT]jáTT E \T dŸ+|ŸPsÁ¿£ ¿ÃD²\T nsTTq x qT ¿£qT>=q+&.
8. »»dŸ+|ŸPsÁ¿£ ¿ÃD²\ ÈÔáýË, ÿ¿£ ¿ÃD+ KºÌÔá+>± n~ó¿£ ¿ÃDyîT® –+&†*µµ n“ €ç¥Ôá #î|ξ +~. B““ úeÚ
n+^¿£]kÍïy? ¿±sÁD+ Ôî\Î+&.
9. ÿ¿£ ¿ÃD+ <‘“ dŸ+|ŸPsÁ¿£ ¿ÃD+ ¿£+fñ 400 mÅ£”Øe, nsTTq € ¿ÃD²“• ¿£qT>=q+&?
10. »»Âs+&ƒT n~ó¿¿£ ÃD²\T dŸ+|ŸPsÁ¿£ ¿ÃD²\ ÈÔá ¿±ýñeÚµµ n“ çoqT nH•&ƒT. B““ MTsÁT n+^¿£]kÍïs? MT
dŸeÖ<ó‘H“• dŸeT]œ+#á+&.
4.2 €dŸq•¿ÃD²\T :
ç|Ÿ¿Ø£ qTq• “eTˆ¿±jáT eTT¿£Ø ºçԐ“• #áÖ&ƒ+&. MTsÁT @$T >·eT“+#sÁT? eT<ó«Š ýË
nHû¿£ ¿ÃD²\T ÿ¿£<‘“ |Ÿ¿£Øq eTs=¿£{ì @sÁÎ&ƒ&†“• MTsÁT >·eT“+#áe#áTÌ.
AOB eT]jáTT BOC ¿ÃD²\qT >·eT“+#á+&. AOB jîTT¿£Ø osÁü+
@$T{ì? BOC jîTT¿£Ø osÁü+ @$T{ì? s +&ƒT ¿ÃD²\ jîTT¿£Ø –eTˆ& osÁü+
JJJG
‘O’ n“ >·eT“+#á+&. Âs+&ƒT ¿ÃD²\ jîTT¿£Ø –eTˆ& uó„TÈ+ OB n“
JJJG
>·eT“+#á+& eT]jáTT Âs+&ƒT ¿ÃD²\T –eTˆ& uó„TÈ+ OB ¿ì #îsà yîÕ|ŸÚq ¿£\eÚ
n“ >·eT“+#á+&. ‚$ ÿ¿£ ÈÔá €dŸq• ¿ÃD²\T.
»»Âs+&ƒT ¿ÃD²\T –eTˆ& osÁü+, –eTˆ& uóT
„ È+ ¿£*Ð € s
 +&ƒT ¿ÃD²\T –eTˆ& uóT
„ È+¿ì #îsÃyî|
Õ ÚŸ q –q•#à y{ì“
€dŸq•¿ÃD²\T n+{²sÁT.µµ
‚³Te+{ì €dŸq• ¿ÃD²\qT ç¿ì+~ ‚eNj&ƒ¦ ºçԐýË¢ >·eT“+#á+&:
m<ŠT\Ý ‹+& #áç¿£eTT
>·&jáÖsÁeTT
|Ÿ³+ýË, KPM eT]jáTT MPN jîTT¿£Ø –eTˆ& osÁü+ P eT]jáTT
JJJJG
–eTˆ& uó„TÈ+ PM KPM eT]jáTT MPN \T –eTˆ& uó„TÈ+
JJJJG
PM ¿ì #îs=¿£ yî|
Õ ÚŸ q –H•sTT. ¿£qT¿£ KPM eT]jáTT MPN \T
€dŸq• ¿ÃD²\T neÚԐsTT.
>·eT“¿£: 1. €dŸq• ¿ÃD²\ n+ÔásÁ+\ýË –eTˆ& _+<ŠTeÚ\T
–+&ƒeÚ.
2. €dŸq• ¿ÃD²\T |ŸPsÁ¿£ ¿ÃD²\T >±“, dŸ+|ŸPsÁ¿£
¿ÃD²\T >±“ ¿±e\d¾q nedŸsÁ+ ýñ<ŠT.
7e ÔásÁ>·Ü >·DìÔáeTT
148
s¹ K\T eT]jáTT ¿ÃD²\T
1.
A
In the figure, AOB and BPC are not adjacent
angles. Why? Give reason.
O
In the figure, AOB and
P
COD have common vertex O.
But AOB, COD are not adjacent angles.
Why? Give reason.
2.
3.
4.2.1
C
In the figure, POQ and POR have common vertex O and
common arm OP but POQ and POR are not adjacent
angles. Why? Give reason.
Linear pair of angles:
Q
P
B
Q
R
Look at the picture beside. Observe the angles formed on
the PR. What are the adjacent angles here? POQ and
QOR are adjacent angles. Observe that these angles are
formed at a point on the line on the same side. What is the
sum of these two angles? Their sum is 1800. These are linear
pair of angles.
“A pair of adjacent angles whose sum is 1800 are called linear pair of angles.”
Observe the linear pair of angles in the given pictures:
Electrical pole
Tree
Pen stand
HJJG
HJJG JJJG
In the figure, DF is a straight line and O is a point on DF . OE is a ray.
DOE, EOF is a linear pair of angles.
If EOF = x, then DOE = 1800 – x
If EOF = 400 then DOE = 1800 – 400 = 1400
If DOE = 830 then EOF = 1800 – 830 = 970
VII Class Mathematics
149
Lines and Angles
€ý˺<‘Ý+!
A
1. |Ÿ³+ýË AOB eT]jáTT BPC \T €dŸq• ¿ÃD²\T ¿±eÚ. m+<ŠTÅ£”?
¿±sÁD+ Ôî\Î+&.
|Ÿ³+ýË, AOB eT]jáTT COD\Å£” O
–eTˆ& osÁü+ O>± ¿£\<ŠT.
2.
AOB, COD €dŸq• ¿ÃD²\T ¿±<ŠT.
m+<ŠTÅ£”? ¿±sÁD+ Ôî\Î+&.
B
P
C
3. |Ÿ³+ýË, POQ eT]jáTT POR\Å£” –eTˆ& osÁü+ O eT]jáTT –eTˆ&
uó„TÈ+ OP>± ¿£\eÚ. ¿±ú POQ eT]jáTT POR\T €dŸq• ¿ÃD²\T ¿±eÚ.
m+<ŠTÅ£”? ¿±sÁD+ Ôî\Î+&.
4.2.1. s¹ FjáT ¿ÃD²\ ÈÔá :
ç|Ÿ¿£ØqTq• ºçԐ“• #áÖ&ƒ+&. PR ™|Õ @sÁÎ&q ¿ÃD²\qT >·eT“+#á+&.
M{ìýË €dŸq• ¿ÃD²\T @$T{ì? POQ eT]jáTT QOR\T €dŸq•
¿ÃD²\T. ‡ s +&ƒT ¿ÃD²\T dŸsÞÁ s¹ø K ™|Õ ÿ¹¿ yîÕ|ŸÚq ÿ¹¿ _+<ŠTeÚ e<ŠÝ
@sÁÎ&†¦jáT“ >·eT“+#á+&. ‡ Âs+&ƒT ¿ÃD²\ yîTTÔáï+ m+Ôá –+³T+~?
y{ì yîTTÔáï+ 1800. ‚$ ÿ¿£ s¹ FjáT¿ÃD²\ ÈÔá.
Q
P
Q
R
»»Âs+&ƒT €dŸq• ¿ÃD²\ yîTTÔá+
ï
180
0
nsTTq#à y{ì“ s
¹ FjáT¿ÃD²\ ÈÔá ýñ<‘ s
¹ FjáT<ŠÇjáT+ n“ n+{²sÁT.µµ
‚³Te+{ì ¹sFjáT¿ÃD²\ ÈÔá\qT ç¿ì+~ ‚eNj&ƒ¦ ºçԐýË¢ >·eT“+#á+&:
™|HŽ
kͼ+&Ž
HJJG
DF
¿£s
Â
+³T
dŸ+
œ
‹+
ç|Ÿ¿Ø£ |Ÿ³+ ýË,
ÿ¿£ dŸsÁÞø ¹sK eT]jáTT dŸsÁÞø ¹sK
‚¿£Ø&ƒ DOE, EOF \T s¹ FjáT ¿ÃD²\ ÈÔá.
EOF = x nsTTq DOE = 1800 – x
EOF = 400 nsTTq DOE = 1800 – 400 = 1400
DOE = 830nsTTq EOF = 1800 – 830 = 970
7e ÔásÁ>·Ü >·DìÔáeTT
HJJG
DF
150
#î³T¼
HJJG
™|Õ O ÿ¿£ _+<ŠTeÚ. OE ÿ¿£ ¿ìsÁD+.
s¹ K\T eT]jáTT ¿ÃD²\T
1.
HJJG
In the figure PR is a straight line and O is a
JJJG
point on the line. OQ is a ray.
If QOR= 500, then what is POQ?
i)
ii) If QOP = 1020, then what is QOR?
Note:
1.
The non-common arms of the linear pair of angles are two opposite rays and they
lie on a straight line.
2.
Linear pair of angles are supplementary.
1.
A linear pair of angles must be adjacent but adjacent angles need not be
linear pair. Do you agree? Draw a figure to support your answer.
2.
Mahesh said that the sum of two angles 300 and 1500 is 1800 hence they
are linear pair. Do you agree? Draw a figure to support your answer.
Example 3:
Find the linear pair of angles which are equal?
Solution:
Let the equal linear pair of angles are x0 and x0
x0 + x0 = 1800 (why?)
2x0 =1800
x0 =
180D
2
x0 = 900
Hence, each angle = 900
i)
HJJG
In the given figure AB is a straight line, O is
HJJG JJJG
a point on AB . OC is a ray. Take a point D
in the interior of AOC, join OD.
Find AOD +DOC + COB?
HJJG
ii) In the given figure, AG is a straight line. Find
the value of 1 + 2 + 3 +4 + 5 + 6?
F
Note : i) ‘the sum of the angles at a point on the same side of the line is 180o.’
ii) ‘the sum of all the angles at a point is 360o.’
VII Class Mathematics
151
Lines and Angles
ú |ç >Ÿ Ü· “
Ôî\TdŸT¿Ã
HJJG
HJJG
1. ç|Ÿ¿Ø£ |Ÿ³+ PR ýË ÿ¿£ dŸsÁÞø ¹sK eT]jáTT dŸsÁÞø ¹sK PR ™|Õ O ÿ¿£ _+<ŠTeÚ.
JJJG
OQ
ÿ¿£ ¿ìsÁD+
i)
QOR= 500 nsTTq POQ $\Te
m+Ôá?
ii)) QOP = 1020 nsTTq QOR $\Te m+Ôá?
>·eT“¿£: 1. ¹sFjáT <ŠÇjáT+ jîTT¿£Ø –eTˆ& uó„TÈ+ ¿±“ uó„TC²\T Âs+&ƒT e«Ü¹s¿£ ¿ìsÁD²\T eT]jáTT n$ ÿ¹¿
dŸsÁÞø¹sK™|Õ –+{²sTT.
2. ¹sFjáT ¿ÃD²\ ÈÔá dŸ+|ŸPsÁ¿±\T.
1. s¹ FjáT<ŠÇjáT+ m\¢|ڟ Î&ƒÖ €dŸq•¿ÃD²\T neÚԐsTT.¿±ú €dŸq•¿ÃD²\T m\¢|ڟ Î&ƒÖ
s¹ FjáT<ŠÇjáT+ ¿±e\d¾q nedŸs+Á ýñ<TŠ . MTsÁT n+^¿£]kÍïs? MT dŸeÖ<ó‘H“•
dŸeT]Æ+#áT³Å£” ÿ¿£ |Ÿ{²“• ^jáT+&.
nHûÇw¾<‘Ý+
2. Âs+&ƒT ¿ÃD²\T 300 eT]jáTT 1500\ yîTTÔáï+ 1800 ¿£qT¿£ n$ ¹sFjáT<ŠÇjáT+
neÚԐsTT n“ eTV²wt #îbÍÎ&ƒT. B““ úeÚ n+^¿£]kÍïy? MT dŸeÖ<ó‘H“•
dŸeT]Æ+#á+&.
–<‘VŸ²sÁD 3: ÿ¿£<‘“Ôà eTs=¿£{ì dŸeÖq+>± –+&û ¹sFjáT ¿ÃD²\ ÈÔáqT ¿£qT>=q+&?
kÍ<óŠq:
dŸeÖq+>± >·\ ¹sFjáT¿ÃD²\ ÈÔáqT x0 eT]jáTT x0 n“ nqT¿=qTeTT.
x0 + x0 = 1800 (m+<ŠTÅ£”?)
2x0 =1800
180D
x =
2
0
x = 900
0
¿£qT¿£, ÿ¿=Ø¿£Ø ¿ÃD+ = 900
€ý˺<‘Ý+!
HJJG
HJJG
i)
ç|Ÿ¿£Ø |Ÿ³+ýË
ÁÞø¹sK, dŸsÁÞø¹sK AB ™|Õ ÿ¿£
JJJG AB ÿ¿£ dŸsJJJG
_+<ŠTeÚ OC ÿ¿£ ¿ìsÁD+. OC n+ÔásÁ+ýË _+<ŠTeÚ D “
rdŸT¿=“, OD ¿£\|Ÿ+&. AOD +DOC + COB?
$\Te ¿£qT>=q+&?
ii)
ç|Ÿ¿£Ø |Ÿ³+ýË AG ÿ¿£ dŸsÁÞø¹sK, 1 + 2 + 3 +4 + 5 + 6
$\Te ¿£qT>=q+&?
HJJG
F
»»dŸsÁÞø¹sK™|Õ ÿ¿£ _+<ŠTeÚ e<ŠÝ € ¹sKÅ£” ÿ¹¿ yîÕ|ŸÚq >·\ ¿ÃD²\ yîTTÔáï+ 1800 neÚÔáT+~.µµ
ii) »ÿ¿£ _+<ŠTeÚ e<ŠÝ >·\ n“• ¿ÃD²\ yîTTÔáï+ 3600µ.
>·eT“¿£: i)
7e ÔásÁ>·Ü >·DìÔáeTT
152
s¹ K\T eT]jáTT ¿ÃD²\T
HJJG
Example 4 : In the given figure, PS is a straight line, find x0.
Solution:
From the given figure, POQ = 600
QOR = x0
60
ROS = 47
0
0
But POQ + QOR + ROS = 1800 (why?)
600 + x0 + 470 = 1800
x0 + 1070 = 1800
x0 = 1800 – 1070
x0 = 730
Exercise - 4.2
Q
1.
Observe the given figure and write any 2 linear pairs of angles.
K
P
2.
Draw a pair of adjacent angles which are complementary to each other.
3.
Draw a pair of adjacent angles which are supplementary to each other.
4.
Give any two examples of adjacent angles in your surroundings.
5.
Observe the figure and write the adjacent angles.
6.
Is it possible for the following pair of angles to form a linear pair? If yes, draw them. If not,
give reason.
i)
7.
1200, 600
L
M
ii) 980, 1020
Draw the following angles as linear pair.
Write the straight line and common arm.
8.
9.
HJJG
In the given figure, AB is a straight line.
HJJG
O is a point on AB . Find the value of x.
Teacher asked students to check whether 400
and 1400 form a linear pair or not by drawing
angles. The students have drawn as follows.
Who will get correct answer?
VII Class Mathematics
153
Lines and Angles
HJJG
–<‘VŸ²sÁD 4: ‚ºÌq |Ÿ³+ýË PS ÿ¿£ dŸsÁÞø¹sK nsTTq x0 ¿£qT>=q+&.
kÍ<óŠq:
‚ºÌq |Ÿ³+ qT+&, POQ = 600
QOR = x0
60
0
ROS = 470
POQ + QOR + ROS = 1800 (m+<ŠTÅ£”?)
600 + x0 + 470 = 1800
x0 + 1070 = 1800
x0 = 1800 – 1070
x0 = 730
nuó²«dŸ+ ` 4.2
Q
1.
2.
3.
4.
‚ºÌq |Ÿ{²“• >·eT“+#á+& eT]jáTT 2 ¹sFjáT ¿ÃD²\ ÈÔá\qT sjáT+&.
ÿ¿£<‘“¿=¿£{ì |ŸPsÁ¿£ ¿ÃD²\jûT« €dŸq• ¿ÃD²\ ÈÔáqT ^jáT+&.
ÿ¿£<‘“¿=¿£{ì dŸ+|ŸPsÁ¿£ ¿ÃD²\jûT« €dŸq• ¿ÃD²\ ÈÔáqT ^jáT+&.
MT |Ÿ]dŸs\ýË úeÚ >·eT“+#û €dŸq• ¿ÃD²\Å£” dŸ+‹+~ó+º @yîHÕ  s +&ƒT
–<‘VŸ²sÁD\T ‚eÇ+&.
K
P
L
M
5. |Ÿ³+qT |Ÿ]o*+#á+&. M\jûT« €dŸq• ¿ÃD²\ ÈÔá\qT sjáT+&.
6. ‚ºÌq ¿ÃD²\ ÈÔá ¹sFjáT <ŠÇjáT+ njûT« ne¿±Xø+ eÚ+<‘? ÿ¿£yûÞø neÚqT nsTTÔû, y{ì“ ^jáT+&.
ÿ¿£yûÞø ýñq³¢sTTÔû, ¿±sÁD+ ‚eÇ+&.
i)
1200, 600
ii) 980, 1020
7. ç¿ì+~ ¿ÃD²\qT ¹sFjáT<ŠÇjáT+>± ^jáT+&. n+<ŠTýË
>·\ dŸsÁÞø¹sKqT eT]jáTT –eTˆ&uó„TC²“• çyjáT+&.
HJJG
HJJG
8. ‚ºÌq |Ÿ³+ýË AB ÿ¿£ dŸsÁÞø¹sK. AB ™|Õ O ÿ¿£ _+<ŠTeÚ. x $\Te
¿£qT¿ÃØ+&.
9. 400 eT]jáTT 1400 ¿ÃD²\T ¹sFjáT <ŠÇjáT+qT
msÁÎsÁTkÍïjîÖ ýñ<à |Ÿ³+ ^º dŸ]#áÖ&ƒeT“ ÿ¿£
{¡#sá TÁ Ôáq $<‘«sÁT\Æ Å£” #î™|ÎqT. € $<‘«sÁT\Æ T ç¿ì+~
$<óŠ+>± |Ÿ³+ ^#îqT. nsTTq me]¿ì dŸ]jî®Tq
dŸeÖ<ó‘q+ e#áTÌqT?
7e ÔásÁ>·Ü >·DìÔáeTT
154
sÁV²Ó yŽT
yûT¯
sÃw¾Ôá
s¹ K\T eT]jáTT ¿ÃD²\T
4.3 Vertically opposite angles:
Look at the picture of scissors. Observe the angles marked in the picture.
How many angles do you observe in the scissors? What are they? The
angles are 1, 2, 3 and 4.
Is there any angle which is not linear pair to 1? Yes, it is 3, which is
exactly opposite to 1. These are vertically opposite angles.
‘When two straight lines intersect each other, the pair of opposite
angles formed at the point of intersection are called as vertically
opposite angles.’
Here, in scissors two lines ‘l’ and ‘m’
intersect each other at point O. 1, 3 is a
pair of vertically opposite angles. And 2, 4
is another pair of vertically opposite angles.
Observe the pair of vertically opposite angles
in the given pictures.
Glass Table
Let us learn an important property of vertically opposite angles through an activity.
Take a white paper. Draw 3 distinct pairs of intersecting lines on this
paper. Measure the angles so formed and fill the table.
S.No.
Figure on white paper
Vertically
opposite angles
B
1
C
D
A
1 =
3 =
2 =
4 =
1 =
L
2
Equal or not
equal
3 =
N
2 =
M
K
1 =
R
Q
3
P
4 =
3 =
2 =
S
4 =
From the above table, we observe that “vertically opposite angles are equal.”
VII Class Mathematics
155
Lines and Angles
4.3 osô_óeTTK¿ÃD²\T:
ç|Ÿ¿£ØqTq• ¿£ÔîïsÁ jîTT¿£Ø ºçԐ“• #áÖ&ƒ+&. |Ÿ³+ýË >·T]ï+ºq ¿ÃD²\qT
>·eT“+#á+&. ¿£Ôsïî ýÁ Ë m“• ¿ÃD²\qT MTsÁT >·eT“+#sÁT? n$ @$T{ì? n$
1, 2, 3, 4 ¿ÃD²\T.
1 Å£”
s¹ FjáT <ŠÇjáT+ ¿±“ ¿ÃD+ @<îHÕ  –q•<‘? neÚqT eÚ+~. n~ 1 Å£”
dŸ]>±Z m<ŠTsÁT>± >·\ ¿ÃD+ 3. ‚$ ÿ¿£ ÈÔá osô_óeTTK ¿ÃD²\T neÚԐsTT.
»»Âs+&ƒT dŸsÞÁ s¹ø K\T ÿ¿£<‘“¿=¿£{ì K+&+#áT¿=q•|ŸÚÎ&ƒT, € K+&ƒq _+<ŠTeÚ e<ŠÝ
@sÁÎ&q n_óeTTK¿ÃD²\ ÈÔáqT osô_óeTTK ¿ÃD²\T n“ n+{²sÁT.µµ
¿£ÔîïsÁ |Ÿ³+ýË ‘l’ eT]jáTT ‘m’ nHû s +&ƒT s¹ K\T
ÿ¿£<‘“Ôà eTs=¿£{ì O _+<ŠTeÚ e<ŠÝ K+&+#áT¿=q•$.
‚¿£Ø&ƒ 1, 3 \T ÿ¿£ ÈÔá osô_óeTTK ¿ÃD²\T.
2, 4 nHû~ eTsà ÈÔá osô_óeTTK ¿ÃD²\T.
‚³Te+{ì osô_óeTTK ¿ÃD²\ ÈÔá\qT ç¿ì+~ ‚eNj&ƒ¦
ºçԐýË¢ >·eT“+#á+&.
¿£Ôá«+
ç¿£.dŸ+.
Ôî\¢ ¿±ÐÔá+™|Õ |Ÿ³+
osô_óeTTK ¿ÃD²\T
B
C
D
A
n&ƒ¦ sÃ&ƒT¢
dŸeÖqeÖ? ¿±<‘?
1 =
3 =
2 =
4 =
1 =
L
2
n&ƒT¦ ^Ôá\ >·\ ¿ì{ì¿ì
‚|ŸÚÎ&ƒT eTq+ osô_óeTTK ¿ÃD²\ jîTT¿£Ø ÿ¿£ eTTK«yîTq® <ósŠ ˆ“• ÿ¿£ ¿£Ôá«+ <‘Çs
HûsÁTÌÅ£”+<‘+.
ÿ¿£ Ôî\¢ ¿±ÐÔá+ rdŸT¿Ã+&. ‡ ¿±ÐÔá+™|Õ 3 $_óq• ÈÔá\ K+&ƒq ¹sK\qT ^jáT+&. ný²
@sÁÎ&q ¿ÃD²\qT ¿=*º |Ÿ{켿£ýË “+|Ÿ+&.
‚~ #û<‘Ý+!
1
n<ŠÝ+ >·\ fñ‹T\T
3 =
N
2 =
M
K
1 =
R
Q
3
P
4 =
3 =
2 =
S
4 =
™|Õ |Ÿ{켿£ qT+º »»osô_óeTTK ¿ÃD²\T dŸeÖq+µµ n“ >·eT“+#á>·\sÁT.
7e ÔásÁ>·Ü >·DìÔáeTT
156
s¹ K\T eT]jáTT ¿ÃD²\T
We can prove this property as follows.
In the figure, lines l and m intersect at O.
Let the angles formed be 1, 2, 3 and 4
1 + 4 = 1800 (why?)
3 + 4 = 1800 (why?)
1 + 4 = 3 + 4
 1 = 3, Similarly, 2 = 4
Hence, we conclude that the pair of vertically opposite angles are equal.
Example 5: Observe the figure, then find x, y and z.
Solution:
From the figure, x = 1100 (vertically opposite angles are equal)
y + 1100 = 1800 (why?)
y = 1800 – 1100
= 700
z = y (why?)
z = 700
Hence x = 1100, y = 700 and z = 700
In the figure three lines p, q and r intersect at a
point O. Observe the angles in the figure. Write
answers to the following.
i)
What is the vertically opposite angle to 1?
ii)
What is the vertically opposite angle to 6?
iii) If 2 = 500, then what is 5?
Exercise - 4.3
1.
Name three pairs of vertically opposite angles in the figure.
If AOB =450. then find DOE.
2.
HJJG
In the given figure PQ is a straight line. Check
whether x and y are vertically opposite angles
or not? Give reason.
3.
Write any three examples for vertically opposite angles in your surroundings.
4.
In the given figure, the lines l and m intersect at point P.
Observe the figure and find the values of x, y and z.
VII Class Mathematics
157
Lines and Angles
‡ <óŠsˆ“• eTq+ ‡ ç¿ì+~ $<óŠ+>± “sÁÖ|¾+#áe#áTÌ.
|Ÿ³+ýË dŸsÁÞø¹sK\T l eT]jáTT m \T _+<ŠTeÚ O e<ŠÝ K+&+#áT#áTq•$. n“
nqT¿=qTeTT.
n|ŸÚÎ&ƒT @sÁÎ&q ¿ÃD²\T 1, 2, 3 eT]jáTT4 nqT¿=qTeTT.
1 + 4 = 1800 (m+<ŠTÅ£”?)
3 + 4 = 1800 (m+<ŠTÅ£”?)
1 + 4 = 3 + 4
 1 = 3
‚<û $<óŠ+>±, 2 = 4
¿£qT¿£, osô_óeTTK ¿ÃD²\T dŸeÖq+ n“ eTq+ “sÝ]+#áe#áTÌ.
–<‘VŸ²sÁD 5: ç|Ÿ¿£ØqTq• |Ÿ{²“• >·eT“+º x, y eT]jáTT z $\Te\qT ¿£qT>=q+&.
kÍ<óŠq:
|Ÿ³+ qT+&, x = 1100 (osô_óeTTK ¿ÃD²\T dŸeÖq+)
y + 1100 = 1800 (m+<ŠTÅ£”?)
y = 1800 – 1100
= 700
z = y (m+<ŠTÅ£”?)
z = 700
¿£qT¿£, x = 1100, y = 700 (eT]jáTT) z = 700
ú |ç >Ÿ Ü· “
Ôî\TdŸT¿Ã
|Ÿ³+ýË eTÖ&ƒT dŸsÁÞø¹sK\T p, q eT]jáTT r \T ÿ¿£
_+<ŠTeÚ O e<ŠÝ K+&+#áT¿=“q$. |Ÿ³+ýË ¿ÃD²\qT
|Ÿ]o*+#á+&. ç¿ì+~ ç|ŸX•ø \Å£” dŸeÖ<ó‘H\T sjáT+&.
i) 1 ¿ì osô_óeTTK¿ÃD+ @~?
ii) 6 ¿ì osô_óeTTK¿ÃD+ @~?
iii) 2 = 500 nsTTq 5 $\Te m+Ôá?
nuó²«dŸ+ ` 4.3
1. |Ÿ³+ qT+& eTÖ&ƒT ÈÔá\ osô_óeTTK ¿ÃD²\ ÈÔá\ |sÁ¢qT |s=Øq+&.
|Ÿ³+ýË AOB =450 nsTTq DOE ¿£qT>=q+&.
HJJG
2. ‚ºÌq |Ÿ³+ýË PQ ÿ¿£ dŸsÁÞø¹sK. x eT]jáTT y
\T osô_ó e TTK ¿ÃD²\T neÚԐjî Ö ýñ < Ã
dŸ]#áÖ&ƒ+&. ¿±sÁD+ Ôî\Î+&.
3. MT |Ÿ]dŸs\ýË osô_óeTTK ¿ÃD²\TÅ£” eTÖ&ƒT –<‘VŸ²sÁD\qT sjáT+&.
4. ‚ºÌq |Ÿ ³ +ýË dŸ s Á Þ ø ¹ s K\T l eT]já T T m \T _+<Š T eÚ P e<Š Ý
K+&+#áT¿=qT#áTq•$. |Ÿ{²“• |Ÿ]o*+º x, y eT]jáTT z \T $\Te\qT
¿£qT>=q+&.
7e ÔásÁ>·Ü >·DìÔáeTT
158
s¹ K\T eT]jáTT ¿ÃD²\T
5.
6.
HJJG
HJJG
In the given figure, two lines AD and EC intersects at O.
Name two pairs of vertically opposite angles in the given
figure.
HJJG
HJJG
Two lines PS and QT intersect at M. Observe the figure and find x?
4.4 Angles made by a transversal :
While you enter into the school, do you observe that the school entrance gate often designed
with iron bars that are arranged in horizontal and vertical order, which looks beautiful? You
might have seen the railway track, the cement bars arranged vertical to the railway track. In
ladder, the wooden bars are arranged horizontally to the vertical poles. In construction of
building, the beams which are made up of iron and cement are arranged in the same way.
These situations explain the concept of a Transversal.
Look at the following pictures to understand Transversal concept.
A transversal is a straight line that intersect two or more straight lines at distinct points.
Observe the following figures.
In figure i) the line t intersects the lines m and n at two distinct points.
In figure ii) the line t intersects three lines j, k and l at three distinct points.
In figure iii) the line t intersects four lines p, q, r and s at four distinct points.
In all the above three cases, we say that the line t is a transversal.
In figure iv) the lines l, m and n intersect at one point O.
Hence, there is no transversal in figure iv.
VII Class Mathematics
159
Lines and Angles
HJJG
HJJG
5. ‚ºÌq |Ÿ³+ýË Âs+&ƒT dŸsÁÞø¹sK\T AD eT]jáTT EC \T _+<ŠTeÚ O
e<ŠÝ K+&+#áT¿=q•~. ‚ºÌq |Ÿ³+ qT+& Âs+&ƒT ÈÔá\ osô_óeTTK
¿ÃD²\ |sÁq¢ T |s=Øq+&.
HJJG
HJJG
6. Âs+&ƒT dŸsÁÞø¹sK\T PS eT]jáTT QT \T _+<ŠTeÚ M e<ŠÝ
K+&+#áT¿=q•$. |Ÿ{²“• |Ÿ]o*+º x qT ¿£qT>=q+&.
4.4 ÜsÁ«>´s¹ K#û @sÁÎ&Tƒ ¿ÃD²\T:
MTsÁT bÍsÄXÁ æ\ ý˓¿ì ç|Ÿy¥û +#û³|ŸÚÎ&ƒT, bÍsÄXÁ æ\ ç|ŸyXû ø <‘ÇsÁ+ ‚qT|Ÿ¿&£ \¦ž Ôà sÁÖ|Ÿ¿\£ Îq #ûd¾ ¿ì܌ È dŸeÖ+Ôás+Á >±
eT]jáTT “\TeÚ>± ç¿£eT+ýË neTsÁ̋& n+<Š+>± ¿£q‹&ƒT³ >·eT“+#s? MTsÁT ÂsÕýñÇ ç{²¿ù >·eT“+ºq#Ã, ÂsÕýñÇ
ç{²¿ùÅ£” n&ƒ¦+>± neT]Ìq d¾yîT+{Ù ~eTˆ\qT MTsÁT #áÖd¾ –+{²sÁT. “#îÌqýË Å£L&† “\TeÚ dŸï+uó²\Å£” #î¿£Ø
¿£&\¦ž T n&ƒ+¦ >± neTsÁ̋&eÚ+{²sTT. n<û $<ó+Š >± uóe„ q “sˆD+ýË Å£L&† ‚qT|Ÿ }#á\T, d¾yTî +{Ù\Ôà ÔájÖá sÁT#ûdq¾
<ŠÖý²\qT (;yŽT) “\TeÚ>± eT]jáTT n&ƒ+¦ >± “]ˆ+#á&†“• >·eT“+#s? ‡ dŸ+<ŠsÒÛ\“•jáTT ÜsÁ«>´s¹ K uó²eqqT
$e]kÍïsTT.
ÜsÁ«>´¹sK uó²eqqT nsÁœ+ #ûdŸT¿Ãe&ƒ+ ¿=sÁÅ£” ~>·Te ºçԐ\qT #áÖ&ƒ+&.
»»Âs+&ƒT ýñ<‘ n+Ôá¿+
£ fñ mÅ£”Øe dŸsÞ
Á s
¹ø K\qT $_óq• _+<ŠTeÚ\ e<ŠÝ K+&+#û dŸsÞ
Á s
¹ø KqT ÜsÁ«ç¹>K n+{²sÁT.µµ
ç¿ì+~ |Ÿ{²\qT >·eT“+#á+&.
|Ÿ³+ i) ýË dŸsÁÞø¹sK\T m eT]jáTT n \qT eTsà dŸsÁÞø¹sK t Âs+&ƒT $_óq• _+<ŠTeÚ\ e<ŠÝ K+&+ºq~.
|Ÿ³+ ii) ýË dŸsÁÞø¹sK\T j, k eT]jáTT l \qT eTsà dŸsÁÞø¹sK t eTÖ&ƒT $_óq• _+<ŠTeÚ\ e<ŠÝ K+&+ºq~.
|Ÿ³+ iii) ýË dŸsÁÞø¹sK\T p, q, r eT]jáTT s \qT eTsà dŸsÁÞø¹sK t H\T>·T $_óq• _+<ŠTeÚ\ e<ŠÝ K+&+ºq~.
™|Õq |s=Øq• eTÖ&ƒT dŸ+<ŠsÒÛýË¢qÖ dŸsÁÞø¹sK t “ ÜsÁ«>´s¹ K n“ n+{²eTT.
|Ÿ³+ iv) ýË dŸsÁÞø¹sK\T l, m eT]jáTT n ¹sK\T ÿ¹¿ _+<ŠTeÚ O e<ŠÝ K+&+#áT¿=qT#áTq•$.
¿£qT¿£ ‡ dŸ+<ŠsÁÒÛ+ýË ÜsÁ«>´¹sK ýñ<ŠT n“ n+{²eTT.
7e ÔásÁ>·Ü >·DìÔáeTT
160
s¹ K\T eT]jáTT ¿ÃD²\T
i)
In the figure, the line l intersects other
two lines m and n at A and B respectively.
Hence l is a transversal. Is there any other
transversal in the figure? Give reason.
ii) How many transversals can be drawn for
the given pair of lines?
Angles made by a transversal:
When a transversal p intersects two lines m and n at two distinct
points, then 8 angles will be formed as shown in the figure.
In the figure, let the angles formed by the transversal be 1, 2,
3, 4, 5, 6,7 and 8.
Now let us learn about various pairs of angles formed by a
transversal.
S.No.
Types of angles
Figure
Angles in the figure
1
Interior angles : Angles in between
the lines m and n
3, 4, 5, 6
2
Exterior angles : Angles that are
not in between the lines m and n
1, 2, 7, 8
Corresponding angles: Two angles
which are on the same side of
transversal, one interior and other
exterior but not adjacent angles are
pair of corresponding angles.
Four pairs
(1, 5),
(4, 8),
(2, 6),
(3, 7)
Alternate interior angles :
Two pairs
(4, 6),
(3,5)
3
4
5
Interior angles on either side of the
transversal but not adjacent angles
are alternate interior angles.
Two pairs
(1, 7),
(2,8)
Alternate exterior angles :
Exterior angles on either side of the
transversal but not adjacent angles
are alternate exterior angles.
VII Class Mathematics
161
Lines and Angles
€ý˺<‘Ý+!
i)
|Ÿ³+ýË dŸsÁÞø¹sK l $TÐ*q Âs+&ƒT dŸsÁÞø¹sK\T m, n
\qT esÁd>Ÿ ± A eT]jáTT B _+<ŠTeÚ\ e<ŠÝ K+&+ºq~.
n+<ŠTe\¢ l ÿ¿£ ÜsÁ«>´¹sK neÚÔáT+~. |Ÿ³+ýË ‚+¿±
@yîT®H ÜsÁ«>´¹sK\T –H•jáÖ? ¿±sÁD+ Ôî\Î+&.
ii) ÿ¿£ ÈÔá dŸsÁÞø¹sK\Å£” m“• ÜsÁ«>´¹sK\qT ^jáTe#áTÌ?
ÜsÁ«>´¹sK#û @sÁÎ&ƒT ¿ÃD²\T:
s +&ƒT dŸsÞÁ s¹ø K\T m eT]jáTT n \qT ÿ¿£ ÜsÁ«>´s¹ K p, s +&ƒT $_óq• _+<ŠTeÚ\
e<ŠÝ K+&+ºq|ŸÚÎ&ƒT, |Ÿ³+ýË #áÖ|¾+ºq $<ó+Š >± 8 ¿ÃD²\T @sÁÎ&ƒTqT. |Ÿ³+ýË
ÜsÁ«>´¹sK#û @sÁÎ&q ¿ÃD²\T 1, 2, 3, 4, 5, 6,7 eT]jáTT 8
n“ nqT¿=qTeTT.
‚|ŸÚÎ&ƒT, eTq+ ÜsÁ«>´s¹ K#û @sÁÎ&q $$<óŠ ¿ÃD²\ ÈÔá\qT >·T]+º HûsTÁ ÌÅ£”+<‘+.
ç¿£.dŸ+.
1
2
3
4
5
¿ÃD²\ÈÔá\T sÁ¿e
£ TT
|Ÿ³eTT
|Ÿ³+ýË ¿ÃD²\T
n+ÔásÁ¿ÃD²\T:
3, 4, 5, 6
u²VŸ²«¿ÃD²\T :
1, 2, 7, 8
dŸ<ŠXø¿ÃD²\T:
H\T>·T ÈÔá\T
dŸsÁÞø¹sK\T m eT]jáTT n \ eT<óŠ« qTq•
¿ÃD²\T
dŸsÁÞø¹sK\T m eT]jáTT n eT<óŠ« ýñ“
¿ÃD²\T
ÜsÁ « >´ ¹ s KÅ£ ” ÿ¹ ¿ yî Õ | Ÿ Ú q eÚ+& , ÿ¿£ { ì
n+ÔásÁ+ýË eTs=¿£{ì u²VŸ²«+>± –+³Ö,
€dŸq• ¿ÃD²\T ¿±“ y{ì“ dŸ<Š Xø¿ÃD²\T
n+{²sÁT.
(1, 5),
(4, 8),
(2, 6),
(3, 7)
@¿±+ÔásÁ¿ÃD²\T:
Âs+&ƒT ÈÔá\T
@¿£u²VŸ²«¿ÃD²\T:
Âs+&ƒT ÈÔá\T
(4, 6),
(3,5)
ÜsÁ«>´¹sKÅ£” #îsÿ£ yîÕ|ŸÚq eÚ+³Ö, €dŸq•
¿ÃD²\T ¿±“, n+ÔásÁ ¿ÃD²\qT @¿±+ÔásÁ
¿ÃD²\T n+{²sÁT.
(1, 7),
(2,8)
ÜsÁ«>´¹sKÅ£” #îs=¿£ yîÕ|ŸÚq eÚ+³Ö,
€dŸq• ¿ÃD²\T ¿±“, u²VŸ²«¿ÃD²\qT
@¿£ u²VŸ²«¿ÃD²\T n+{²sÁT.
7e ÔásÁ>·Ü >·DìÔáeTT
162
s¹ K\T eT]jáTT ¿ÃD²\T
S.No
Types of angles
Figure
Angles in the figure
6
Co-interior angles : Interior angles
on same side of transversal are cointerior angles.
Two pairs
(3, 6),
(4,5)
7
Co-exterior angles:
Exterior angles on same side of
transversal are co-exterior angles.
Two pairs
(2, 7),
(1,8)
Observe the figures
Figure Transversal
(i)
(ii)
Exterior
angles
i) and ii) then fill the table.
Pairs of
Interior
corresponding
angles
angles
Pairs of
alternative
interior
angles
Pairs of
alternative
exterior
angles
a, e
b, f
c, g
d, h
n
1, 4,
5, 8
4.4.1 Transversal on parallel lines :
Two coplanar lines which do not intersect at any point are called as parallel
lines.
In the picture, the horizontal bars of window represent parallel lines and
the vertical bar which intersect horizontal bars represent a transversal.
VII Class Mathematics
162
Lines and Angles
ç¿£.dŸ+.
|Ÿ³eTT
¿ÃD²\ÈÔá\T sÁ¿e
£ TT
dŸV²Ÿ n+ÔásY ¿ÃD²\T :
6
7
Âs+&ƒT ÈÔá\T
ÜsÁ«>´s¹ KÅ£” ÿ¹¿ yî|Õ ÚŸ q >·\ n+ÔásÁ ¿ÃD²\T
dŸV²Ÿ n+ÔásÁ ¿ÃD²\T n>·TqT.
(3, 6),
(4,5)
Âs+&ƒT ÈÔá\T
dŸV²Ÿ u²VŸ²« ¿ÃD²\T :
(2, 7),
(1,8)
ÜsÁ«>´s¹ KÅ£” ÿ¹¿ yî|Õ ÚŸ q >·\ u²VŸ²« ¿ÃD²\T
dŸV²Ÿ u²VŸ²« ¿ÃD²\T n>·TqT.
ú |ç >Ÿ Ü· “
Ôî\TdŸT¿Ã
|Ÿ³+ (i) eT]jáTT (ii) \qT >·eT“+º |Ÿ{켿£qT |ŸP]+#á+&.
|Ÿ³+ (i)
|Ÿ³+
ÜsÁ«>´¹sK
(i)
n
(ii)
|Ÿ³+ýË ¿ÃD²\T
‹VŸ²«¿ÃD²\T
|Ÿ³+ (ii)
dŸ<Š Xø ¿ÃD²\
ÈÔá\T
n+ÔásÁ
¿ÃD²\T
@¿±+ÔásÁ
¿ÃD²\ ÈÔá\T
@¿£u²VŸ²«
¿ÃD²\ ÈÔá\T
a, e
b, f
c, g
d, h
1, 4,
5, 8
4.5 dŸeÖ+ÔásÁ s¹ K\™|Õ ÜsÁ«>´s¹ K :
ÿ¹¿ Ôá\+ý˓ Âs+&ƒT dŸsÁÞø¹sK\T K+&+#áT ¿=q“#à y{ì“ dŸeÖ+ÔásÁ¹sK\T n“ n+{²sÁT.
ç|Ÿ¿£ØºçÔá+ýË, ¿ì{ì¿¡¿ì n&ƒ¦+>± neT]Ìq }#á\T dŸeÖ+ÔásÁ¹sK\qT dŸÖºdï y{ì“ K+&dŸÖï
“\TeÚ>± –q• }#á, ÜsÁ«>´¹sKqT dŸÖºdŸTï+~.
7e ÔásÁ>·Ü >·DìÔáeTT
163
s¹ K\T eT]jáTT ¿ÃD²\T
Now let us know the properties of pair of angles formed when a transversal intersect a pair of
parallel lines.
1.
Property of corresponding angles:
Now let us know the property of corresponding angles through an activity.
Take a graph sheet. Draw a pair of parallel lines m and n
and draw a transversal p which intersects ‘m’ and ‘n’ at
two distinct points on a graph sheet. Let the angles formed
be 1, 2, 3, 4, 5, 6,7 and 8 as shown in the
figure.
Take a small paper and cut it into the shape of 1, then
keep that piece at 5.
What do you observe? You observe that paper coincides
with 5 as shown in the figure.
Take another small paper and cut it in the shape of 2.
Now keep that piece at 6. What do you observe?
You observe that it coincides with 6. Similarly, by
continuing this activity you observe that 4 coincides
with 8 and 3 coincides with 7.
From the above activity, we conclude that, ‘when a transversal intersects a pair of parallel
lines then the corresponding angles are equal.’
Here, in the above figure if 1 = 1020, then 5 = 1020,
if 8 = 780, then 4 = 780
JJJG JJJG
JJJG
Example 6: In the given figure AB || CD and AE is transversal.
Solution:
If BAC =1200, then find x and y?
JJJG JJJG
JJJG
In the given figure, AB || CD and AE is transversal
BAC = 1200
ACD = x
DCE = y
BAC = DCE (correpsonding angles are equal)
 y = 1200
x + y = 1800 (Linear pair of angles are supplementary)
x + 1200 = 1800
x =1800 – 1200
 x = 600
Hence x = 600, y =1200.
VII Class Mathematics
164
Lines and Angles
ÿ¿£ ÈÔá dŸeÖ+ÔásÁ¹sK\qT ÜsÁ«>´¹sK K+&+ºq|ŸÚÎ&ƒT @sÁÎ&û ¿ÃD²\ ÈÔá\ \¿£ŒD²\qT >·Ö]Ì eTq+ ‚|ŸÚÎ&ƒT
Ôî\TdŸTÅ£”+<‘+.
1. dŸ<ŠXø¿ÃD²\ <óŠsÁˆ+ :
‚|ŸÚÎ&ƒT eTq+ dŸ<ŠXø¿ÃD²\ jîTT¿£Ø <óŠsÁˆ+ >·Ö]Ì ç¿ì+~ ¿£Ôá«+ <‘Çs Ôî\TdŸTÅ£”+<‘+.
‚~ #û<‘Ý+!
¿£Ôá«+
ÿ¿£ ç>±|˜t ¿±ÐÔá+ rdŸT¿Ã+&. <‘“™|Õ ÿ¿£ ÈÔá dŸeÖ+ÔásÁ ¹sK\T m
eT]jáTT n ^jáT+&. dŸeÖ+ÔásÁ ¹sK\T »mµ eT]jáTT »nµ \qT Âs+&ƒT
$_óq• _+<ŠTeÚ\ e<ŠÝ K+&+#û $<óŠ+>± ÿ¿£ ÜsÁ«>´¹sK p “ |Ÿ³+ýË
#áÖ|¾q³T¢ ^jáT+&. n|ŸÚÎ&ƒT ç|Ÿ¿£Ø |Ÿ³+ýË #áÖ|¾q $<óŠ+>± @sÁÎ&q
¿ÃD²\qT 1, 2, 3, 4, 5, 6, 7 eT]jáTT 8 n“
nqT¿=qTeTT..
ÿ¿£ ºq• ¿±ÐԐ“• rdŸT¿=“ <‘““ 1 €¿±sÁ+ýË¿ì ¿£Üï]+#á+&.
¿£Üï]+ºq € eTT¿£ØqT 5 e<ŠÝ –+#á+&.
úeÚ @$T >·eT“+#eÚ? |Ÿ³+ýË #á֓ $<ó+Š >± ¿±ÐÔá+ eTT¿£Ø 5ÔÃ
@¿¡u$
„ó dŸT+ï <Š“ MTsÁT >·eT“kÍïsTÁ . eTsà ºq• ¿±ÐԐ“• rdŸTÅ£”“ <‘““
2 €¿±sÁ+ýË ¿£Üï]+#á+&. ¿£Üï]+ºq € eTT¿£ØqT 6 e<ŠÝ
–+#á + & . úeÚ @$T >· e T“+#eÚ? ¿±ÐÔá + eTT¿£ Ø 6 ÔÃ
@¿¡u$„ó dŸT+ï <Š“ MTsÁT >·eT“kÍïsTÁ . ‚<û$<ó+Š >± ¿£Ôá«+ ¿=qkÍÐ+#á&+ƒ <‘Çs ¿ÃD²\T 4, 8 \T @¿¡u$„ó kÍïjTá “
eT]jáTT ¿ÃD²\T 3, 7 \T @¿¡uó„$kÍïjáT“ >·eT“+#á+&.
™|Õ ¿£Ôá«+ qT+&, »ÿ¿£ ÈÔá dŸeÖ+ÔásÁ s¹ K\ qT ÜsÁ«>´¹sK K+&+ºq|ŸÚÎ&ƒT @sÁÎ&q dŸ<Š Xø ¿ÃD²\T dŸeÖq+µ n“
Ôî\TdŸTï+~.
‚¿£Ø&ƒ ™|Õ|³
Ÿ +ýË 1 = 1020 nsTTÔû 5 = 1020 neÚÔáT+~.
ÿ¿£yÞû ø 8 = 780 nsTTÔû 4 = 780 neÚÔáT+~.
–<‘VŸ²sÁD 6: ‚ºÌq |Ÿ³+ýË
JJJG JJJG
AB || CD
JJJG
eT]jáTT AE ÿ¿£ ÜsÁ«>´¹sK.
BAC =1200 nsTTÔû x eT]jáTT y\qT ¿£qT>=q+&?
kÍ<óŠq:
JJJG JJJG
JJJG
‚ºÌq |Ÿ³+ýË AB || CD eT]jáTT AE ÿ¿£ ÜsÁ«>´¹sK
BAC = 1200
ACD = x
DCE = y
BAC = DCE (dŸ<ŠXø¿ÃD²\T dŸeÖq+)
 y = 1200
x + y = 1800 (¹sFjáT <ŠÇjáT+ m|Ÿð&ƒT dŸ+|ŸPsÁ¿±\T)
x + 1200 = 1800
x =1800 – 1200
 x = 600
¿£qT¿£ x = 600, y =1200.
7e ÔásÁ>·Ü >·DìÔáeTT
165
s¹ K\T eT]jáTT ¿ÃD²\T
In the figure, p || q and t is a transversal.
Observe the angles formed.
i) If 1 = 1000, then what is 5?
ii) If 8 = 800, then what is 4?
iii) If 3 = 1450, then what is 7?
iv) If 6=300, then what is 2?
2)
Property of alternate interior angles:
Let us find the relation between alternate interior
angles formed when a transversal intersect a pair
of parallel lines.
Let p || q and r is a transversal.
From figure 1 = 3 (why?)
1 = 5 (why?)
Hence 3 = 5
Similarly 4 = 6
Hence, we conclude that, ‘when a transversal intersect a pair of parallel lines, then the
alternate interior angles are equal.’
In the above figure if 3 = 1100 then 5 = 1100.
If 4 = 700 then 6 = 700.
JJJG JJJG
Example 7: In the given figure, BA || CD and BC is transversal. Find x.
JJJG JJJG
Solution:
In the given figure, BA || CD and BC is transversal
C = x + 350 and B = 600
C = B (' alternate interior angles are equal)
x + 350 = 600
x = 600 – 350
 x = 250
What is the relation between alternate exterior angles formed by a
transversal on parallel lines?
3)
Property of Co- interior angles:
Let us find the relation between co interior angles formed when a transversal intersect a
pair of parallel lines.
Let n || m and l is a transversal. From the figure,
2 + 3 = 1800 (why?)
2 = 6 (why?)
Hence, 3 +6= 1800
similarly, 4 +5 = 1800
VII Class Mathematics
166
Lines and Angles
ú |ç >Ÿ Ü· “
Ôî\TdŸT¿Ã
|Ÿ³+ýË p || q eT]jáTT t ÿ¿£ ÜsÁ«>´¹sK.
|Ÿ³+ýË ¿ÃD²\qT |Ÿ]o*+º ç¿ì+~ y“¿ì Èy‹T\T sjáT+&.
i) If 1 = 1000 nsTTq 5 $\Te m+Ôá?
ii) If 8 = 800 nsTTq 4 $\Te m+Ôá?
iii) If 3 = 1450 nsTTq 7 $\Te m+Ôá?
iv) If 6=300 nsTTq 2 $\Te m+Ôá?
2) @¿±+ÔásÁ¿ÃD²\ <óŠsÁˆ+:
ÿ¿£ ÈÔá dŸeÖ+ÔásÁ s¹ K\qT ÜsÁ«>´s¹ K K+&+ºq|ŸÚÎ&ƒT @sÁÎ&û @¿±+Ôás¿Á ÃD²\
eT<óŠ« >·\ dŸ+uó„+<Š+ >·Ö]Ì eTq+ ‚|ŸÚÎ&ƒT Ôî\TdŸTÅ£”+<‘+.
p || q eT]jáTT r ÿ¿£ ÜsÁ«>´s
¹ K nqT¿=+<‘+.
(m+<ŠTÅ£”?)
|Ÿ³+ qT+&, 1 = 3
1 = 5
(m+<ŠTÅ£”?)
3 = 5
¿£qT¿£,
‚<û $<óŠ+>± 4 = 6
¿£qT¿£,
eTq+ “sÆ]+#áe#áTÌqT.
™|Õ |Ÿ³+ýË 3 = 1100 nsTTq#à 5 = 1100 neÚÔáT+~.
ÿ¿£yûÞø 4 = 700 nsTTq#Ã 6 = 700. neÚÔáT+~.
»»ÿ¿£
ÈÔá
dŸeÖ+ÔásÁ
–<‘VŸ²sÁD 7 : ‚ºÌq |Ÿ³+ýË,
kÍ<óŠq:
‚ºÌq |Ÿ³+ýË,
¹sK\qT
ÜsÁ«>´¹sK
JJJG JJJG
BA || CD
JJJG JJJG
BA || CD
K+&+ºq|ŸÚÎ&ƒT
@sÁÎ&û
@¿±+ÔásÁ
¿ÃD²\T
dŸeÖq+µµ
n“
eT]jáTT BC ÿ¿£ ÜsÁ«>´s¹ K nsTTq xqT ¿£qT>=q+&?
eT]jáTT BC ÿ¿£ ÜsÁ«>´¹sK
C = x + 350 eT]jáTT B = 600
C = B (' @¿±+ÔásÁ
¿ÃD²\T dŸeÖq+)
x + 350 = 600
x = 600 – 350
 x = 250
€ý˺<‘Ý+!
ÿ¿£ ÈÔá dŸeÖ+ÔásÁ s¹ K\qT ÜsÁ«>´s¹ K K+&+ºq|ŸÚÎ&ƒT @sÁÎ&û @¿£ u²VŸ²«¿ÃD²\
eT<ó«Š >·\ dŸ+‹+<ó+Š @$T{ì?
3) ÜsÁ«>´¹sK ÿ¹¿ yîÕ|ŸÚq >·\ n+ÔásÁ¿ÃD²\ <óŠsÁˆ+:
ÿ¿£ ÈÔá dŸeÖ+ÔásÁ ¹sK\ qT ÜsÁ«>´¹sK K+&+ºq|ŸÚÎ&ƒT @sÁÎ&q, ÜsÁ«>´¹sKÅ£” ÿ¹¿ yîÕ|ŸÚq >·\ n+ÔásÁ¿ÃD²\
eT<óŠ« >·\ dŸ+‹<óŠ+ >·Ö]Ì eTq+ ‚|ŸÚÎ&ƒT Ôî\TdŸTÅ£”+<‘+.
Âs+&ƒT dŸsÁÞø¹sK\T n || m eT]jáTT l ÿ¿£ ÜsÁ«>´¹sK .
|Ÿ³+ qT+&,
2 + 3 = 1800 (m+<ŠTÅ£”?)
2 = 6 (m+<ŠTÅ£”?)
¿£qT¿£ 3 +6= 1800
‚<û$<óŠ+>±, 4 +5 = 1800
7e ÔásÁ>·Ü >·DìÔáeTT
167
s¹ K\T eT]jáTT ¿ÃD²\T
Hence, we conclude that, ‘when a transversal intersect a pair of parallel lines, then the interior
angles on the same side of transversal are supplementary.’
In the above figure,
if 3 = 1050 then 6 = 750
if 4 = 750 then 5 = 1050
JJJJG JJJG
Example 8: In the figure MN || KL and MK is transversal. Find x.
JJJJG JJJG
Solution :
From the figure, MN || KL and MK is transversal
M = 2x and K = x + 300
M +K = 1800 (' co-interior angles are supplementary)
2x + x + 300 = 1800
3x + 300 = 1800
3x = 1800 – 300
3x = 1500
x=
150D
3
 x = 500
Example 9 : In the figure AB || DE and C is a point in between them. Observe the figure, then
find x, y and BCD.
Solution :
In the figure AB || DE and C is a point in between them.
Draw a parallel line CF to AB through C.
AB || CF , and BC is a transversal.
x + 1030 = 1800 (why?)
x = 1800 – 1030
x = 770
From the figure,
DE || CF and CD is a transversal.
y + 1030 = 1800 (why?)
y = 1800 – 1030
y = 770
and BCD = x + y
= 770 + 770
= 1540
VII Class Mathematics
168
Lines and Angles
¿£qT¿£,
»»ÿ¿£ ÈÔá dŸeÖ+ÔásÁ s
¹ K\qT ÜsÁ«>´s
¹ K K+&+ºq|ŸÚÎ&ƒT @sÁÎ&q, ÜsÁ«>´s
¹ KÅ£” ÿ¹¿ yî|
Õ ÚŸ q >·\ n+Ôás¿
Á ÃD²\T
dŸ+|ŸPsÁ¿±\Tµµ
|Ÿ³+ýË
n“ eTq+ “sÆ]+#áe#áTÌqT.
3 = 1050 nsTTq 6 = 750 neÚÔáT+~.
4 = 750 nsTTq 5 = 1050 neÚÔáT+~.
–<‘VŸ²sÁD 8 : |Ÿ³+ýË
JJJJG JJJG
MN || KL
kÍ<óŠq :
JJJJG JJJG
eT]jáTT MK ÿ¿£ ÜsÁ«>´¹sK
nsTTq x qT ¿£qT¿ÃØ+&?
|Ÿ³+ýË MN || KL eT]jáTT MK ÿ¿£ ÜsÁ«>´¹sK
|Ÿ³+ qT+&, M = 2x eT]jáTT K = x + 300
M +K = 1800 (ÜsÁ«>´¹sKÅ£”
ÿ¹¿ yîÕ|ŸÚq >·\ n+ÔásÁ¿ÃD²\ ÈÔá dŸ+|ŸPsÁ¿±\T)
2x + x + 300 = 1800
3x + 300 = 1800
3x = 1800 – 300
3x = 1500
x=
150D
3
 x = 500
–<‘VŸ²sÁD 9 : ‚ºÌq ºçÔá+ýË
eT]jáTT
kÍ<óŠq :
AB || DE
BCD $\Te\qT
eT]jáTT y{ì eT<óŠ«ýË ÿ¿£ _+<ŠTeÚ C. |Ÿ{²“• |Ÿ]o*+º x, y
¿£qT>=q+&?
|Ÿ³+ qT+& AB || DE eT]jáTT y{ì eT<óŠ«ýË ÿ¿£ _+<ŠTeÚ C.
AB
¿ì dŸeÖ+ÔásÁ ¹sK CF qT C qT+& ^jáT+&.
AB || CF eT]jáTT BC
ÿ¿£ ÜsÁ«>´¹sK
x + 1030 = 1800 (m+<ŠTÅ£”?)
x = 1800 – 1030
x = 770
|Ÿ³+ qT+&,
eT]jáTT CD ÿ¿£ ÜsÁ«>´¹sK
y + 1030 = 1800 (m+<ŠTÅ£”?)
DE || CF
y = 1800 – 1030
y = 770
eT]jáTT BCD = x + y
= 770 + 770
= 1540
7e ÔásÁ>·Ü >·DìÔáeTT
169
s¹ K\T eT]jáTT ¿ÃD²\T
1.
In the figure, m || n and t is a transversal.
i)
If 3 = 1160 then what is 5?
ii) If 4 = 510 then what is 5?
iii) If 1 = 1230 then what is 7?
iv) If 2 = 660 then what is 7?
What is the relation between co-exterior angles, when a transversal cuts
a pair of parallel lines?
Now, let us find whether the two straight lines are parallel or not in the following cases:
Case 1: When the corresponding angles are equal: Let us find whether the two straight lines
are parallel or not through an activity.
Take a white paper and draw a pair of non-parallel lines p and q and a
transversal r shown in the figure 1. Measure the corresponding angles and
fill the table. Now, take a graph sheet and draw a pair of parallel lines m, n
and a transversal t as shown in the figure 2. Measure the pair of corresponding
angles and fill the table.
S.No
Figure
1
Corresponding angles
1 =
5 =
2 =
6 =
3 =
Equal or not
7 =
4 =
8 =
2
1 =
5 =
2 =
6 =
3 =
7 =
4 =
8 =
What do you observe from the above table? If the pair of corresponding angles are not
equal, then what do you say about the pair of lines? If the pair of corresponding angles are equal,
then what do you say about the pair of lines? Which pair of lines are parallel? Hence, we conclude
that ‘when a transversal intersects two lines and a pair of corresponding angles are equal
then, the two lines are parallel.’
VII Class Mathematics
170
Lines and Angles
ú |ç >Ÿ Ü· “
Ôî\TdŸT¿Ã
1. |Ÿ³+ýË m || n eT]jáTT t ÿ¿£ ÜsÁ«>´¹sK.
i) 3 = 1160 nsTTq 5 $\Te m+Ôá neÚÔáT+~?
ii) 4 = 510nsTTq 5 $\Te m+Ôá neÚÔáT+~?
iii) 1 = 1230 nsTTq 7 $\Te m+Ôá neÚÔáT+~?
iv) 2 = 660 nsTTq 7 $\Te m+Ôá neÚÔáT+~?
€ý˺<‘Ý+!
ÿ¿£ ÈÔá dŸeÖ+ÔásÁ¹sK\qT ÜsÁ«>´¹sK K+&+ºq|ŸÚÎ&ƒT @sÁÎ&û dŸVŸ²u²VŸ²«¿ÃD²\ eT<óŠ«
>·\ dŸ+‹+<óŠ+ @$T{ì?
‚|ŸÚÎ&ƒT eTq+ ç¿ì+~ dŸ+<ŠsÒÛ\ýË s +&ƒT dŸsÞÁ s¹ø K\T dŸeÖ+Ôás+Á >± eÚ+{²jáÖ ýñ<‘ nHû $wŸjÖá “• Ôî\TdŸTÅ£”+<‘+:
dŸ+<ŠsÒÁ +Û 1. dŸ<Š Xø¿ÃD²\T dŸeÖq+ nsTTq|ŸÚÎ&ƒT. s +&ƒT dŸsÞÁ s¹ø K\T dŸeÖ+Ôás+Á >± –+{²jáÖ ýñ<‘ nHû $wŸjÖ
á “•
eTq+ ÿ¿£ ¿£Ôá«+ #ûd¾ Ôî\TdŸTÅ£”+<‘+
‚~ #û<‘Ý+!
ÿ¿£ Ôî\¢ ¿±ÐԐ“• rdŸT¿=qTeTT. |Ÿ³+ 1 ýË #áÖ|¾q $<óŠ+>± <‘“™|Õ dŸeÖ+ÔásÁ+>± ýñ“ Âs+&ƒT
dŸsÁÞø¹sK\T p eT]jáTT q \qT ^jáT+&. y{ì“ K+&+#û³³T¢ ÿ¿£ ÜsÁ«>´¹sK r qT ^jáT+&. n|ŸÚÎ&ƒT
@sÁÎ&q dŸ<ŠXø¿ÃD²\T ¿=*º |Ÿ{켿£qT |ŸP]+#á+&. ‚|ŸÚÎ&ƒT ÿ¿£ ç>±|˜ŸÚ ¿±ÐÔá+ rdŸT¿=“ <‘“™|Õ
¿£Ôá«+
|Ÿ³+ 2 ýË #áÖ|¾q $<óŠ+>± ÿ¿£ ÈÔá dŸeÖ+ÔásÁ¹sK\T m , n eT]jáTT y{ì™|Õ ÿ¿£ ÜsÁ«>´¹sK t “
^jáT+&. ‡ dŸ+<ŠsÁÒÛ+ýË @sÁÎ&q dŸ<ŠXø ¿ÃD²\T ¿=*º |Ÿ{켿£ýË çyjáT+&.
ç¿£.dŸ+.
dŸ<ŠXø¿ÃD²\T
|Ÿ³+
1
1 =
5 =
2 =
6 =
3 =
dŸeÖqeÖ?¿±<‘?
7 =
4 =
8 =
2
1 =
5 =
2 =
6 =
3 =
7 =
4 =
8 =
™|Õ |Ÿ{켿£ qT+& MTsÁT @$T >·eT“+#sÁT? dŸ<ŠXø¿ÃD²\T dŸeÖq+>± ýñq|ŸÚÎ&ƒT, dŸsÁÞø¹sK\ ÈÔá >·T]+º MTsÁT
@$T #î|ŸÎ>·\sÁT? dŸ<ŠXø¿ÃD²\T dŸeÖq+>± –q•|ŸÚÎ&ƒT, dŸsÁÞø¹sK\ ÈÔá >·T]+º MTsÁT @$T #î|ŸÎ>·\sÁT? @
dŸsÁÞø¹sK\ ÈÔá dŸeÖ+ÔásÁ+>± ¿£\eÚ?
n“ #î|Ο e#áTÌ.
»»ÿ¿£ ÈÔá dŸsÁÞø¹sK\qT ÜsÁ«>´¹sK K+&+ºq|ŸÚ&ƒT @sÁÎ&q dŸ<ŠXø¿ÃD²\ÈÔá
dŸeÖq+ nsTTÔû € s
 +&ƒT dŸsÞ
Á s
¹ø K\T dŸeÖ+Ôás+
Á >± –+{²sTT.µµ
7e ÔásÁ>·Ü >·DìÔáeTT
171
s¹ K\T eT]jáTT ¿ÃD²\T
Example 10: In the figure transversal p intersects two lines m and n.
Observe the figure, check whether m || n or not.
Solution:
In the given figure, it is given that each angle in the pair
of corresponding angles is 450. So they are equal.
Since a pair of corresponding angles are equal the lines are parallel.
Hence, m || n.
Case 2. When the alternate interior angles are equal.
Let m and n are two straight lines and r is a transversal.
alternate interior angles are equal
Let 3 = 5
But 3 = 1 (why?)
 1 = 5
Since corresponding angles are equal, the lines are parallel.
Hence, m || n.
Case 3. When the co interior angles are supplementary.
Let p, q are two straight lines and r is a transversal.
co-interior angles are supplementary.
Let 4 + 5 = 180 0
But 1 +4 = 180 0 (why?)
4 +5 = 1 + 4
 1 =5
Since corresponding angles are equal, the lines are parallel.
Hence, p || q.
From the figure, state which property that is used
in each of the following.
i)
If 3 = 5 then p || q.
ii) If 3 + 6 = 1800 then p || q.
iii) If 3 = 8 then p || q.
iv) If p || q then 1 = 8.
1.
When a transversal intersects two lines and a pair of alternate exterior angles are equal,
What can you say about the two lines?
2.
When a transversal intersects two lines and a pair of co exterior angles are supplementary,
what can you say about the two lines?
VII Class Mathematics
172
Lines and Angles
–<‘VŸ²sÁD 10: ‚ºÌq |Ÿ³+ýË m, n \T Âs+&ƒT dŸsÁÞø¹sK\T. p ÿ¿£ ÜsÁ«>´¹sK. |Ÿ{²“•
|Ÿ]o*+º m || n neÚÔáT+<à ýñ<à ¿£qT¿ÃØ+&?
‚ºÌq |Ÿ³+ýË ÿ¿£ ÈÔá dŸ<ŠXø ¿ÃD²\T ÿ¿=Ø¿£Ø{ì 450 >± ‚eNj&+~.
‚$ dŸeÖH\T. ÿ¿£ ÈÔá dŸsÞÁ s¹ø K\qT ÜsÁ«>´s¹ K K+&+ºq|ŸÚ&ƒT @sÁÎ&q
dŸ < Š  Xø ¿ÃD²\ ÈÔá dŸ e Öq+ nsTTq#à €  s +&ƒ T dŸ s Á Þ ø ¹ s K\T
dŸeÖ+ÔásÁ+>± –+{²sTT. ¿£qT¿£, m || n neÚÔáT+~.
kÍ<óŠq:
dŸ+<ŠsÁÒÛ+ 2.
@¿±+ÔásÁ ¿ÃD²\T dŸeÖq+ nsTTq|ŸÚÎ&ƒT.
m, n \T
Âs+&ƒT dŸsÁÞø¹sK\T eT]jáTT r ÿ¿£ ÜsÁ«>´¹sK.
ÿ¿£ ÈÔá @¿±+ÔásÁ ¿ÃD²\T dŸeÖq+
3 = 5 nqT¿=qTeTT.
¿±ú, 3 = 1 (m+<ŠTÅ£”?)
 1 = 5
dŸ<ŠXø ¿ÃD²\ ÈÔá dŸeÖq+ nsTTq#à € Âs+&ƒT dŸsÁÞø¹sK\T
dŸeÖ+ÔásÁ+>± –+{²sTT.
¿£qT¿£, p || q .
dŸ+<ŠsÁÒÛ+ 3.
ÜsÁ«>´s¹ K ÿ¹¿ yî|
Õ ÚŸ q >·\ n+ÔásÁ ¿ÃD²\T dŸ+|ŸPsÁ¿±\T nsTTq|ŸÚÎ&ƒT.
p, q \T
Âs+&ƒT dŸsÁÞø¹sK\T eT]jáTT r ÿ¿£ ÜsÁ«>´¹sK>± rdŸTÅ£”H•+
ÜsÁ«>´¹sKÅ£” ÿ¹¿ yîÕ|ŸÚq >·\ n+ÔásÁ ¿ÃD²\T dŸ+|ŸPsÁ¿±\T ¿£qT¿£
4 + 5 = 1800 nqT¿=qTeTT.
¿±ú 1 +4 = 1800 (m+<ŠTÅ£”?)
4 +5 = 1 + 4
 1 =5
dŸ<ŠXø ¿ÃD²\ ÈÔá dŸeÖq+ ¿£qT¿£ € Âs+&ƒT dŸsÁÞø¹sK\T dŸeÖ+ÔásÁ+>± –+{²sTT.
¿£qT¿£, p || q.
ú |ç >Ÿ Ü· “
Ôî\TdŸT¿Ã
|Ÿ{²“• |Ÿ]o*+º ç¿ì+<Š ‚ºÌq ç|ŸÜ<‘“ýË –|ŸjÖî Ð+ºq H«já֓• çyjáT+&.
i) 3 = 5 nsTTq p || q.
ii) 3 + 6 = 1800 nsTTq p || q.
iii) 3 = 8 nsTTq p || q.
iv) p || q nsTTq 1 = 8.
nHûÇw¾<‘Ý+
1. ÿ¿£ ÈÔá dŸsÁÞø¹sK\qT ÜsÁ«>´¹sK K+&+ºq|ŸÚ&ƒT @sÁÎ&q @¿£ u²VŸ²«¿ÃD²\ ÈÔá dŸeÖq+ nsTTq €
Âs+&ƒT dŸsÁÞø¹sK\T >·Ö]Ì úyû$T #î|ŸÎ>·\eÚ?
2. ÿ¿£ ÈÔá dŸsÁÞø¹sK\qT ÜsÁ«>´¹sK K+&+ºq|ŸÚ&ƒT @sÁÎ&q dŸV²Ÿ u²VŸ²«¿ÃD²\ ÈÔá dŸ+|ŸPsÁ¿±\T nsTTq#Ã
€ Âs+&ƒT dŸsÁÞø¹sK\T >·Ö]Ì úyû$T #î|ŸÎ>·\eÚ?
7e ÔásÁ>·Ü >·DìÔáeTT
173
s¹ K\T eT]jáTT ¿ÃD²\T
Exercise - 4.4
1.
In the given figure, two lines p || q and r is transversal.
If 3 =1350, then find the remaining angles.
HJJJ HJJJ
In the given figure, AB || CD
JJJG
and DE is a transversal. Find x.
2.
3.
In the given figure, m || n and p is transversal.
Find x and y.
JJJG JJJG JJJG
In the given figure, AB || CD || FE .
4.
Find x, y and AEC.
5.
In the given figure, a transversal t intersects two lines p and q. Check
whether p || q or not.
6.
7.
In the given figure if l || m, then find x, y and z.
In the given figure p, q, r and s are parallel lines
and t is a transversal. Find x, y and z.
8.
JJJG JJJG
In the given figure AB || CD and E is a point in
between them. Find x + y + z.
JJJG
(Hint : Draw a parallel line to AB through E)
VII Class Mathematics
174
Lines and Angles
nuó²«dŸ+ ` 4.4
1. ‚ºÌq |Ÿ³+ýË p || q eT]jáTT r ÿ¿£ ÜsÁ«>´¹sK 3 =1350 nsTTq#Ã
$TÐ*q n“• ¿ÃD²\qT ¿£qT>=q+&?
JJJG
HJJJ HJJJ
2. ‚ºÌq |Ÿ³+ýË, AB || CD eT]jáTT DE
ÿ¿£ ÜsÁ«>´s¹ K nsTTq x $\Te ¿£qT>=q+&.
3. ‚ºÌq |Ÿ³+ýË, m || n eT]jáTT p ÿ¿£ ÜsÁ«ç¹>K nsTTq x eT]jáTT
y $\Te\qT ¿£qT>=q+&.
JJJG JJJG JJJG
4. ‚ºÌq |Ÿ³+ýË, AB || CD || FE nsTTq
x, y eT]já T T AEC $\Te\qT
¿£qT>=q+&.
5. ‚ºÌq |Ÿ³+ýË Âs+&ƒT dŸsÁÞø¹sK\T p eT]jáTT q \qT ÿ¿£ ÜsÁ«>´¹sK t
K+&+#îqT. p || q neÚÔáT+<à ýñ<à dŸ]#áÖ&ƒ+&?
6. ‚ºÌq |Ÿ³+ýË l || m nsTTq x, y
eT]jáTT z $\Te\qT ¿£qT>=q+&?
7. ‚ºÌq |Ÿ³+ýË p, q, r eT]jáTT s \T dŸeÖ+Ôáss¹Á K\T eT]jáTT
t ÿ¿£ ÜsÁ«>´s
¹ K nsTTq x, y eT]jáTT z $\Te\qT ¿£qT>=q+&.
JJJG JJJG
8. ‚ºÌq |Ÿ³+ýË AB || CD eT]jáTT y{ì eT<Š«ýË ÿ¿£ _+<ŠTeÚ E
nsTTq x + y + z $\TeqT ¿£qT>=q+&.
JJJG
(dŸÖ#áq : E >·T+&† AB ¿ì ÿ¿£ dŸeÖ+ÔásÁ¹sKqT ^jáT+&).
7e ÔásÁ>·Ü >·DìÔáeTT
175
s¹ K\T eT]jáTT ¿ÃD²\T
9.
Identify the pair of parallel lines in the given figure and write them.
1.
Find the complementary, supplementary and conjugate angle of 360.
2.
Observe the figure and write any 4 pairs of adjacent angles.
3.
4.
In the given figure the lines l and m intersect at O. Find x.
HJJG
In the given figure AE is a straight line. If the ratio of angles
1 ,2, 3, 4 in the given figure is 1:2:3:4, then find the
angles.
5.
Write any two examples for linear pair of angles in your surroundings.
6.
Mani said, “Two obtuse angles can form a pair of conjugate angles.” Do you agree.
Justify your answer.
7.
Draw a pair of adjacent angles which are not supplementary to each other.
8.
In the figure, if l || m, t is a transversal. Find 1 and 2.
9.
A line p intersects two lines l and m at two distinct points. Observe the
figure and fill in the blanks:
i)
The line ‘p’ is known as ......, ...........
ii)
1 and 5 is a pair of ..........angles.
iii)
4 and 6 is a pair of ..........angles.
iv)
3 and 6 is a pair of ......... angles.
VII Class Mathematics
176
Lines and Angles
9. ç¿ì+~ ‚ºÌq |Ÿ³+ýË dŸeÖ+ÔásÁ ¹sK\ ÈÔá\qT >·T]ï+º çyjáT+&.
jáT֓{Ù nuó²«dŸ+
1. 360 jîTT¿£Ø |ŸPsÁ¿£, dŸ+|ŸPsÁ¿£ eT]jáTT dŸ+jáTT>·ˆ ¿ÃD² \qT ¿£qT>=q+&?
2. ç|Ÿ¿Ø£ |Ÿ{²“• |Ÿ]o*+#á+&. |Ÿ³+ýË @yîHÕ  4 ÈÔá\ €dŸq• ¿ÃD² \qT sjáT+&.
3. ‚ºÌq |Ÿ³+ýË dŸsÁÞø¹sK\T l eT]jáTT m \T O nHû
_+<ŠTeÚ e<ŠÝ K+&+ºq x $\TeqT ¿£qT>=q+&.
HJJG
4. ‚ºÌq |Ÿ³+ýË AE ÿ¿£ dŸsÞÁ s¹ø K. ¿ÃD²\T 1 , 2, 3, 4 \
“wŸÎÜï 1:2:3:4 nsTTq € ¿ÃD²\TqT ¿£qT>=q+&.
5. s¹ FjáT ¿ÃD²\ ÈÔáÅ£” MT |Ÿ]dŸs\ qT+& Âs+&ƒT –<‘VŸ²sÁD\T sjáT+&.
6. »»Âs+&ƒT n~ó¿£ ¿ÃD²\T ÿ¿£ ÈÔá dŸ+jáTT>·ˆ ¿ÃD²\qT @sÁÎsÁTkÍïsTTµµ n“ eTDì #î™|ÎqT. B““ MTsÁT n+^¿£]kÍïs?
MT dŸeÖ<ó‘H“• dŸeT]Æ+#áTeTT.
7. ÿ¿£<‘“¿=¿£{ì dŸ+|ŸPsÁ¿±\T ¿±“ ÿ¿£ ÈÔá €dŸq• ¿ÃD²\Å£” |Ÿ{²“• ^jáT+&.
8. ‚ºÌq |Ÿ³+ýË l || m eT]jáTT t ÿ¿£ ÜsÁ«>´¹sK. 1 eT]jáTT 2 \qT
¿£qT>=q+&.
9. Âs+&ƒT dŸsÁÞø¹sK\T l eT]jáTT m \qT eTsà dŸsÁÞø¹sK p Âs+&ƒT $_óq• _+<ŠTeÚ\ e<ŠÝ
K+&+ºq~. |Ÿ{²“• |Ÿ]o*+º U²°\qT |ŸP]+#á+&.
i) dŸsÁÞø¹sK p “ .................n“ n+{²sÁT.
ii) 1 eT]jáTT 5 \T ................¿ÃD²\ ÈÔá.
iii) 4 eT]jáTT 6 \T ................¿ÃD²\ ÈÔá.
iv) 3 eT]jáTT 6 \T ................¿ÃD²\ ÈÔá.
7e ÔásÁ>·Ü >·DìÔáeTT
177
s¹ K\T eT]jáTT ¿ÃD²\T
HJJG JJJG JJJG
10. In the given figure CF || BD, BE is transversal.
CAE = 1350, then find ABD.
1.
2.
3.
4.
5.
6.
7.
8.
9.
If the sum of two angles is 900, then the angles are called as complementary angles to each
other.
If the sum of two angles is 1800, then the angles are called as supplementary angles to each
other.
If the sum of the two angles is 3600, then the angles are called as conjugate angles to each
other.
Two angles are said to be adjacent angles, if they have a common vertex, common arm and
lie on either side of the common arm.
A pair of adjacent angles whose sum is 1800 are called as linear pair of angles.
When two straight lines intersect, the opposite angles at the point of intersection are called
as vertically opposite angles.
Vertically opposite angles are equal.
A transversal is a straight line that intersects two or more straight lines at distinct points.
When a transversal intersects a pair of parallel lines p and q,
let the angles formed be 1,  2,  3, 4, 5, 6, 7 and 8.
VII Class Mathematics
178
Lines and Angles
HJJG
JJJG
JJJG
JJJG
10. ‚ºÌq |Ÿ³+ýË CF || BD, BE eT]jáTT BE ÿ¿£
ÜsÁ«>´¹sK CAE = 1350 nsTTq ABD $\Te
¿£qT>=q+&.
>·TsÁTï+#áT¿Ãy*àq
n+Xæ\T
1.
2.
3.
4.
5.
6.
7.
8.
9.
Âs+&ƒT ¿ÃD²\ yîTTÔáï+ 900 nsTTÔû, € s +&ƒT ¿ÃD²\qT ÿ¿£<‘“¿=¿£{ì |ŸPsÁ¿£ ¿ÃD²\T n+{²sÁT.
Âs+&ƒT ¿ÃD²\ yîTTÔáï+ 1800 nsTTÔû, € s +&ƒT ¿ÃD²\qT ÿ¿£<‘“¿=¿£{ì dŸ+|ŸPsÁ¿£ ¿ÃD²\T n+{²sÁT.
Âs+&ƒT ¿ÃD²\ yîTTÔáï+ 3600 nsTTÔû, € s +&ƒT ¿ÃD²\qT ÿ¿£<‘“¿=¿£{ì dŸ+jáTT>·ˆ ¿ÃD²\T n+{²sÁT.
s +&ƒT ¿ÃD²\T –eTˆ& osÁü+, –eTˆ& uóT„ È+ ¿£*Ð € s +&ƒT ¿ÃD²\T –eTˆ& uóT„ È+ ¿ì #îsÃyî|Õ ÚŸ q –q•#Ã
y{ì“ €dŸq•¿ÃD²\T n+{²sÁT.
Âs+&ƒT €dŸq• ¿ÃD²\ yîTTÔáï+ 1800 nsTTq#à y{ì“ s¹ FjáT¿ÃD²\ ÈÔá ýñ<‘ s¹ FjáT <ŠÇjáT+ n“ n+{²sÁT.
Âs+&ƒT dŸsÁÞø¹sK\T ÿ¿£<‘“¿=¿£{ì K+&+#áT¿=q•|ŸÚÎ&ƒT, € K+&ƒq _+<ŠTeÚ e<ŠÝ @sÁÎ&q n_óeTTK¿ÃD²\
ÈÔáqT osô_óeTTK ¿ÃD²\T n“ n+{²sÁT.
osô_óeTTK ¿ÃD²\T dŸeÖq+.
Âs+&ƒT ýñ<‘ n+Ôá¿£+fñ mÅ£”Øe dŸsÁÞø¹sK\qT $_óq• _+<ŠTeÚ\ e<ŠÝ K+&+#û dŸsÁÞø¹sKqT ÜsÁ«>´¹sK n+{²sÁT.
ÿ¿£ ÈÔá dŸeÖ+ÔásÁ ¹sK\T p, q \qT ÿ¿£ ÜsÁ«>´¹sK r K+&+ºq|ŸÚÎ&ƒT, @sÁÎ&q ¿ÃD²\T 1, 2, 3,
4, 5, 6, 7 eT]jáTT 8 n“ nqT¿=qTeTT.
7e ÔásÁ>·Ü >·DìÔáeTT
179
s¹ K\T eT]jáTT ¿ÃD²\T
10. When a transversal intersects two lines,
Euclid (323-283BC) was a Greek mathematician. He is well known
as ‘Father of Geometry.’ He wrote a book named ‘Elements’ which
was a collection of 13 volumes, dealing with geometry. ‘Elements’
was most influenced work in the history of mathematics. In his
books he deduced many theorems using axioms. These theorems
are being used in the study of Euclidean geometry.
Logical Venn Diagrams
A Venn diagram is an illustration that uses circles to show the relationships among things or
finite groups of things. Circles that overlap have a commonality while circles that do not overlap
do not share those traits.
Venn diagrams help to visually represent the similarities and differences between three or more
concepts.
S.No Group
1
Venn diagram
a
VII Class Mathematics
Use of venn
diagram
Example
Each class is totally 1. Shark, Whale sea turtle
different from the 2. Iron, Copper and Brass
3. Apple, Mango, Orange
other.
4. Book, Pen, Eraser
5. Fruits, Vegetables,
Flowers.
180
Lines and Angles
10. ÿ¿£ ÜsÁ«ç¹>K ÿ¿£ ÈÔá dŸsÁÞø¹sK\qT K+&+ºq|ŸÚ&ƒT,
#]çÔá¿£ n+XøeTT
jáTÖ¿ì¢&Ž (323-283BC) ÿ¿£ ç^Å£” >·DìÔá XæçdŸïyûÔáï. €jáTq »»¹sU²>·DìÔá
|¾ÔeTVŸQ&ƒTµµ>± ç|Ÿd~¾ Æ #î+<‘&ƒT. €jáTq »»m*yîT+{Ù൵ nHû |sÁTÔà ÿ¿£ |ŸÚdŸ¿ï ±“•
sÁº+#îqT. ‚~ ¹sU²>·DìԐ“• >·T]+º $e]+#û 13 |ŸÚdŸï¿±\ dŸeÖVŸäsÁ+. >·DìÔá
XæçdŸï #á]çÔáýË »»m*yîT+{Ù൵ nÔá«+Ôá ç|Ÿuó²$ÔáyîT®q sÁ#áq. €jáTq Ôáq |ŸÚdŸï¿±ýË¢
nHû¿£ dÓÇ¿£Ԑ\qT –|ŸjÖî Ð+º nHû¿£ d¾<‘Æ+Ԑ\qT $e]+#sÁT. ‡ d¾<‘Æ+Ԑ\T
jáTÖ¿ì¢&jáTHŽ C²«$TÜ n<óŠ«jáTq+ýË ‚|ŸÎ{ì¿¡ –|ŸjîÖÐ+#á‹&ƒTÔáTH•sTT.
Ԑ]Ø¿£ yîHŽ ºçԐ\T
edŸTïeÚ\ eT<óŠ« dŸ+‹+<ó‘\qT ýñ<‘ edŸTïeÚ\ jîTT¿£Ø ºq• dŸeTÖVŸä\ eT<óŠ« dŸ+‹+<ó‘\qT #áÖ|¾+#á&†“¿ì
–|ŸjîÖÐ+#û eԐïýñ yîHŽ ºçÔá+.zesY ý²«|t njûT« eԐï\ \¿£ŒD²\T –eTˆ&>± –+³Ö, zesY ý²«|t ¿±“ eԐï\T
y{ì \¿£ŒD²\qT |Ÿ+#áT¿ÃeÚ. yîHŽ ºçԐ\T eTÖ&ƒT ýñ<‘ n+Ôá¿£+fñ mÅ£”Øe uó²eq\ eT<óŠ« –+&û bþ*¿£\T eT]jáTT
Ôû&†\qT <ŠXø«|ŸsÁ+>± çbÍܓ<óŠ«+ eV¾²+#á&†“¿ì kÍjáT|Ÿ&ƒÔsTT.
ç¿£.dŸ+. dŸeTÖVŸ²+
1
a
7e ÔásÁ>·Ü >·DìÔáeTT
Ԑ]Ø¿£ yîHŽ ºçÔáeTT
yîHŽ ºçÔáeTT –|ŸjÖ
î >·eTT
–<‘VŸ²sÁD
ç|Ÿ Ü Ôá s Á > · Ü $TÐ\q 1. kþsÁ#|û ,Ÿ çÜ$T+>·\eTT, Ԑuñ\T.
Ôás>Á Ô· Tá \Ôà ¿£\e Å£”+&† 2. ‚qTeTT, sÐ eT]jáTT ‚Ôá&ï 
$_óq•+>± –+³T+~. 3. €|¾ýÙ, eÖ$T&, ¿£eTý
4. |ŸÚdŸ¿ï e£ TT, ™|qT•, sÁ‹ÒsÁT
5. |Ÿ+&ƒT,¢ Å£LsÁ>±jáT\T, |ŸÚeÚÇ\T
181
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S.No Group
Venn diagram
Use of venn diagram
Example
2
b
Each class is totally 1. Musician, Instrumentalist,
Violinist.
inserted in another, but
2. Carrot, Food, Vegetables
not mixed.
3. Andhra Pradesh, India,
Asia.
4. Ring, Ornaments, Golden
ring.
3
c
One class is partly 1. Lawyer, Woman, Doctor
related to other two but 2. Dog, Cat, Pet
these two are no ware
related.
4
d
All the three are partly
related to each other.
5
e
One class is completely 1. Sun, Moon, Stars.
inserted other and one 2. Pen, Stationary, Soap.
is completely different 3. Bird, Owl, Elephant.
from other two.
6
f
Two are totally inserted 1. Lotus, Flower and Rose
in one but these two are 2. Dog, Cat, Animal
3. Food, Milk, Fruits
no where related.
VII Class Mathematics
182
1. Athletes, Footballers &
Cricketers.
2. Boys, Students, 7th Class
Boys
3. Tennis fans, Cricket. fans,
Kabaddi fans.
Lines and Angles
ç¿£.dŸ+. dŸeTÖVŸ²+
Ԑ]Ø¿£ yîHŽ ºçÔáeTT
yîHŽ ºçÔáeTT –|ŸjÖ
î >·eTT
–<‘VŸ²sÁD
2
b
ç|ŸÜ ÔásÁ>·Ü Å£L&† |ŸP]ï>± 1. dŸ+^Ôá $<‘Ç+dŸT\T, y<Š«¿±sÁT\T,
eTsà <‘“ýË #=|¾ Î +#á ejîÖ*HŽ $<‘«+dŸT&ƒT.
‹&ƒ T Ôá T +~ ¿±ú ¿£ \ |Ÿ 2. ¿±«Âs{Ù, €VŸäsÁ+, Å£LsÁ>±jáT\T
‹&ƒýñ<ŠT.
3. €+ç<óçŠ |Ÿ<Xû Ù, uó²sÁÔ<á Xû +ø , €d¾jÖá
4. –+>·s+Á , €uós„ D
Á ²\T, ‹+>±sÁ|ڟ
–+>·s+Á .
3
c
ç|ŸÜ ÔásÁ>·Ü Å£L&† eTsà 1. ý²jáTsY, €&ƒyÞø—ß, yîÕ<ŠT«\T
<‘“¿ì |Ÿ P ]ï > ± _ó q •+>± 2. Å£”¿£Ø, |¾*¢, ™|+|ŸÚ&ƒT È+ÔáTeÚ
–+³T+~.
4
d
5
e
6
f
7e ÔásÁ>·Ü >·DìÔáeTT
‡ eTÖ&ƒ Ö ÿ¿£ < ‘“Ôà 1. ç¿¡&†¿±sÁT\T, |˜ÚŸ {Ù u²ýÙ ç¿¡&†¿±sÁT\T
eT]jáTT ç¿ì¿Â ³sÁT
eTs=¿£ <‘“¿ì bͿ쌿£+>± 2. u²\TsÁT, $<‘«sÁTœ\T, @&ƒe ÔásÁ>·Ü
dŸ+‹+<óŠ+ ¿£*Ð eÚH•sTT. u²\TsÁT
3. fÉ “ •dt n_ó e ÖqT\T, ç¿ì  ¿ {Ù
n_óeÖqT\T, ¿£‹&ž¦ n_óeÖqT\T.
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|Ÿ P ]ï > ± #=|¾ Î +#á ‹ & + ~ 2.d¼wŸq¯, ¿£\eTT (™|HŽ), ™|sÁT>·T
eT]jáTT ÿ¿£ ÔásÁ>·Ü |ŸP]ï>± 3.|Ÿ¿ìŒ, >·T&ƒ¢>·Ö‹, @qT>·T
$TÐ*q s +&+{ì¿ì _óq•yîTq® ~.
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¿±“ ‡ Âs+&ƒT ÔásÁ>·ÔáT\Å£” 3. €VŸäsÁ+, bÍ\T, |Ÿ+&ƒT¢
dŸ+‹+<óŠ+ ýñ<ŠT.
183
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Practice problems :
Indicate the group (a, b, c, d, e, f, g, h) to which given below belongs to
1.
State, District, Mandal
16.
Fruits, Mango, Onions
2.
Boys, Girls, Artistists
17.
School, Teacher, Students
3.
Hours, Days, Minutes
18.
Rivers, Oceans, Springs
4.
Women, Teacher, Doctor
19.
India, Andhra Pradesh, Visakhapatnam
5.
Food, Curd, Spoon
20.
Animals, Cows, Horses
6.
Himans, Dancer, Player
21.
Fish, Tiger, snakes
7.
Bulding, Brick, Bridge
22.
Flowers, Jasmine, Banana
8.
Tree, Branch, Leaf
23.
Authors, teachers, Men
9.
Gold, Silver, Jewelary
24.
Dog, Fish, Parrot
10. Bulbs, Mixtures, Electricals
25.
Rose, Flower, Apple
11. Women, Illiteracy, Men
26.
School, Benches, Class Room
12. Medicine, Tablets, Syrup
27.
Pen, Stationary, Powder
13. Carrots, Oranges, Vegetables
28.
Crow, Pigeon, Bird
14. Female, Mothers, Sisters
29.
Mammals, Elephants, Dinosaurs
15. Table, Furniture, Chair
30.
Writers, Teachers, Researchers
VII Class Mathematics
184
Lines and Angles
ç|ŸXø•\qT çbÍ¿¡¼dt #ûjáT+& :
™|Õ |Ÿ{켿£ý˓ @ dŸeTÖVŸ²+ (A, B, C, D, E, F), ç¿ì+<Š ‚ºÌq ç|ŸXø•\Å£” #î+~q<à dŸÖº+#á+&.
1. sçwŸ¼eTT, ›ý²¢, eT+&ƒ\+
16. |Ÿ+&ƒT¢, eÖ$T&, –*¢
2. u²\TsÁT, u²*¿£\T, ¿£Þ²¿±sÁT\T
17. bÍsÄÁXæ\, –bÍ<ó‘«jáTT&ƒT, $<‘«sÁTœ\T
3. >·+³\T, sÃE\T, “$TcÍ\T
18. nsÁ{,ì #=¿±Ø, ‹ýÙÒ
4. eTV¾²Þø\T, –bÍ<ó‘«jáTT&ƒT, yîÕ<ŠT«&ƒT
19. uó²sÁÔá<ûXø+, €+ç<óŠç|Ÿ<ûXÙ, $XæK|Ÿ³¼D+
5. €VŸäsÁeTT, ™|sÁT>·T, #î+#
20. È+ÔáTeÚ\T, €eÚ\T, >·Tçs\T
6. eÖqeÚ\T, H³«¿±sÁT&ƒT, €³>±&ƒT
21. #û|Ÿ, |ŸÚ*, bÍeTT\T
7. uó„eqeTT, ‚³T¿£, e+Ôîq
22. |ŸP\T, eTýÉ¢|ŸP\T, nsÁ{ì
8. #î³T¼, ¿=eTˆ, €Å£”
23. sÁ#ásTTÔá\T, –bÍ<ó‘«jáTT\T, |ŸÚsÁTwŸ§\T
9. ‹+>±sÁ+, yî+&, €uó„sÁD²\T
24. Å£”¿£Ø, #û|Ÿ, º\T¿£
10. ‹\TÒ\T, d¾Ç#Y\T, m\ç¿ì¼¿£ýÙà
25. >·Tý²_, |ŸÚwŸÎ+, €|¾ýÙ
11. eTV¾²Þø\T, “sÁ¿£ŒsdŸ«Ôá, |ŸÚsÁTwŸ§\T
26. dŸÖØ\T, ¿±¢dt sÁÖyŽT, uÉ+N\T
12. eT+<ŠT\T, eÖçÔá\T, d¾sÁ|t
27. ¿£\eTT, d¼wŸq¯, bå&ƒsY
13. ¿±«Âs³T¢, €Âs+CÙ\T, Å£LsÁ>±jáT\T
28. ¿±¿ì, bÍeÚsÁ+, |ŸÅ£Œ”\T
14. È>´, |ŸÚdŸï¿£+, Å£”+&ƒ
29. ¿¡ŒsÁ<‘\T, @qT>·T\T, &îÕHÃkÍsÁT¢
30. sÁ#ásTTÔá\T, –bÍ<ó‘«jáTT\T, |Ÿ]XË<óŠÅ£”\T
15. ‹\¢, |˜Ÿ]•#ásY, Å£”¯Ì
7e ÔásÁ>·Ü >·DìÔáeTT
185
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5
TRIANGLES
Learning Outcomes
Content Items
The learner is able to,
l
measure the sides and angles of a triangle.
l
classify the triangles based on sides and angles.
l
find unknown measurements in a triangle by using
the properties of triangles.
l
explain the propertes of interior and exterior angles
of a triangle.
l
verify the properties of angles and sides of a triangle.
l
understand and differentiate the concurrent points
to a triangle.
l
establish congruence criterion for the pairs of
triangles such as SSS, SAS, ASA and RHS.
l
identify the suitable congruency criterion in pair of
triangles.
5.0 Introduction
5.1 Classification of triangles
5.2 Interior angles sum property
5.3 Exterior angle property
5.4 Properties of sides of a triangle
5.5 Concurrent points of a
triangles
5.6 Congruence of triangles
5.0 Introduction:
We have learnt about triangle and it’s parts in the previous class. Triangle is one of the basic
shape in geometry. We use triangle shapes in several situations of our daily life. It is used in bridge
construction like long hanging bridges and other structures for more strength and stability. They
are also used in high tension electric towers, buildings, land surveying, signal towers, roof
constructions, Astronomy, Navigation, Physics etc. because this shape absorbs the pressure and
remain rigid. Look at the pictures and speak about them. What do you observe?
VII Class Mathematics
188
Triangles
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nuó„«dŸq |˜Ÿ*Ԑ\T
$wŸjáÖ+Xæ\T
nuó²«dŸÅ£”&ƒT :
l
l
l
l
l
l
l
l
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¿£qT>=q>·\&ƒT.
çÜuó„TÈeTTý˓ n+ÔásÁ¿ÃD, u²VŸ²«¿ÃDeTT\ <óŠsÁˆeTT\qT
$e]+#á>·\&ƒT.
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5.0 |Ÿ]#ájTá +
5.1 çÜuóT„ C²\e¯Z¿s£ D
Á
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5.3 çÜuóT„ È u²VŸ²«¿ÃD<ósŠ ˆÁ eTT
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5.6. çÜuóT„ C²\ dŸsÇÁ dŸeÖqÔáÇ+.
5.0 |Ÿ]#ájáT+ :
ç¿ì+~ Ôás>Á Ü· ýË çÜuóT„ C²\T eT]jáTT y{ì uó²>±\T >·T]+º HûsTÁ ÌÅ£”H•eTT. çÜuóT„ ÈeTT s¹ U²>·DÔì +á ýË ÿ¿£
çbÍ<óŠ$T¿£ €¿±sÁ+. eTq “Ôá« J$Ôá dŸ+<ŠsÒÛ\ýË çÜuó„TC²¿±s\qT nHû¿£ dŸ+<ŠsÒÛ\ýË –|ŸjîÖÐdŸTïH•+.
çÜuóT„ C²¿±s\qT ç_&ž\¨ “sˆD\Å£” eTTK«+>± nÜ bõ&ƒyqÕî çyûý²&û e+Ôîq\T eT]jáTT <óŠ &ƒyTî q® eT]jáTT d¾sœ yÁ Tî q®
“sˆD²\Å£” –|ŸjîÖÐdŸTïH•+. n~ó¿£ n~ó¿£ $<ŠT«ÔYqT ç|ŸkÍsÁ+ #ûd dŸÇuó²\T >·\ $<ŠT«ÔY ³esYà? uó„eH\T,
uó„ÖeTT\dŸ¹sÇ, d¾>·•ýÙ ³esYà, ™|Õ¿£|ŸÚÎ\T “sˆD²\ýË, H$¹>wŸHŽ eT]jáTT uó Ü¿£XæçdŸïeTT yîTT<Š\>·T y{ìýË Å£L&†
çÜuóT„ È €¿±s\qT –|ŸjÖî ÐkÍïsTÁ . m+<ŠT¿£q>± ‡ €¿±sÁ+ ÿÜï&“ ç>·V²¾ +º d¾sœ +Á >± –+³T+~. ç¿ì+~ ºçԐ\qT
#áÖd¾, y{ì >·T]+º #á]Ì+#á+&? MTsÁT @$T >·eT“+#sÁT?
7e ÔásÁ>·Ü >·DìÔáeTT
189
çÜuóT„ C²\T
We can observe so many triangles in the above pictures, they are in different sizes. To know
their properties, it is needed to learn more about triangles. In this chapter, we are going to learn
about classification of triangles, properties of triangles, concurrent points and congruence of
triangles. Before going to discuss these concepts in detail, let us recall and review the previous
concepts what we have learnt in previous class.
A triangle is a simple closed plane figure formed by three line segments.
A triangle has three vertices, three sides and three angles. A triangle can be named in either
clockwise or anti clockwise direction, i.e. ACB or ABC.
Look at the adjacent figure. The elements in ABC are:
i)
three vertices: A, B and C
ii)
three angles: A, B and C
iii)
three sides: AB , BC and CA
The opposite side to vertex A is BC and similarly, the opposite sides to vertices B and C are AC
and AB respectively.
1.
2.
3.
Mark any three non collinear points A, B and C in your note book, join them to make a triangle
and name it.
Observe the given triangle and answer the following:
i) Write the interior points of the triangle.
ii) Write the points marked on the triangle.
iii) Write the exterior points of the triangle.
Observe the given triangle and answer the following:
i) The opposite side to vertex L is _______
ii) The opposite side to K is ______
iii) The opposite angle to KL is _____
iv) The opposite vertex to LM is ______
VII Class Mathematics
190
Triangles
™|qÕ #áÖ|¾q |Ÿ³eTT\ýË eTqeTT #ý² sÁ¿±\ çÜuóT„ ÈeTT\qT >·eT“+#áe#áTÌ, n$ $$<óŠ sÁ¿±\ |Ÿ]eÖD²\ýË
–H•sTT. çÜuóT„ ÈeTT\ <ósŠ ˆÁ eTT\qT ne>±VŸ²q #ûdTŸ ¿=qT³Å£” y{ì >·T]+º eT]¿=“• n+Xæ\T Ôî\TdŸT¿=qT³
nedŸs+Á . ‡ n<ó‘«jáT+ýË çÜuóT„ ÈeTT\ e¯Z¿s£ DÁ , çÜuóT„ ÈeTT\ <ósŠ ˆ\T, $T[Ôá _+<ŠTeÚ\T, çÜuóT„ C²\ dŸsÇÁ dŸeÖqÔáÇ+
yîTT<Š\>·T $wŸjÖá \ >·T]+º eTq+ HûsTÁ ÌÅ£”+<‘+. ‡ $wŸjÖá \qT #á]Ì+#áT³Å£” eTT+<ŠT>± ç¿ì+~ Ôás>Á Ô· Tá \ýË
HûsTÁ ÌÅ£”q• çÜuóT„ C²“¿ì dŸ+‹+~ó+ºq |ŸPsÁÇC²ãH“• >·TsÁTÅï ”£ Ôî#Tá ÌÅ£”+<‘+.
eTÖ&ƒT s¹ U² K+&†\#û @sÁÎ&q dŸsÞÁ ø dŸ+eÔá |Ÿ³eTTqT çÜuóT„ È+ n+{²sÁT.
çÜuóT„ ÈeTTqÅ£” eTÖ&ƒT osÁüeTT\T, eTÖ&ƒT uóT„ ÈeTT\T eT]jáTT eTÖ&ƒT ¿ÃDeTT\T ¿£\eÚ. çÜuóT„ ÈeTTqÅ£” dŸe«
ýñ<‘ n|ŸdŸe« ~Xø\ýË |sÁT ™|³¼e#áTÌ. nq>± ACB ýñ<‘ ABC
ç|Ÿ¿Ø£ |Ÿ³eTT qT+&, çÜuóT„ ÈeTT ABC ý˓ n+XøeTT\T @eTq>±,
i) eTÖ&ƒT osÁüeTT\T: A, B eT]jáTT C
ii) eTÖ&ƒT ¿ÃDeTT\T: A, B eT]jáTT C
iii) eTÖ&ƒT uóT„ ÈeTT\T: AB , BC eT]jáTT CA .
osÁüeTT AÅ£” m<ŠTsÁT>± –q• uóT„ ÈeTT BC , €<û$<óŠ+>± B eT]jáTT C \Å£” m<ŠTsÁT>± –q• uó„TÈeTT\T
esÁTdŸ>± AC eT]jáTT AB .
|ŸÚq]ÇeTsÁô nuó²«dŸ+
1. @yîHÕ  eTÖ&ƒT dŸs¹ FjáÖ\T ¿±“ _+<ŠTeÚ\T A, B eT]jáTT C \qT ¿±ÐÔáeTT ™|Õ >·T]ï+#áTeTT y{ì“ ¿£*|¾
çÜuó„TÈeTT ^d¾ |sÁT ™|³T¼eTT.
2. ‚ºÌq çÜuóT„ ÈeTTqT |Ÿ]o*+º, ç¿ì+~ y{ì¿ì Èy‹T çyjáT+&.
i)
çÜuóT„ ÈeTT jîTT¿£Ø n+ÔásÁ _+<ŠTeÚ\T sjáT+&.
ii)
çÜuóT„ ÈeTT ™|qÕ –q• _+<ŠTeÚ\T sjáT+&.
iii)
çÜuóT„ ÈeTT jîTT¿£Ø u²VŸ²« _+<ŠTeÚ\T sjáT+&.
3. ‚ºÌq çÜuóT„ ÈeTTqT |Ÿ]o*+º, ç¿ì+~ y{ì¿ì dŸeÖ<ó‘q+ sjáT+&.
i)
osÁüeTT L qÅ£” m<ŠTsÁT>± –q• uóT„ ÈeTT _______
ii)
K Å£”m<ŠTsÁT>± –q• uóT„ ÈeTT _______
iii)
KL qÅ£” m<ŠTsÁT>± –q• ¿ÃDeTT _______
iv)
LM qÅ£” m<ŠTsÁT>± –q• osÁüeTT _______.
7e ÔásÁ>·Ü >·DìÔáeTT
191
çÜuóT„ C²\T
4.
Classify the following angles into acute, obtuse and right angles :
200, 500, 1020, 470, 1250, 650, 360, 900, 950 and 1100.
5.
Identify the intersecting point and concurrent
point in the adjust figure.
5.1 Classification of triangles :
Based on the length of the sides, the triangles are classified into three types.
They are:
1.
Equilateral Triangle: A triangle with three equal sides is called an
Equilateral Triangle.
Ex: i) In ABC, AB = BC = CA = 3cm
ii)
In PQR, PQ = QR = RP
Do you know?
The sides with equal length are denoted
with same number of strokes similiarly for
equal angles too.
2.
Isosceles Triangle : A triangle with two equal sides is called an Isosceles Triangle.
Ex :
In KLM, KM = ML = 2.5 cm
i)
3.
ii)
In XYZ, XY = XZ
Scalene Triangle: A triangle with no two sides are equal is called a Scalene Triangle.
Ex :
i)
In DEF, DE  EF  FD
VII Class Mathematics
ii)
192
In STU, ST  TU  US
Triangles
4. ç¿ì+<Š ‚ºÌq ¿ÃD²\qT n\Î, n~ó¿£ eT]jáTT \+‹¿ÃDeTT\T>± e¯Z¿]£ +#á+&.
200, 500, 1020, 470, 1250, 650, 360, 900, 950 eT]jáTT 1100.
5. ç|Ÿ¿Ø£ |Ÿ³eTT qT+& K+&ƒq _+<ŠTeÚ eT]jáTT $T[Ôá
_+<ŠTeÚqT çyjáT+&.
5.1 çÜuó„TC²\ e¯Z¿£sÁD (çÜuó„TÈeTT\ýË sÁ¿£eTT\T) :
uóT„ C²\ ¿=\Ôá\ €<ó‘sÁ+>±, çÜuóT„ ÈeTT\qT eTÖ&ƒT sÁ¿±\T>± e¯Z¿]£ +#áe#áTÌ. n$ :
1. dŸeTu²VŸQ çÜuóT„ È+ : n“• uóT„ ÈeTT\T dŸeÖqeTT>± –q• çÜuóT„ ÈeTTqT dŸeTu²VŸQ
çÜuóT„ È+ n+{²sÁT.
–<‘VŸ²sÁD : i) ABC ýË, AB = BC = CA = 3 ™d+.MT.
ii)
PQR ýË PQ = QR = RP.
MTÅ£” Ôî\TkÍ?
dŸeÖq uóT„ ÈeTT\qT dŸeÖq dŸ+K« >·\ ^Ôá\ÔÃ
dŸÖºkÍïsTÁ . n<û$<ó+Š >± dŸeÖq ¿ÃD²\qT Å£L&†
dŸeÖq dŸ+K« >·\ ^Ôá\Ôà dŸÖºkÍïsTÁ .
2. dŸeT~Çu²VŸQ çÜuóT„ È+ : s +&ƒT uóT„ ÈeTT\T dŸeÖqeTT>± –q• çÜuóT„ ÈeTTqT dŸeT~Çu²VŸQçÜuóT„ È+ n+{²sÁT.
–<‘ :
i)
KLM ýË KM = ML = 2.5 ™d+.MT.
ýË XY = XZ.
3. $wŸeTu²VŸQ çÜuóT„ È+ : @ s +&ƒT uóT„ ÈeTT\T dŸeÖqeTT>± ýñ“ çÜuóT„ ÈeTTqT $wŸeTu²VŸQ çÜuóT„ ÈeTT n+{²sÁT.
–<‘ :
i)
DEF ýË DE  EF  FD.
7e ÔásÁ>·Ü >·DìÔáeTT
ii)
XYZ
ii)
193
STU ýË ST  TU  US.
çÜuóT„ C²\T
Classify the following triangles according to the length of their sides:
Z
U
R
Q
P
X
Y
T
S
Based on the measure of the angles, triangles are classified into three types. They are :
1.
Acute angled triangle: A triangle with all acute angles is called an Acute angled triangle.
Ex :
2.
Right angled triangle: A triangle with one right angle is called a right angled triangle.
Ex :
Do you know?
In a right angled triangle, the
side opposite to right angle is
called ‘Hypotenuse’.
3.
Obtuse angled triangle: A triangle with one obtuse angle is called an obtuse angled triangle.
Ex :
F
A
C D
B
VII Class Mathematics
194
E
Triangles
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
ç¿ì+<Š ‚ºÌq çÜuóT„ ÈeTT\qT uóT„ ÈeTT\ ¿=\Ôá\ €<ó‘sÁ+>± e¯Z¿]£ +#á+&.
Z
U
R
Q
P
X
Y
T
S
¿ÃD²\ ¿=\Ôá\T €<ó‘sÁ+>± çÜuóT„ ÈeTT\qT eTÖ&ƒT sÁ¿±\T>± e¯Z¿]£ +#áe#áTÌ. n$ :
1. n\οÃD çÜuóT„ È+: çÜuóT„ ÈeTT jîTT¿£Ø n“• ¿ÃDeTT\T n\οÃDeTT\T>± >·\ çÜuóT„ ÈeTTqT n\οÃD çÜuóT„ È+
n+{²sÁT.
–<‘ :
2. \+‹¿ÃD çÜuóT„ È+ : çÜuóT„ ÈeTTýË ÿ¿£ ¿ÃD+ \+‹¿ÃDeTT >·\ çÜuóT„ ÈeTTqT \+‹¿ÃD çÜuóT„ È+ n+{²sÁT.
–<‘ :
MTÅ£” Ôî\TkÍ?
\+‹¿ÃD çÜuóT„ ÈeTTýË
\+‹¿ÃD²“¿ì m<ŠTsÁT>± –+&û
uóT„ C²“• »¿£se’Á TTµ n+{²sÁT.
3. n~ó¿¿£ ÃD çÜuóT„ È+ : çÜuóT„ ÈeTTýË ÿ¿£ ¿ÃD+ n~ó¿¿£ ÃDeTT>± >·\ çÜuóT„ ÈeTTqT n~ó¿¿£ ÃD çÜuóT„ È+ n+{²sÁT.
–<‘ :
F
A
B
7e ÔásÁ>·Ü >·DìÔáeTT
C D
195
E
çÜuóT„ C²\T
Classify the following triangles according to the measure of their angles.
M
J
G
E
F
I
H
K
L
Exercise - 5.1
1.
Classify the following triangles based on the length of their sides:
A
X
E
D
B
F
C
Z
Y
Z
S
Q
P
2.
R T
Classify the following triangles based on the measure of angles:
T
a)
b)
U
3.
V
U
Y
X
I
c)
Q
S
R
K
J
Classify the following triangles based on their sides and also on their angles:
a)
T
L
I
b)
J
VII Class Mathematics
K
c)
M
N
196
U
V
Triangles
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
¿ì+<Š ‚ºÌq çÜuóT„ ÈeTT\qT ¿ÃDeTT\ €<ó‘sÁ+>± e¯Z¿]£ +#á+&.
M
J
G
E
I
H
F
K
L
nuó²«dŸ+ ` 5.1
1. ç¿ì+<Š ‚ºÌq çÜuóT„ ÈeTT\qT uóT„ ÈeTT\ €<ó‘sÁ+>± e¯Z¿]£ +#á+&.
A
D
B
X
E
F
C
Z
Y
Z
S
Q
P
R
U
T
2) ç¿ì+<Š ‚ºÌq çÜuóT„ ÈeTT\qT ¿ÃDeTT\ €<ó‘sÁ+>± e¯Z¿]£ +#á+&.
T
a)
Y
X
b)
I
c)
Q
U
V
S
R
K
J
3) ç¿ì+<Š ‚ºÌq çÜuóT„ ÈeTT\qT ¿ÃDeTT\T, uóT„ ÈeTT\ €<ó‘sÁ+>± e¯Z¿]£ +#á+&.
T
L
I
a)
b)
c)
T
L
J
7e ÔásÁ>·Ü >·DìÔáeTT
K
M
N
197
U
V
çÜuóT„ C²\T
5.2 Interior Angles sum Property:
1.
On a white paper, draw a triangle
ABC. Using colour pencils,
colour its angles as shown in the
figure.
2.
Using a scissors, cut out the three angular regions.
3.
Draw a line PQ and mark a point ‘O’ on it.
P
4.
O
Q
Paste the three angular cut outs adjacent to each other to form one angle at ‘O’ as shown in the
figure given below.
You will find that three angles now constitute a straight angle.
Thus, the sum of three interior angles of a triangle is equal to 1800.
Proof of the interior angle-sum property of a triangle :
Statement: The sum of the interior angles of a triangle is 1800.
Given : ABC is a triangle
To prove :A + B + C = 1800
HJJG
Construction : Draw a line PQ , through ‘A’ parallel to BC
Proof :
Mark the angles as indicated in the figure:
2 = 5
3 = 4
(i)
(ii)
(why?)
(why?)
(i) + (ii)  2 +3 = 5 + 4
1 + 2 + 3 = 1 + 5 + 4
(adding 1 on both sides)
1 + 5 +4 = 1800
(linear angles)
Therefore, 1 + 2 + 3 = 1800
A + B + C = 1800. Hence proved.
VII Class Mathematics
198
Triangles
5.2 n+ÔásÁ ¿ÃDeTT\ yîTTÔáïeTT ` <óŠsÁˆeTT :
‚~ #û<‘Ý+!
¿£Ôá«+
1. ÿ¿£ Ôî\¢¿±ÐÔá+ ™|Õ ABC çÜuóT„ ÈeTTqT
^jáT+&. sÁ+>·T ™|“àýÙ Ôà ¿ÃDeTT\Å£”
|Ÿ³eTTýË #áÖ|¾q³T¢ sÁ+>·T\T yûjTá +&.
2. ¿£Ôsïî ÔÁ à eTÖ&ƒT ¿ÃD uó²>·eTT\qT ¿£Ü]ï +#á+&.
3.
PQ nqT s
¹ KqT ^jáT+& <‘“™|Õ ‘O’ _+<ŠTeÚqT >·T]ï+#á+&.
P
O
Q
4. ç¿ì+~ |Ÿ³+ýË #áÖ|¾q³T¢ ‘O’ e<ŠÝ ÿ¹¿ ¿ÃD+ @sÁÎ&ƒTq³T¢ ¿£Ü]ï +ºq ¿ÃD uó²>·eTT\qT |Ÿ¿Ø£ |Ÿ¿Ø£ Hû nÜ¿ì+#á+&.
@$T >·eT“+#sÁT?
eTÖ&ƒT ¿ÃDeTT\ uó²>·eTT\T dŸsÞÁ s¹ø K™|Õ 1800 ¿ÃD+ @sÁÎsÁ#&á †“ MTsÁT >·eT“+#áe#áTÌ. ¿£qT¿£ çÜuóT„ ÈeTTý˓
eTÖ&ƒT n+ÔásÁ ¿ÃD²\ yîTTÔáeï TT 1800 n“#î|Ο e#áTÌ.
çÜuóT„ ÈeTTýË n+ÔásÁ ¿ÃDeTT\ yîTTÔáeï TT 1800 nqT <ósŠ ˆÁ eTTqÅ£” ÿ¿£ “sÁÖ|ŸD :
ç|Ÿe#áqeTT : ÿ¿£ çÜuóT„ ÈeTTýË n+ÔásÁ ¿ÃDeTT\ yîTTÔáeï TT 1800.
<ŠÔï+XøeTT : ABC ÿ¿£ çÜuóT„ ÈeTT.
kÍs+XøeTT : A + B + C = 1800.
“sˆDeTT :
osÁüeTT ‘A’ >·T+&†
BC uó„TÈeTTqÅ£”
HJJG
¹ UqT ^jáT+&.
PQ nqT s
dŸeÖ+ÔáseÁ TT>±
“sÁÖ|ŸD : |Ÿ³+ýË #áÖ|¾q $<ó+Š >± ¿ÃDeTT\qT >·T]ï+#á+&.
(i)
2 = 5
(m+<ŠTÅ£”?)
(ii)
3 = 4
(m+<ŠTÅ£”?)
(i) + (ii)  2 +3 = 5 + 4
1 + 2 + 3 = 1 + 5 + 4
1 + 5 +4 = 1800
¿±eÚq, 1 + 2 + 3 = 1800
(‚sÁT yîÕ|ŸÚý² 1qT Å£L&ƒ>±)
(dŸsÞÁ s¹ø K™|Õ ÿ¿£ _+<ŠTeÚ e<ŠÝ ¿ÃD+)
A + B + C = 1800 “sÁÖ|¾+#á‹&+~.
7e ÔásÁ>·Ü >·DìÔáeTT
199
çÜuóT„ C²\T
Example 1 : In a triangle, two of the angles are 430 and 570. Find the third angle.
Solution :
Given two angles of a triangle are 430 and 570
Sum of these two angles = 430 + 570 = 1000
In a triangle the sum of the interior angles is equal to 1800.
Third angle = 1800 – 1000 = 800
Example 2 : Find the value of ‘x’ in the following triangles:
Fig. (i)
Fig. (iii)
Fig. (ii)
Solution: From the fig. (i)
A +B +C = 1800
(' Sum of three angles in triangle is 1800)
 1250 + x = 1800
650 + 600 + x = 1800
x = 1800 – 1250 = 550
From the fig. (ii)
P + Q + R = 1800
(why?)
 2x + 800 = 1800
x + x + 800 = 1800
100D
 50D  x = 500
2x = 180 – 80 = 2x = 100  x 
2
0
0
0
From the fig. (iii)
X + Y + Z = 1800
(why?)
x + 900 + 420 = 1800  x + 1320 = 1800
x = 1800 – 1320
VII Class Mathematics
 x = 480
200
Triangles
–<‘VŸ²sÁD 1 : ÿ¿£ çÜuóT„ È+ ý˓ s +&ƒT ¿ÃDeTT\T 430 eT]jáTT 570 nsTTq eTÖ&ƒe ¿ÃDeTTqT ¿£qT>=q+&.
kÍ<óŠq :
<ŠÔï+Xø+ qT+& çÜuóT„ ÈeTTý˓ s +&ƒT ¿ÃDeTT\T 430 eT]jáTT 570
‡ s +&ƒT ¿ÃDeTT\ yîTTÔáeï TT R 430 G 570 R 1000
çÜuóT„ ÈeTTýË n+ÔásÁ ¿ÃDeTT\ yîTTÔáeï TT 1800
 eTÖ&ƒe¿ÃD+ R 1800 - 1000 R 800
–<‘VŸ²sÁD 2 : ç¿ì+<Š ‚ºÌq çÜuóT„ ÈeTT\ýË ‘x’ $\TeqT ¿£qT>=q+&.
|Ÿ³eTT (i)
|Ÿ³eTT (iii)
|Ÿ³eTT (ii)
kÍ<óŠq : |Ÿ³eTT (i) qT+&
A +B +C = 1800
(' çÜuóT„ ÈeTTýË n+ÔásÁ ¿ÃDeTT\ yîTTÔáeï TT 1800)
 1250 + x = 1800
650 + 600 + x = 1800
x = 1800 – 1250 = 550
|Ÿ³eTT (ii) qT+&
P + Q + R = 1800
x + x + 800 = 1800
(m+<ŠTÅ£”?)
 2x + 800 = 1800
2x = 1800 – 800 = 2x = 1000  x 
100D
 50D  x = 500
2
|Ÿ³eTT (iii) qT+&
(m+<ŠTÅ£”?)
X + Y + Z = 1800
x + 900 + 420 = 1800
x + 900 + 420 = 1800  x + 1320 = 1800
 x = 480
x = 1800 – 1320
7e ÔásÁ>·Ü >·DìÔáeTT
201
çÜuóT„ C²\T
Example 3: Find the values of x and y in the given triangle.
Solution :
A
In the DABC,
ACD +ACB = 1800 (linear pair of angles)
1100 + y = 1800
y = 1800 – 1100
y = 700
(i)
BAC + ACB + CBA = 180
(Why?)
0
x + y + 600 = 1800
x + 700 + 600 = 1800
x + 130 = 180
0
(From (i))
0
x = 500
Example 4 : In a right angled triangle one acute angle is 440. Find the other angle.
Solution :
We know that, sum of the two acute angles in a right angled triangle is 900.
Given that, in the right angled triangle one of the acute angles = 440
Other acute angle in right angled triangle = 900 – 440 = 460
Example 5 : The angles of a triangle are (x + 10)0, (x – 20)0 and (x + 40)0. Find the value of x and
the measure of the angles.
Solution :
Given that the angles of the triangle are (x + 10)0, (x – 20)0 and (x + 40)0.
(x + 10)0 + (x – 20)0 + (x + 40)0 = 1800
(why?)
x + 100 + x – 200 + x + 400 = 1800
3x + 300 = 1800
 3x = 1800 – 300
3x = 1500
150D
 x
 50D
3
Each of the angles are,
x + 100 = 500 + 100 = 600
x – 200 = 500 – 200 = 300
x + 400 = 500 + 400 = 900
 Measure of the angles are 600, 300 and 900
N
A
G
S
L
From the given figure, find the sum of
A, N, G, L, E and S.
E
VII Class Mathematics
202
Triangles
–<‘VŸ²sÁD 3 : ‚ºÌq çÜuóT„ ÈeTTýË x eT]jáTT y $\Te\T ¿£qT>=q+&.
kÍ<óŠq:
DABC ýË,
ACD +ACB = 180 (¹sFjáT
1100 + y = 1800
y = 1800 – 1100
y = 700
0
A
¿ÃD²\ <ŠÇjáT+)
(i)
(m+<ŠTÅ£”?)
BAC + ACB + CBA = 1800
x + y + 600 = 1800
x + 700 + 600 = 1800
x + 1300 = 1800
((i) qT+&)
x = 500
–<‘VŸ²sÁD 4 : \+‹¿ÃD çÜuóT„ ÈeTTýË ÿ¿£ n\οÃD+ 440 nsTTq s +&ƒe n\οÃD+ ¿£qT>=q+&.
kÍ<óŠq :
\+‹¿ÃD çÜuóT„ ÈeTT ýË s +&ƒT n\Î ¿ÃDeTT\ yîTTÔá+ï 900\T n“ eTqÅ£” Ôî\TdŸT.
<ŠÔï+Xø ç|Ÿ¿±sÁ+ \+‹¿ÃD çÜuóT„ ÈeTT ýË ÿ¿£ n\οÃD+ R 440.
\+‹¿ÃD çÜuóT„ ÈeTTýË s +&ƒe n\Î ¿ÃD+ R 900 - 440 R 460
–<‘VŸ²sÁD 5 : çÜuó„TÈ+ý˓ ¿ÃDeTT\T (x + 10)0, (x – 20)0 eT]jáTT (x + 40)0. € ¿ÃDeTT\ýË x $\TeqT
¿£qT>=q+&. € ¿ÃDeTT\ ¿=\Ôá\qT ¿£qT>=q+&.
kÍ<óŠq :
<ŠÔï+Xø ç|Ÿ¿±sÁ+ çÜuóT„ È+ý˓ ¿ÃDeTT\T (x + 10)0, (x – 20)0 eT]jáTT (x + 40)0
(x + 10)0 + (x – 20)0 + (x + 40)0 = 1800
(m+<ŠTÅ£”?)
x + 100 + x – 200 + x + 400 = 1800
3x + 300 = 1800
 3x = 1800 – 300
3x = 1500
 x
€ ¿ÃD²\T,
150D
 50D
3
x + 100 = 500 + 100 = 600
x – 200 = 500 – 200 = 300
x + 400 = 500 + 400 = 900

¿ÃD²\ jîTT¿£Ø ¿=\Ôá\T 600, 300 eT]jáTT 900
nHûÇw¾<‘Ý+
ç|Ÿ¿Ø£ |Ÿ³+ qT+& A, N, G, L, E eT]jáTT S ¿ÃD²\
yîTTԐ ¿£qT>=q+&?
N
A
G
S
L
E
7e ÔásÁ>·Ü >·DìÔáeTT
203
çÜuóT„ C²\T
Example 6 : Find the sum of the interior angles in a pentagon.
Solution :
A
Divide the pentagon into three triangles as shown in the fig.
Sum of the angles in DABC = 180
0
E
3
2
B
1
Sum of the angles in DADC = 1800
D
Sum of the angles in DADE = 1800
C
Sum of the interior angles in the pentagon = Sum of angles in DABC + Sum of the angles in DADC
+ Sum of the angles in DADE
= 1800 + 1800 + 1800 = 5400.
Can you find a triangle in which each angle is less than 600?
Exercise - 5.2
1.
Which of the following angles form a triangle?
a)
600, 700, 800
d)
600, 300, 900
b)
650, 450, 700
e)
380, 1020, 400
c)
400, 500, 600
f)
1000, 300, 450
2.
Sum of two interior angles of a triangle is 1050. Find the third angle.
3.
In DPQR, P=650,Q=500. Find R.
4.
Find the missing angles in each of the following triangles.
a)
5.
b)
c)
Find the value of ‘x’ in each of the given triangles.
a)
b)
E
VII Class Mathematics
204
Triangles
–<‘VŸ²sÁD 6 : |Ÿ+#áuT„ó › jîTT¿£Ø n+ÔásÁ ¿ÃDeTT\ yîTTÔá+ï ¿£qT>=q+& .
kÍ<óŠq:
|Ÿ³eTTýË #áÖ|¾q³T¢ |Ÿ+#áuT„ó ›“ eTÖ&ƒT çÜuóT„ ÈeTT\T >± $uó›„ +#áTeTT
A
E
DABC ýË
3
B
1
2
¿ÃDeTT\yîTTÔá+ï R 1800
DADC ýË¿ÃDeTT\ yîTTÔá+
ï R1800
C
D
DADE ýË ¿ÃDeTT\ yîTTÔá+
ï R 1800
|Ÿ+#áuT„ó › jîTT¿£Ø n+ÔásÁ ¿ÃDeTT\ yîTTÔá+ï R DABCýË ¿ÃDeTT\ yîTTÔá+ï G DADCýË ¿ÃDeTT\ yîTTÔá+ï
G DADEýË ¿ÃDeTT\ yîTTÔá+ï
0
0
0
R 180 G 180 G 180 R 5400.
€ý˺<‘Ý+!
ç|Ÿr ¿ÃDeTT 60 &ç^\ ¿£H• ÔáŔ£ Øe –q• çÜuóT„ ÈeTT kÍ<ó«Š eÖ?
nuó²«dŸ+ ` 5.2
1. ç¿ì+<Š ‚ºÌq ¿ÃDeTT\ýË @$ çÜuóT„ ÈeTTqT @sÁÎsÁ#Tá qT?
a)
600, 700, 800
d)
600, 300, 900
b)
650, 450, 700
e)
380, 1020, 400
c)
400, 500, 600
f)
1000, 300, 450
2. çÜuóT„ È+ý˓ s +&ƒT ¿ÃDeTT\ yîTTÔáeï TT 1050 nsTTq eTÖ&ƒe¿ÃD+ ¿£qT>=q+&.
3. DPQRýË P=650,Q=500 nsTTq R qT ¿£qT>=q+&.
4. ç¿ì+~ çÜuóT„ ÈeTT\ýË $TÐ*q ¿ÃDeTTqT ¿£qT>=q+&.
a)
b)
c)
5. ç¿ì+<Š ‚ºÌq çÜuóT„ ÈeTT\ýË ‘x’ $\TeqT ¿£qT>=q+&.
a)
b)
E
7e ÔásÁ>·Ü >·DìÔáeTT
205
çÜuóT„ C²\T
6.
Find the value of ‘x’ and ‘y’ in each of the following triangles.
a)
b)
680
7.
In a right angled triangle one acute angle is 370. Find the other acute angle.
8.
If the three angles of a triangular signboard are 2x0, (x – 10)0 and (x + 30)0 respectively. Then
find it’s angles.
9.
If one angle of a triangle is 800, find the other two angles which are equal.
10. State TRUE or FALSE for each of the following statements and give the reasons for the
FALSE statement.
i)
A triangle can have two right angles.
ii)
A triangle can have two acute angles.
iii) A triangle can have two obtuse angles.
11. The angles of a triangle are in the ratio 2 : 4 : 3, then find the angles.
12. Find the sum of interior angles of an hexagon.
5.3 Exterior Angle Property :
Draw DABC and extended BC upto D.
Observe the ACD formed at point C. This angle lies
in the exterior of DABC. We call it is the exterior angle of
DABC formed at vertex C.
Clearly BCA is an adjacent angle to ACD. The remaining
two angles of the triangle namely BAC or A and CBA
or B are called the two interior opposite angles of ACD.
Now cut out (or make trace copies of) A and B and place
them adjacent to each other at ‘C’ as shown in the Figure.
Do these two pieces together entirely cover ACD?
Can you say that ACD = A + B?
From the above activity, we can say that an exterior angle of a triangle is equal to the sum of its
opposite interior angles.
VII Class Mathematics
206
Triangles
6. ‚ºÌq çÜuóT„ ÈeTT\ýË ‘x’ eT]jáTT ‘y’ $\Te\T ¿£qT>=q+&.
a)
b)
680
7. ÿ¿£ \+‹¿ÃD çÜuóT„ ÈeTTýË ÿ¿£ n\Î ¿ÃDeTT 370 nsTTq s +&ƒe n\Î ¿ÃD+ ¿£qT>=q+&.
8. çÜuóT„ C²¿±sÁ+ýË –q• ™dHÕ Ž uËsÁT¦ jîT¿£Ø eTÖ&ƒT n+ÔásÁ ¿ÃDeTT\T esÁdTŸ >± 2x0, (x – 10)0, (x + 30)0
nsTTq € ¿ÃDeTT\T ¿£qT>=q+&.
9. çÜuóT„ È+ jîTT¿£Ø ÿ¿£ ¿ÃD+ 800 eT]jáTT $TÐ*q s +&ƒT ¿ÃD²\T dŸeÖq+ nsTTq y{ì“ ¿£qT>=qTeTT.
10. ç¿ì+~ ç|Ÿe#áH\T dŸÔ«á yîÖ, ndŸÔ«á yîÖ çyd¾ ndŸÔ«á ç|Ÿe#áH\qT dŸ¿±sÁD+>± $e]+#á+&.
i) çÜuóT„ ÈeTT s
 +&ƒT \+‹ ¿ÃDeTT\T ¿£*Ð –+&ƒe#áTÌ.
ii) çÜuóT„ ÈeTT s
 +&ƒT n\Î ¿ÃDeTT\T ¿£*Ð –+&ƒe#áTÌ.
iii) çÜuóT„ ÈeTT s
 +&ƒT n~ó¿£ ¿ÃDeTT\T ¿£*Ð –+&ƒe#áTÌ.
11. çÜuóT„ È+ jîTT¿£Ø ¿ÃDeTT\T 2:4:3 “wŸÎÜïýË –q•, € ¿ÃDeTT\T ¿£qT>=q+& .
12. wŸ&Tƒ Òۛ jîTT¿£Ø n+ÔásÁ ¿ÃD²\ yîTTÔáeï TT ¿£qT>=q+&.
5.3 u²VŸ²«¿ÃD <óŠsÁˆeTT
‚~ #û<‘Ý+!
¿£Ôá«+
ÿ¿£ ¿±ÐÔá+™|Õ ABC çÜuóT„ ÈeTT ^jáT+& BC uóT„ ÈeTTqT D esÁŔ£
bõ&Ð+#á+&.
C e<ŠÝ @sÁÎ&q ACD ¿ÃDeTTqT |Ÿ]o*+#á+&.
‡ ¿ÃDeTT ABC çÜuóT„ ÈeTTqT ‹jáT³–q•~. B“Hû ABC çÜuóT„ ÈeTTqÅ£” osÁüeTT C e<ŠÝ u²VŸ²«¿ÃDeTT n+{²sÁT.
BCA, ACDÅ£” €dŸq•¿ÃD+ nHû~ dŸÎwŸ+
¼ .
çÜuóT„ ÈeTTý˓ $TÐ*q s +&ƒT ¿ÃDeTT\T BAC ýñ<‘ A eT]jáTT
CBA ýñ<‘ B\T, u²VŸ²«¿ÃDeTT ACDÅ£” n+Ôás_óeTTK ¿ÃDeTT\T
n+{²sÁT.
‚|ŸÚÎ&ƒT ¿ÃDeTT A, ¿ÃD+ B\qT ¿£Ü]ï +º (çfñdt #ûdq¾ eTT¿£ØýÉHÕ )
u²VŸ²«¿ÃDeTT ACD™|Õ |Ÿ³eTTýË #áÖ|¾q³T¢ |Ÿ¿Ø£ |Ÿ¿Ø£ Hû nÜ¿ì+#á+&.
‡ s +&ƒT ¿ÃDeTT\T ¿£*d¾ u²VŸ²«¿ÃDeTT ACDÅ£” dŸ]bþsTTqy
ACD = A + B n“ #î|ŸÎ>·\s?
™|Õ ¿£Ôá«eTT qT+& ÿ¿£ çÜuóT„ È+ý˓ u²VŸ²«¿ÃDeTT <‘“ jîTT¿£Ø n+Ôásn
Á _óeTTK ¿ÃDeTT\ yîTTÔáeï TTqÅ£” dŸeÖqeT“
#î|Ο e#áTÌ.
7e ÔásÁ>·Ü >·DìÔáeTT
207
çÜuóT„ C²\T
Proof of Exterior angle property :
We can prove this property as follows.
Statement :
An exterior angle of triangle is equal to the sum of its interior opposite angles.
Given :
In DPQR, PRS is an exterior angle at R, P and Q are interior opposite angles.
To prove :
PRS = P + Q
Construction: Through point‘R’, Draw RT parallel to PQ
Proof :
eq. (i) + (ii)
1 = 3
(i) (alternative angles)
2 = 4
(ii) (corresponding angles)
1 + 2 = 3 + 4
PRS = P + Q
Thus, the exterior angle of a triangle is equal to sum of it’s interior opposite angles.
Example 7 : In the DABC, exterior angle at C = 1050 and A=650. Find the other interior opposite
angle.
Solution :
Given that, exterior angle at C =1050.
One of the interior opposite angle A = 650.
The other interior opposite angle is B.
A + B =1050 (' exterior angle of a triangle is equal to sum of the interior opposite
angles) 650 + B = 1050
 B =1050 – 650 = 400
Example 8 : Find the exterior angle of the given triangle.
Solution :
From the figure, P = 300, Q = 350
Exterior angle at R = P + Q (' exterior angle property)
= 300 + 350 = 650
Find the value of ‘x’ in the given figure.
VII Class Mathematics
208
Triangles
u²VŸ²« ¿ÃD <óŠsÁˆeTT “sÁÖ|ŸD :
ç|Ÿe#áq+
<ŠÔï+XøeTT
:
:
çÜuóT„ ÈeTTý˓ u²VŸ²«¿ÃDeTT n+Ôás_óeTTK ¿ÃDeTT\ yîTTÔáeï TTqÅ£” dŸeÖqeTT.
DPQR ýË R e<ŠÝ u²VŸ²«¿ÃDeTT PRS. P eT]jáTT Q nHû$ y{ì n+Ôás_óeTTK
¿ÃD²\T.
kÍs+XøeTT
“sˆDeTT
:
:
PRS = P + Q
R qT+& PQ Å£”
1 = 3
2 = 4
(i), (ii)\
dŸeÖ+Ôás+Á >± RT qT ^jáT+&.
(i) (@¿±+ÔásÁ ¿ÃDeTT\T)
(ii) (dŸ<ŠXø« ¿ÃDeTT\T)
qT+& 1 + 2 = 3 + 4
PRS = P + Q
¿£qT¿£ çÜuóT„ ÈeTTý˓ u²VŸ²«¿ÃDeTT, n+Ôás_óeTTK ¿ÃDeTT\ yîTTÔáeï TTqÅ£” dŸeÖqeTT.
–<‘VŸ²sÁD 7 : DABCýË C e<ŠÝ u²VŸ²«¿ÃDeT R 1050 eT]jáTT A=650 nsTTq s +&ƒe n+Ôás_óeTTK ¿ÃD+
¿£qT>=q+&.
kÍ<óŠq :
<ŠÔï+XøeTT qT+& C e<ŠÝ u²VŸ²«¿ÃDeTT R 1050.
ÿ¿£ n+Ôás_óeTTK ¿ÃDeTT A = 650.
s +&ƒe n+Ôás_óeTTK ¿ÃD+ B.
A + B =1050 ('
dŸeÖqeTT)
çÜuóT„ ÈeTTý˓ u²VŸ²«¿ÃDeTT n+Ôás_óeTTK ¿ÃDeTT\ yîTTÔáeï TTqÅ£”
650 + B = 1050
 B =1050 – 650 = 400
–<‘VŸ²sÁD 8 : ç¿ì+~ çÜuóT„ ÈeTTýË u²VŸ²«¿ÃDeTT ¿£qT>=q+&.
kÍ<óŠq :
|Ÿ³eTT qT+&, P = 30 , Q=35
0
R e<ŠÝ
0
u²VŸ²«¿ÃDeTT = P + Q (' u²VŸ²«¿ÃD <ósŠ ˆÁ eTT)
= 300 + 350 = 650
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
7e ÔásÁ>·Ü >·DìÔáeTT
‚ºÌq |Ÿ³+ýË ‘x’ $\Te ¿£qT>=q+&.
209
çÜuóT„ C²\T
Exercise - 5.3
1.
Write the exterior angles of DXYZ.
2.
Find the exterior angles in each of the following triangles:
60 0
a)
b)
730
3.
R
F
Find the value of ‘x’ in the following figures.
P
Fig (ii)
Fig (i)
4.
5.
If the exterior angle of a triangle is 1100 and it’s interior opposite angles are x0 and (x + 10)0. Find
the value of ‘x’.
Find the values of ‘x’ and ‘y’ in each of the following figures.
A
Q
Fig (i)
VII Class Mathematics
P
210
Fig (ii)
Triangles
nuó²«dŸ+ ` 5.3
1.
DXYZ jîTT¿£Ø
u²VŸ²«¿ÃD²\T çyjáT+&.
2. ‚ºÌq çÜuóT„ C²\ýË u²VŸ²«¿ÃD²\qT ¿£qT>=qTeTT.
a)
b)
600
730
R
3. ‚ºÌq çÜuóT„ C²\ýË ‘x’ jîTT¿£Ø $\Te ¿£qT>=qTeTT.
F
P
|Ÿ³+ (ii)
|Ÿ³+ (i)
4. ÿ¿£ çÜuóT„ ÈeTTý˓ u²VŸ²«¿ÃDeTT 1100 eT]jáTT n+Ôás_óeTTK ¿ÃDeTT\T x0, (x + 10)0 nsTTq ‘x’ $\Te
¿£qT>=q+&.
5. ‚ºÌq çÜuóT„ ÈeTT\ýË ‘x’ eT]jáTT‘y’ $\Te\T ¿£qT>=q+&.
A
Q
|Ÿ³+ (i)
7e ÔásÁ>·Ü >·DìÔáeTT
P
211
|Ÿ³+ (ii)
çÜuóT„ C²\T
5.4 Properties of sides of a triangle :
Measure the length of the sides of ABC and fill the
following table.
Name of the Triangle
ABC
Length of the Sides Sum of the lengths of two sides
AB =
BC =
CA =
BC + CA =
CA + AB =
AB + BC =
Relation
True/False
BC + CA > AB
CA + AB > BC
AB + BC > CA
From the above table, we can conclude that the sum of the lengths of any two sides of a triangle
is greater than the length of the third side.
Take the same measurements of the previous activity and note the results as given below.
Name of the
triangle
Length of the
side
Difference of
lengths
Relation
ABC
AB =
BC – CA =
BC – CA < AB
BC =
CA – AB =
CA – AB < BC
CA =
AB – BC =
AB – BC < CA
True/False
From the above table we can conclude that the difference of the lengths of any two sides of a
triangle is less than the length of the third side.
590
760
450
Fill the following table as per measurements given in the above triangles:
Name of the
triangle
Measure of
each angles
Lenth of each
side
ABC
A=__, B=__, C=__
AB =__, BC =__, CA =__
PQR
P=__,Q=__, R=__
PQ =__,QR =__, RP =__
Smallest Shortest Biggest Longest
angle
side
angle
side
What do you observe from the last four columns in above table?
v In any triangle the opposite side to the biggest angle is bigger than the other two sides.
v In any triangle the opposite side to the smallest angle is smaller than the other two sides.
VII Class Mathematics
212
Triangles
5.4 Üuó„TÈeTT jîTT¿£Ø uó„TC²\ <óŠsˆ\T :
‚~ #û<‘Ý+!
ABC, PQR, XYZ \ýË
|Ÿ{¿¼ì ý£ Ë qyîÖ<ŠT #ûjTá +&.
¿£Ôá«+
çÜuóT„ ÈeTT |sÁT
ABC
ç|ŸÜ uóT„ ÈeTT\qT ¿=*º ç¿ì+~
ÿ¿£ uóT„ È+ bõ&ƒeÚ s +&ƒT uóT„ ÈeTT\ yîTTÔáeï TT
AB =
BC =
CA =
BC + CA =
CA + AB =
AB + BC =
dŸ+‹+<óeŠ TT
dŸÔ«á eTT/ndŸÔ«á eTT
BC + CA > AB
CA + AB > BC
AB + BC > CA
™|Õ |Ÿ{¿¼ì £ qT+&, ÿ¿£ çÜuóT„ È+ý˓ @yîHÕ  s +&ƒT uóT„ ÈeTT\ yîTTÔáeï TT, eTÖ&ƒe uóT„ È+ ¿£H• mÅ£”Øe n“ “sÆ]+#áe#áTÌ.
‚+ÔáŔ£ eTT+<ŠT ¿£Ôá«+ýË d¿£]+ºq nyû uóT„ ÈeTT\ ¿=\Ôá\qT ç¿ì+~ |Ÿ{¿¼ì ý£ Ë qyîÖ<ŠT #ûjTá +&.
çÜuóT„ ÈeTT
|sÁT
ABC
ÿ¿£ uóT„ È+ bõ&ƒeÚ
s +&ƒT uóT„ ÈeTT\ uó<ñ eŠó TT
dŸ+‹+<óeŠ TT
AB =
BC – CA =
BC – CA < AB
BC =
CA – AB =
CA – AB < BC
CA =
AB – BC =
AB – BC < CA
dŸÔ«á eTT /
ndŸÔ«á eTT
™|Õ |Ÿ{¿¼ì £ qT+&, ÿ¿£ çÜuóT„ È+ý˓ @yîHÕ  s +&ƒT uóT„ ÈeTT\ uñ<eŠó TT , eTÖ&ƒe uóT„ È+ ¿£H• ÔáŔ£ Øe n“ “sÆ]+#áe#áTÌ.
‚~ #û<‘Ý+!
¿£Ôá«+
590
760
450
™|Õ çÜuó„TC²\ ¿=\Ôá\qT ‹{ì¼ ç¿ì+~ |Ÿ{켿£qT |ŸP]+#á+&.
çÜuóT„ ÈeTT
uóT„ ÈeTT\ ¿=\Ôá\T
¿ÃDeTT\ ¿=\Ôá\T
|sÁT
ABC
A=__, B=__, C=__
AB =__, BC =__, CA =__
PQR
P=__,Q=__, R=__
PQ =__,QR =__, RP =__
v
v
nÜ ºq• nÜ ºq• nÜ ™|<ŠÝ nÜ ™|<ŠÝ
¿ÃDeTT uóT„ ÈeTT ¿ÃDeTT uóT„ ÈeTT
@ çÜuóT„ ÈeTTýË nsTTq nܺq• ¿ÃDeTTqÅ£” m<ŠTsÁT>± –q• uóT„ ÈeTT $TÐ*q s +&ƒT uóT„ ÈeTT\ ¿£H• ºq•~.
@ çÜuóT„ ÈeTTýË nsTTq n~ó¿£ ¿ÃDeTTqÅ£” m<ŠTsÁT>± –q• uóT„ ÈeTT $TÐ*q s +&ƒT uóT„ ÈeTT\ ¿£H• ™|<Š~Ý .
7e ÔásÁ>·Ü >·DìÔáeTT
213
çÜuóT„ C²\T
1260
7 cm
We can observe that ‘In an Isosceles triangle, the angles which are opposite to equal sides are also
equal’.
Similarly the sides which are opposite to equal angles are also equal’.
What are the measurements of angles of an equilateral triangle?
Example 9 : Find the values of ‘x’ and ‘y’ in the following figure.
L
Solution : LNM + 1100 = 1800 (Linear pair angles)
LNM = 1800 – 1100 = 700
LMN = LNM
xo
(Angles opposite to equal sides)
y = 700
x + y = 1100
(Exterior angle property)
x + 700 = 1100
1100
yo
x = 1100 – 700 = 400
M
N
O
Exercise - 5.4
1.
Two sides of a triangle are 5cm and 4cm respectively. Write any three possible measurements
that suit for the third side.
2.
The lengths of line segments are 3cm, 5cm, 6cm and 9cm.
3.
i)
From the above measurements which group of the line segments can form a triangle.
ii)
Which group of line segments can not form a triangle, give the reason.
Find the value of ‘x’ in the following triangles:
Fig. (i)
VII Class Mathematics
Fig. (ii)
214
Triangles
‚~ #û<‘Ý+!
¿£Ôá«+
1260
7 cm
ÿ¿£ dŸeT~Çu²VŸQ çÜuóT„ È+ýË dŸeÖq uóT„ ÈeTT\Å£” m<ŠTsÁT>± –q• ¿ÃDeTT\T dŸeÖq+ n<û$<ó+Š >± dŸeÖq
¿ÃDeTT\Å£” m<ŠTsÁT>± –q• uóT„ ÈeTT\T dŸeÖqeTT n“ eTq+ >·eT“+#áe#áTÌ.
€ý˺<‘Ý+!
dŸeTu²VŸQ çÜuóT„ È+ýË ¿ÃD²\ ¿=\Ôá\T ¿£qT>=qTeTT?
–<‘VŸ²sÁD 9 : ç¿ì+~ |Ÿ³+ qT+& ‘x’ eT]jáTT ‘y’ $\Te\T ¿£qT¿ÃØ+&.
kÍ<óŠq :
LNM + 110 = 180 (¹sU² K+&ƒ+™|Õ ¿ÃDeTT\T)
0
L
0
LNM = 1800 – 1100 = 700
xo
(dŸeÖq uóT„ C²\Å£”
m<ŠTsÁT>± –+&û ¿ÃD²\T)
LMN = LNM
y = 700
(u²VŸ²«¿ÃD <ósŠ ˆÁ eTT)
x + y = 1100
x + 70 = 110
0
y
1100
o
0
x = 110 – 70 = 40
0
0
M
0
N
O
nuó²«dŸ+ ` 5.4
1. çÜuóT„ ÈeTTý˓ s +&ƒT uóT„ ÈeTT\T 5™d+.MT. eT]jáTT 4 ™d+.MT. nsTTq eTÖ&ƒe uóT„ È+ uóT„ ÈeTTqÅ£” kÍ<ó«Š eTjûT«
3 ¿=\Ôá\T sjáT+&.
2. s¹ U² K+&ƒeTT\ bõ&ƒeÚ\T 3 ™d+.MT., 5 ™d+.MT., 6 ™d+.MT. eT]jáTT 9™d+.MT.
i) ™|Õ ¿=\Ôá\ qT+& @ eTÖ&ƒT¿=\Ôá\T çÜuóT„ ÈeTTqT @sÁÎsÁ#Tá Hà #î|Ο +&.
ii) @ ¿=\Ôá\T çÜuóT„ ÈeTTqT @sÁÎsÁ#e
á Ú? ¿±sÁD+ #î|Ο +&.
|Ÿ³+ (i)
7e ÔásÁ>·Ü >·DìÔáeTT
|Ÿ³+ (ii)
215
çÜuóT„ C²\T
4.
ABC is an isosceles triangle in which AB = AC. If A = 800 then find B and C.
5.
Find the values of ‘x’ in each of the beside triangles.
6.
Which of the following statements are true in the following diagram.
i)
OY< OT
ii)
TY< TO
iii)
Y < T
Do you Know?
The perpendicular line segment from
any vertex to it’s opposite side is
called ‘Altitude.’
iv) TY < OY
5.5 Concurrent points:
Ortho Centre:
v
v
Make a paper cut of ABC, from vertex A fold the triangle on the side BC
as shown in Fig. (i). This folded line represents the altitude (orthogonal) of
the triangle. Name it as AD.
In the same way you can get two other altitudes. Name them BE and CF respectively as shown in
Fig. (ii) and (iii).
You can observe that these three altitudes are orthogonals concurrent.
v
The concurrent point of the orthogonals of a triangle is called Orthocenter. It is denoted
by ‘O’.
Note :
1.
2.
In right angled triangle Orthocentre lies on the vertex at the right angle.
It lies on the exterior of the obtuse angled triangle.
VII Class Mathematics
216
Triangles
4. ABC ÿ¿£ dŸeT~Çu²VŸQ çÜuó„TÈ+, AB = AC. A = 800, nsTTq B eT]jáTT C ¿£qT>=q+&.
5. ç|Ÿ¿Ø£ q ‚eNj&q çÜuóT„ C²\ý˓ ‘x’ $\Te\T ¿£qT¿ÃØ+&.
6. ç¿ì+<Š ‚eNj&q |Ÿ³eTT qT+& ‡ ç¿ì+<Š ‚eNj&q y¿£«eTT\T @$ dŸÔ«á eTT\T?
i)
OY< OT
ii)
TY< TO
MTÅ£” Ôî\TkÍ?
çÜuóT„ ÈeTTýË @<îHÕ  ÿ¿£ osÁüeTT qT+&
<‘“ m<ŠT{ì uó„TC²“¿ì ^jáT‹&q
\+‹¹sKqT –q•Ü n+{²sÁT.
iii) Y < T
iv) TY < OY
5.5 $T[Ôá _+<ŠTeÚ\T
\+‹¹¿+ç<ŠeTT
‚~ #û<‘Ý+!
v
¿£Ôá«+
¿±ÐÔá+™|Õ ABC çÜuóT„ ÈeTTqT ¿£Ü]ï +#áTeTT. osÁü+ A qT+& |Ÿ³+ 1ýË #áÖ|¾q³T¢
BC uóT„ ÈeTT eT&ƒe+&.
|Ÿ³eTTýË #áÖ|¾q³T¢, ‡ eT&ºq s¹ K yî+‹& >·\ s¹ U² K+&ƒeTT çÜuóT„ ÈeTT jîTT¿£Ø mÔáTqï T dŸÖº+#áTqT. B“¿ì
AD n“|sÁT™|³¼+&.
v
n<û $<ó+Š >± $TÐ*q s +&ƒT –q•ÔáT\qT bõ+<Še#áTÌ. ‡ s +&ƒT –q•Ôá\Å£” |Ÿ³eTT 2, 3ýË #áÖ|¾q³T¢.
CF n“ |sÁT™|³¼+&.
BE,
eTÖ&ƒT –q•ÔáT\T $T[Ôá s¹ K\“ >·eT“+#áe#áTÌ.
v
çÜuóT„ È+ýË –q•ÔáT\T $T[Ôá _+<ŠTeÚqT \+‹¹¿+ç<ŠeTT n+{²sÁT. B““ ‘O’ n¿£sŒ +Á Ôà dŸÖºkÍïeTT.
>·eT“¿£ :
1. \+‹¿ÃD çÜuóT„ C+ýË \+‹¹¿+ç<ŠeTT \+‹¿ÃDeTT ¿£*ÐjáTTq• osÁüeTT e<ŠÝ eÚ+³T+~.
2. ‚~ n~ó¿¿£ ÃD çÜuóT„ È+ýË u²VŸ²« _+<ŠTeÚ>± eÚ+³T+~.
7e ÔásÁ>·Ü >·DìÔáeTT
217
çÜuóT„ C²\T
Incentre:
Do you Know?
A line which divides given angle into two equal
parts is called angle bisector.
1.
Make a paper cut of ABC Fold the triangle as shown in Fig. (i) by coinciding the sides AB and
AC. This folded line represents the angle bisector of A.
2.
In the same way, you can get two more angle bisectors of B and C respectively as shown in
Fig. (ii) and (iii).
You observe that these three angle bisectors are concurrent.
3.
The concurrent point of the angle bisectors is called ‘Incentre’ of the triangle and it is denoted
by ‘I’.
S
I = Incentre, IS = radius of the Incircle
Note : Incentre of a triangle is equidistant from all the three sides.
Circumcentre :
Do you Know?
A perpendicular line passing through mid point of
any side is called perpendicular bisector.
u
u
Make a paper cutof ABC. Fold the triangle along the side, by coinciding the vertices B and C
each other. The folded line represents the perpendicular bisector of side BC.
In the same way you can get two more perpendicular bisectors to the sides AC and AB respectively.
You observe that these three perpendicular bisectors are concurrent.
u
The concurrent point of perpendicular bisectors of a triangle is called ‘Circumcentre’ of
that triangle. It is denoted by ‘S’.
VII Class Mathematics
218
Triangles
n+ÔáseY Ôáï ¿¹ +ç<ŠeTT
‚~ #û<‘Ý+!
¿£Ôá«+
MTÅ£” Ôî\TkÍ?
‚ºÌq ¿ÃD²“• s +&ƒT dŸeÖq uó²>±\T>± $uó›„ +#û
s¹ KqT ¿ÃD dŸeT~ÇK+&ƒq s¹ K n+{²sÁT.
1. ¿±ÐÔá+™|Õ ABC çÜuóT„ ÈeTTqT ¿£Ü]ï +#áTeTT. |Ÿ³+ 1ýË #áÖ|¾q³T¢ çÜuóT„ ÈeTTqT AB eT]jáTT AC
uóT„ ÈeTT\T @¿¡u$„ó +#áTq³T¢>± eT&ƒTeÚeTT. ‡ $<ó+Š >± eT&ºq s¹ K A ¿ì ¿ÃD dŸeT~ÇK+&ƒq s¹ K neÚÔáT+~.
2. ‚<û $<óŠ+>± B eT]jáTT C osÁüeTT\ e<ŠÝ s +&ƒT ¿ÃD dŸeT~ÇK+&ƒq s¹ K\qT |Ÿ³eTT 2 eT]jáTT 3ýË
#áÖ|¾q³T¢ ^jáTe#áTÌ. ‡ eTÖ&ƒT ¿ÃD dŸeT~ÇK+&ƒq s¹ K\T $T[Ԑ\T n“ >·eT“+#áe#áTÌ.
3. eTÖ&ƒT ¿ÃD dŸeT~ÇK+&ƒq s¹ K\ $T[Ôá _+<ŠTeÚqT n+ÔáseY Ôá ¿¹ï +ç<ŠeTT n+{²sÁT. B““ I nqT n¿£sŒ +Á ÔÃ
dŸÖºkÍïsTÁ .
S
I R n+ÔásYeÔáï ¹¿+ç<ŠeTT, IS R n+ÔásYeÔáï ¹¿+ç<Š y«kÍsÁœ+
>·eT“¿£ : n+ÔásÁ eÔáï ¿¹ +ç<ŠeTT çÜuóT„ CeTT jîTT¿£Ø eTÖ&ƒT uóT„ C²\Å£” dŸeÖq <ŠÖsÁ+ýË –+³T+~.
|Ÿ]eÔáï ¹¿+ç<ŠeTT
MTÅ£” Ôî\TkÍ?
@<îÕH uó„TÈeTT jîTT¿£Ø eT<óŠ« _+<ŠTeÚ
>·T+&† bþjûT \+‹ s¹ KqT \+‹
dŸeT~ÇK+&ƒq s¹ K n+{²sÁT.
‚~ #û<‘Ý+!
¿£Ôá«+
u
u
u
¿±ÐÔá+™|Õ ABC çÜuóT„ ÈeTTqT ¿£Ü]ï +#áTeTT. ¿£Ü]ï +ºq ‡ çÜuóT„ ÈeTTqT B eT]jáTT C osÁüeTT\T
@¿¡u$„ó +ºq³T¢>± eT&ƒe+&. eT&ƒÔá yî+‹& s¹ K \+‹ dŸeT~ÇK+&ƒq s¹ KqT
n<û $<ó+Š >± |Ÿ³eTT 2 eT]jáTT 3ýË #áÖ|¾q³T¢ AC eT]jáTT AB uóT„ ÈeTT\Å£” \+‹ dŸeT~ÇK+&ƒq s¹ K\qT
bõ+<Še#áTÌ.
‡ eTÖ&ƒT \+‹ dŸeT~ÇK+&ƒq s¹ K\T $T[ÔáeTT\T n“ >·eT“+#á>\· eÚ.
ÿ¿£ çÜuóT„ È+ýË \+‹ dŸeT~ÇK+&ƒq s¹ K\ $T[Ôá _+<ŠTeÚqT |Ÿ]eÔΌï +ç<ŠeTT n+{²sÁT. B““ ‘S’ nHû
n¿£sŒ +Á Ôà dŸÖºkÍïsTÁ .
7e ÔásÁ>·Ü >·DìÔáeTT
219
çÜuóT„ C²\T
Note :
l
l
l
Circumcentre is equidistant from all three vertices of the triangle.
It lies on the midpoint of the hypotenuse in right angled triangle.
It lies on the exterior of the obtuse angled triangle.
Centroid :
Do you Know?
A line segment joining any vertex to mid point of
it’s opposite side of triangle is called median
n
n
Make paper cut out of ABC, now fold triangle in such a way that the vertex B falls on
vertex C. The line along which the triangle has been folded will intersect at midpoint of side
BC as shown in the figure. The point of intersection is the mid-point of the side BC, name it
as S. Draw a line joining vertex A to the mid-point S. This line is called the median of the
triangle on BC from A.
In the same way find the mid points to AB and AC. Name the mid points as ‘U’ and ‘T’
respectively. Join UC and TB (medians).
You observe that these three medians are concurrent.
n
The concurrent point of medians of a triangle is called ‘Centroid’ and it is denoted by ‘G’.
2
1
Note : Centroid ‘G’ divides each median in the ratio of 2:1 from the vertex of a triangle.
Write the location of the concurrent points in different triangles.
Type of triangle
Orthocentre (O)
Incentre (I)
Acute angled
triangle
VII Class Mathematics
Centroid (G)
Mid point of the
hypotenuse
Right angled
triangle
Obtuse angled
triangle
Circumcentre (S)
Interior
Exterior
220
Triangles
>·eT“¿£ :
l
l
l
ÿ¿£ çÜuóT„ È+ýË |Ÿ]eÔáï ¿¹ +ç<ŠeTT eTÖ&ƒT osÁüeTT\Å£” dŸeÖq <ŠÖsÁ+ýË –+³T+~.
\+‹¿ÃD çÜuóT„ È+ýË |Ÿ]eÔáï ¿¹ +ç<ŠeTT ¿£se’Á TT jîTT¿£Ø eT<ó«Š _+<ŠTeÚ neÚÔáT+~.
n~ó¿£ ¿ÃD çÜuóT„ ÈeTTýË |Ÿ]eÔáï ¿¹ +ç<ŠeTT çÜuóT„ ÈeTTqÅ£” ‹jáT³ –+³T+~.
>·TsÁTÔáǹ¿+ç<ŠeTT :
MTÅ£” Ôî\TkÍ?
‚~ #û<‘Ý+!
çÜuóT„ ÈeTTýË @<îHÕ  osÁüeTT qT+& <‘“ m<ŠT³ uóT„ ÈeTT jîTT¿£Ø
eT<ó«Š _+<ŠTeÚqT ¿£*| s¹ U²K+&†“• eT<ó«Š >·Ôá s¹ K n+{²sÁT.
¿£Ôá«+
n
n
ÿ¿£ ¿±ÐÔá+™|Õ ABC ¿£Ü]ï +#áT+&.‡ çÜuóT„ ÈeTTqT B osÁüeTT, C osÁüeTT\T @¿¡u$„ó +#áTq³T¢>± eT&ƒTeÚeTT.‡
eT&ƒÔá |Ÿ³+ 1ýË #áÖ|¾q³T¢ BC “ eT<ó«Š _+<ŠTeÚ e<ŠÝ K+&+#áTqT. € _+<ŠTeÚqT S nqT¿=qTeTT. eT<ó«Š _+<ŠTeÚ
S eT]jáTT osÁüeTT AqT ¿£\T|ŸÚeTT. AS çÜuó„TÈeTTqÅ£” eT<óŠ«>·ÔáeTT n>·TqT.
‚<û $<óŠ+>± AB eT]jáTT AC \Å£” eT<óŠ« _+<ŠTeÚ\T >·T]ï+#áTeTT. y{ì“ esÁTdŸ>± U eT]jáTT T nqTeTT.
UC eT]jáTT TB\qT (eT<óŠ«>·ÔáeTT\T) ¿£\T|ŸÚeTT.
‡ eTÖ&ƒT eT<ó«Š >·Ôá s¹ K\T $T[ÔáeTT\T n“ >·eT“+#áe#áTÌ.
n
‡ eT<ó«Š >·Ôá s¹ K\ $T[Ôá_+<ŠTeÚqT >·TsÁTÔáÇ¿¹ +ç<Š+ n+{²sÁT. B““ G n¿£sŒ +Á Ôà dŸÖºkÍïsTÁ .
2
1
>·eT“¿£ :
>·TsÁTÔáǹ¿+ç<ŠeTT ‘G’ ç|ŸÜ eT<ó«Š >·Ôá s¹ KqT <‘“ osÁüeTT qT+& 2:1 “wŸÎÜïýË $uó›„ dŸT+ï ~.
nHûÇw¾<‘Ý+
$$<óŠ çÜuóT„ C²\ýË $T[Ôá_+<ŠTeÚ\ jîTT¿£Ø k͜H“• çyjáT+&.
çÜuóT„ È sÁ¿e£ TT
\+‹ ¹¿+ç<ŠeTT (O) n+ÔásÁ eÔáï ¿¹ +ç<eŠ TT (I) |Ÿ]eÔáï ¿¹ +ç<ŠeTT (S) >·TsÁTÔáÇ¿¹ +ç<ŠeTT (G)
\+u¿ÃD
çÜuóT„ È+
¿£se’Á TT jîTT¿£Ø
eT<óŠ« _+<ŠTeÚ
n\οÃD
çÜuóT„ È+
n~ó¿£ ¿ÃD
çÜuóT„ È+
7e ÔásÁ>·Ü >·DìÔáeTT
n+Ôás+Á ýË
u²VŸ²«+ýË
221
çÜuóT„ C²\T
Euler Line:
Draw a triangle. Draw medians, perpendicular bisectors, angle bisectors
and altitudes in that triangle. Join centroid, circumcentre, incentre and
orthocentre.
What do you observe?
In a triangle, the centroid ‘G’, circumcentre ‘S’, incentre ‘I’ and orthocenter ‘O’ lie
on a straight line (Collinear points). This line is called the ‘Euler line’.
Note :
1.
2.
3.
In Isoscles triangle ‘G’, ‘I’, ‘O’ and ‘S’ are Collinear.
In scalene triangle ‘G’, ‘O’ and ‘S’ are Collinear.
In Equilateral triangle ‘G’, ‘I’, ‘O’ and ‘S’ are coincide to each other.
Exercise - 5.5
1.
State TRUE or FALSE. If it is FALSE write the correct statement.
i)
Centroid (G) of a triangle always lies in the interior of that triangle.
ii)
Orthocenter (O) of a right angled triangle is the mid point of it’s hypotenuse.
iii) Circumcentre (S) is equidistant from all the sides of a triangle.
iv) The concurrence of medians of a triangle is centroid.
v)
The point of concurrence of Perpendicular bisectors of sides is Incentre.
2.
In PQR, QT = TR, PS  QR, then what do you name PT and PS.
3.
Fill the following table with correct answers:
Concurrent lines in triangle
i)
Medians
ii)
Altitudes
Concurrent point
iii) Angle bisectors
iv) Perpendicular bisectors
VII Class Mathematics
222
Triangles
‚~ #û<‘Ý+!
¿£Ôá«+
€jáT\sY ¹sK :
ÿ¿£ çÜuóT„ C²“• ^jáT+&. çÜuóT„ C²“¿ì eT<ó«Š >·Ôá s¹ K\T, \+‹ dŸeT~ÇK+&ƒq s¹ K\T, ¿ÃD
dŸeT~ÇK+&ƒq s¹ K\T eT]jáTT \+‹ s¹ K\T ^jáT+&.
MTsÁT @$T >·eT“+#sÁT?
€sTT\sY ¹sK : çÜuóT„ È+ýË >·TsÁTÔáÇ¿¹ +ç<Š+ (G), |Ÿ]eÔáï ¿¹ +ç<Š+ (S), n+Ôás¿¹Á +ç<Š+ (I) eT]jáTT
\+‹ ¿¹ +ç<Š+(O)\T ÿ¹¿ s¹ K™|Õ –+&ƒTqT.€ s¹ KHû €sTT\sY s¹ K n+{²sÁT.
>·eT“¿£ : 1. dŸeT~Çu²VŸQ çÜuó„TÈ+ýË ‘G’, ‘I’, ‘O’ eT]jáTT ‘S’ \T dŸ¹sFjáÖ\T.
2. $wŸeTu²VŸQ çÜuóT„ È+ýË ‘G’ ‘O’ eT]jáTT ‘S’ \T dŸ¹sFjáÖ\T.
3. dŸeTu²VŸQ çÜuóT„ È+ýË ‘G’, ‘I’, ‘O’ eT]jáTT ‘S’ \T ÿ¹¿ _+<ŠTeÚ e<ŠÝ @Fuó$„ +#áTqT.
nuó²«dŸ+ ` 5.5
1. ç¿ì+~ y¿±«\T dŸÔ«á yîÖ, ndŸÔ«á yîÖ Ôî\|Ÿ+&. ndŸÔ«á yîÖ nsTTÔû dŸs qÕ y¿±«\qT sjáT+&.
i) çÜuó„TÈ+ jîTT¿£Ø >·TsÁTÔáÇ ¹¿+ç<Š+ (G) m\¢|ڟ Î&ƒÖ çÜuóT„ È+ ýË|Ÿ* uó²>·+ýË –+³T+~.
ii) \+‹¿ÃD çÜuóT„ È+ jîTT¿£Ø \+‹ ¿
¹ +ç<Š+ (O) <‘“ ¿£sÁ’+ jîTT¿£Ø eT<óŠ«_+<ŠTeÚ.
iii) |Ÿ]eÔáï ¿
¹ +ç<Š+ (S) çÜuóT„ È+ uóT„ C²\ qT+& dŸeÖq <ŠÖsÁeTTýË –+³T+~.
iv) çÜuó„TÈ+ jîTT¿£Ø eT<óŠ«>·Ôá ¹sK\ $T[Ôá _+<ŠTeÚ >·TsÁTÔáǹ¿+ç<Š+.
v) uóT„ C²\ \+‹dŸeT~ÇK+&ƒq s
¹ K\ $T[Ôá _+<ŠTeÚ n+Ôás¿¹Á +ç<Š+.
2. PQR ýË, QT = TR, PS  QR, nsTTq PT eT]jáTT PS\ |sÁT¢ @$T{ì?
3. ç¿ì+~ ‚ºÌq |Ÿ{¿¼ì q£ T dŸs qÕ Èy‹T\Ôà |ŸP]+#á+&.
$T[Ôá s¹ K\T
i) eT<ó«Š >·Ôe
á TT\T
ii) –q•ÔáT\T
iii) ¿ÃDdŸeT~ÇK+&ƒq s
¹ K\T
iv) \+‹dŸeT~ÇK+&ƒq s
¹ K\T
7e ÔásÁ>·Ü >·DìÔáeTT
223
$T[Ôá_+<ŠTeÚ
çÜuóT„ C²\T
4.
5.
6.
In ABC, AD is median and G is centroid write the following ratios:
i) AG : GD ii) AG : AD iii) AD : GD iv) GD : AG
In RST, If RX, SY and TZ are medians of a triangle.
If RX = 12cm, SY = 15cm then find,
i) GX
ii) RG
iii) SG
iv) GY
Which centre is equidistant from all the sides of a triangle.
5.6 Congruence of triangles:
Look at the following figures. What do you observe in them?
Let us observe congruence of some geometrical figures.
Name of the
figure
Diagram
B P
Line segments
A
Symbolic
representation
Equal in length
AB PQ
Equal measure
of angles
A  B
Equal radii
C1  C2
Q
Angles
C2
C1
Circles
Squares
Rule for
congruency
A
B
P
Q
D
C
R
S
Equal sides
ABCD 
PQRS
What do you think of rules for congruence of triangles?
Congruence of Triangles:
We can check congruence of plane figures by method of superimposition. After superimposition
if they coincide each other exactly then we say that they are congruent.
A
P
B
C
R
Q
Here ABC and PQR have the same size and the same shape. So we can represent this as
ABC PQR
This means that, when you place PQR on ABC, P coincide with A, Q coincide with B and R
coincide with C, PQ also coincide with AB , QR coincide with BC and PR coincide with AC . The
coinciding parts are called as corresponding parts. In two congruent triangles, the corresponding parts
(i.e., angles and sides) that match each other are equal.
VII Class Mathematics
224
Triangles
4.
5.
ABCýË AD nHû~
i)
AG : GD
i)
GX
eT<óŠ«>·ÔáeTT, G >·TsÁTÔáÇ¿¹ +ç<Š+ nsTTq ç¿ì+~ “wŸÎÔáT\ï qT sjáT+&.
ii) AG : AD
iii) AD : GD
RST, If RX, SY çÜuó„TÈ+ jîTT¿£Ø eT<óŠ«>·Ôá¹sK\T eT]jáTT
RX = 12 ™d+.MT., SY = 15™d+.MT. nsTTq M{ì“ ¿£qT¿ÃØ+&.
ii) RG
iii) SG
iv) GD : AG
iv) GY
6. çÜuóT„ È+ýË uóT„ C²\Å£” dŸeÖq <ŠÖsÁ+ýË –q• $T[Ôá _+<ŠTeÚ @~?
5.6 çÜuó„TC²\ dŸsÁÇdŸeÖqÔáÇ+ :
ç¿ì+<Š ‚ºÌq |Ÿ{²\qT >·eT“+#á+&. y{ìýË MTsÁT @$T >·eT“+#sÁT?
¿=“• C²«$TrjáT |Ÿ{²\ dŸsÇÁ dŸeÖqÔáÇ+ |Ÿ]o*<‘Ý+.
dŸsÇÁ dŸeÖqÔáÇ “jáTeT+
|Ÿ³+ |sÁT
ºçÔá+
B P
s¹ U² K+&†\T
A
Q
¿ÃD²\T
A
#áÔTá sÁçkÍ\T
C2
C1
eԐï\T
B
P
dŸeÖq bõ&ƒeÚ\T
AB PQ
dŸeÖq ¿ÃD²\T
A  B
dŸeÖq y«kÍsœ\T
C1  C2
Q
dŸeÖq uóT„ C²\T,
D
C
R
kÍ+¹¿Ü¿£yTî q® çbÍܓ<ó«Š +
ABCD 
PQRS
S
çÜuóT„ C²\ jîTT¿£ØdŸsÇÁ dŸeÖqÔáÇ “jáTeÖ\ >·T]+º MTsÁT @eTqTÅ£”+³TH•sÁT?
çÜuó„TC²\ jîTT¿£Ø dŸsÁÇdŸeÖqÔáÇeTT:
€<ó‘«sÃ|ŸD(dŸÖ|ŸsY ‚+bþ›wŸH)Ž $<ó‘q+ <‘Çs eTq+ dŸeTÔá\ |Ÿ{²\ jîTT¿£Ø dŸsÇÁ dŸeÖqԐǓ•
dŸ]#áÖ&ƒe#áTÌ. €<ó‘«sÃ|ŸD ÔásTÁ yÔá n$ ÿ¿£<‘“Ôà eTs=¿£{ì dŸ]>±Z @¿¡u$„ó dï n$ dŸsÇÁ dŸeÖqeTT\T.
A
P
C
R
B
Q
‚¿£Ø&ƒ ABC eT]jáTT PQR\T ÿ¹¿ €¿±sÁ+ eT]jáTT ÿ¹¿ |Ÿ]eÖD+ ¿£*Ð –H•sTT. n|ŸÚÎ&ƒT eTq+ B““
‚ý² Ôî*jáTCñkÍï+. ABC PQR
nq>± PQRqT ABC™|Õ –+ºq|ŸÚ&ƒT P nHû~ AÔà @¿¡uó„$dŸTï+~. Q nHû~ BÔà @¿¡uó„$dŸTï+~ eT]jáTT
R osÁü+ CÔÃ
@¿¡uó„$dŸTï+~. PQ nHû~ Å£L&† AB Ôà @¿¡uó„$dŸTï+~. QR nHû~ BC Ôà @¿¡uó„$dŸTï+~.
PR nHû~ AC ÔÃ @¿¡u$
„ó dŸT+ï ~. @¿¡u$„ó +#û uó²>±\qT dŸ<Š Xø uó²>±\T n“ n+{²sÁT. s +&ƒT dŸsÇÁ dŸeÖqÔáÇeTT
çÜuó„TC²ýË¢ dŸ<ŠXø« uó²>±\T (n+fñ, ¿ÃD²\T eT]jáTT uó„TC²\T) dŸeÖq+.
7e ÔásÁ>·Ü >·DìÔáeTT
225
çÜuóT„ C²\T
Statement
Corresponding vertices Corresponding sides Corresponding angles
ABC  PQR
AP
AB = PQ
A = P
BQ
BC = QR
B = Q
CR
AC = PR
C = R
DEF  RTS
DR
DE = RT
D = R
ET
EF = TS
E = T
FS
DF = RS
F = S
If EFG  KLM, then write the corresponding congruent sides,
congruent angles and congruent vertices?
Conditions for congruence of triangles:
We know that for two triangles to be congruent, it is necessary that three corresponding sides and
three corresponding angles of the two triangles must be equal. Now, we can understand that all the six
conditions are not necessarily required to be verified for establishing the congruence of two triangles.
SSS (Side-Side-Side) Congruence criterion: If three sides of a triangle are equal to the
corresponding three sides of the other triangle, then the two triangles are congruent.
Ex : In ABC and PQR,
Corresponding sides
AB = PQ,
BC = QR,
CA = RP
By SSS congruency, ABC and PQR are congruent. ABC  PQR. It is read as triangle ABC
is congruent to triangle PQR.
You can observe that the two triangles coincide each other exactly.
ABC FED
SAS (Side-Angle-Side) Congruence criterion :
If two sides and their included angle of one triangle are equal to the corresponding two
sides and their included angle of the other triangle, then the two triangles are congruent.
Ex :
In ABC and PQR
Corresponding sides
AB = PQ
BC = QR
Corresponding anglesB = Q
By SAS congruency, ABC and PQR are congruent.
ABC PQR
VII Class Mathematics
226
Triangles
ç|Ÿe#áq+
dŸ<Š Xø« osÁüeTT
ABC  PQR
dŸ<Š Xø« osÁüeTT
AP
BQ
CR
DR
ET
FS
DEF  RTS
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
AB = PQ
BC = QR
AC = PR
DE = RT
EF = TS
DF = RS
dŸ<Š Xø« ¿ÃD²\T
A = P
B = Q
C = R
D = R
E = T
F = S
EFG  KLM nsTTÔû
dŸ<ŠXø« uó„TC²\T, dŸ<ŠXø« ¿ÃD²\T eT]jáTT
dŸ<Š Xø« osü\qT sjáT+&.
çÜuó„TC²\ dŸsÁÇ dŸeÖqÔáÇ “jáTeÖ\T :
s +&ƒT çÜuóT„ C²\ eTÖ&ƒT nqTsÁÖ|Ÿ uóT„ C²\T eT]jáTT eTÖ&ƒT nqTsÁÖ|Ÿ ¿ÃD²\T dŸeÖqeTÔû n$ dŸsÇÁ dŸeÖqeT“
eTqÅ£” Ôî\TdŸT. ¿±ú çÜuóT„ C²\T dŸsÇÁ dŸeÖqeT“ “sÁÖ|¾+#á<‘“¿ì €sÁT nqTsÁÖ|Ÿ uó²>±\T dŸeÖq+ ¿±qedŸs+Á
ýñ<ŠT.
uóT„ È+ - uóT„ È+ - uóT„ È+ dŸsÇÁ dŸeÖqÔáÇ “jáTeTeTT:
ÿ¿£ çÜuóT„ È+ý˓ eTÖ&ƒT uóT„ C²\T esÁTdŸ>± s +&ƒe çÜuóT„ ÈeTTý˓ y{ì dŸ<Š Xø uóT„ C²\Å£” dŸeÖqyîTÔ® û €
s +&ƒT çÜuóT„ C²\T dŸsÇÁ dŸeÖqeTT.
–<‘VŸ²sÁD: ABC eT]jáTT PQR
dŸ<ŠXø« uó„TC²\T AB = PQ,
BC = QR,
CA = RP
uóT„ ÈeTT-uóT„ ÈeTT-uóT„ ÈeTT dŸsÇÁ dŸeÖq “jáTeT+ qT+& ABC eT]jáTT PQR\T dŸsÁÇdŸeÖH\T.
B““
ABC  PQR
çÜuó„TÈ+ ABC , çÜuó„TÈ+ PQR\T dŸsÇÁ dŸeÖq+ n“ #á<TŠ eÚԐ+.
MTsÁT @$T >·eT“+#sÁT? s +&ƒT çÜuóT„ C²\T dŸ]>±Z ÿ¿£<‘“Ôà eTs=¿£{ì m¿¡u$„ó dŸTHï •jáT“ MTsÁT >·eT“+#áe#áTÌ.
ABC FED
uóT„ ÈeTT - ¿ÃDeTT - uóT„ ÈeTT dŸsÇÁ dŸeÖqÔáÇ “jáTeTeTT :
ÿ¿£ çÜuóT„ È+ý˓ s +&ƒT uóT„ C²\T y{ì eT<ó«Š ¿ÃD+ esÁTdŸ>± s +&ƒe çÜuóT„ ÈeTTý˓ y{ì dŸ<Š Xø s +&ƒT
uóT„ C²\T, y{ì eT<ó«Š ¿ÃD²“¿ì dŸeÖqyîTÔ® û € s +&ƒT çÜuóT„ C²\T dŸsÇÁ dŸeÖqeTT.
–<‘VŸ²sÁD :
ABC eT]jáTT PQR \ýË
dŸ<ŠXø« uó„TC²\T:
AB = PQ
BC = QR
dŸ<ŠXø« ¿ÃD²\T:
B = Q
uóT„ ÈeTT-¿ÃDeTT-uóT„ ÈeTT dŸsÇÁ dŸeÖqÔá #ûÔá ABC eT]jáTT PQR\T dŸsÁÇdŸeÖH\T.
ABC PQR
7e ÔásÁ>·Ü >·DìÔáeTT
227
çÜuóT„ C²\T
ASA (Angle-Side-Angle) Congruence criterion:
If two angles and their included side of one triangle is equal to the corresponding two
angles and their included side of the another triangle, then the two triangles are congruent.
Ex : In ABC and PQR
Here, corresponding angles B = Q, C = R,
Corresponding side BC = QR
By ASA congruency, ABC and PQR are congruent.
ABC  PQR
RHS (Right angle-Hypotnuse-Side) Congruence criterion :
If the hypotenuse and one side of a right angled triangle are equal to the corresponding hypotenuse
and side of the other right-angled triangle, then the two triangles are congruent.
Ex : In ABC and PQR
Corresponding sides BC = QR
Corresponding hypotenuses AC = PR
Corresponding angles B = Q = 900
By RHS congruency, ABC and PQR are congruent.
ABC PQR
Note : In two triangles, if the three angles of one triangle are respectively equal to the three angles
of the other, then the triangles not necessarily be congruent.
Exercise - 5.6
1.
2.
Given that ABC DEF.
i)
List out all the corresponding congruent sides.
ii)
List out all the corresponding congruent angles.
ABC and FGE are congruent, determine whether the given pair of sides and angles are
corresponding sides and corresponding angles or not.
i)
AB and FG
ii)
iii)
BC and EF
iv) A and F
v)
B and F
vi) C and E.
VII Class Mathematics
AC and GE
228
Triangles
¿ÃDeTT - uóT„ ÈeTT - ¿ÃDeTT dŸsÇÁ dŸeÖqÔáÇ “jáTeTeTT :
ÿ¿£ çÜuóT„ ÈeTTý˓ s +&ƒT ¿ÃDeTT\T y{ì –eTˆ& uóT„ ÈeTT esÁTdŸ>± s +&ƒe çÜuóT„ ÈeTTý˓ s +&ƒT dŸ<Š Xø
¿ÃDeTT\T y{ì –eTˆ& uóT„ ÈeTTqÅ£” dŸeÖqyîTÔ® û € s +&ƒT çÜuóT„ C²\T dŸsÇÁ dŸeÖqeTT.
–<‘VŸ²sÁD: ABC eT]jáTT PQRý\Ë
‚¿£Ø&ƒ dŸ<ŠXø«¿ÃD²\T B = Q, C = R,
dŸ<ŠXø« uó„TC²\T BC = QR
¿ÃDeTT-uóT„ ÈeTT-¿ÃDeTT dŸsÇÁ dŸeÖqÔá “jáTeT ç|Ÿ¿±sÁ+
ABC eT]jáTT PQR\T dŸsÇÁ dŸeÖH\T.
ABC  PQR
\+‹¿ÃDeTT - ¿£s+’Á - uóT„ ÈeTT dŸsÇÁ dŸeÖqÔáÇ “jáTeTeTT :
ÿ¿£ \+‹¿ÃD çÜuóT„ ÈeTTý˓ ¿£s+’Á , ÿ¿£ uóT„ È+ esÁTdŸ>± s +&ƒe \+‹¿ÃD çÜuóT„ È+ýË ¿£se’Á TT, dŸ<Š Xø«
uóT„ C²“¿ì dŸeÖqyîTÔ® û € çÜuóT„ C²\T dŸsÇÁ dŸeÖqeTT.
–<‘: ABC eT]jáTT PQR\ýË
dŸ<ŠXø« uó„TC²\T : BC = QR
dŸ<ŠXø« ¿£s’\T : AC = PR
dŸ<ŠXø« ¿ÃD²\T : B = Q = 900
\+‹¿ÃDeTT-¿£se’Á TT-uóT„ ÈeTT dŸsÇÁ dŸeÖqÔá “jáTeT+ ç|Ÿ¿±sÁ+ ABC eT]jáTT
PQR\T dŸsÁÇdŸeÖH\T.
ABC PQR
>·eT“¿£ : s +&ƒT çÜuóT„ C²ýË¢, ÿ¿£ çÜuóT„ È+ jîTT¿£Ø eTÖ&ƒT ¿ÃD²\T esÁd>Ÿ ± eTsà çÜuóT„ È+ jîTT¿£Ø eTÖ&ƒT
¿ÃD²\Å£” dŸeÖq+>± –q•³¢sTTÔû, n|ŸÚÎ&ƒT çÜuóT„ C²\T Ôá|Ο “dŸ]>± dŸsÖÁ |ŸÔá ¿£*Ð –+&ƒeÚ.
nuó²«dŸ+ ` 5.6
‚#ÌsÁT.
i) dŸ<ŠXø« dŸsÁÇdŸeÖq uó„TÈeTT\qT çyjáT+&.
ii) dŸ<ŠXø« dŸsÁÇdŸeÖq ¿ÃDeTT\T çyjáT+&.
2. ABC eT]jáTT FGE s +&ƒT çÜuóT„ C²\T dŸsÇÁ dŸeÖqeTT, nsTTq ‚ºÌq uóT„ ÈeTT\ ÈÔá\T eT]jáTT ¿ÃDeTT\
ÈÔá\T dŸ<ŠXø« uó„TÈeTT\T eT]jáTT dŸ<ŠXø« ¿ÃDeTT\T neÚԐjîÖ ýñ<à “sÆ]+#á+&.
1.
ABC DEF n“
eT]jáTT
i)
AB
iii)
eT]jáTT EF
B eT]jáTT F
v)
BC
7e ÔásÁ>·Ü >·DìÔáeTT
FG
ii)
AC
eT]jáTT GE
iv) A eT]jáTT F
vi) C eT]jáTT E.
229
çÜuóT„ C²\T
3.
Find the three pairs of corresponding parts to ensure SSS congruence criterion in the given pair of
triangles. Write congruence of triangles in symbolic form.
4.
Find the three pairs of corresponding parts to ensure SAS congruence criterion in the given pair of
triangles. Write congruence of triangles in symbolic form.
5.
Find the three pairs of corresponding parts to ensure ASA congruence criterion in the given
pair of triangles. Write congruence of triangles in symbolic form.
VII Class Mathematics
230
Triangles
3. ‚ºÌq çÜuó„TC²\ ÈÔá\Å£” uó„T.uó„T.uó„T dŸsÁÇdŸeÖqÔáÇ+ “jáTeTeTT e]ï+|Ÿ#ûjáTT³Å£”, ¿±e\d¾q eTÖ&ƒT ÈÔá\
dŸ<Š Xø uó²>·eTT\T ¿£qT>=q+&. çÜuóT„ C²\ dŸsÇÁ dŸeÖqÔáÇeTT >·TsÁTqï T –|ŸjÖî Ð+ºsjáT+&.
4. ‚ºÌq çÜuóT„ C²\ ÈÔá\Å£” uóT„ .¿Ã.uóT„ . dŸsÇÁ dŸeÖqÔáÇ+ “jáTeTeTT e]ï+|Ÿ#j
û Tá T³Å£”, ¿±e\d¾q eTÖ&ƒT ÈÔá\
dŸ<Š Xø uó²>·eTT\T ¿£qT>=q+&. çÜuóT„ C²\ dŸsÇÁ dŸeÖqÔáÇeTTqT >·TsÁTï –|ŸjÖî Ð+º sjáT+&.
5. ‚ºÌq çÜuóT„ C²\ ÈÔá\Å£” ¿Ã.uóT„ .¿Ã dŸsÇÁ dŸeÖqÔáÇ+ “jáTeTeTT e]ï+|Ÿ#j
û Tá T³Å£”, ¿±e\d¾q eTÖ&ƒT ÈÔá\
dŸ<Š Xø« uó²>·eTT\T ¿£qT>=q+&. çÜuóT„ C²\ dŸsÇÁ dŸeÖqÔáÇeTTqT >·TsÁTqï T –|ŸjÖî Ð+º sjáT+&.
7e ÔásÁ>·Ü >·DìÔáeTT
231
çÜuóT„ C²\T
6.
Find the three pairs of corresponding parts to ensure RHS congruence criterion in the given
pair of triangles. Write congruence of triangles in symbolic form.
1.
2.
How many right angles exist in a triangle?
Which is the longest side in XYZ having right angle at ‘Z’?
3.
Is the sum of any two angles of a triangle always greater than the third angle? Give examples
to justify your answer.
4.
Choose any three measures from the following to make three different triangular wooden
frames.
11m, 9m, 3m, 7m and 5m.
5.
Write any two possible measurements to be suitable for the following triangles.
i)
Right angled triangle.
ii)
Obtuse angled triangle.
iii) Acute angled Triangle.
6.
What is Euler line?
7.
In which ratio the centroid ‘G’ divides the median?
8.
Which concurrent points lie outside the obtuse angled triangle?
9.
What are the two concurrent points that lies on the right angled triangle?
10. Find the value of ‘x’ and ‘y’ in the adjacent figure.
VII Class Mathematics
232
Triangles
6. ‚ºÌq çÜuóT„ C²\ ÈÔá\Å£” \+.¿£.uóT„ dŸsÇÁ dŸeÖqÔáÇ+ “jáTeTeTT e]ï+|Ÿ#j
û Tá T³Å£”, ¿±e\d¾q eTÖ&ƒT ÈÔá\
dŸ<Š Xø« uó²>·eTT\T ¿£qT>=q+&. çÜuóT„ C²\ dŸsÇÁ dŸeÖqÔáÇeTTqT >·TsÁTqï T –|ŸjÖî Ð+º sjáT+&.
jáT֓{Ù nuó²«dŸ+
1. ÿ¿£ çÜuóT„ È+ýË m“• \+‹¿ÃD²\T –+{²sTT?
2. ‘Z’ e<ŠÝ \+‹¿ÃDeTT >·\ XYZ ýË nܙ|<ŠÝ uó„TÈ+ @~?
3. çÜuóT„ È+ýË @ s +&ƒT ¿ÃD²\ yîTTÔá+ï nsTTH eTÖ&ƒe¿ÃD+ ¿£+fñ mÅ£”Øe>± –+³T+<‘? –<‘VŸ²sÁD\ÔÃ
$e]+#á+&.
4. ~>·Te ‚ºÌq ¿=\Ôá\ýË @yû“ eTÖ&ƒT ¿=\Ôá\qT mqT•¿=“ eTÖ&ƒT $uóq„ • #î¿Ø£ çÜuóT„ È¿±s\qT ÔájÖá sÁT
#ûjTá +&.
11 ™d+.MT, 9 ™d+.MT, 3 ™d+.MT, 7 ™d+.MT eT]jáTT 5 ™d+.MT.
5. ~>·Te |s=Øq• çÜuóT„ C²\Å£” dŸ]|Ÿ&Tƒ s +&ƒT sÁ¿±ýÉqÕ ¿=\Ôá\qT |s=Øq+&.
i) \+‹¿ÃD çÜuóT„ È+
ii) n~ó¿¿
£ ÃD çÜuóT„ È+
iii) n\Î ¿ÃD çÜuóT„ È+
6. €sTT\sY s¹ K nq>± Hû$T?
7. eT<óŠ«>·Ôá ¹sKqT >·TsÁTÔáǹ¿+ç<Š+ ‘G’ @ “wŸÎÜïýË $uó„›dŸTï+~?
8. n~ó¿¿£ ÃD çÜuóT„ È+ýË çÜuóT„ ÈeTTqÅ£” ‹jáT³ –+&û $T[Ôá _+<ŠTeÚ\T @$?
9. \+‹¿ÃD çÜuóT„ È+™|Õ –+&û s +&ƒT $T[Ôá _+<ŠTeÚ\qT Ôî\|Ÿ+&.
10. ç|Ÿ¿Ø£ |Ÿ³eTTý˓ ‘x’ eT]jáTT ‘y’ $\Te\qT ¿£qT>=qTeTT.
7e ÔásÁ>·Ü >·DìÔáeTT
233
çÜuóT„ C²\T
11. In ABC, A is four times to B and C is five times to B. Find the three angles.
12. Ladder was faced to a wall. One end of the ladder was making 700 with the floor. Find the angle
of the other end of the ladder with the wall.
13. Write the possible measurements of angles in the following table. One example is given for you.
Type of the triangle
Acute angled triangle
Scalene triangle
400, 600, 800
Equilateral triangle
Isosceles triangle
Right angled triangle
Obtuse angled triangle
14. Write the congruency law in sybolic form for the following triangles.
I
K
J
1.
A triangle is a simple closed plane figure made up of three line segments.
2.
Based on the sides triangles are of three types.
i)
Equilateral triangle
ii)
Isosceles triangle
iii) Scalene triangle
3.
4.
5.
6.
Based on the angles, triangles are of three types.
i) Acute angled triangle
ii) Obtuse angled triangle iii) Right angled triangle
The sum of the three interior angles in a triangle is 180o.
The exterior angle of a triangle is equal to the sum of it’s opposite interior angles.
Properties of the lengths of the sides of a triangle:
l
l
The sum of the lengths of any two sides of a triangle is greater than the length of the third
side.
The difference between the lengths of any two sides of a triangle is smaller than the
length of the third side.
VII Class Mathematics
234
Triangles
11. çÜuó„TÈ+ ABCýË ¿ÃDeTT A, ¿ÃDeTT BÅ£” H\T>·Ts ³T¢ eT]jáTT ¿ÃDeTT C, ¿ÃDeTT BÅ£”
×<ŠTs ³T¢ nsTTq, eTÖ&ƒT ¿ÃD²\qT ¿£qT>=qTeTT.
12. >Ã&ƒÅ”£ €“+º “\uÉ{q¼ì ÿ¿£ “#îÌq ÿ¿£ yî|Õ ÚŸ Hû\Ôà 70 &ç^\T ¿ÃD+ #ûjTá T#áTq•, “#îÌq s +&ƒe yî|Õ ÚŸ >Ã&ƒÔÃ
#ûjTá T ¿ÃD+ m+Ôá?
13. ~>·Te Ôî*|¾q |Ÿ{¿¼ì q£ T dŸs qÕ ¿=\Ôá\Ôà |ŸP]+#á+&. ÿ¿£ –<‘VŸ²sÁD ‚eNj&+~.
$wŸeT u²VŸQ çÜuóT„ È+ dŸeT~Çu²VŸQ çÜuóT„ È+
dŸeTu²VŸQ çÜuóT„ È+
0
0
0
n\οÃD çÜuó„TÈ+
40 , 60 , 80
\+‹¿ÃD çÜuóT„ È+
n~ó¿£ ¿ÃD çÜuó„TÈ+
14. ç¿ì+<Š ‚ºÌq çÜuóT„ C²\ dŸsÇÁ dŸeÖqÔáÇ|ŸÚ “jáTeTqT dŸ+¿¹ Ôá sÁÖ|Ÿ+ýË sjáT+&.
I
K
J
>·TsÁTï+#áT¿Ãy*àq
n+Xæ\T
1. eTÖ&ƒT dŸsÞÁ ø s¹ K\#û @sÁÎ&q dŸsÞÁ ø dŸ+eÔá dŸeTÔá\ |Ÿ³eTTqT çÜuóT„ È+ n+{²sÁT.
2. uóT„ ÈeTT\ €<ó‘sÁ+>±, çÜuóT„ ÈeTT\T eTÖ&ƒT sÁ¿e£ TT\T n$ :
i) dŸeTu²VŸQ çÜuóT„ È+
ii) dŸeT~Çu²VŸQ çÜuóT„ È+
iii) $wŸeTu²VŸQ çÜuóT„ È+.
3. ¿ÃDeTT\T €<ó‘sÁ+>± çÜuóT„ ÈeTT\T eTÖ&ƒT sÁ¿e£ TT\T n$ :
i) n\οÃD çÜuóT„ È+
ii)
n~ó¿¿£ ÃD çÜuóT„ È+
iii)
\+‹¿ÃD çÜuóT„ È+.
4. çÜuóT„ ÈeTTýË n+ÔásÁ ¿ÃDeTT\ yîTTÔáeï TT 1800.
5. ÿ¿£ çÜuóT„ È+ýË u²VŸ²«¿ÃD+ <‘“ n+Ôás_óeTTK ¿ÃDeTT\ yîTTÔáeï TTqÅ£” dŸeÖqeTT.
6. çÜuó„TÈeTTýË uó„TÈeTT\ ¿=\Ôá\ <óŠsˆ\T.
ÿ¿£ çÜuóT„ È+ýË @yîHÕ  s +&ƒT uóT„ ÈeTT\ yîTTÔáeï TT, eTÖ&ƒe uóT„ È+ ¿£H• mÅ£”Øe.
ÿ¿£ çÜuóT„ ÈeTTýË @ s +&ƒT uóT„ ÈeTT\ uó<ñ yŠ Tî q® eTÖ&ƒe uóT„ È+ ¿£H• ÔáŔ£ Øe.
l
l
7e ÔásÁ>·Ü >·DìÔáeTT
235
çÜuóT„ C²\T
7.
The line segment joining a vertex of a triangle to the mid-point of its opposite side is called
median of the triangle. The concurrent point of the medians of triangle is called the Centroid.
Centroid ‘G’ divides the median in the ratio of 2:1 from the vertex.
8.
The perpendicular line segment from any Vertex to its opposite side is called altitude.
Concurrent point of altitudes or orthogonals of a triangle is called an Orthocentre ‘O’.
9.
The line segment which divides the angle into the equal parts is called angle bisector.
Concurrent point of angle bisectors in a triangle is called Incentre ‘I’.
10. In a triangle perpendicular by sector or concurrent on the concurrent is called circumcentre.
11. In a triangle, the Centroid ‘G’, Circumcentre ‘S’, Incentre ‘I’ and Orthocenter ‘O’ are Collinear.
This line is called the Euler line.
12. Congruent objects or figures have the same shape and the same size.
13. Two congruent line segments have the same length. Two congruent angles have the same
measure.
14. Superimposition method is used for enhancing of congruent plane figures.
15. For the Congruence criterion of triangles, The necessary and sufficient conditions.
SSS congruency rule: If three sides of one triangle are equal to the corresponding three sides
of the another triangle then the triangle are congruent
SAS congruency rule: If two sides and their included angle of one triangle are equal to the
corresponding two sides and their included angle of the other triangle, then the triangles are
congruent.
ASA congruency rule: If two angles and their included side of one triangle are equal to the
corresponding two angles and their included side of the other triangle then the triangles are
congruent.
RHS congruency rule: If the hypotenuse and the side of one right-angled triangle are equal
to the corresponding hypotenuse and the side of the another right-angled triangle, then the
triangles are congruent.
Find the interior angles of all triangles by using given clues.
VII Class Mathematics
236
Triangles
7. ÿ¿£ çÜuóT„ ÈeTTýË osÁüeTT qT+& <‘“ m<ŠT{ì uóT„ È+ jîTT¿£Ø eT<ó«Š _+<ŠTeÚqÅ£” ¿£\T|ŸÚÔáÖ ^ºq s¹ KqT
eT<ó«Š >·Ôeá TT n+{²sÁT. ÿ¿£ çÜuóT„ È+ýË eT<ó«Š >·Ôs¹á K\T $T[ÔáeTT\T. $T[Ôá _+<ŠTeÚqT >·TsÁTÔáÇ¿¹ +ç<ŠeTT. (G)
n+{²sÁT. >·TsÁTÔáÇ¿¹ +ç<ŠeTT eT<ó«Š >·Ôeá TTqT osÁüeTT qT+& 2:1 “wŸÎÜï ýË $uó›„ +#áTqT.
8. osÁüeTT qT+& m<ŠT{ìuT„ó ÈeTT™|¿Õ ì ^ºq \+‹¹sKqT mÔáTï ýñ<‘ –q•Ü n+{²sÁT. çÜuóT„ È+ýË –q•ÔáT\
$T[Ôá_+<ŠTeÚqT \+‹¹¿+ç<ŠeTT (O) n+{²sÁT.
9. ‚ºÌq ¿ÃD²“• s +&ƒT dŸeÖq uó²>±\T>± $uó›„ +#û s¹ KqT ¿ÃD dŸeT~Ç K+&ƒq s¹ K n+{²sÁT. çÜuóT„ È+ýË
¿ÃDdŸeT~ÇK+&ƒq s¹ K\T $T[ÔáeTT\T. ¿ÃDdŸeT~Ç K+&ƒq s¹ K\ $T[Ôá _+<ŠTeÚqT n+Ôás¿¹Á +ç<ŠeTT (I)
n+{²sÁT.
10. çÜuóT„ ÈeTTýË uóT„ C²\ \+‹dŸeT~ÇK+&ƒq s¹ K\T $T[ÔáeTT\T. ‡ $T[Ôá_+<ŠTeÚqT |Ÿ]eÔáï ¿¹ +ç<ŠeTT (S)
n+{²sÁT.
11. çÜuó„TÈeTTýË >·TsÁTÔáǹ¿+ç<ŠeTT (G), n+Ôás¿¹Á +ç<ŠeTT (I), |Ÿ]eÔáï ¹¿+ç<ŠeTT(S) eT]jáTT \+‹¹¿+ç<ŠeTT(O)
dŸs¹ FjáTeTT\T.‡ s¹ KqT €sTT\sY s¹ K n+{²sÁT.
12. dŸsÇÁ dŸeÖq edŸTeï Ú\T ýñ<‘ |Ÿ{²\T ÿ¹¿ €¿±sÁ+ eT]jáTT ÿ¹¿ |Ÿ]eÖD+ ¿£*Ð –+{²sTT.
13. s +&ƒT dŸsÇÁ dŸeÖq s¹ U²K+&†\T ÿ¹¿ bõ&ƒeÚqT ¿£*Ð –+{²sTT. s +&ƒT dŸsÇÁ dŸeÖq ¿ÃD²\T ÿ¹¿ ¿=\ÔáqT ¿£*Ð
–+{²sTT.
14. dŸeTÔá\ |Ÿ{²\ jîTT¿£Ø dŸsÇÁ dŸeÖqÔáÇ+ ™|+bõ+~+#á&†“¿ì dŸÖ|ŸsY ‚+bþ›wŸHŽ |Ÿ<ÜƊ (€<ó‘«sÃ|ŸD) <‘«s
#ûjTá e#áTÌ.
15. çÜuóT„ C²\ jîTT¿£Ø dŸsÇÁ dŸeÖqÔáÇ+.
€eXø«¿£-|Ÿs«|Ÿï “jáTeÖ\T:
uóT„ .uóT„ .uóT„ .dŸsÇÁ dŸeÖqÔáÇ “jáTeTeTT: s +&ƒT uóT„ C²\ýË ÿ¿£ çÜuóT„ È+ý˓ eTÖ&ƒT uóT„ C²\ ¿=\Ôá\T esÁTdŸ>± s +&ƒe
çÜuó„TÈeTTý˓ dŸ<ŠXø« uó„TC²\ ¿=\Ôá\Å£” dŸeÖqyîT®Ôû € Âs+&ƒT çÜuó„TC²\T dŸsÁÇdŸeÖqeTT
uó„T.¿Ã.uó„T. dŸsÁÇdŸeÖqÔáǓjáTeTeTT: s +&ƒT çÜuóT„ C²\ýË ÿ¿£ çÜuóT„ È+ý˓ s +&ƒT uóT„ C²\T y{ì eT<ó«Š ¿ÃD+
esÁTdŸ>± s +&ƒe çÜuóT„ ÈeTTý˓ dŸ<Š Xø«uóT„ C²\T y{ì eT<ó«Š ¿ÃD+eTTqÅ£” dŸeÖqyîTÔ® û € çÜuóT„ C²\T dŸsÇÁ dŸeÖqeTT.
¿Ã.uóT„ .¿Ã.dŸsÇÁ dŸeÖqÔáÇ “jáTeTeTT : s +&ƒTçÜuóT„ C²\ýË, ÿ¿£ çÜuóT„ ÈeTTý˓ s +&ƒT ¿ÃDeTT\T y{ì –eTˆ& uóT„ ÈeTT
esÁTdŸ>± s +&ƒe çÜuóT„ ÈeTTý˓ dŸ<Š Xø«¿ÃDeTT\T y{ì –eTˆ& uóT„ ÈeTTqÅ£” dŸeÖqyîTÔ® û € çÜuóT„ C²\T dŸsÇÁ dŸeÖqeTT
\+.¿£.uó„T.dŸsÁÇdŸeÖqÔáÇ “jáTeTeTT : s +&ƒT \+‹¿ÃD çÜuóT„ ÈeTTýË, ÿ¿£ çÜuóT„ ÈeTTý˓ ¿£s+’Á , ÿ¿£ uóT„ ÈeTT
esÁTdŸ>± s +&ƒe çÜuóT„ È+ýË ¿£s+’Á , dŸ<Š Xø« uóT„ C²“¿ì dŸeÖqyîTÔ® û € çÜuóT„ C²\T dŸsÇÁ dŸeÖqeTT.
|Ÿ›ýÙ fÉ®yŽT
7e ÔásÁ>·Ü >·DìÔáeTT
ç¿ì+<Š ‚ºÌq dŸÖ#áq\qT –|ŸjÖî Ð+º #áÔTá sÁçkÍ¿±sÁ |Ÿ³eTTý˓ çÜuóT„ ÈeTT\
n+Ôás¿Á ÃDeTT\qT ¿£qT>=q+&.
237
çÜuóT„ C²\T
6
DATA HANDLING
Content Items
Learning Outcomes
The learner is able to
l
l
6.0 Introduction
explain the concept of ‘Data’.
6.1 Arithmetic Mean
understand the concepts arithmetic mean, mode and
median of ungrouped data and can use these concepts
in daily life to solve the problems.
6.2 Mode
l
construct and interprets double bar graphs.
l
construct and interprets pie charts.
l
represent and visualise the data in the form of graphs
and pie charts.
6.3 Median
6.4 Visual Representation of Data
(Double bar graph and Pie
chart)
6.0 Introduction:
One day Kiran and Bindu visited the nearby
‘GRAMA SACHIVALAYAM’ along with
their father.
Kiran :
Dad, what are the details about,
those are on the board?
Father : It is about the welfare schemes
implemented by AP government and
the number of beneficiaries.
Bindu :
Kiran, could you tell me what is the
population of our village?
Kiran :
3254 people are residing in our
village.
Father : Bindu, how many ‘AMMAVODI’
beneficiaries are there in our village?
Bindu :
In our village there are 185
‘AMMAVODI’ beneficiaries.
The information exhibited on the board is "basic data" of the village.
VII Class Mathematics
248
Data Handling
<ŠÔï+Xø “sÁÇVŸ²D
nuó²«dŸÅ£”\T :
l
l
l
l
l
nuó„«dŸq |˜Ÿ*Ԑ\T
$wŸjáÖ+Xæ\T
<ŠÔï+Xø uó²eqqT $e]+#á>\· sÁT.
n+¿£>D· Ôì á dŸ>³· T, u²VŸQÞø¿e£ TT, eT<ó«Š >·Ôeá TT nHû ne¯Z¿£ Ôá
<ŠÔï+Xø|ŸÚ uó²eq\qT ne>±VŸ²q #ûdŸT¿Ã>·\&ƒT eT]jáTT
<îqÕ +~q J$Ôá+ýË ‡ uó²eq\qT –|ŸjÖî Ð+º dŸeTdŸ«\qT
kÍ~ó+#á>\· sÁT.
Âs+&ƒT esÁTdŸ\ ¿£MTˆ ºçԐ\T (&ƒ‹TýÙ u²sY ç>±|˜t) ^jáT>·\&ƒT
eT]jáTT y«U²«“+#á>\· sÁT.
eÔáï¹sU² ºçԐ\T (™|Õ ºçԐ\T) ^jáT>·\&ƒT eT]jáTT
y«U²«“+#á>\· sÁT
<ŠÔï+XøeTTqT ç>±|˜\t T eT]jáTT eÔás¹ï U² ºçԐ\ sÁÖ|Ÿ+ýË
ç|Ÿ<]Š ô+#á>\· sÁT.
6.0 |Ÿ]#ájTá +
6.1 n+¿£>·DìÔá dŸ>·³T
6.2 u²VŸQÞø¿e£ TT
6.3 eT<ó«Š >·Ôeá TT
6.4 <ŠÔï+XøeTTqT <ŠXø« sÁÖ|ŸeTTýË
ç|Ÿ<Š]ô+#áT³ (Âs+&ƒT esÁTdŸ\ ¿£MTˆ
ºçÔá+ eT]jáTT eÔás¹ï U² ºçÔá+)
6.0. |Ÿ]#ájáT+ :
¿ìsDÁ Y eT]jáTT _+<ŠT\T ÿ¿£sÃE ÔáeT Ôá+ç&Ôà ¿£*d¾ dŸMT|Ÿ+ý˓
»»ç>±eT dŸºy\jáT+µµ qT dŸ+<Š]ô+#sÁT.
¿ìsDÁ Y : Hq•>±sÁT, uËsÁT™¦ |Õ >·\ $es\T <û“¿ì
dŸ+‹+~ó+ºq$?
Ôá+ç& : €+ç<óçŠ |Ÿ<Xû Ù ç|ŸuT„ó ÔáÇ+ neT\T #ûdTŸ qï •
dŸ+¿¹ eŒ T |Ÿ<¿Š¸ ±\T eT]jáTT \_Æ<‘sÁT\
dŸ+K«Å£” dŸ+‹+~ó+ºq $es\T
¿£\eÚ.
_+<ŠT : ¿ìsÁDY, eTq ç>±eT ÈHuó² m+ÔÃ
#î|Ο >·\y?
¿ìsDÁ Y : eTq ç>±eT+ýË 3,254 eT+~
“ed¾dTŸ Hï •sÁT.
Ôá+ç& : _+<ŠT, eTq ç>±eT+ýË »»neTˆ ÿ&µµ
\_Æ<‘sÁT\T m+ÔáeT+~ –H•sÁT?
_+<ŠT : eTq ç>±eT+ýË 185 eT+~ »»neTˆ
ÿ&µµ \_Æ<‘sÁT\T –H•sÁT.
uËsÁT™¦ |Õ ç|Ÿ<]Š ô+#á‹&ƒ¦ dŸeÖ#s“• ç>±eTeTT jîTT¿£Ø »»çbÍ<ó$Š T¿£ <ŠÔï+XøeTTµµ n+{²eTT.
7e ÔásÁ>·Ü >·DìÔáeTT
249
<ŠÔï+Xø “sÁÇVŸ²D
The information which collected either in the form of numbers, words or pictures is called
‘Data’. Always we can see several kinds of information or data through newspapers, magazines,
television and other sources of media in our daily life.
Let’s observe the following charts and pictures. What are they representing?
Auto
Riksha
IYER
The above charts and pictures are showing certain information or data, which is related to the top
five batsmens and bowlers in IPL - 2020, pass percentages of school students and mode of transport
of students which make us to understand easy.
Collection of data places a very important role in the statistical analysis. Data helps us to understand
well and identify in which aspect we are good and in which aspect we need improvement. The method
of collecting information is divided into two different types, namely ‘Primary data’ and ‘Secondary
data’.
1.
Primary data : The data colected directly through personal experiences,
interviews, direct observations, physical testing etc. is called Primary data.
It is also described as raw data or first-hand information.
Example : Marks noted in the personal marks register of a teacher etc.
2.
Secondary data : Secondary data is the information which has been
collected in the past by someone else but used by the investigator for his
own purpose. The sources of secondary data are books, journals,
newspapers, websites etc.
Example : Collection of village population details from census register.
Dr. Calyampudi Radhakrishna Rao (C.R.Rao) - India
A wellknown statistician, famous for his theory of estimations.
He worked on Cramer - Rao Inequality and Fisher - Rao theorem.
His schooling was completed in Gudur, Nuzvid, Nandigama and
Visakhapatnam.
He studied M.A (Mathematics) in Andhra University.
He obtained a Ph. D. Degree under the guidance of Sir R.A. Fisher.
VII Class Mathematics
250
Data Handling
dŸ+K«\T, |Ÿ<‘\T ýñ<‘ ºçԐ\ sÁÖ|Ÿ+ýË d¿£]+ºq dŸeÖ#sÁeTTHû »»<ŠÔï+XøeTTµµ n+{²sÁT. eTq+ “Ôá«
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Measures of Central Tendency:
We have come across several situations where we use the term ‘Average’ in our day-to-day life.
Consider the following statements.
l
The average marks scored by class VI students in mathematics summative test is 74.
l
The average temperature at Vijayawada in the month of May is 40°C.
l
Jyothika’s average study time is 4 hours per day.
Children! consider the example,
‘The average marks scored by class VI students in mathematics summative test is 74’. Does it
mean that every student has scored 74 marks in the class? No, certainly not. Some students may got
more than 74 marks and some students may got less than 74 marks. Average is the value that represents
the general performance of class VI students in mathematics test. Similarly, 40°C is the representative
temperature of Vijayawada in the month of May.
We can say average is a representative or Central Tendency Value of the group of data. By using
a ‘Central Tendency Value’ we can summarize the given data. Different forms of data need different
forms of representative values or Central Tendency values to describe it. We study three types of
measures of Central Tendency of data namely Arithmetic Mean, Mode and Median.
6.1 Average (or) Arithmetic Mean:
Kishore :
Hai Samreen, how are you studying?
Samreen :
I am studying well Kishore.
Kishore :
How many marks have you got in .
mathematics in FA-2?
Samreen :
I got 49 marks and stood first in
mathematics in my class.
Kishore :
Congratulations! So you are the first
ranker in FA-2 exam.
Samreen :
No Kishore! I got 2nd rank, after calculating of
‘Average’ on all subjects.
The most common Measure of Central Tendency of a group of data is average or Arithemetic Mean.
To understand it in a better way, let us observe at the following case.
Suppose marks secured by two students in different subjects are
Samreen : 32, 37, 25, 49, 36, 31
Sujatha : 45, 42, 22, 34, 36, 43 then who pereformed well ?
Now Average or Arithmetic Mean of marks of Samreen and Sujatha will be calculated by
Total of marks in all subjects
Number of subjects
VII Class Mathematics
252
Data Handling
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· T,
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6.1 dŸsdŸ] (ýñ<‘) n+¿£>D· Ôì á dŸ>³· T:
¿ìcþsY : VŸäjYT dŸçMTHŽ, mý² #á<ŠTeÚ ÔáTH•eÚ?
dŸçMTHŽ : u²>± #á<TŠ eÚÔáTH•qT ¿ìcþsY.
¿ìcþsY : FA-2ýË >·DÔì á |Ÿ¯¿£ýŒ Ë úÅ£”
m“• eÖsÁTØ\T e#ÌsTT?
dŸçMTHŽ : HûqT 49 eÖsÁTØ\T kÍ~ó+º, Ôás>Á Ü· ýË >·DÔì á
dŸuÅ¨É ”£ ý¼ Ë yîTT<Š{ì k͜q+ýË “*#qT.
¿ìcþsY : n_óq+<Šq\T, ¿±‹{ì¼ úeÚ FA-2
|Ÿ¯¿£ýŒ Ë yîTT<Š{ì s«+¿£sY nq•eÖ³.
dŸçMTHŽ : ¿±<ŠT ¿ìcþsY, HÅ£” n“• dŸuɨţ”¼\™|Õ dŸ>·³T
ýÉ¿Øì +ºq ÔásTÁ yÔá s +&ƒe s«+Å£” eºÌ+~.
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· Ôì á dŸ>³
· TqT <ŠÔï+XøeTT jîTT¿£Ø ¿¹ +çBjáT k͜q eÖ|ŸqeTT>±
–|ŸjÖî ÐkÍïeTT.
B““ dŸ>³
· TqT eT]+Ôá yîTsÁT>±Z nsÁ+œ #ûdTŸ ¿=qT³Å£” ¿=sÁŔ£ ç¿ì+~ dŸ+<ŠsÒۓ• |Ÿ]o*<‘Ý+.
$_óq• dŸuÅ¨É ”£ ý¼ Ë¢ ‚<ŠsÝ TÁ $<‘«sÁT\œ T kÍ~ó+ºq eÖsÁTØ\T ç¿ì+<Š ‚eNj&†¦sTT.
dŸçMTHŽ : 32, 37, 25, 49, 36, 31
dŸTC²Ôá : 45, 42, 22, 34, 36, 43 mesÁT eT+º ç|ŸÜuóq„ T ¿£q‹]#sÁT?
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dŸuÅ¨É ”£ \¼ dŸ+K«
l
l
l
7e ÔásÁ>·Ü >·DìÔáeTT
253
<ŠÔï+Xø “sÁÇVŸ²D
Arthmetic Mean of Samreen marks =
32  37  25  49  36  31 210

 35
6
6
45  42  22  34  36  43 222

 37
6
6
So, Sujatha’s performance is better than Samreen.
Arithmetic means of Sujatha marks =
‘Arithmetic Mean’(Average) is a number or value that represents or shows the
Central Tendency of a group of observations of data.
Average or Arithmetic Mean can be defind as follows:
Sum of observations
Arithmetic Mean = Number of observations
Example 1 : Mid Day Meal (MDM) taken by the students in six days are 132, 164, 145,182, 163
and 114. Find Arithmetic Mean of students who took MDM per day?
Solution : Given, number of students who took MDM 132, 164, 145, 182, 163, 114
Sum of observations
Number of observations
Arithmetic Mean =

132  164  145  182  163  114 900

 150
6
6
Where does the Arithmetic Mean lie? :
The marks obtained by Sarala, Bindu, Geeta and Rekha in Telugu, Hindi and English are given below.
Students
Telugu
Hindi
English
Sarala
20
16
15
Bindu
14
10
20
Geeta
16
12
18
Rekha
14
10
15
Now let us calculate the average marks obtained by the students in each subject.
Telugu
Hindi
English
20  14  16  14
4
16  10  12  10
4
15  20  18  15
4
64
4
=
=
= 16
=
=
Arithmetic Mean =

Highest marks
Least marks
20
14
What do you observe from the above table?
Does the mean lies between maximum and minimum values in each case?
VII Class Mathematics
254
Data Handling
dŸçMTHŽ bõ+~q eÖsÁTØ\ n+¿£>D
· Ôì á dŸ>³
· T=
32  37  25  49  36  31 210

 35
6
6
dŸTC²Ôá bõ+~q eÖsÁTØ\ n+¿£>D
· Ôì dá >Ÿ ³
· T=
45  42  22  34  36  43 222

 37
6
6
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· Ôì á dŸ>³
· TqT ‡ ç¿ì+~ $<óeŠ TT>± “sÁǺ+#áe#áTÌ.
sXø—\ yîTTÔáeï TT
n+¿£>·DìÔá dŸ>·³T R sXø—\ dŸ+K«
–<‘VŸ²sÁD 1 : ÿ¿£ bÍsÄXÁ æ\ýË eT<ó‘«VŸ²• uóËÈq |Ÿ<¿Šó e£ TTýË 6 sÃE\bͳT uóT„ ›+ºq $<‘«sÁT\Æ dŸ+K« esÁTdŸ>±
kÍ<óŠq:
132, 164, 145,182, 163 eT]jáTT114 nsTTq eT<ó‘«VŸ²•+ uóËÈq+ #ûdq¾ $<‘«sÁT\Æ n+¿£>D· Ôì á
dŸ>³
· T ¿£qT>=qTeTT?
eT<ó‘«VŸ²• uóËÈq |Ÿ<¿Šó e£ TTýË 6 sÃE\bͳT uóT„ ›+ºq $<‘«sÁT\Æ dŸ+K« esÁTdŸ>± 132, 164,
145, 182, 163, 114
sXø—\ yîTTÔáeï TT
n+¿£>·DìÔá dŸ>·³T R sXø—\ dŸ+K«

132  164  145  182  163  114 900

 150
6
6
n+¿£ dŸ>³· T @ $\Te\ eT<ó«Š –+³T+~?
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V¾²+B
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20
16
15
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14
10
20
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16
12
18
s¹ K
14
10
15
ç|Ÿr dŸu¿¨É ý¼ù Ë $<‘«sÁT\Æ T bõ+~q dŸsdŸ] eÖsÁTØ\T >·Dq #û<‘Ý+.
Ôî\T>·T
V¾²+B
‚+^¢wt
n+¿£>·DìÔá dŸ>·³T R
20  14  16  14
4
16  10  12  10
4
15  20  18  15
4
64
4
=
=
= 16
=
=

nÔá«~ó¿£ eÖsÁTØ\T
20
nÔá«\Î eÖsÁTØ\T
14
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7e ÔásÁ>·Ü >·DìÔáeTT
255
<ŠÔï+Xø “sÁÇVŸ²D
Note: ‘Arithmetic Mean’ of given data always lies between the highest and
lowest observations of the data.
Find the Arithmetic Mean of first three multiples of 5?
Collect the information about the weights of any ten students of your class in
kilograms and answer the following :
1.
What are the greatest and smallest weights?
2.
Find Arithmetic mean of collected data.
3.
Verify whether Arithmetic Mean lies between greatest and smallest
observations or not .
Exercise - 6.1
1.
Find Arithmetic Mean of the following .
i)
4, 5, 11, 8
ii) 10, 15, 21, 12, 17
iii)
1 1 3 3 5
, , , ,
4 2 4 2 4
2.
Amounts donated by eight students to ‘NIVAR’ cyclone effected people are `300, `450, `700,
`650, `400, `750, `900 and `850. Find the Arithmetic Mean of amounts donated.
3.
The number of passengers who travelled in APSRTC bus
from Eluru to Rangapuram in 5 trips in a day
are 35, 42, 28, 41 and 44. What is the average
of number of passengers travelled per trip?
4.
Find Arithmetic mean of factors of 24.
5.
Find the Arithmetic Mean of x, x + 1 and x + 2.
sÁ+>±|ŸÚsÁ+
Range of the data
The difference between maximum and minimum values of data is its ‘Range’.
Example : If the observations of data are 15, 22, 9, 45, 54 and 36 then
Range = Maximum value – Minimum value
= 54 – 9 = 45
What is the range of first ten whole numbers?
VII Class Mathematics
256
Data Handling
>·eT“¿£: n+¿£>D· Ôì á dŸ>³· T m\¢|ڟ Î&ƒÖ nÔá«\Î eT]jáTT nÔá«~ó¿£ |Ÿ]o\H
$\Te\ eT<ó«Š –+³T+~.
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
nHûÇw¾<‘Ý+
5 jîTT¿£Ø yîTT<Š{ì eTÖ&ƒT >·TDìC²\ n+¿£ >·DÔì á dŸ>³· T ¿£qT>=qTeTT?
Ôás>Á Ü· ý˓ 10 eT+~ $<‘«sÁT\Æ ‹sÁTeÚ\qT (¿ìýËç>±eTT\ýË) d¿£]+#á+& ç¿ì+~
ç|ŸX•ø \Å£” dŸeÖ<ó‘H\T sjáT+&.
1. nÔá«~ó¿£ eT]jáTT nÔá«\Î ‹sÁTeÚ\T @$ ?
2. ™d¿£]+ºq <ŠÔï+XøeTTqÅ£” n+¿£>D
· Ôì á dŸ>³
· T ¿£qT>=qTeTT.
3. n+¿£>·DìÔá dŸ>·³T, nÔá«~ó¿£ eT]jáTT nÔá«\Î |Ÿ]o\H $\Te\ eT<óŠ«
–q•<‘ ýñ<à >·eT“+#á+&.
nuó²«dŸ+ ` 6.1
1. ç¿ì+~ <ŠÔï+XøeTT\ n+¿£>D
· Ôì á dŸ>³
· T ¿£qT>=qTeTT.
i)
4, 5, 11, 8
ii) 10, 15, 21, 12, 17
iii)
1 1 3 3 5
, , , ,
4 2 4 2 4
2. »úesYµ ÔáTb˜ÍqT u²~óÔTá \Å£” 8 eT+~ $<‘sÁT\Æ T `300, `450, `700, `650, `400, `750, `900 eT]jáTT
`850\T <óq
Š dŸVäŸ jáT+ #ûkÍsÁT. nsTTq $<‘«sÁT\Æ T dŸVäŸ jáT+ #ûdq¾ dŸ>³
· T <óqŠ + m+Ôá?
3. APSRTC ‹dŸTàýË @\ÖsÁT qT+& sÁ+>±|ŸÚsÁ+qÅ£” ÿ¿£ sÃEýË ×<ŠT
kÍsÁT¢ ç|ŸjÖá Dì+ºq ç|ŸjÖá DìŔ£ \ dŸ+K« 35, 42, 28, 41
eT]jáTT 44. nsTTq € ‹dŸTàýË, ç{ì|ڟ ÎÅ£” ç|ŸjÖá Dì+ºq dŸ>³
· T
sÁ+>±|ŸÚsÁ+
ç|ŸjÖá DìŔ£ \ dŸ+K« m+Ôá?
4. 24 jîTT¿£Ø ¿±sÁD²+¿£eTT\ n+¿£>D
· Ôì á dŸ>³
· T ¿£qT>=qTeTT.
5. x, x + 1, x + 2 \ n+¿£>D
· Ôì á dŸ>³
· T ¿£qT>=qTeTT.
<ŠÔï+XøeTT jîTT¿£Ø y«|¾ï
‚ºÌq <ŠÔï+XøeTTý˓ >·]wŸ¼ eT]jáTT ¿£“wŸ¼ $\Te\ eT<óŠ« uóñ<‘Hû• »y«|¾ïµ n+{²sÁT.
–<‘VŸ²sÁD : <ŠÔï+Xøý˓ sXø—\T 15, 22, 9, 45, 54, 36 nsTTq
y«|¾ï R >·]wŸ¼ $\Te - ¿£“wŸ¼ $\Te
R 54 - 9 R 45
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
7e ÔásÁ>·Ü >·DìÔáeTT
yîTT<Š{ì 10 |ŸPs’+¿£eTT\ y«|¾ï ¿£qT>=qTeTT.
257
<ŠÔï+Xø “sÁÇVŸ²D
6.2 Mode :
We have discussed earlier that the arithmetic mean is one of the Measures of Central Tendency.
Depending upon the data and its purpose, other measures of Central Tendency may be used.
Following table represents sale details of different sizes of footwear in a shop for a week.
Size of the footwear (inches)
4"
5"
6"
7"
8"
9"
10"
Number of footwear sets sold
22
30
16
70
32
23
10
Then which size footwear should the shopkeeper has to replenish in his stock at the end of the
week.
Suppose we find the Arithmetic Mean of the footwear sold.
Arithmetic Mean =
=
Sum of observations
Number of observations
203
22  30  16  70  32  23  10
=
= 29
7
7
Average number of footwear sold is 29. This means that the shopkeeper has to get 29 pairs of
footwear in each size. Will it be wise to decide like this? (Why)?
It has to be observed that the maximum purchase falls on the footwear of size 7 inches. So the
shopkeeper has to get more number of footwear of size 7 inches. Hence arithmetic mean does not
suit for this purpose.
Here we need another type of measure of central tendency called ‘Mode’.
The observation which occurs most frequently in the given
data is called ‘Mode’ of the data.
Note :
In general, Mode is used to find the representative value of ‘verbal data’.
a) To know the most preferred flavour of ice-cream to children.
b) To know the most popular toy liked by the kids by a toy factory.
Example 2: Ages of students (in years) are 8, 5, 6, 6, 5, 7, 5, 6, 5, 4, 7, 6, 7, 6, 5, 8 and 6 find the
mode of given data?
Solution : Given, ages of students are 8, 5, 6, 6, 5, 7, 5, 6, 5, 4, 7, 6, 7, 6, 5, 8, 6.
By arranging the numbers with same values together
4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 8, 8
As ‘6’ occurs more frequently than other observations.
Mode = 6
Data having only one mode is known as ‘Unimodal Data’.
VII Class Mathematics
258
Data Handling
6.2 u²VŸQÞø¿£+ :
€e¯Z¿£ Ôá <ŠÔï+XøeTT jîTT¿£Ø ÿ¿£ ¿¹ +çBjáT k͜q eÖ|ŸqeTT>± n+¿£>D
· Ôì á dŸ>³
· T –+&û dŸ+<ŠsÒÛ\qT eTq+
#á]Ì+#eTT. ‚ºÌq <ŠÔï+XøeTT eT]jáTT <‘“ €eXø«¿£Ôá €<ó‘sÁeTT>± ‚ÔásÁ ¹¿+çBjáT k͜q eÖ|ŸqeTT\T
nedŸseÁ T>·TqT.
Õ \ #î|ڟ Î\ $es\T ç¿ì+<Š |Ÿ{¿¼ì ý£ Ë ‚eNj&†¦sTT.
<ŠT¿±D<‘sÁT&ƒT ÿ¿£ ysÁ+ýË n$Tˆq $$<óŠ ™dE
5"
6"
7"
8"
9"
10"
#î|ڟ Î ™dE
Õ (n+>·TÞ²\ýË)
4"
n$Tˆq #î|ŸÚÎ\ ÈÔá\ dŸ+K«
22
30
16
70
32
23
10
nsTTÔû € <ŠT¿±D<‘sÁT&ƒT ysÁ+ÔáseÁ TTýË @ ™dE
Õ >·\ #î|ڟ Î\qT mÅ£”Øe dŸ+K«ýË “\Ç –+#áT¿Ãy*?
n$Tˆq #î|ڟ Î\ ÈÔá\ dŸ+K« jîTT¿£Ø n+¿£>D
· Ôì á dŸ>³
· TqT eTq+ ¿£qT>=+<‘eTT.
n+¿£>·DìÔá dŸ>·³T R
sXø—\ yîTTÔáeï TT
sXø—\ dŸ+K«
= 22  30  16  70  32  23  10 = 203 = 29
7
7
¿±‹{ì¼ <ŠT¿±D<‘sÁT&ƒT dŸ>³· Tq ç|Ÿr ™dE
Õ #î|ڟ Î\T 29 ÈÔá\T neTTˆqT. B“ ç|Ÿ¿±sÁ+ <ŠT¿±D<‘sÁT&ƒT ç|Ÿr ™dE
Õ
#î|ڟ Î\T 29 ÈÔá\T “\Ç –+#*. ‚ý² #ûjTá T³ dŸ]jîT® q “sÁj
’ Tá yûTH? (m+<ŠTÅ£”)
‚ºÌq |Ÿ{¿¼ì £ qT+& 7 n+>·TÞ²\ ™dE
Õ ýË –q• #î|ڟ Î\T >·]wŸ¼ dŸ+K«ýË ¿=qT>Ã\T #ûd¾ –+&ƒ³+ >·eT“+#áe#áTÌ.
¿±‹{ì¼ <ŠT¿±D<‘sÁT&ƒT 7 n+>·TÞ²\ ™dE
Õ ýË –q• #î|ڟ Î\qT mÅ£”Øe dŸ+K«ýË “\Ç –+#*. n+<ŠTe\¢ ‡
dŸ+<ŠsÁÒÛeTTýË n+¿£>·DìÔá dŸ>·³T –|ŸjîÖ>·|Ÿ&ƒ<ŠT.
‡ dŸ+<ŠsÒÁ eÛ TTýË eTqÅ£” u²VŸQÞø¿+£ nHû eTsà ¿¹ +çBjáT k͜q eÖ|ŸqeTT nedŸseÁ T>·TqT.
‚eNj&q <ŠÔï+XøeTTýË mÅ£”Øe kÍsÁT¢ |ŸÚqseÔá+ njûT« s¥“
»u²VŸQÞø¿+£ µ n+{²sÁT.
>·eT“¿£: kÍ<ó‘sÁDeTT>± |Ÿ<Š sÁÖ|ŸýË –q• <ŠÔï+XøeTT jîTT¿£Ø çbÍܓ<ó«Š $\Te>± u²VŸQÞø¿+£ qT –|ŸjÖî ÐkÍïeTT.
1. |¾\¢\T mÅ£”Øe>± ‚wŸ¼|Ÿ&û ×dt ç¿¡+ b˜Í¢esY Ôî\TdŸT¿=qT³Å£”.
2. €³ u¤eTˆ\T ÔájÖá sÁT #ûd ¿£+™|ú ysÁT |¾\\¢ T mÅ£”Øe>± @ €³ u¤eTˆqT ‚wŸ|¼ &Ÿ Tƒ ÔáTH•sÃ
Ôî\TdŸT¿=qT³Å£”.
–<‘VŸ²sÁD 2 : $<‘«sÁT\Æ ejáTdŸT\T (dŸ+eÔáàs\ýË) 8, 5, 6, 6, 5, 7, 5, 6, 5, 4 , 7, 6, 7, 6 , 5, 8 eT]jáTT
6 nsTTq y{ì u²VŸQÞø¿e£ TT m+Ôá?
kÍ<óŠq :
$<‘«sÁT\Æ ejáTdŸT\T 8, 5, 6, 6, 5, 7, 5, 6, 5, 4 , 7, 6, 7, 6 , 5, 8, 6 >± ‚eNj&q~.
‚ºÌq sXø—\ýË ÿ¹¿ $\Te >·\ sXø—\qT ÿ¿£ ç¿£eT |Ÿ<Ü݊ ýË neT]ÌÔû
4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 8, 8
$T>·Ô y{ì¿+£ fñ »6µ mÅ£”Øe kÍsÁT¢ |ŸÚqseÔá+ nsTTq~.
¿±‹{ì¼ u²VŸQÞø¿e£ TT R »6µ.
u²VŸQÞø¿e£ TT>± ÿ¹¿ ÿ¿£ s¥ >·\ <ŠÔï+XøeTTqT »@¿£ u²VŸQÞø¿£ <ŠÔï+XøeTTµ n+{²sÁT.
7e ÔásÁ>·Ü >·DìÔáeTT
259
<ŠÔï+Xø “sÁÇVŸ²D
Example 3 : Find the mode of the data of Grades A, B, E, A, C, E, B, C, D, A, D, C, F, A and C?
Solution :
Given, A, B, E, A, C, E, B, C, D, A, D, C, F, A, C.
By arranging the letters of same type together
A, A, A, A, B, B, C, C, C, C, D, D, E, E, F
As ‘A’ and ‘C’ occurs most frequently in the data
Mode = A and C
Data having two modes is known as ‘Bimodal Data’.
Note :
i)
If each observation in a data is repeated an equal number of times, then the data set has
no mode.
ii)
If no observation in the data is repeated then the given data has no mode.
Find the mode of 10, 9, 12, 10, 8, 7, 6, 10, 9, 7, 8, 5 and 2.
Take a dice, roll it 20 times and record the numbers you
got on its top face. Find the ‘Mode’ of resulting numbers.
The following data shows that the number of hours spent
by students for study. Find the mode :
Number of study hours
1
2
3
4
5
6
Number of students
4
2
1
2
1
0
Exercise - 6.2
1.
Find mode of the following data.
i)
2, 3, 7, 5, 3, 2, 6, 7, 1, 2
ii) K, A , B , C, B, C, D, K, B, D, B, K, A, K
iii)
First ten natural numbers
iv) 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8
2.
20 students were participated in ‘SWATCH BHARAT ABHIYAN’ campaign. The number of
days each student participated were 5, 1, 2, 4, 1, 2, 3, 2, 1, 2, 3, 2, 5, 3, 4, 2, 1, 3, 4 and 5. Find
mode of the data.
3.
The number of goals scored by a football team in different matches are 3, 2, 4, 6, 1, 3, 2, 4, 1 and
6. Find the mode of data.
4.
Find the mode of letters in the adjacent figure.
Verify whether it is Unimodel or Bimodal Data?
VII Class Mathematics
260
Data Handling
–<‘VŸ²sÁD 3 :
kÍ<óŠq :
A, B, E, A, C, E, B, C, D, A, D, C, F, A eT]jáTT C jîTT¿£Ø u²VŸQÞø¿+
£
A, B, E, A, C, E, B, C, D, A, D, C, F, A, C\T ‚eNj&†¦sTT.
m+Ôá?
‚ºÌq sXø—\ýË ÿ¹¿ $<óyŠ Tî q® sXø—\qT ÿ¿£ ç¿£eT |Ÿ<Ü݊ ýË neT]ÌÔû,
A, A, A, A, B, B, C, C, C, C, D, D, E, E, F
$T>·Ô y{ì¿+£ fñ »Aµ eT]jáTT »Cµ \T mÅ£”Øe kÍsÁT¢
¿±‹{ì¼ u²VŸQÞø¿e£ TT R »Aµ eT]jáTT »Cµ
|ŸÚqseÔá+ njáÖ«sTT.
u²VŸQÞø¿e£ TT>± s +&ƒTsXø—\T >·\ <ŠÔï+XøeTTqT »~Çu²VŸQÞø¿£ <ŠÔï+XøeTTµ n+{²sÁT.
>·eT“¿£: i) ÿ¿£ <ŠÔï+XøeTTýË ç|Ÿr s¥ dŸeÖq dŸ+K«ýË |ŸÚqseÔá+ nsTTÔû € <ŠÔï+XøeTTqÅ£” u²VŸQÞø¿e£ TT
–+&ƒ<TŠ .
ii) ÿ¿£ <ŠÔï+XøeTTýË @ ÿ¿£Ø s¥ |ŸÚqseÔá+ ¿±¿£bþÔû € <ŠÔï+XøeTTqÅ£” u²VŸQÞø¿e
£ TT –+&ƒ<TŠ .
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
nHûÇw¾<‘Ý+
€ý˺<‘Ý+!
10, 9, 12, 10, 8, 7, 6, 10, 9, 7, 8, 5 eT]jáTT 2 sXø—\ u²VŸQÞø¿e£ TT m+Ôá?
ÿ¿£ bͺ¿£qT rdŸT¿Ã+&, <‘““ 20 kÍsÁT¢ <=]¢+#á+&.
bͺ¿£ ™|Õ uó²>·+ýË eºÌq n+¿\qT qyîÖ<ŠT #ûjTá +&.
€ n+¿\ u²VŸQÞø¿+£ ¿£qT>=qTeTT.
¿ì+~ |Ÿ{¿¼ì ý£ Ë $<‘«sÁT\œ T sÃEÅ£” #á<TŠ eÚýË yîºÌ+#û dŸeTjáT+ (>·+³\ýË)
‚eNj&q~ nsTTq u²VŸQÞø¿e£ TT ¿£qT>=qTeTT.
#á<TŠ eÚ³ýË yîºÌ+ºq >·+³\ dŸ+K«
$<‘«sÁT\œ T dŸ+K«
1
4
2
2
3
1
4
2
5
1
6
0
nuó²«dŸ+ ` 6.2
1. ç¿ì+~ <ŠÔï+XøeTT\Å£” u²VŸQÞø¿e£ TT ¿£qT>=qTeTT.
i)
2, 3, 7, 5, 3, 2, 6, 7, 1, 2
ii) K, A , B , C, B, C, D, K, B, D, B, K, A, K
yîTT<Š{ì 10 dŸVŸ²È dŸ+K«\T
iv) 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8
2. 20 eT+~ $<‘«sÁT\Æ T »»dŸÇ#áÌ uó²sÁÔY n_ójÖá HŽµµ ¿±sÁ«ç¿£eTeTTýË bÍý¤ZH•sÁT. $<‘«sÁT\Æ T bÍý¤Zq•
sÃE\ dŸ+K« esÁTdŸ>± 5, 1, 2, 4, 1, 2, 3, 2, 1, 2, 3, 2, 5, 3, 4, 2, 1, 3, 4 eT]jáTT 5 nsTTq
u²VŸQÞø¿e£ TT m+Ôá?
3. ÿ¿£ |˜ÚŸ {Ùu²ý٠ȳT¼ $$<óŠ eÖ«#Y\ýË kÍ~ó+ºq >ÃýÙà dŸ+K« 3, 2, 4, 6, 1, 3, 2, 4, 1 eT]jáTT 6.
nsTTq u²VŸQÞø¿+£ ¿£qT>=qTeTT.
4. ç|Ÿ¿Ø£ |Ÿ³eTTý˓ ‚+^¢wt n¿£sŒ \ u²VŸQÞø¿e£ TT ¿£qT>=qTeTT?
‚~ @¿£ u²VŸQÞø¿£ <ŠÔï+XøeÖ ýñ¿£ ~Çu²VŸQÞø¿£ <ŠÔï+XøeÖ?
iii)
7e ÔásÁ>·Ü >·DìÔáeTT
261
<ŠÔï+Xø “sÁÇVŸ²D
6.3 Median:
The following are the salaries (in rupees) earned by the manager and the workers in a production unit.
Manager - `88,000
Worker 1
`6,600
Worker 7
`9,000
Worker 2
`10,000
Worker 8
`9,200
Worker 3
`8,000
Worker 9
`9,000
Worker 4
`8,400
Worker 10
`6,400
Worker 5
`7,600
Worker 11
`6,400
Worker 6
`7,000
Worker 12
`6,400
Will the mean salary or the mode of salaries be a representative value for this data?
Let us calculate the mean of the salaries in the production unit.
Mean of the salary =
=
=
Sum of salaries of all employees
Number of employees
88000  6600  10000  8000  8400  7600  7000  9000  9200  9000  6400  6400  6400
13
1,82, 000
13
= `14,000
Is this salary a representative of the salaries of either the manager or the workers? (Why?) It is
much lesser than the manager’s salary and more greater than the salary of all the workers.
Now, let us consider the mode. From the given data, 6400 is the most frequently occurring value
So, Mode = `6400.
However, it can’t be a representative of the data. (Why?)
So, now, let us use another way of calculating the representative or Central Tendency Value.
First arrange the numbers in ascending order
6400 6400 6400 6600 7000 7600 8000 8400 9000 9000 9200 10000 88000
The middle value of this data is ‘8000’. Clearly it divides employees into 2 equal groups.
6 are earning more than `8,000 and 6 are earning less than `8,000.
This value is called ‘Median’ and as you can see it provides a representative picture or Central
Tendency for all the values of the given data.
The middle most value of the data, when the observations are
arranged in either ascending or descending order is called ‘Median’.
In the above example, the number of observations were 13 i.e. an odd number, thus the median
divides the data into 2 equal groups.
Now what is the Median, if the number of observations were in even number?
VII Class Mathematics
262
Data Handling
6.3 eT<óŠ«>·ÔáeTT :
ÿ¿£ –ԐÎ<Š¿£ dŸ+dŸÆ ýË yûTHûÈsY eT]jáTT ¿±]ˆÅ£”\ jîTT¿£Ø yûÔHá \ (sÁÖbÍjáT\ýË) $es\T ¿ì+<Š ‚eNj&†¦sTT.
yûTHûÈsÁT - `88,000
`6,600
@&à ¿±]ˆÅ£”&ƒT
`9,000
yîTT<Š{ì ¿±]ˆÅ£”&ƒT
s +&ƒe ¿±]ˆÅ£”&ƒT
`10,000
m“$T<à ¿±]ˆÅ£”&ƒT
`9,200
eTÖ&ƒe ¿±]ˆÅ£”&ƒT
`8,000
Ô=$Tˆ<à ¿±]ˆÅ£”&ƒT
`9,000
`8,400
|Ÿ<à ¿±]ˆÅ£”&ƒT
`6,400
H\T>à ¿±]ˆÅ£”&ƒT
×<à ¿±]ˆÅ£”&ƒT
`7,600
|Ÿ<¿Š =+&à ¿±]ˆÅ£”&ƒT
`6,400
€sà ¿±]ˆÅ£”&ƒT
`7,000
|ŸH•î +&ƒe ¿±]ˆÅ£”&ƒT
`6,400
‡ <ŠÔï+X擿ì dŸ>³
· T yûÔqá eTT ýñ<‘ yûÔHá \ u²VŸQÞø¿e£ TTqT çbÍܓ<ó«Š $\Te\T>± rdŸT¿Ãe#áTÌH?
yîTT<Š³>± çbõ&ƒ¿HŒ£ Ž jáT֓{ÙýË yûÔHá \ dŸ>³
· TqT eTq+ ýÉ¿Øì <‘Ý+.
–<ë>·T\+<Š] yûÔáH\ yîTTÔáï+
yûÔáH\ dŸ>·³T R
–<ë>·T\ dŸ+K«
=
=
88000  6600  10000  8000  8400  7600  7000  9000  9200  9000  6400  6400  6400
13
1,82, 000
13
= `14,000
‡ yûÔqá + yûTHûÈsY ýñ<‘ ¿±]ˆÅ£”\ yûÔHá \Å£” çbÍܓ<ó«Š $\Te>± –+³T+<‘? ( m+<ŠTÅ£”? ). ‚~ yûTHûÈsY yûÔqá +
¿£+fñ #ý² ÔáŔ£ Øe>± –+~ eT]jáTT ¿±]ˆÅ£”\ yûÔHá \ ¿£+fñ #ý² mÅ£”Øe>± –+³T+~.
‚|ŸÚÎ&ƒT u²VŸQÞø¿±“• |Ÿ]o*<‘Ý+. ‡ <ŠÔï+XøeTT ýË mÅ£”Øe kÍsÁT¢ |ŸÚqseÔá+ nsTTq $\Te 6400
¿±‹{ì¼ u²VŸQÞø¿+£ R `6400
nsTTq|ŸÎ{ì¿ì, ‚~ <ŠÔï+Xø+ jîTT¿£Ø çbÍܓ<óŠ« $\Te>± –+&ƒ<ŠT (m+<ŠTÅ£”?)
¿£qTq, ‚|ŸÚÎ&ƒT ¿¹ +çBjáT k͜q $\TeqT ýÉ¿Øì +#á&†“¿ì eTsà |Ÿ<ÜƊ “ |Ÿ]o*<‘Ý+.
dŸ+dŸýœ ˓ –<ë>·T\ yûÔHá \Å£” €sÃVŸ²D ç¿£eT+ýË neTsÁÌ>±
6400 6400 6400 6600 7000 7600 8000 8400 9000 9000 9200 10000 88000
‡ <ŠÔï+XøeTT jîTT¿£Ø eT<ó«Š eT $\Te 8000 n>·TqT. ‚~ –<ë>·+ s +&ƒT dŸeÖq dŸ+K« >·\ dŸeTÖVŸä\T
>± $uó„›dŸTï+~.
` 8,000 ¿£+fñ mÅ£”Øe dŸ+bÍ~+#û €sÁT>·TsÁT –<ë>·T\T eT]jáTT ` 8,000 ¿£+fñ ÔáŔ£ Øe dŸ+bÍ~+#û €sÁT>·TsÁT
–<ë>·T\T>± $uó„›dŸTï+~.
‡ $\TeHû eT<óŠ«>·Ôá+ n+{²sÁT. ‡ dŸ+dŸýœ ˓ –<ë>·T\ yûÔHá \Å£” ‚~ çbÍܓ<ó«Š $\Te ýñ<‘ ¿¹ +çBjáT
k͜q $\Te>± –+³T+<Š“ >·eT“+#áe#áTÌ.
<ŠÔï+XøeTTý˓ sXø—\qT €sÃVŸ²D ýñ<‘ nesÃVŸ²D ç¿£eTeTTýË neTsÁÌ>±,
€ neT]¿£ý˓ eT<ó«Š eT $\TeqT eT<ó«Š >·Ô+á n+{²sÁT.
™|Õ –<‘VŸ²sÁDýË |Ÿ]o\q\ dŸ+K« 13 ÿ¿£ uñd¾ dŸ+K«. ¿±‹{ì¼ eT<ó«Š >·Ô+á <ŠÔï+XøeTTqT s +&ƒT dŸeÖq
uó²>±\T>± $uó„›dŸTï+~.
|Ÿ]o\q\ dŸ+K« dŸ]dŸ+K« nsTTq n|ŸÚÎ&ƒT eT<óŠ«>·Ôá+ m$T{ì?
7e ÔásÁ>·Ü >·DìÔáeTT
263
<ŠÔï+Xø “sÁÇVŸ²D
Let us take the example of the production unit again.
If a new worker joined and earning `8,200 in the production unit, then what will be the Median?
Arrange the salaries in ascending order :
6400
6400 6400 6600 7000 7600 8000 8200 8400 9000 9000 9200 10000 88000
Here both 8000 and 8200 lie in the middle of the data.
Here the median will be calculated by finding the average of these two values.
Thus, the median salary =
8000  8200 16200

= `8,100/2
2
Example 4 : Find the median of data 32, 43, 25, 67, 46, 71 and 182.
Solution :
Given 32, 43, 25, 67, 46, 71, 182
Arrange the given observations in ascending order
25, 32, 43, 46, 67, 71, 182
In ‘7’ observations the 4th obsrevation is the middle most value.
 Median = 46
th
 n 1
If the number of observations(n) is odd then median is 
 observation.
 2 
Example 5 : The monthly incomes of 8 members are `8000, `9000, `8200, `7900, `8500, `8600,
`7700 and `60000. Find the median income.
Solution :
Given that, the monthly incomes of 8 members are:
`8000, `9000, `8200, `7900, `8500, `8600, `7700, `60000
Arrange the given observations in ascending order:
7700, 7900, 8000, 8200, 8500, 8600, 9000, 60000
Here, we have two middle most values 8200 and 8500.
In this case, Median = Average of two middle most values

8200  8500 16700

= `8,350
2
2
If number of observations(n) is even then the median is
th
th
n
n 
Average of   and   1 observations
2
2 
What is the median of first 7 prime numbers?
Collect 10th class pass percentage for last six years of your
school (or) your near by school. Find Median of the data.
VII Class Mathematics
264
Data Handling
™|Õ –ԐÎ<Š¿£ dŸ+dŸœ –<‘VŸ²sÁDHû eTsÁý² rdŸTÅ£”+<‘+.
sÁÖ.8,200 yûÔqá +>± >·\ eTsà e«¿ìï –ÔÎ<Š¿£ dŸ+dŸýœ Ë ¿=Ôá>ï ± #ûs&ƒT nqT¿=+<‘eTT? n|ŸÚÎ&ƒT ¿=Ôáï eT<ó«Š >·Ô+á m+Ôá?
‚|ŸÚÎ&ƒT 14 eT+~ yûÔHá \qT €sÃVŸ²D ç¿£eT+ ýË neTsÁTÌ<‘+.
6400 6400 6400 6600 7000 7600 8000 8200 8400 9000 9000 9200 10000 88000
‡ <ŠÔï+XøeTTýË 8000 , 8200 nHû s +&ƒT $\Te\T eT<ó«Š eT $\Te\T>± –H•sTT.
‚ý²+{ì dŸ+<ŠsÒÛ\ýË ‡ s +&ƒT $\Te\T dŸsdŸ] ¿£qT¿ÃØe³+ <‘Çs eT<ó«Š >·Ô“• >·Dkì ÍïeTT.
8,000 G 8,200 16,200 R `8,100/¿±‹{ì¼ eT<ó«Š >·Ôá yûÔqá + R
R 2
2
–<‘VŸ²sÁD 4: 32, 43, 25, 67, 46, 71 eT]jáTT 182 \ eT<ó«Š >·Ôeá TT ¿£qT>=qTeTT.
kÍ<óŠq :
32, 43, 25, 67, 46, 71, 182
‚ºÌq sXø—\T €sÃVŸ²D ç¿£eTeTTýË neT]ÌÔû
25, 32, 43, 46, 67, 71, 182
<ŠÔï+XøeTTý˓ @&ƒT sXø—\ýË 4e s¥ eT<ó«Š |Ÿ<+Š n>·TqT
 eT<ó«Š >·Ôe
á TT R 46
‚ºÌq <ŠÔï+XøeTTýË sXø—\ dŸ+K«(n) uñd¾ dŸ+K« nsTTq eT<óŠ«>·ÔáeTT
 n 1 

 2 
e
|Ÿ<eŠ TT n>·TqT.
–<‘VŸ²sÁD 5 : 8 eT+~ Hî\dŸ] €<‘jáÖ\T `8000, `9000, `8200, `7900, `8500, `8600, `7700
eT]jáTT `60000 nsTTq y] eT<ó«Š >·Ôá €<‘já֓• ¿£qT>=qTeTT.
8 Å£”³T+u²\ Hî\dŸ] €<‘jáÖ\T
`8000, `9000, `8200, `7900, `8500, `8600, `7700, `60000
€<‘jáÖ\qT €sÃVŸ²D ç¿£eTeTTýË neT]ÌÔû
7700, 7900, 8000, 8200, 8500, 8600, 9000, 60000
eT<ó«Š eT |Ÿ<‘\T 8200 eT]jáTT 8500
eT<ó«Š >·Ô+á , 8200, 8500\ dŸsdŸ] n>·TqT
8200 G 8500 16700 R `8,350
eT<óŠ«>·Ôá+ R
R 2
2
‚ºÌq <ŠÔï+XøeTTýË sXø—\ dŸ+K«(n) dŸ] dŸ+K« nsTTq
kÍ<óŠq:
eT<óŠ«>·ÔáeTT
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
nHûÇw¾<‘Ý+
7e ÔásÁ>·Ü >·DìÔáeTT
n
 e
2
eT]jáTT
n 
 1 e
2 
|Ÿ<‘\ dŸsdŸ] n>·TqT.
yîTT<Š{ì 7 ç|Ÿ<‘ó q dŸ+K«\ eT<ó«Š >·Ôeá TT ¿£qT>=qTeTT.
MT bÍsÄXÁ æ\ ýñ<‘ MTÅ£” dŸMT|Ÿ+ýË >·\ bÍsÄXÁ æ\ jîTT¿£Ø >·Ôá 6 dŸ+eÔáàs\
|Ÿ<eŠ Ôás>Á Ü· –rïsԒÁ  XæÔáeTT\qT qyîÖ<ŠT #ûd¾ eT<ó«Š >·Ôeá TT ¿£qT>=qTeTT.
265
<ŠÔï+Xø “sÁÇVŸ²D
Exercise - 6.3
1.
Find the Median of the following :
i)
7, 3, 15, 0, 1, 71, 19, 4, 17
ii) 12, 23, 11, 18, 15, 20, 86, 27
2.
The number of pages in text books of different subjects are 421, 175, 128, 117,150, 145, 147
and 113 find Median of given data.
3.
The weekly sales of motor bikes in a showroom for the past 14 weeks are 10, 6, 8, 3, 5, 6, 4, 7,
12, 13, 16, 10, 4 and 7. Find the Median of the data.
4.
Find the Median of 0.3, 0.25, 0.32, 0.147, 0.19, 0.2 and 7.1
5.
If the Median of observations 2x, 3x, 4x, 5x, 6x, (x > 0) is 28 then find the value of ‘x’?
Visit any vegetable market along with your parent. Collect the information about the costs of
different vegetables. By using this data fill the following table.
i) Arithmetic Mean =
ii) Mode =
iii) Median =
Measure of Central Tendency
Useful situation
Arithmetic Mean
When data set has no extreme values.
Mode
When the data has many identical values and for quick
calculation.
Median
When the data set has extreme values and there are no big gaps
in the middle of the data.
VII Class Mathematics
266
Data Handling
nuó²«dŸ+ ` 6.3
1. ç¿ì+~ <ŠÔï+XøeTT\Å£” eT<óŠ«>·ÔáeTT ¿£qT>=qTeTT
i) 7, 3, 15, 0, 1, 71, 19, 4, 17
ii) 12, 23, 11, 18, 15, 20, 86, 27
2. $_óq• dŸuÅ¨É ”£ \¼ Å£” dŸ+‹+~ó+ºq bÍsÄ«Á |ŸÚdŸ¿ï e£ TT\ýË >·\ |J\ dŸ+K«\T 421, 175, 128,117, 150, 145,
147 eT]jáTT 113 nsTTq ‡ <ŠÔï+XøeTTqÅ£” eT<ó«Š >·Ôeá TT ¿£qT>=qTeTT.
3. ÿ¿£ yîÖ{²sÁT yVŸ²qeTT\ <ŠT¿±DeTTýË >·Ôá 14 ysÁeTT\ýË n$Tˆq yVŸ²qeTT\ dŸ+K« esÁTdŸ>± 10, 6,
8, 3, 5, 6, 4, 7, 12, 13, 16, 10, 4 eT]jáTT 7 nsTTq y{ì eT<ó«Š >·Ôeá TT ¿£qT>=qTeTT.
4. 0.3, 0.25, 0.32, 0.147, 0.19, 0.2, 7.1 \ eT<ó«Š >·Ô“• ¿£qT>=q+&.
5. 2x, 3x, 4x, 5x, 6x, (x > 0) sXø—\ eT<ó«Š >·Ôeá TT 28 nsTTq x $\Te m+Ôá?
çbÍCÉÅ£”¼ |Ÿ“
MT Ôá*/¢ Ôá+ç&Ôà bͳT <Š>sZ· ýÁ ˓ Å£LsÁ>±jáT\ eÖÂsØ{ÙqT dŸ+<Š]ô+º, $$<óŠ Å£LsÁ>±jáT\ <ósŠ \Á T d¿£]+#á+&.
ç¿ì+~ |Ÿ{¿¼ì q£ T |ŸP]+º ¿¹ +çBjáT k͜q $\Te\T ¿£qT>=qTeTT.
Å£LsÁ>±jáT |sÁT¢
1 ¹¿› <óŠsÁ
n+¿£eT<ó«Š eT+
i)
uÉ+&ƒ¿±jáT
n+¿£>·DìÔá dŸ>·³T R
³eÖ³
¿±«Âs{Ù
ii)
u²VŸQÞø¿e£ TT R
iii)
eT<ó«Š >·Ô+á R
;{ÙsÖÁ {Ù
|ŸºÌ$TsÁ|¿Ÿ ±jáT
e+¿±jáT
–*¢bÍjáT
¿¹ +çBjáT k͜q eÖ|ŸqeTT
n+¿£>D· Ôì á dŸ>³· T
u²VŸQÞø¿+£
eT<ó«Š >·Ô+á
7e ÔásÁ>·Ü >·DìÔáeTT
nÔá«+Ôá –|ŸjÖî >·¿s£ yÁ Tî q® dŸ+<ŠsÒÁ eÛ TT
<ŠÔï+XøeTTý˓ sXø—\ $\Te\ eT<ó«Š e«Ô«dŸ+ ÔáŔ£ Øe>± –q•|ŸÚÎ&ƒT.
<ŠÔï+XøeTTýË ÿ¹¿ $<óyŠ Tî q® nHû¿£ $\Te\T –q•|ŸÚÎ&ƒT eT]jáTT yû>e· TT>±
>·D+ì #áT³Å£”.
<ŠÔï+XøeTTý˓ sXø—\ $\Te\ eT<ó«Š e«Ô«dŸ+ u²>± mÅ£”Øe>±
–q•|ŸÚÎ&ƒT eT]jáTT <ŠÔï+Xø+ eT<ó«Š $\TeýË ™|<ŠÝ e«Ô«dŸ+ ýñq|Ÿð&ƒT
267
<ŠÔï+Xø “sÁÇVŸ²D
6.4 Visual Representation of the Data:
You have already learnt how to present data as a visual representation in the form of pictographs
and bar graphs in the previous class.
Prasanna wanted to buy a mobile phone. He selected two mobile phones of different
companies having same features. Now he would like to know which mobile phone has good
performance. He collected the information of star rating from different magazines and news papers.
1.
What information is given in the table?
2.
Is the above information being useful to Prasanna?
3.
Which mobile phone you will suggest to Prasanna?
Data represented in the form of pictures, graphs or charts is Visual Representation. Visual
Representation makes the data clear to understand. Data visualisation is important in data analysis
because it is able to make the data easily and quickly understood.
Bar graph:
Samuel :
Shabreen, did you observe our
school display board.
Shabreen :
Yes, it was about our school
S.S.C pass percentage.
Samuel :
Can you tell our school pass
percentage in 2018 - 19.
Shabreen :
Yes, the pass percentage
was 70.
Samuel :
Can you remember what type
Shabreen :
of visual representation it is.
Yes, It is ‘Bar graph’.
Representation of numerical data by using bars of uniform width is called ‘Bar graph’. In
bar graphs the width of all rectangles (bars) is equal and the height (length) of each bar is
proportional to the data that it represents.
VII Class Mathematics
268
Data Handling
6.4dŸeÖ#sÁeTTqT <ŠXø« sÁÖ|ŸeTTýË ç|Ÿ<Š]ô+#áT³ :
<ŠÔï+XøeTTqT |Ÿ³ ºçÔáeTT eT]jáTT ¿£MTˆ¹sU² ºçÔá sÁÖ|ŸeTTýË ç|Ÿ<]Š ô+#áT³qT ‚+ÔáŔ£ eTT+<ŠT Ôás>Á Ü· ýË
MTsÁT HûsTÁ ÌÅ£”H•sÁT.
ç|ŸdqŸ • ÿ¿£ yîTTuÉýÕ Ù b˜þHŽ ¿=H\qTÅ£”H•&ƒT. nÔáqT ÿ¹¿ \¿£D
Œ ²\T (kå¿£s«\T) >·\ s +&ƒT $_óq• ¿£+™|ú\
yîTTuÉýÕ Ù b˜þq¢qT m+|¾¿£ #ûdTŸ Å£”H•&ƒT. ‡ s +&ƒT yîTTuÉýÕ Ù b˜þqTýË¢ @~ yîTsÁT> qÕ <à nÔáqT Ôî\TdŸT¿Ãy\qTÅ£”H•&ƒT.
nÔáqT $$<óŠ |ŸçÜ¿£\T eT]jáTT eÖ«>·CqÕÉ T¢ qT+& ‡ ç¿ì+~ dŸeÖ#sÁeTTqT d¿£]+#&ƒT.
1. ™|Õ |Ÿ{켿£ýË >·\ dŸeÖ#sÁ+ <û““ dŸÖºdŸTï+~?
2. |Ÿ{¿¼ì ý£ ˓ dŸeÖ#sÁ+ ç|ŸdqŸ •Å£” –|ŸjÖî >·|&Ÿ Tƒ ÔáT+<‘?
3. úyîÔÕ û ç|ŸdqŸ • Å£”, @ yîTTuÉýÕ Ù b˜þHŽ qT dŸÖºkÍïeÚ?
‚ºÌq dŸeÖ#s“• ºçԐ\T, ç>±|˜ÚŸ \T ýñ<‘ #sÁT\¼ sÁÖ|ŸeTTýË ç|Ÿ<]Š ô+#áT³Hû <ŠXø« sÁÖ|Ÿ ç|Ÿ<sŠ ôÁ q n+{²sÁT.
‚ºÌq <ŠÔï+XøeTTqT <ŠXø« sÁÖ|ŸeTTýË ç|Ÿ<]Š ô+#áT³ e\q dŸeÖ#sÁeTTqT nsÁ+Æ #ûdTŸ ¿=qT³ dŸT\uó+„ n>·TqT.
<ŠÔï+XøeTT jîTT¿£Ø <ŠXø« sÁÖ|Ÿ ç|Ÿ<sŠ ôÁ q ‚ºÌq dŸeÖ#sÁeTTqT $Xâw¢ +¾ #áT³Å£” eT]jáTT Ôû*¿±>± ÔáÇ]Ôá>Ü· q
ne>±VŸ²q #ûdTŸ ¿=qT³Å£” eTTK«yîTq® ~.
¿£MTˆ ºçÔáeTT :
XæeTÖ«ýÙ : wŸç;HŽ, eTq bÍsÄXÁ æ\ý˓ ç|Ÿ<sŠ ôÁ q (display)
uËsÁTq¦ T >·eT“+#y?
wŸç;HŽ
: neÚqT, n~ |Ÿ<eŠ Ôás>Á Ü· –rïsԒÁ  XæÔáeTTqT
10e ÔásÁ>·Ü –rïsÁ’Ԑ XæÔáeTT
Ôî*jáT#ûdTŸ +ï ~ ¿£<‘!
›.|Ÿ. –q•Ôá bÍsÄX
Á æ\
XæeTÖ«ýÙ : 2018 - 19 $<‘« dŸ+eÔáàsÁeTTýË eTq
bÍsÄXÁ æ\ jîTT¿£Ø –rïsԒÁ  XæÔáeTT m+ÔÃ
#î|Ο >·\y?
wŸç;HŽ
: €eÚqT. –rïsԒÁ á XæÔá+ 70.
XæeTÖ«ýÙ : ‡ $<óŠyîT®q <ŠXø« sÁÖ|Ÿ ç|Ÿ<ŠsÁôqqT @eT“
|¾\TkÍïsÃ, >·TsÁT+ï <‘?
wŸç;HŽ
: >·TsÁT+ï ~, B““ ¿£MTˆ ºçÔá+ n+{²sÁT.
‚ºÌq dŸ+U²« <ŠÔï+XøeTTqT dŸeÖq yî&\ƒ TÎ >·\ ¿£MTˆ\ sÁÖ|Ÿ+ýË ç|Ÿ<]Š ô+#á{²Hû• »»¿£MTˆ ºçÔá+µµ n+{²sÁT. ¿£MTˆ
ºçÔá+ý˓ ¿£MTˆ\ yî&\ƒ TÎ dŸeÖq+>± –+&ƒTqT eT]jáTT ¿£MTˆ mÔáTï (bõ&ƒeÚ) n~ dŸÖº+#û n+XøeTT jîTT¿£Ø
$\TeÅ£” nqTýËeÖqTbÍÔáeTTýË –+³T+~.
7e ÔásÁ>·Ü >·DìÔáeTT
269
<ŠÔï+Xø “sÁÇVŸ²D
Number of People
Observe the adjacent bar graph and answer the following:
1.
Which fruit most of the people like?
2.
How many people likes banana?
Fruits
40
35
30
25
20
15
10
Apple Orange Banana
Kiwi
6.4.1 Double bar graph:
Observe the following two bar graphs which shows the sales of two mobile phone companies A
and B in various years:
y - axis
y - axis
Sales of Mobile Company ‘A’
Sales of Mobile Company ‘B’
Sale of Mobiles in lakhs
80
Sale of Mobiles in lakhs
80
60
60
40
40
20
20
2016
2017
2018
Years
2019
2016
x - axis
2017
2018
Years
2019
x - axis
From the above bar graphs, we can study the individual sale capability of A and B. But we want
to compare sales of both companies in every year, we need a different visual representation.
Observe the following graph.
Mobile phone- A
y-axis
Mobile phone- B
Sale of mobiles in lakhs
80
60
60
50 50
40
40
40
20
70
70
20
2016
2017
Years
2018
2019
x - axis
By using the above graph, it is easy to compare the sales of both companies.
These type of graphs are called ‘Double bar graphs’.
VII Class Mathematics
270
Data Handling
ú |ç >Ÿ Ü
· “
Ôî\TdŸT¿Ã
|Ÿ+&ƒT¢
e«Å£”\ï dŸ+K«
40
35
30
25
20
15
10
5
0
yîTTuÉýÕ Ù b˜þq¢ neTˆ¿±\T \¿£\Œ ýË
yîTTuÉýÕ Ù b˜þq¢ neTˆ¿±\T \¿£\Œ ýË
ç|Ÿ¿Ø£ ¿£MTˆ ºçÔá+qT >·eT“+º ¿ì+~ ç|ŸX•ø \Å£” dŸeÖ<ó‘H\T sjáTTeTT.
1. mÅ£”Øe eT+~ ‚wŸ|¼ &Ÿ û |Ÿ+&ƒT @~?
2. nsÁ{ì |Ÿ+&ƒTqT ‚wŸ|¼ &Ÿ û y] dŸ+K« m+Ôá?
€|¾ýÙ H]+È nsÁ{ì ¿ì$
6.4.1 s +&ƒT esÁTdŸ\ ¿£MTˆ ºçԐ\T (&ƒ‹TýÙ u²sY >ç ±|˜)t
ç¿ì+~ ¿£MTˆ ºçԐ\qT >·eT“+#á+&. ‚$ $$<óŠ dŸ+eÔáàs\ýË s +&ƒT yîTTuÉýÕ Ù b˜þHŽ ¿£+™|ú\T A, B \
neTˆ¿±\qT dŸÖºdŸTHï •sTT.
y - n¿£+
Œ
y - n¿£+
Œ
yîTTuÉýÕ Ù ¿£+™|ú A neTˆ¿±\T
yîTTuÉýÕ Ù ¿£+™|ú B neTˆ¿±\T
80
60
40
20
2016
2017
2018
2019
80
60
40
20
2016
2017
2018
2019
x - n¿£+
Œ
x - n¿£+
Œ
dŸ+eÔáàs\T
dŸ+eÔáàs\T
eTq+ ™|Õ ¿£MTˆ ºçԐ\ qT+& ç|ŸÜ yîTTuÉýÕ Ù b˜þHŽ ¿£+™|ú\T A, B\ e«¿ì>ï Ô· á neTˆ¿£|ڟ kÍeTsÆ«“• $Xâw¢ +¾ #áe#áTÌ.
¿±“ s +&ƒT ¿£+™|ú\ neTˆ¿£|ڟ kÍeTsÆ«“• dŸ]bþ\T̳ţ” ç|ŸÜ dŸ+eÔáàsÁ+ eTqÅ£” $_óq• <ŠXø«sÁÖ|Ÿ ç|Ÿ<sŠ ôÁ q
nedŸs+Á n>·TqT.
ç¿ì+~ ºçԐ“• >·eT“+#á+&.
yîTTuÉÕýÙ b˜þHŽ - A
y-n¿£+
Œ
yîTTuÉÕýÙ b˜þHŽ - B
yîTTuÉýÕ Ù b˜þq¢ neTˆ¿±\T \¿£\Œ ýË
80
60
60
50 50
40
40
40
20
70
70
20
2016
2017
2018
dŸ+eÔáàs\T
2019
x-
n¿£+Œ
™|Õ ºçԐ“• –|ŸjÖî Ð+º eTq+ s +&ƒT ¿£+™|ú\ neTˆ¿£|ڟ kÍeTsÆ«“• dŸ]bþ\T̳ dŸT\uóe„ T>·TqT.
‚³Te+{ì ¿£MTˆ ºçÔáeTTHû s +&ƒT esÁTdŸ\ ¿£MTˆ ºçÔá+ (&ƒ‹TýÙ u²sY ç>±|˜)t n+{²eTT.
7e ÔásÁ>·Ü >·DìÔáeTT
271
<ŠÔï+Xø “sÁÇVŸ²D
Observe the above double bar graph and answer the following:
1. In which year the sales of both mobile phone companies are equal?
2. In 2017, which mobile phone company has more sales?
So, ‘Double bar graph’ is used to display two sets of data on the same graph. With the help
of ‘Double bar graph’, we can compare the two group of observations in a single look.
6.4.2 Construction of Double bar graph:
Example 6 : The number of CFL bulbs and LED bulbs by a seller every month from March to
August are given Below. Draw by vertical double bar graph.
MONTH
CFL BULBS
LED BULBS
March
70
75
April
35
30
May
65
75
June
90
100
July
22
35
August
50
50
Solution: Steps in drawing a Double bar graph:
1. Draw x-axis (horizontal line) and y - axis (vertical line) on the graph paper and mark their intersection
point as ‘O’.
2. Take ‘months’ on x - axis.
3. Take number of CFL and LED bulbs on y - axis.
4. Take an appropriate scale on y - axis so that number of both the bulbs can be shown easily. Here
the maximum value to be plotted on y - axis as 100, so let us take 1cm = 10 bulbs on y - axis.
5. Find the length of each bar by dividing the value by 10 (as scale is 1cm = 10 bulbs).
Eg : Length of bar represents 70 CFL bulbs =
70
= 7 cm
10
75
= 7.5 cm
10
Draw the bars of uniform width representing ‘CFL bulb’ and ‘LED bulb’ side by side of every month.
Length of bar represents 75 LED bulbs =
6.
y - axis
CFL Bulbs
100
90
LED Bulbs
80
70
60
50
40
30
20
10
March
VII Class Mathematics
April
272
May
June
July
Aug
x - axis
Data Handling
™|qÕ ‚eNj&q s +&ƒT esÁTdŸ\ ¿£MTˆ ºçÔá+qT |Ÿ]o*+º ç¿ì+~ ç|ŸX•ø \Å£” dŸeÖ<ó‘H\T sjáT+&.
1. @ dŸ+eÔáàsÁeTT ýË s +&ƒT yîTTuÉýÕ Ù b˜þHŽ ¿£+™|ú\ neTˆ¿±\T dŸeÖq+?
2. 2017 e dŸ+eÔáàsÁeTT ýË @ yîTTuÉýÕ Ù b˜þHŽ ¿£+™|ú neTˆ¿±\T mÅ£”Øe?
¿±‹{ì,¼ s +&ƒT $_óq• <ŠÔï+XøeTT\qT ÿ¹¿ ç>±|˜ÚŸ ¿±ÐÔáeTT ™|Õ ç|Ÿ<]Š ô+#áT³Å£” s +&ƒT esÁTdŸ\ ¿£MTˆ ºçÔá+qT
–|ŸjÖî ÐkÍïeTT. s +&ƒT esÁTdŸ\ ¿£MTˆ ºçÔá+qT –|ŸjÖî Ð+º s +&ƒT $_óq• <ŠÔï+XøeTT\qT ÿ¹¿ kÍ] dŸ]bþ\Ìe#áTÌ.
6.4.2 s +&ƒT esÁTdŸ\ ¿£MTˆ ºçԐ\qT “]ˆ+#áT³ :
–<‘VŸ²sÁD 6 : ÿ¿£ <ŠT¿±D<‘sÁT&ƒT eÖ]Ì qT+& €>·w§Ÿ ¼ esÁŔ£ ç|Ÿr Hî\ýË n$Tˆq CFL ‹\TÒ\T eT]jáTT LED
‹\TÒ\ neTˆ¿±\ $es\T ç¿ì+~|Ÿ{¿¼ì ý£ Ë ‚eNj&†¦sTT nsTTq ç¿ì+~ <ŠÔï+XøeTTqÅ£” s +&ƒT esÁTdŸ\
¿£MTˆ ºçԐ“• “]ˆ+#á+&.
Hî\
CFL ‹\TÒ\T
LED ‹\TÒ\T
eÖ]Ì
70
75
@ç|¾ýÙ
35
30
yûT
65
75
pHŽ
90
100
pýÉÕ
22
35
€>·w§Ÿ ¼
50
50
kÍ<óŠq : s +&ƒT esÁTdŸ\ ¿£MTˆ ºçÔá+ “sˆDeTTýË kþbÍH\T:
1. ç>±|˜ŸÚ ¿±ÐÔáeTT ™|Õ x - n¿£+Œ (¿ì܌ È s¹ K), y - n¿£+Œ (“\TeÚ s¹ K) ^jáT+&. y{ì K+&ƒq _+<ŠTeÚqT »Oµ>±
>·T]ï+#á+&.
2. x - n¿£Œ+™|Õ Hî\\ |sÁT¢ rdŸT¿Ã+&.
3. y - n¿£+Œ ™|Õ CFL ‹\TÒ\dŸ+K«, LED ‹\TÒ\dŸ+K«qT rdŸT¿Ã+&.
4. Âs+&ƒT sÁ¿±\ ‹\TÒ\ dŸ+K« ç>±|˜t ¿±ÐÔáeTT™|Õ >·T]ï+#áT³Å£” M\T>± dŸÂsÕq dØ\TqT y - n¿£+Œ ™|Õ rdŸT¿Ã+&.
y - n¿£+
Œ ™|Õ >·T]ï+#áe\d¾q >·]wŸ¼ $\Te 100. ¿±‹{ì¼ 1™d+.MT R 10 ‹\TÒ\T>± rdŸT¿Ãe#áTÌ.
5. ‚ºÌq $\Te\qT 10Ôà uó²Ð+#áT³ <‘Çs ¿£MTˆ bõ&ƒeÚ “sÆ]+#á+&. (dŸÖº¿£ _óq•+ 1 ™d+.MT R 10 ‹\TÒ\T).
70
–<‘ : 70 CFL ‹\TÒ\qT dŸÖº+#áT ¿£MTˆ bõ&ƒeÚ R 10 = 7 ™d+.MT
75 LED ‹\TÒ\qT dŸÖº+#áT ¿£MTˆ bõ&ƒeÚ R
75
=
10
7.5 ™d+.MT
5. ç|Ÿr Hî\ýË n$Tˆq CFL ‹\TÒ\ eT]jáTT LED ‹\TÒ\ dŸ+K«qT dŸeÖq yî&\ƒ TÎ >·\ ¿£MTˆ\ sÁÖ|ŸeTTýË
ç|Ÿ¿£Ø ç|Ÿ¿£Øq ^jáT+&.
y - n¿£Œ+
CFL
‹\TÒ\T
LED
‹\TÒ\T
100
90
80
70
60
50
40
30
20
10
eÖ]Ì @ç|¾ýÙ
7e ÔásÁ>·Ü >·DìÔáeTT
273
yûT
pHŽ pýÉÕ €>·dTŸ ¼
x-
n¿£Œ+
<ŠÔï+Xø “sÁÇVŸ²D
6.4.3 Pie charts :
Observe the adjacent figure :
1. Largest portion of the circle is shaded by which colour?
2. Are the portions of circle shaded by blue and pink are in same size?
3. Smallest portion of the circle is shaded by which colour?
In the above figure, you can observe that the circle is divided in to some component parts. Have
you observed that each component part is bounded by an arc and two radii of circle?
These parts of the circle are called ‘sectors’ of the circle.
Children, observe the following figures. Have you observe data represented in the circular form?
This type of chart, which display information as parts of a circle (sector) is called a ‘Pie chart’.
The ‘Pie chart’ represents each item as a portion of the circle, as how much part of ‘the total item’ is
shared by each item.
‘Pie chart’ is the visual representation of the numerical data by sectors of the circle such that
angle of each sector (area of sector) is proportional to value of the data that it represents.
Observe the below picture which shows various expenses of Manasa’s family :
i)
For which item highest amount spent?
ii)
For which items same amount spent?
iii)
For which item least amount spent?
VII Class Mathematics
274
Data Handling
6.4.3 eÔáï s¹ U² ºçÔáeTT (™|Õ ºçÔáeTT) :
ç|Ÿ¿Ø£ |Ÿ³+qT >·eT“+#á+&.
1. eÔáeï TTý˓ n~ó¿£ uó²>·eTT @ sÁ+>·TÔà w&Ž #ûjTá ‹&q~?
2. ú\+ sÁ+>·T, |¾+¿ù sÁ+>·T uó²>±\T dŸeÖq |Ÿ]eÖDeTTýË –H•jáÖ?
3. eÔáeï TT ý˓ nÔá«\Î uó²>·eTT @ sÁ+>·TÔà w&Ž #ûjTá ‹&q~?
™|Õ |Ÿ³eTTýË eÔá+ï ¿=“• uó²>±\T>± $uó›„ +#á‹& –+&ƒ&+ƒ eTq+ >·eT“+#áe#áTÌ. ç|Ÿr uó²>·eTT, s +&ƒT y«kÍsœ\T
eT]jáTT eÔáï #|ŸeTT\Ôà €e]+#á‹& –+&ƒT³ >·eT“+#s?
eÔáeï TTý˓ ‡ uó²>±\Hû »»™d¿±¼sTÁ µ¢ µ n+{²sÁT.
|¾\\¢ Ö, ç¿ì+~ |Ÿ{²\qT |Ÿ]o*+#á+&. MTsÁT <ŠÔï+Xø+qT eÔáï sÁÖ|Ÿ+ýË dŸÖº+#á³qT >·eT“+#s?
48 |ŸsTÁ >·T\T
4’S
36 |ŸsTÁ >·T\T
T
TT\
·‘s¡e
Ç
T· \T
Ÿ| TÁ >
s
16
25¾+%& T
| sœ\
|Ÿ<‘
6’S
dŸeTÔáT\« €VŸäsÁ+
Å£L|s Ÿ+&ƒT¢ Ê
Á>
40%±jáT\T
10%
|Ÿ<翑s=eÚÇ
œ\T
Ôás>Á Ü
· y¯ $<‘«sÁT\œ T
çbþ 25%
{¡q
T¢
dü∫Hé kÕ~Û+∫q |üs¡T>∑T\T
yÓTT‘·Ô+ |üs¡T>∑T\T ` 100
‡ $<óeŠ TT>± eÔáeï TTqT uó²>±\T (™d¿±¼sTÁ )¢ >± $uó›„ +#áT³ <‘Çs <ŠÔï+XøeTTqT ç|Ÿ<]Š ô+#áT³Hû ™|Õ ºçÔá+ ýñ<‘ eÔáï
s¹ U² ºçÔá+ n+{²eTT. ™|Õ ºçÔáeTT eÔáeï TTý˓ ç|ŸÜ uó²>·eTTqT dŸÖºdŸT+ï ~. yîTTÔá+ï n+Xæ\ýË ç|ŸÜ n+XøeTT m+Ôá
uó²>·+ |Ÿ+#áTÅ£”+³T+<à dŸÖºdŸT+ï ~.
¿±‹{ì¼ ‚ºÌq dŸeÖ#s“• eÔáeï TTqT ™d¿±¼sTÁ >¢ ± $uó›„ +#áT³ <‘Çs <ŠXø« sÁÖ|ŸeTTýË ç|Ÿ<]Š ô+#áT³qHû ™|Õ ºçÔá+ ýñ<‘
eÔáï s¹ U² ºçÔá+ n+{²eTT. ç|Ÿr ™d¿±¼sTÁ ¿¹ +ç<Š+ e<ŠÝ #ûd ¿ÃDeTT (™d¿±¼sTÁ yîXÕ æ\«eTT) n~ dŸÖº+#û n+Xø $\TeÅ£”
nqTýËeÖqTbÍÔá+ýË –+³T+~.
¿ì+~ |Ÿ³eTTýË eÖqdŸ Å£”³T+‹+ jîTT¿£Ø $$<óŠ KsÁTÌ\ $es\T #áÖ|Ÿ‹&†¦sTT.
i) n~ó¿£ uó²>·+ <û“ ¿=sÁÅ£” KsÁTÌ ™|³¼‹&+~?
€VŸäsÁ+
ii) dŸeÖq yîTTÔá+
ï ýË KsÁTÌ #ûjTá ‹&q n+Xæ\T @$?
<ŠTdŸT\ï T
iii) nÔá«\Î uó²>·+ <û“ ¿=sÁÅ£” KsÁTÌ ™|³¼‹&+~?
dŸVäŸ jáT+
#á<TŠ eÚ
‚ÔáseÁ TT\T
7e ÔásÁ>·Ü >·DìÔáeTT
275
<ŠÔï+Xø “sÁÇVŸ²D
6.4.4 Interpretation of Pie chart :
Observe the adjacent Pie chart:
It is about the time spent by a child during a day.
The size of each sector is proportional to information it represents
and also each sector shows the relation between whole and its part.
i)
Proportion of sector for hours spent in sleeping
=
Number of sleeping hours
Whole day
8 hours
= 24 hours
=
1 rd
of circle
3
As, total angle at the center of circle = 3600
So, angle of the sector represents sleeping =
ii)
Proportion of sector for hours spent in playing =
1
 3600 = 1200
3
Number of sleeping hours
Whole day
3 hours
= 24 hours
1
= 8 th of circle
Angle of the sector represents playing =
So from the above,
Angle of sector =
1
 3600 = 450
8
Value of the item
 360D
Sum of the values of all items
6.4.4 Construction of Pie chart:
Now let us learn about how data is presented on a pie chart.
Example 7 :
In a school, there are 100 students in class VII and each one is a member of any one of the club.
The following table shows the number of students in various clubs, then.
VII Class Mathematics
276
Data Handling
6.4.4 eÔáï ¹sU² ºçÔá+ (™|Õ ºçÔá+) - y«U²«“+#áT³ :
\T
6 >∑+≥
“ç<ŠýË >·&| q¾ dŸeTjáT+ ¿=sÁŔ£ ™d¿±¼sÁT jîTT¿£Ø uó²>·eTT R
R
ìÁ<ä
i)
≥\T
8 >∑+
|Ÿ¿Ø£ ™|Õ ºçÔá+ qT |Ÿ]o*+#á+&. ‚~ ÿ¿£ sÃEýË $<‘«]Æ >·&| q¾ dŸeTjáT+
>·T]+º Ôî*jáT#ûdTŸ +ï ~. ç|ŸÜ ™d¿±¼sTÁ jîTT¿£Ø |Ÿ]eÖD+ n~ çbÍܓ<ó«Š + eV¾²+#û
dŸeÖ#s“¿ì nqTýËeÖqTbÍÔá+ýË –+³T+~.
ç|ŸÜ ™d¿±¼sTÁ , n+Xø+ yîTTÔá+ï eT]jáTT n~ çbÍܓ<ó«Š + eV¾²+#û n+X擿ì eT<ó«Š
dŸ+‹+<󑓕 #áÖ|ŸÚÔáT+~.
Ê\
bÕsƒ¡X
T
∑+≥\
3>
TT\T
·‘s¡e
Ç
Ä≥\T 3 >∑∑+≥\T
Ç+{Ï |üì
\T
∑+≥
4>
“ç<ŠýË >·&| q¾ dŸeTjáT+
ÿ¿£ |ŸP]ï ~q+
8 >·+³\T
24 >·+³\T
1
3
R eÔá+ï ýË e uó²>·eTT
eÔáï ¿¹ +ç<Š+ e<ŠÝ yîTTÔá+ï ¿ÃD+ R 3600
¿±‹{ì,¼ “ç<ŠýË >·&| q¾ dŸeTjá֓¿ì çbÍܓ<ó«Š + eV¾²+#û ™d¿±¼sÁT ¿ÃD+ R
1
 3600 = 1200
3
€³\ýË >·&| q¾ dŸeTjáT+
ii) €³\ýË >·&|
 q¾ dŸeTjáT+ ¿=sÁŔ£ ™d¿±¼sÁT jîTT¿£Ø uó²>·eTT R
ÿ¿£ |ŸP]ï ~q+
3 >·+³\T
R 24 >·+³\T
1
R eÔá+ï ýË 8 e uó²>·+
€³\ýË >·&| q¾ dŸeTjá֓¿ì çbÍܓ<ó«Š + eV¾²+#û ™d¿±¼sÁT ¿ÃD+ R
1
 3600 = 450
8
™|Õ y{ì qT+&
n+XøeTT $\Te
™d¿±¼sTÁ ¿ÃD+ R n“• n+XøeTT\ $\Te\ yîTTÔáeï TT
 3600
6.4.5 eÔáï ¹sU² ºçÔáeTT “sˆDeTT (™|Õ ºçÔá+) :
eÔáï s¹ U² ºçÔá sÁÖ|ŸeTTýË dŸeÖ#sÁ+ mý² ç|Ÿ<]Š ôkÍïyÖî ‚|ŸÚÎ&ƒT HûsTÁ ÌÅ£”+<‘+.
–<‘VŸ²sÁD 7: ÿ¿£ bÍsÄXÁ æ\ý˓ 7e Ôás>Á Ü· ýË 100 eT+~ $<‘«sÁT\Æ T ¿£\sÁT. 7e Ôás>Á Ü· ý˓ ç|Ÿr $<‘«]œ @<Ã
ÿ¿£ ¿£¢uÙýË dŸuó„T«\T>± –H•sÁT.¿ì+~ |Ÿ{켿£ $$<óŠ ¿£¢uÙ\ý˓ $<‘«sÁTœ\ dŸ+K«qT #áÖ|ŸÚÔáT+~, nsTTq
7e ÔásÁ>·Ü >·DìÔáeTT
277
<ŠÔï+Xø “sÁÇVŸ²D
Construct the pie chart to following data :
Club
Number of members
Mathematics
50
Science
30
Social Studies
40
English
40
Arts
20
Solution: The angle of each sector will depend on the ratio between the number of students in club
and total number of students.
Value of the item
0
Angle of sector = Sum of the values of all items  360
Total
180
3600
Steps of construction:
1. Draw a circle with any convenient radius and mark it’s centre as ‘O’.
2. Mark a point A, somewhere on the circumference and join OA.
3. Construct AOB = 100º to represent angle of the sector for Maths club.
4. Construct BOC = 60º to represent angle of the sector for Science club.
5. Construct COD = 80º to represent angle of the sector for Social studies club.
6. Construct DOE = 80º to represent angle of the sector for English club.
7. Now EOA = 40º represents the angle of the sector for Arts club.
VII Class Mathematics
278
Data Handling
|Ÿ{¿¼ì ý£ ˓ dŸeÖ#s“¿ì ™|Õ ºçԐ“• ^jáT+&.
¿£u¢ Ù
dŸuT„ó «\ dŸ+K«
>·DÔì +á
50
kÍeÖq« XæçdŸ+ï
30
kÍ+|˜T¾ ¿£ XæçdŸ+ï
40
‚+^¢w§Ÿ
40
¿£Þ\ø T
20
kÍ<óŠq: ™d¿±¼sTÁ jîTT¿£Ø ¿ÃD+ ¿£u¢ Ùý˓ $<‘«sÁT\Æ dŸ+K« eT]jáTT yîTTÔá+ï $<‘«sÁT\œ dŸ+K«Å£” >·\ “wŸÎÜï™|Õ €<ó‘sÁ|&Ÿ Tƒ qT.
n+XøeTT $\Te
™d¿±¼sTÁ ¿ÃD+ R n“• n+XøeTT\ $\Te\ yîTTÔáeï TT
¿£u¢ Ù
dŸuT„ó «\ dŸ+K«
 3600
n+XøeTT $\Te
™d¿±¼sTÁ ¿ÃD+ R n“• n+XøeTT\ $\Te\ yîTTÔáïeTT  3600
>·DÔì +á
kÍeÖq« Xæg+
kÍ+|˜T¾ ¿£ Xæg+
‚+^¢w§Ÿ
¿£Þ\ø T
yîTTÔáeï TT
180
3600
“sˆD kþbÍH\T:
1. @<û“ y«kÍsÁ+œ Ôà eԐ ^º, <‘“ ¿¹ +ç<‘“• »Oµ >± >·T]ï+#á+&
2. eÔáï |Ÿ]~ó ™|Õ @<îÕH ÿ¿£ _+<ŠTeÚqT »Aµ >± >·T]ï+#á+&. OAqT ¿£\|Ÿ+&.
3. >·DÔì á ¿£u¢ Ù ™d¿±¼sTÁ qT dŸÖº+#áTq³T¢ AOB =100º “ “]ˆ+#á+&.
4. kÍeÖq« XæçdŸ+ï ¿£u¢ Ù ™d¿±¼sTÁ qT dŸÖº+#áTq³T¢ BOC = 60º“ “]ˆ+#á+&.
5. kÍ+|˜T¾ ¿£ XæçdŸ+ï ¿£u¢ Ù ™d¿±¼sTÁ qT dŸÖº+#áTq³T¢ COD = 80º“ “]ˆ+#á+&.
6. ‚+^¢w§Ÿ ¿£u¢ Ù ™d¿±¼sTÁ qT dŸÖº+#áTq³T¢ DOE = 80º“ “]ˆ+#á+&.
7. EOA = 40º nHû ™d¿±¼sTÁ ¿ÃD+ ¿£Þ\ø ¿£u¢ ÙqT dŸÖºdŸT+ï ~.
k
Xæç ÍeÖq
d+ïŸ «
T
¿Þ£ \ø
kÍ
Xæçd+|T¾˜ ¿£
+ïŸ
+á
>D· Ôì
‚+^¢w§Ÿ
7e ÔásÁ>·Ü >·DìÔáeTT
279
<ŠÔï+Xø “sÁÇVŸ²D
Exercise - 6.4
In the adjacent Double bar graph, number of students of a class
are shown according to academic year.
Answer the following questions on the basis
of this Double bar graph.
90
i)
In which academic year, number of girls
were more than number of boys in the
school?
ii)
In which academic year, number of both
girls and boys in the school were equal?
iii)
What was the total number of students in
Number of Students
1.
Boys
Girls
80
70
60
50
40
30
20
10
2013-14 2014-15 2015-16 2016-17 2017-18
Academic Years
the school in the academic year 2013-14 ?
2.
The marks in the Mathematics and Science of five students of class 7 are given in the table.
Exhibit these by the vertical Double bar graph by given information.
3.
4.
Name of student
Mathematics
Science
Praveena
50
60
Venkat
80
75
Uzma
75
70
Joseph
65
75
Ramya
75
60
The adjacent Pie chart gives the expenditure on various items during a month for a family. In the
figure angles made by each sector at the center are given then answer the following:
i)
if the expenditure on rent is `3000 then how much
amount they saved?
ii)
on which item the expenditure is minimum?
iii)
on which item the expenditure is maximum?
The following data shows the number of students opting different subjects in college.
Subjects
Number of
students
Botany
45
Mathematics Physics Chemistry Economics Commerce
60
20
30
10
15
Construct a Pie diagram to represent the above data.
VII Class Mathematics
280
Data Handling
nuó²«dŸ+ ` 6.4
$<‘«sÁTÆ\ dŸ+K«
1. ç|Ÿ¿Ø£ s +&ƒT esÁTdŸ\ ¿£MTˆ ºçÔá+ýË $$<óŠ $<‘« dŸ+eÔáàs\ýË bÍsÄXÁ æ\ý˓ ÿ¿£ Ôás>Á Ü· ý˓ $<‘«sÁT\Æ
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u²\TsÁT
sjáT+&.
u²*¿£\T
i) @ $<‘« dŸ+eÔáàsÁ+ýË bÍsÄX
Á æ\ýË u²\TsÁ dŸ+K« ¿£+fñ 90
80
70
u²*¿£\T dŸ+K« n~ó¿£eTT>± ¿£\<ŠT.
60
50
ii) @ $<‘« dŸ+eÔáàsÁ+ýË bÍsÄX
Á æ\ýË u²\TsÁ dŸ+K«, 40
u²*¿£\ dŸ+K«Å£” dŸeÖqeTT?
30
20
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$<‘«sÁT\Æ dŸ+K« m+Ôá?
á $<‘« dŸ+eÔáàs\T
2013-14 2014-15 2015-16 2016-17 2017-18
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yî+¿£{Ù
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sÁeT«
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80
75
65
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kÍeÖq« XæçdŸ+ï
60
75
70
75
60
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i)
n<îÝ ¿=sÁŔ£ ™|{ìq¼ KsÁTÌ sÁÖ.3000, nsTTq
€ Hî\ýË bõ<ŠT|ŸÚ m+Ôá?
ii) ÔáŔ£ Øe KsÁTÌ #ûjTá ‹&q n+Xø+ @~?
iii) n~ó¿£ KsÁTÌ #ûjTá ‹&q n+Xø+ @~?
€VŸäsÁ+ `
bõ<ŠT|ŸÚ `
#á<TŠ eÚ `
n<îÝ `
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dŸuÅ¨É ”£ ¼
u˳ú
>·DÔì +á
|˜›¾ ¿ùà
¿Â $TçdÓ¼
m¿£H$T¿ùà
$<‘«sÁTÆ\ dŸ+K«
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60
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30
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15
™|qÕ ‚eNj&q dŸeÖ#s“¿ì –|ŸjÖî Ð+º ™|Õ ºçԐ“• “]ˆ+#á+&.
7e ÔásÁ>·Ü >·DìÔáeTT
281
<ŠÔï+Xø “sÁÇVŸ²D
1.
2.
3.
4.
5.
Runs made by two bats men in 3 matches are given below.
Kohli : 49, 98, 72
Rohit : 64, 45, 83 then find average of runs scored by Kohli and Rohit. Whose average is
higher?
Find mode of 38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38 and 47? Verify whether
it is Unimodel or Bimodal data?
The temperature in different places are 0, –5, 7, 10, 13, –1 and 41 in degree celsius. Find the
Median? If another observation, ‘40C’ is added to the given data, is there any change in value
of Median? Explain?
If the range of observation 7x, 5x, 3x, 2x, x (x > 0) is 12, then find value of ‘x’ and express all
the observations in numerical form?
Birth and death rates of different states in 2015 are given below. Approximately Draw a
double bar graph for the given data.
State
6.
Birth rate (per 1000)
Death rate (per 1000)
Andhra Pradesh
17
7
Arunachal Pradesh
19
6
Tamilnadu
15
7
Jharkhand
24
6
Gujarath
20
6
Odisha
19
8
The following data relates to the cost of construction of a house.
Items
Expenditure (%)
Cement
Steel
Bricks
Timber
30
10
10
15
Labour Miscellaneous
20
15
Draw a Pie diagram to represent the above data.
Collect the data presented in the form of Bar graph, Double bar graph and Pie charts in magazines,
newspapers etc. paste them on wall magazine present in your class.
1.
The information either in form of pictures, numbers or words is called ‘Data’.
2.
Based on the method of collecting information the data is divided into two
different types namely ‘Primary data’ and ‘Secondary data’.
3.
The difference between maximum and minimum values of data is its ‘Range’.
VII Class Mathematics
282
Data Handling
jáT֓{Ù nuó²«dŸ+
1. 3 eÖ«#Y\ýË ¿ÃVÓ²¢ eT]jáTT sÃV¾²ÔY\T #ûdq¾ |ŸsTÁ >·T\T ç¿ì+<Š ‚eNj&†¦sTT.
¿ÃVÓ²¢ : 49, 98, 72
sÃV¾²ÔY: 64, 45, 83. ¿ÃVÓ²¢ eT]jáTT sÃV¾²ÔY\T #ûdq¾ |ŸsTÁ >·T\ dŸ>³
· T ¿£qT>=qTeTT. me] dŸ>³
· T mÅ£”Øe?
2. 38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38 eT]jáTT 47\ u²VŸQÞø¿e£ TT ¿£qT>=qTeTT?
‚~ @¿£ u²VŸQÞø¿£ <ŠÔï+XøeÖ ýñ¿£ ~Çu²VŸQÞø¿£ <ŠÔï+XøeÖ?
3. $_óq• çbÍ+Ԑ\ýË –cþ’ç>·Ô\á T 0, -5, 7, 10, 13, -1 eT]jáTT 41 &ç^ ™d*àjáTdt (0C)\ýË >·\eÚ. nsTTq
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$\Te eÖsÁTÔáT+<‘? $e]+#á+&?
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sÁÖ|Ÿ+ýË e«¿£|ï sŸ #Á Tá eTT.
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¿£MTˆ ºçԐ“• ^jáT+&.
sçwŸe¼ TT
ÈqH\ s¹ ³T (ç|Ÿr 1000¿ì) eTsÁD²\ s¹ ³T (ç|Ÿr 1000¿ì)
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19
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$wŸjTá eTT
d¾yTî +{Ù
dÓý¼ Ù
‚³T¿£\T ‚+{ì ¿£\|Ÿ Å£L*
‚ÔáseÁ TT\T
KsÁTÌ(%)
30
10
10
15
20
15
™|qÕ ‚eNj&q dŸeÖ#s“• –|ŸjÖî Ð+º ™|Õ - ºçԐ“• ^jáT+&.
çbÍCÉÅ£”¼ |Ÿ“
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>Ã&ƒ|çŸ Ü¿£™|Õ ç|Ÿ<]Š ô+#á+&.
>·TsÁTï+#áT¿Ãy*àq
n+Xæ\T
1. dŸ+K«\T, |Ÿ<‘\T ýñ<‘ ºçԐ\ sÁÖ|Ÿ+ýË d¿£]+ºq dŸeÖ#sÁyTû »»<ŠÔï+XøeTTµµ.
2. dŸeÖ#s“• d¿£]+#û |Ÿ<ÜƊ “ ‹{ì¼ <ŠÔï+X擕 »»çbÍ<ó$Š T¿£ <ŠÔï+XøeTTµµ eT]jáTT »»>šD
<ŠÔï+XøeTTµµ nHû s +&ƒT sÁ¿±\T>± $uó›„ +#áe#áTÌ.
3. ‚ºÌq <ŠÔï+XøeTTý˓ >·]wŸ¼ eT]jáTT ¿£“wŸ¼ $\Te\ eT<óŠ« uóñ<‘Hû• »y«|¾ïµ n+{²sÁT.
7e ÔásÁ>·Ü >·DìÔáeTT
283
<ŠÔï+Xø “sÁÇVŸ²D
4.
The most common representative or Central Tendency Value of a grouped data is average or
Arithmetic Mean
5.
Arithmetic Mean =
6.
Arithmetic Mean of given data always lies between the lowest and highest observations in
the data.
The observation which occurs most frequently in the given data is ‘Mode’.
Data having only one mode is Unimodal data and data having two modes is Bimodal data.
The middle most value of the data when the observations are arranged in either ascending or
descending order is called Median.
7.
8.
9.
Sum of observations
Number of observations
 n  1  th
10. If number of observations(n) is odd then Median is 
 observations.
 2 
n
n 
11. If number of observations(n) is even then median is average of   th and   1  th
2
2 
observations.
12. Representation of numerical data by using bars of uniform width is Bar graph.
13. Representation of two sets of numerical data on the same graph by using bars of uniform
width is Double bar graph.
14. A Pie chart is the representation of the numerical data by sectors of the circle such that angle
of each sector (area of sector) is proportional to value of the given data.
15. Angle of sector =
Value of the item
 3600 .
Sum of the values of all items
Letter Series
Letter series is a logical arrangement of English alphabetical letters arranged
in a specific pattern. In this series of letters, groups of letters, combination of letters
and numbers are given. Each letter or group is called term. The terms of a series are arranged in a
particular order or pattern. We have to identify the pattern, and find the missing term (next term)
from the alternatives, which will satisfy the pattern. Assigning numbers to the alphabets is very
useful to practice letter series.
Example 1:
B, D, F, H, --1) I 2) K 3) J 4) L
Example 2:
A, B, D, E, G, --1) H 2) I 3) K 4) F
Answer : J
Answer : H
4. dŸsÇÁ kÍ<ó‘sÁD+>± çbÍܓ<ó«Š $\Te ýñ<‘ ¿¹ +çBjáT k͜q $\Te>± »»dŸ>³
· Tµµ ýñ<‘ »»n+¿£>D
· Ôì á dŸ>³
· Tµµ qT
–|ŸjÖî ÐkÍïsTÁ .
sXø—\ yîTTÔáeï TT
5. n+¿£>·DìÔá dŸ>·³T R sXø—\ dŸ+K«
6. n+¿£>·DìÔá dŸ>·³T m\¢|ŸÚÎ&ƒÖ nÔá«\Î eT]jáTT nÔá«~ó¿£ |Ÿ]o\H $\Te\ eT<óŠ« –+³T+~.
7. <ŠÔï+XøeTTýË mÅ£”Øe kÍsÁT¢ |ŸÚqseÔá+ njûT« s¥“ »»u²VŸQÞø¿+£ µ n+{²sÁT.
8. ÿ¹¿ ÿ¿£ s¥ u²VŸQÞø¿e£ TT>± >·\ <ŠÔï+XøeTTqT »»@¿£ u²VŸQÞø¿£ <ŠÔï+XøeTTµµ n“ eT]jáTT s +&ƒT sXø—\T
u²VŸQÞø¿e£ TT>± >·\ <ŠÔï+XøeTTqT »»~Ç u²VŸQÞø¿£ <ŠÔï+XøeTTµµ n+{²sÁT.
9. <ŠÔï+XøeTTý˓ sXø—\qT €sÃVŸ²D ýñ<‘ nesÃVŸ²D ç¿£eTeTTýË neTsÁÌ>±, € neT]¿£ý˓ eT<ó«Š eT $\TeqT
»»eT<ó«Š >·Ô+á µµ n+{²sÁT.
10. <ŠÔï+XøeTTýË sXø—\ dŸ+K«(n) uñd¾ dŸ+K« nsTTq eT<ó«Š >·Ôeá TT
 n 1 

e
 2 
11. ‚ºÌq <ŠÔï+XøeTTýË sXø—\ dŸ+K«(n) dŸ] dŸ+K« nsTTq eT<ó«Š >·Ôeá TT,
|Ÿ<eŠ TT n>·TqT.
n
 e
2
eT]jáTT
n 
  1 e
2 
|Ÿ<‘\
dŸsdŸ].
12. <ŠÔï+XøeTTqT dŸeÖq yî&\ƒ TÎ>·\ –q• ¿£MTˆ\ sÁÖ|ŸeTTýË ç|Ÿ<]Š ô+#áT³Hû ¿£MTˆ ºçÔá+ n+{²sÁT.
13. @¿£¯Ü yî&\ƒ TÎ ¿£*Ðq ¿£MTˆ\qT –|ŸjÖî Ð+#á&+ƒ <‘Çs ÿ¹¿ ç>±|˜t MT<Š s +&ƒT $_óq• <ŠÔï+XøeTT\qT
ç|Ÿ<]Š ô+#áT³qT &ƒ‹TýÙ u²sY ç>±|˜t ýñ<‘ s +&ƒT esÁTdŸ\ ¿£MTˆ ºçÔá+ n+{²sÁT.
14. eÔá+ï qT ™d¿±¼sTÁ >¢ ± $uó›„ +#áT³ <‘Çs <ŠÔï+XøeTTqT ç|Ÿ<]Š ô+#áT³Hû eÔás¹ï U² ºçÔáeTT ýñ<‘ ™|Õ ºçÔáeTT
n+{²eTT. ™|Õ ºçÔá+ýË ç|Ÿr ™d¿±¼sTÁ eÔáï ¿¹ +ç<Š+ e<ŠÝ #ûd ¿ÃDeTT (™d¿±¼sTÁ yîXÕ æ\«eTT) n~ dŸÖº+#û n+Xø
$\TeÅ£” nqTýËeÖqTbÍÔá+ýË –+³T+~.
n+XøeTT $\Te
15. ™d¿±¼sTÁ ¿ÃD+ R n“• n+Xø
eTT\ $\Te yîTTÔáeï TT
 3600
n¿£ŒsÁ çXâDì
n¿£sŒ çÁ XâDì nHû~ ÿ¿£ “]Æw¼Ÿ ç¿£eT+ýË neTsÁ̋&ƒ¦ ‚+^¢wt n¿£sŒ eÁ Ö\ jîTT¿£Ø Ԑ]Ø¿£ neT]¿£.
M{ìýË n¿£Œs\çXâDì (n¿£Œs\T), n¿£Œs\ dŸeTÖVŸä\T ýñ<‘ n¿£Œs\T eT]jáTT dŸ+K«\ ¿£\sTT¿£
‚eNj&+~. ç|ŸÜ n¿£sŒ +Á ýñ<‘ n¿£sŒ \ dŸeTÖVŸ²+qT |Ÿ<+Š n“ n+{²sÁT. çXâDýì ˓ |Ÿ<‘\T
ÿ¿£ “]Ýw¼Ÿ ç¿£eT+ýË ýñ<‘ qeTÖHýË neTsÁ̋&†¦sTT. eTq+ çXâD“ì >·T]ï+º U²°ýË, € çXâD“ì dŸ+Ôá|Ÿï |Ÿs#Á û
|Ÿ<eŠ TT (ÔásTÁ yÔá |Ÿ<+Š ) € ç|ŸÔ«e֕jáÖ\ qT+& ¿£qT>=H*. n¿£sŒ \çXâD“ì kÍ<óqŠ #ûjTá &†“¿ì n¿£sŒ \Å£”
Hî+‹sÁT¢ ¿¹ {²sTT+#á&+ƒ m+Ôà –|ŸjÖî >·¿s£ +Á >± –+³T+~.
–<‘VŸ²sÁD 1:
B, D, F, H, --1) I 2) K 3) J 4) L
–<‘VŸ²sÁD 1:
A, B, D, E, G, --1) H 2) I 3) K 4) F
Cy‹T 'J'
Cy‹T 'H'
Example 3:
Z, X, U, Q
Answer : L
1) M 2) K 3) N 4) L
Example 4:
QPO, NML, KJI,
EDC
This series consists of letters in a reverse
1) KL 2) GHI 3) CAB 4) HGF alphabetical order so, answer should be ‘HGF’.
Example 5:
AB, DE, HI, MN,
1) TV 2) TU 3) ST 4) RS
Example 6:
So, in the series next word is 'ST'
AB, EF, IJ, MN,
1) QR 2) OP 3) XY 4) PQ
Example 7:
So, in the series next word is 'QR'
B2, D4, F6, H8, J10,
.
Alternate letters with their assigned
1) L12 2) K11 3) N14 4) M13 numbers. So, answer is L12
Example 8:
AFK, BGL, CHM, DIN
Each group next letter is forwarding 5th letter.
1) GJO 2) FIO 3) EJO 4) GJN So, answer is ‘EJO’
Model Problems :
In the following alphabetical series, a term (next term) is missing. Choose the missing term from the
options.
1.
B, F, J, N, R, V,
a)
Z
b) W
c) X
d) W
2.
A, C, E, G, I, K,
a) P
b) O
c) N
d) M
3.
M, O, R,T, ..
a) W
b) U
c) V
d) Q
4.
U, S, P, L, ....
a) F
b) G
c) H
d) I
5.
ZA, YB, XC, WD, ...
a) UE
b) EV
c) VE
d) SH
6.
AM, BO, CQ, DS, EU,
a) WF
b) FU
c) GV
d) FW
7.
ZY, XV, UR, QM,
a) LG
b) LI
c) LH
d) KJ
8.
AC, DF, GI, JL,
a) NO
b) MO
c) MN
d) NP
9.
DN, EM, FL, GK, HJ,
a) IK
b) GI
c) IJ
d) II
10. CBA, STU, FED, VWX,
a) IHG
b) GHI
c) IJK
d) YZA
11. AZC, DYF, GXI, JWL,
a) OVM
b) UNV
c) MVO
d) MNO
12. ABK, CDL, EFM, GHN,
a) JIO
b) IJO
c) MNO
d) ONM
13. A2C, D5F, G8I, J11L,
a) M14O
b) M12O
c) N15P
d) N12P
14. A, CD, HIJ, PQRS,
a) ZABCD
b) ZYXW
c) ABCDE
d) RSTUV
15. A, BC, DEF, GHIJ,
a) KLMNP b) LMNOP c) KLMNO d) JKLMN
–<‘VŸ²sÁD 3:
Z, X, U, Q
Cy‹T 'L'
1) M 2) K 3) N 4) L
–<‘VŸ²sÁD 4:
QPO, NML, KJI,
‚ºÌq çXâDìý˓ n¿£Œs\T ‚+^¢wt n¿£ŒsÁeÖ\ jîTT¿£Ø e«Ü¹s¿£
(n|ŸdŸe«) ~Xø sjáT‹&†¦sTT ¿±‹{ì¼ Èy‹T »HGFµn>·TqT.
EDC
1) KL 2) GHI 3) CAB 4) HGF
–<‘VŸ²sÁD 5:
¿±‹{ì ¼ , çXâ D ì ý ˓
ÔásTÁ yÔá e#ûÌ |Ÿ<+Š
»STµ
¿±‹{ì,¼ çXâDýì ˓ ÔásTÁ yÔá
e#ûÌ |Ÿ<Š+ »QRµ
AB, DE, HI, MN,
1) TV 2) TU 3) ST 4) RS
–<‘VŸ²sÁD 6:
–<‘VŸ²sÁD 7:
–<‘VŸ²sÁD 8:
AB, EF, IJ, MN,
1) QR 2) OP 3) XY 4) PQ
B2, D4, F6, H8, J10,
.
1) L12 2) K11 3) N14 4) M13
AFK, BGL, CHM, DIN
1) GJO 2) FIO 3) EJO 4) GJN
n¿£sŒ \T eT]jáTT y{ì¿ì ¿¹ {²sTT+ºq dŸ+K«\T (ÿ¿£{ì
$&º ÿ¿£{)ì ¿±‹{ì¼ Èy‹T »L12µ
ç|Ÿr dŸeTÖVŸ²+ýË n¿£sŒ +Á eT]jáTT <‘“ ÔásTÁ yÔá e#ûÌ
jîTT¿£Ø 5 e n¿£sŒ +Á ¿±‹{ì¼ Èy‹T »EJOµ
eÖ~] dŸeTdŸ«\T:
~>·Te n¿£sŒ çÁ XâDýì Ë, U²°\ýË –+&ƒe\d¾q |Ÿ<+Š (ÔásTÁ yÔá|<Ÿ +Š )qT ‚ºÌq ׺̿±\qT+&m+#áT¿=“ |ŸP]+#á+&.
1.
B, F, J, N, R, V,
a)
Z
b) W
c) X
d) W
2.
A, C, E, G, I, K,
a) P
b) O
c) N
d) M
3.
M, O, R,T, ..
a) W
b) U
c) V
d) Q
4.
U, S, P, L, ....
a) F
b) G
c) H
d) I
5.
ZA, YB, XC, WD, ...
a) UE
b) EV
c) VE
d) SH
6.
AM, BO, CQ, DS, EU,
a) WF
b) FU
c) GV
d) FW
7.
ZY, XV, UR, QM,
a) LG
b) LI
c) LH
d) KJ
8.
AC, DF, GI, JL,
a) NO
b) MO
c) MN
d) NP
9.
DN, EM, FL, GK, HJ,
a) IK
b) GI
c) IJ
d) II
10. CBA, STU, FED, VWX,
a) IHG
b) GHI
c) IJK
d) YZA
11. AZC, DYF, GXI, JWL,
a) OVM
b) UNV
c) MVO
d) MNO
12. ABK, CDL, EFM, GHN,
a) JIO
b) IJO
c) MNO
d) ONM
13. A2C, D5F, G8I, J11L,
a) M14O
b) M12O
c) N15P
d) N12P
14. A, CD, HIJ, PQRS,
a) ZABCD
b) ZYXW
c) ABCDE
d) RSTUV
15. A, BC, DEF, GHIJ,
a) KLMNP b) LMNOP c) KLMNO d) JKLMN
CHAPTER - 1
Review Exercise :
1)
i)
+ `2000
2)
3)
4)
5)
ii)
–350 ft.
–7
–3
iii)
+ 8848 m
0
iv) –14ºC
4
6
i)
Descending order : 0,–1, –6, –9, –10
Asending order : –10, –9, –6, –1, –0
ii)
Descending order : 10, 6, 5, –3, –6, –9
Asending order : –9, –6, –3, 5, 6, 10
iii) Descending order : 2, 0, –15, –20, –35
Asending order : –35, –20, –15, 0, 2
i)
1
ii)
v)
7
4 0C
–8
iii)
14
iv) –5
vi) –103
vii)
53
viii) –10
6)
7)
Loss, `50
–33ft
Exercise 1.1
1)
2)
i)
35
ii)
v)
124
i)
2 × (–5)
v)
(–8) × (–5)
5)
–54
iii)
–36
iv) –56
vi) 84
vii)
–441
viii) –105
ii)
iii)
(–3) × (–21)
iv) 6 × (–16)
–30 inches
6)
Profit, `3000
7) Loss, `2000
iii)
–8
iv) –5
iii)
–3
iv) –2
vii)
–10
viii) –6
4)
9PM, –140C
5)
(–6) × (–7)
4)
–15m
8)
i)
5
ii)
v)
–6
vi) –1
–7
Exercise 1.2
1)
i)
–6
ii)
–2
v)
–4
vi) 6
ix) 2
x)
1
2)
11
3)
`22,000
6)
i) 10 ii) 20
119 Kgs.
Exercise 1.3
1)
i)
Additive commutative property
ii)
Multiplicative identity property
iii) Multiplicative Associative property
iv)
Distributive over addition
v)
vi)
Additive Identity property
Multiplicative Closure property
n<ó‘«jáT+ - 1
|ŸÚq]ÇeTsÁô nuó²«dŸ+:
1)
i)
+ `2000
2)
3)
5)
–350 ft.
–7
–3
iii)
+ 8848 m
0
4
iv) –14ºC
6
iii)
nesÃVŸ²D ç¿£eT+: 0, –1, –6, –9, –10
nesÃVŸ²D ç¿£eT+: 10, 6, 5, –3, –6, –9
nesÃVŸ²D ç¿£eT+: 2, 0, –15, –20, –35
€sÃVŸ²D ç¿£eT+ : –10, –9, –6, –1, –0
€sÃVŸ²D ç¿£eT+ : –9, –6, –3, 5, 6, 10
€sÃVŸ²D ç¿£eT+ : –35, –20, –15, 0, 2
i)
1
ii)
iii)
14
iv) –5
v)
7
vi) –103
vii)
53
viii) –10
6)
7)
qwŸ+¼ , `50
i)
ii)
4)
ii)
4 0C
–8
–33ft
nuó²«dŸ+ - 1.1
1)
2)
i)
35
ii)
v)
124
i)
2 × (–5)
v)
(–8) × (–5)
–54
iii) –36
iv) –56
vi) 84
vii) –441
viii) –105
ii)
(–6) × (–7)
iii) (–3) × (–21)
iv) 6 × (–16)
ý²uó+„ , `3000
4)
–15m
5)
–30 inches
6)
8)
i)
5
ii)
–7
iii) –8
v)
–6
vi) –1
7)
qwŸ+¼ , `2000
iv) –5
nuó²«dŸ+ - 1.2
1)
i)
–6
ii)
–2
v)
–4
vi) 6
ix) 2
x)
1
2)
11
3)
`22,000
6)
i) 10 ii) 20
iii) –3
iv) –2
vii) –10
viii) –6
4)
5)
9PM, –140C
119 Kgs.
nuó²«dŸ+ - 1.3
1)
i)
iii)
v)
dŸ+¿£\q d¾Ôœ «á +ÔásÁ <ósŠ ˆÁ +
>·TD¿±sÁ dŸVŸ²#ásÁ <óŠsÁˆ+
>·TD¿±sÁ dŸ+eÔá <ósŠ ˆÁ +
ii)
iv)
vi)
>·TD¿±sÁ ÔáÔàá eÖ+dŸ+
dŸ+¿£\q+ ™|Õ $uó²>·<óŠsÁˆ+
dŸ+¿£\q ÔáÔàá eÖ+dŸ+
2)
i)
positive
ii)
negative
3)
i)
0
ii)
2
iii)
3, –2
iv) 1
v)
5, 9
i)
False
ii)
False
iii)
True
iv) True
v)
False
5)
i)
–1100
ii)
–150
6)
Integers are not associative under subtraction
4)
Exercise 1.4
1)
i)
v)
52
8
ii)
0
iii)
17
iv) –10
2)
i)
v)
18
71
ii)
–2
iii)
–9
iv) 7
3)
i)
False
ii)
True
iii)
False
iv) True
4)
i)
–3
ii)
–4
iii)
–30
Unit Exercise
1)
2)
i)
–8
ii)
v)
7
i)
v)
–350
iii)
+120
iv) 148
vi) –2
vii)
7
viii) –7
–3
ii)
–6
iii)
8
iv) –12
–25
vi) –7
vii)
–90
viii) –144
4)
Profit, `25
5)
190 calories
6)
–3125
7)
i)
0
ii)
–43
iii)
42
8)
i)
700
ii)
150
iii)
150
v)
10 – p
vi) y – 7
iv) 35
2)
i)
<óHŠ Ôሿ£+
ii)
‹TTD²Ôሿ£+
3)
i)
0
ii)
2
iii)
3, –2
iv) 1
v)
5, 9
i)
ii)
Ôá|ðŸ
iii)
ÿ|Ÿð
iv)
v)
Ôá|ðŸ
Ôá|ðŸ
5)
i)
–1100
ii)
–150
6)
|ŸPsÁ’ dŸ+K«\T e«e¿£\q+ ™|Õ dŸV²Ÿ #ásÁ <ósŠ ˆÁ + bÍ{ì+#áeÚ
4)
ÿ|Ÿð
nuó²«dŸ+ - 1.4
1)
i)
v)
52
8
ii)
0
iii)
17
iv) –10
2)
i)
v)
18
71
ii)
–2
iii)
–9
iv) 7
3)
i)
Ôá|ðŸ
ii)
ÿ|Ÿð
iii)
Ôá|ðŸ
iv)
4)
i)
–3
ii)
–4
iii)
–30
ÿ|Ÿð
jáT֓{Ù nuó²«dŸ+
1)
2)
i)
–8
ii)
v)
7
i)
v)
–350
iii)
+120
iv) 148
vi) –2
vii)
7
viii) –7
–3
ii)
–6
iii)
8
iv) –12
–25
vi) –7
vii)
–90
viii) –144
4)
ý²uó+„ , `25
5)
190 ¿\¯\T
6)
–3125
7)
i)
0
ii)
–43
iii)
42
8)
i)
700
ii)
150
iii)
150
v)
10 – p
vi) y – 7
iv) 35
CHAPTER - 2
Review Exercise :
1)
2)
i)
1
2
v)
19
100
i)
Ascending order
iii) Ascending order
ii)
5 2
,1
3 3
vi)
99
29
, 1
70
70
iii)
1 3 5 9 17
, , , ,
2 2 2 2 2
11 2
,1
9 9
iv)
ii) Ascending order
23
25
5 7 11 6 19
, , , ,
10 10 10 5 5
7 5 8 11 1
, , , ,3
6 3 3 4
4
3)
i)
5
2
ii)
1
5
12
4)
i)
3
4
ii)
5
12
5)
i)
1
4
ii)
3
11
15
iii)
1
27
40
iv) 1
iii)
8
1
28
iv) 8
iii)
6
iv) 1
1
6
59
62
Exercise - 2.1
1)
9kg.
sq.m.
2)
15 cm.
3)
9 km.
4)
1 0 1
5)
`24
6)
50 km.
7)
1
kg or 200gm.
5
8)
60
Exercise - 2.2
1)
2)
3)
4)
i)
140.4
ii) 6131.25
iii)
746.06
iv)
20.16
v)
50.325
i)
2310.4
ii) 1000
iii)
0.02105
iv)
923.4
v)
1000
vi) 5.9001
i)
41.31
ii) 17.06535
iii)
1.1889
iv)
6.4449
v)
5.0256
5) 14.31Sq.cm.
6)
`7836
7)
`213.50
17.5 hours
8)
`68
n<ó‘«jáT+ - 2
|ŸÚq]ÇeTsÁô nuó²«dŸ+:
1)
i)
1
2
ii)
5 2
,1
3 3
vi)
99
29
, 1
70
70
v)
19
100
i)
€sÃVŸ²D ç¿£eT+
iii)
€sÃVŸ²D ç¿£eT+
3)
i)
5
2
ii)
1
4)
i)
3
4
ii)
5
12
5)
i)
1
4
ii)
3
2)
iii)
1 3 5 9 17
, , , ,
2 2 2 2 2
11 2
,1
9 9
€sÃVŸ²D ç¿£eT+
ii)
23
25
iv)
5 7 11 6 19
, , , ,
10 10 10 5 5
7 5 8 11 1
, , , ,3
6 3 3 4
4
5
12
11
15
iii)
1
27
40
iv)
1
iii)
8
1
28
iv)
8
iii)
6
iv)
1
1
6
59
62
nuó²«dŸ+ - 2.1
1)
9¿ì.ç>±.
2)
15 ™d+.MT.
3)
9 ¿ì.MT.
5)
`24
6)
50 ¿ì.MT.
7)
1
kg or 200gm.
5
4)
101 #á.™d+.MT.
8) 60
nuó²«dŸ+ - 2.2
1)
2)
3)
i)
140.4
ii) 6131.25
iii)
746.06
iv)
20.16
v)
50.325
i)
2310.4
ii) 1000
iii)
0.02105
iv)
923.4
v)
1000
vi) 5.9001
i)
41.31
ii) 17.06535
iii)
1.1889
iv)
6.4449
v)
5.0256
5) 14.31#á.™d+.MT. 6)
`7836
7)
`213.50
4)
17.5 >·+³\T
8)
`68
Exercise - 2.3
1)
i)
56.361
ii) 1000
iii) 2016.4
v)
1.23
vi) 1000
ii) 12.8
iii) 9.565
iv) 51.1
v)
0.1401
vi) 3.33
vii) 2
viii) 0.91
ix) 25
x)
9.7
i) 391.2
ii) 3.23
iii) 0.41
iv) 256
v)
0.2
vi) 48.3
vii) 30
viii) 25.4
ix) 4
x)
270.5
4)
i)
ii) 0.72
iii) 4
5)
30.9 Km.
6) 59.5 hours
7)
vi) 590.01
2)
3)
i)
2.755
6.59
`5.15
8)
`2.65
Exercise - 2.4

15
9
ii) 
20
12
1)
i)
3)
2 6 8
5 10 15
  ;

 ;
3 9 12 4 8 12
6)
i)

7)
i)
True
8)
Yes,
2 3

7 7
ii)
2) i)

4
2 0 0
 ;

6
3 3 5
5 5

9 8
ii) False
iii)
16
ii)
36
40
90
2
1 4 2
4)  , 0, , ,
9
3 9 3
13 7

12
6
iii) True
iv) True
3
lies between –1 and –2.
2
Exercise - 2.5
1)
37
14
2) `94
6)
1
7)
3)
`442.75
4)
9.5
5
6
Unit Exercise
1)
i)
d
ii) b
iii) b
iv) d
5)
–5.50C
nuó²«dŸ+ - 2.3
1)
i)
56.361
ii) 1000
iii) 2016.4
v)
1.23
vi) 1000
ii) 12.8
iii) 9.565
iv)
51.1
v) 0.1401
vi) 3.33
vii) 2
viii) 0.91
ix)
25
x) 9.7
i)
ii) 3.23
iii) 0.41
iv)
256
v) 0.2
vi) 48.3
vii) 30
viii) 25.4
ix)
4
x) 270.5
4)
i)
ii) 0.72
iii) 4
5)
30.9 Km.
6) 59.5 hours
7)
8)
`2.65
vi) 590.01
2)
3)
i)
2.755
391.2
6.59
`5.15
nuó²«dŸ+ - 2.4

15
9
ii) 
20
12
1)
i)
3)
2 6 8
5 10 15
  ;

 ;
3 9 12 4 8 12
6)
i)

7)
i)
ÿ|Ÿð
8)
neÚqT,
2 3

7 7
2) i)

4
2 0 0
 ;

6
3 3 5
ii)
5 5

9 8
iii)
13 7

12
6
ii)
Ôá|ðŸ
iii)
ÿ|Ÿð
3
, –1 eT]jáTT –2
2
16
ii)
36
40
90
2
1 4 2
4)  , 0, , ,
9
3 9 3
iv)
ÿ|Ÿð
4)
9.5
iv)
d
\ eT<ó«Š –+³T+~.
nuó²«dŸ+ - 2.5
1)
37
14
2) `94
6)
1
7)
3)
`442.75
5
6
jáT֓{Ù nuó²«dŸ+
1)
i)
d
ii) b
iii) b
5) –5.50C
2
i)
Rational
ii) 0.0121
iii)

3)
i)
13.23
ii) 54.87035
iii)
4)
i)
153.1
ii) 15.6333333
5)
–0.02
6) –
3
7
9)
`1058.125
12) 
13
6
5
2
4
6
or  etc.
6
9
iv)

26.7435
iv)
5.1876
iii)
– 0.52
iv)
14.63636363
7)
40 Km.
8)
`150
iv)
3m – 5 = 11
CHAPTER - 3
Exercise - 3.1
1)
2)
i)
x – 5 = 14
ii) 8y + 3 = –5
iii)
v)
3x = 120
vi) 4a = 14
i)
A number ‘m’ is decreased by 5 is 12
iii) 7 is added to 4 times of ‘x’ is 15
3)
i)
Yes
ii) No
4)
i)
x=1
ii) y = 6
3z
3 7
4
ii) one third of ‘a’ is 4
iv) three times of ‘y’ is subtracted from 2 is 11
iii)
Yes
iv)
No
Exercise - 3.2
1)
i)
m=6
ii) n = 9
iii)
x=7
iv)
y=5
2)
i)
x=6
ii) b = –14
iii)
x=5
iv)
a=5
v)
m = –3
vi) p = –7
vii)
x=
ix) k = 3
16
9
viii) n = 3
x) a = 0
Exercise - 3.3
1)
i)
x=7
ii) y = 4
iii)
a = 3.8
iv)
b= 
7
4
2
i)
n¿£sDÁ j
¡ Tá +
ii) 0.0121
iii)

3)
i)
13.23
ii) 54.87035
iii)
4)
i)
153.1
ii) 15.6333333
5)
–0.02
6) –
3
7
9)
`1058.125
12) 
13
6
5
2
4
6
or  etc.
6
9
iv)

26.7435
iv)
5.1876
iii)
– 0.52
iv)
14.63636363
7)
40 Km.
8)
`150
iv)
3m – 5 = 11
iv)
¿±<ŠT
n<ó‘«jáT+ - 3
nuó²«dŸ+ - 3.1
1)
2)
i)
x – 5 = 14
ii) 8y + 3 = –5
iii)
v)
3x = 120
vi) 4a = 14
i)
‘m’ qT+& 5 rd¾yd
û q¾ |˜*Ÿ Ôá+ 12.
3z
3 7
4
3)
i)
ýË eTÖ&ƒe uó²>·+ 4 edŸT+ï ~.
»xµ jîTT¿£Ø 4 s ³T¢¿ì 7 ¿£*|¾Ôû 15 edŸT+ï ~.
2 qT+& ‘y’ jîTT¿£Ø 3 s ³T¢qT ÔáÐ+Z #á>± 11 edŸT+ï ~.
neÚqT
ii) ¿±<ŠT
iii) neÚqT
4)
i)
x=1
ii)
iii)
iv)
‘a’
ii) y = 6
nuó²«dŸ+ - 3.2
1)
i)
m=6
ii) n = 9
iii)
x=7
iv)
y=5
2)
i)
x=6
ii) b = –14
iii)
x=5
iv)
a=5
v)
m = –3
vi) p = –7
vii)
x=
1)
ix) k = 3
x) a = 0
i)
ii) y = 4
x=7
16
9
viii) n = 3
nuó²«dŸ+ - 3.3
iii)
a = 3.8
iv)
b= 
7
4
v)
p=
60
7
vi) q = 12
ix) k = 3
x) a = 0
2)
ii) q =
i)
p = 10
vi) t = 4vii)
vii)
9
2
k=2
m = –9
viii) n = 30
iii) x = 18
iv)
y = –0.7
v) r = 0
viii) m = –3
ix)
n=2
x) x = 1
Exercise - 3.4
1)
x = 1.7m
2) 38
3)
32
4)
x = 7cm.
5) –40ºF
6)
`49
7) smaller 10, larger 17
8)
16, 18, 20
9) girls 12
iv)
c
v) d
iv)
No solution
v) 8
11)
11years, 8 years
10) Mary age 20 years, Joseph age 12 years
11) `5 notes 80, `10 notes 10
12) `1305, `2695
13) length 11m, breadth 5m
14) White balls 3, Blue balls 6, Red balls 18
15) Cuboid = 54 , Cube = 90
Unit Exercise
1)
i)
c
ii) b
2)
i)
18
ii) transposition iii) 4x = 60
3)
i)
Yes
ii) No
5)
i)
7 more than 2 times of ‘m’ is 21
ii)
one seventh of ‘n’ is 4
6)
i)
x=7
7
i)
a=–
8)
7
4)
k=4
ii) m = 3
4
9
ii) y = 5
9) 4
12) length 35m, breadth 15m
13) rice 24kg, wheat 6kg.
14) 43
iii) a
15) 280 Km.
10) 10m
v)
p=
60
7
vi) q = 12
ix) k = 3
x) a = 0
2)
ii) q =
i)
p = 10
vi) t = 4
vii)
9
2
vii) k = 2
m = –9
viii) n = 30
iii) x = 18
iv)
y = –0.7
v) r = 0
viii) m = –3
ix)
n=2
x) x = 1
nuó²«dŸ+ - 3.4
1)
x = 1.7m
6)
`49
10)
11)
2) 38
3)
32
ºq•~ 10, ™|<ŠÝ~ 17
yûT¯ ejáTdŸTà 20 dŸ+, C˙d|˜t ejáTdŸTà 12 dŸ+
`5 Hó¢ dŸ+K« 80, `10 Hó¢ dŸ+K« 10
7)
4) x = 7 ™d+.MT. 5) –40ºF
8) 16, 18, 20
9)
12 eT+~ u²*¿£\T
12) `1305, `2695
13)
14)
15)
bõ&ƒeÚ 11 MT. yî&\ƒ TÎ 5MT.
Ôî\¢ ‹+ÔáT\ dŸ+K« 3, ú\+ ‹+ÔáT\ dŸ+K« 6, msÁT|ŸÚ ‹+ÔáT\ dŸ+K« 18
BsÁé |˜TŸ q+ R 54, |˜TŸ q+ R 90
jáT֓{Ù nuó²«dŸ+
1)
i)
c
ii) b
2)
i)
18
ii)
3)
i)
Yes
ii) No
5)
i)
‘m’ jîTT¿£Ø 2 s
 ³T¢Å”£ 7 ¿£*|¾q |˜*Ÿ Ôá+ 21 neÚÔáT+~.
|Ÿ¿±Œ +Ôás+Á
ii)
‘n’ ýË 7
6)
i)
x=7
7
i)
a=–
8)
7
12)
bõ&ƒeÚ 35 MT. yî&\ƒ TÎ 15MT.
_jáT« 24 ¿Â .›\T, >Ã<óTŠ eT\T 6 ¿Â .›.\T
13)
14) 43
iii) a
iv)
c
v) d
iii) 4x = 60
iv)
kÍ<óŠq ýñ<ŠT.
v) 8
11)
11 dŸ+eÔáàs\T, 8 dŸ+eÔáàs\T
4)
k=4
uó²>·+ 4 neÚÔáT+~.
ii) m = 3
4
9
ii) y = 5
9) 4
15) 280 Km.
10) 10m
CHAPTER - 4
Review Exercise :
1)
Points A, B, C, D, E ; Line segments BC , CD, AB, BD, DE , AD, AE , BE
Rays BA, DA BE , DE , AE , EA ; Lines AE
4) POQ, POR, POS, QOR, QOS, ROS
2)
l, p ; l, m, n
5)
Acute angle, Right angle, Obtuse angle, Straight angle, Reflex angle
6)
90º
8) l||m ; l  n, m  n
9)
AOB = 40º
Exercise - 4.1
1)
Complementary angles (ii) ; Supplementary angles (iv), (v)
2)
36º, 54º
8)
No, one angle need not be obtuse angle
3) 6º
4)
45º, 45º
6)
Sujatha
9)
70º
7) 30º
10) Yes, sum of two obtuse angles are greater than 180º
Exercise - 4.2
1)
KOQ, QOP;
QOP, POM;MOL, LOK;
5)
AOC, COD;
COD, DOB;
6)
i)
8)
50º
Yes
LOK, KOQ
AOC, COB;
AOD, DOB
ii) No, the sum of angles 98º + 102º = 200º
9) Rositha
Exercise - 4.3
1)
AOB, DOE; BOC, EOF;
COD, AOF;
DOE = 45º
2)
No, SOR is not a straight angle
4)
160º, 20º, 160º
5)
AOE, COD; AOC, DOE
6)
65º
Exercise - 4.4
1)
1 = 135º, 2 = 45º, 4 = 45º, 5 = 135º, 6 = 45º, 7 = 45º, 8 = 135º
2)
65º
6)
60º, 80º, 40º
9)
AL||EH; BK||DI; CJ||FG
3)
120º, 75º
4)
20º, 33º ; AEC = 53º
5) Yes
7)
100º, 80º, 80º
8) 360º
Unit Exercise
1)
54º, 144º, 324º
2)
AOB, BOC; BOC, COD; COD, DOE; DOE, EOF
3)
80º
6)
I don’t agree, sum of two obtuse angles is less than 360º.
4) 18º, 36º, 54º, 72º
n<ó‘«jáT+ - 4
|ŸÚq]ÇeTsÁô nuó²«dŸ+:
1)
_+<ŠTeÚ\T A, B, C, D, E ; s¹ U² K+&†\T
¿ìsÁD²\T
BA, DA BE , DE , AE , EA ;
BC , CD, AB, BD, DE , AD, AE , BE
¹sK\T
AE
4) POQ, POR, POS, QOR, QOS, ROS
2)
l, p ; l, m, n
5)
n\οÃDeTT, \+‹¿ÃDeTT, n~ó¿¿£ ÃDeTT, dŸsÞÁ ¿ø ÃDeTT, |ŸsesÁqï ¿ÃDeTT
6)
90º
8) l||m ; l  n, m  n
9)
AOB = 40º
6)
dŸTC²Ôá
70º
nuó²«dŸ+ - 4.1
1)
|ŸPsÁ¿£ ¿ÃD²\T (ii) dŸ+|ŸPsÁ¿±\T (iv), (v)
2)
36º, 54º
8)
¿±<ŠT, ÿ¿£ ¿ÃD+ n~ó¿¿£ ÃD+ ¿±qedŸs+Á ýñ<TŠ .
neÚqT, s +&ƒT n~ó¿£ ¿ÃD²\T yîTTÔá+ï 1800 ¿£+fñ mÅ£”Øe
9)
1)
KOQ, QOP;
QOP, POM;MOL, LOK;
LOK, KOQ
5)
AOC, COD;
COD, DOB;
6)
i)
8)
50º
10)
3) 6º
4)
45º, 45º
7) 30º
nuó²«dŸ+ - 4.2
neÚqT
ii)
9)
AOC, COB;
¿±<ŠT, ¿ÃD²\T yîTTÔá+ï 98º + 102º
sÃw¾Ôá
AOD, DOB
= 200º
nuó²«dŸ+ - 4.3
1)
AOB, DOE; BOC, EOF;
COD, AOF;
DOE = 45º
2)
¿±<ŠT, SOR dŸsÞÁ ¿ø ÃD+ ¿±<ŠT
4)
160º, 20º, 160º
5)
AOE, COD; AOC, DOE
6)
65º
nuó²«dŸ+ - 4.4
1)
1 = 135º, 2 = 45º, 4 = 45º, 5 = 135º, 6 = 45º, 7 = 45º, 8 = 135º
2)
65º
6)
60º, 80º, 40º
9)
AL||EH; BK||DI; CJ||FG
3)
120º, 75º
neÚqT
4)
20º, 33º ; AEC = 53º
5)
7)
100º, 80º, 80º
8) 360º
jáT֓{Ù nuó²«dŸ+
1)
54º, 144º, 324º
2)
AOB, BOC; BOC, COD; COD, DOE; DOE, EOF
3)
80º
6)
HûqT n+^¿£]+#áqT s +&ƒT n~ó¿¿£ ÃD²\T yîTTÔá+ï 3600 ¿£+fÉ ÔáŔ£ Øe.
4) 18º, 36º, 54º, 72º
8)
110º, 70º
9)
i)
transversal
10) 45º
ii) corresponding iii) alternate interior iv) co-interior
11) 40º
CHAPTER - 5
Review Exercise :
2)
i)
J. D, C
3)
X, Y, Z ; XY, YZ, ZX ; X, Y, Z
4)
i)
6)
Acute angles : 20º, 36º, 47º, 50º, 65º
KM ,
ii) P, A, R, B, G, E, A
iii) M
ii) LM ,
Obtuse angles : 95º, 102º, 110º, 125º
Right angle : 90º
7)
Intersecting points P, Q; concurrent point Q.
Exercise - 5.1
1)
fig (i) and fig (vi) are scalene triangles
fig (ii) and fig (v) are equilateral triangles.
fig (iii) and fig (iv) are isosceles triangles.
2)
fig (ii) and fig (vi) are acute angle triangles.
fig (iii) and fig (v) are obtuse angle triangles.
fig (i) and fig (iv) are right angle triangles
3)
iii) and (v) are scalene triangles
(ii) and (vi) are equilateral triangles
(i) and (iv) are isosceles triangles
iii)
iv)
K
F, H, I
8)
110º, 70º
9)
i)
ÜsÁ«ç¹>K
10) 45º
ii)
nqTsÁÖ|Ÿ
iii)
i)
3)
X, Y, Z ; XY, YZ, ZX ; X, Y, Z
4)
i)
6)
n\οÃD²\T : 20º, 36º, 47º, 50º, 65º
n~ó¿£ ¿ÃD²\T : 95º, 102º, 110º, 125º
\+‹¿ÃD+ : 90º
K+&ƒq _+<ŠTeÚ\T P, Q, $T[Ôá _+<ŠTeÚ Q,
J. D, C
KM ,
ii) P, A, R, B, G, E, A
ii) LM ,
iii) M
nuó²«dŸ+ - 5.5
2)
3)
dŸV²Ÿ n+ÔásÁ ¿ÃH\T
n<ó‘«jáT+ - 5
2)
1)
iv)
11) 40º
|ŸÚq]ÇeTsÁô nuó²«dŸ+:
7)
@¿±+ÔásÁ ¿ÃD²\T
|Ÿ³+ (i) eT]jáTT |Ÿ³+ (vi) \T $wŸeTu²VŸQ çÜuóT„ C²\T
|Ÿ³+ (ii) eT]jáTT |Ÿ³+ (v) \T dŸeTu²VŸQ çÜuó„TC²\T
|Ÿ³+ (iii) eT]jáTT |Ÿ³+ (iv) \T dŸeT~Çu²VŸQ çÜuó„TC²\T
|Ÿ³+ (ii) eT]jáTT |Ÿ³+ (vi) \T n\οÃD çÜuó„TC²\T
|Ÿ³+ (iii) eT]jáTT |Ÿ³+ (v) \T n~ó¿ÃD çÜuó„TC²\T
|Ÿ³+ (i) eT]jáTT |Ÿ³+ (iv) \T \+‹¿ÃD çÜuóT„ C²\T
(iii) eT]jáTT (v) \T $wŸeTu²VŸQ çÜuóT„ C²\T
(ii) eT]jáTT (vi) \T dŸeTu²VŸQ çÜuó„TC²\T
(i) eT]jáTT (iv) \T n~ó¿ÃD çÜuó„TC²\T
iii)
iv)
K
F, H, I
4)
(i), (iv) and (v) are acute angle triangles
(iii) and (vi) are obtuse angle triangles
(ii) is right angle triangle
5)
fig (i) is Obtuse angle triangle and scalene triangle
fig (ii) is right angle triangle and scalene triangle
fig (iii) is obtuse angle triangle and equilateral triangle
Exercise - 5.2
1)
b, d and e measurements forms triangles
4)
S = 500, N = 200, P = 550
6)
fig(i) x = 680 and y = 520
7)
530 8) 800, 300, 700
11) 400, 800, 600
2) 750
3) 650
5) CUT x = 1100 in NET x = 510
In fig(ii), x = 360 and y = 1440
9) 500 , 500
10)
i) False ii) True iii) False
12) 10800
Exercise - 5.3
1)
YXP, ZYQ and XZR
2) ACD = 1330 and DAR = 1200
3)
350, 170
5)
In fig (i) x = 1000, y = 1450; in fig (ii) x = 1300, y = 500
4) 500
Exercise - 5.4
2)
3)
i)
3, 5, 6, and 5, 6, 9 groups forms triangles. Why because sum of any two sides is greater
than the third side in this group.
ii)
3, 6, 9 does not form a triangle . Why because sum of 3cm and 6cm is not greater than
9cm.
650, 4.3 cm.
4) 500, 500
5)
740 , 550
6)
(i), (iii) is true
Exercise - 5.5
1)
i)
True
ii)
False, orthocentre of right angled triangle lies at the vertex at the right angle
iii) True
iv) True
v)
False, The point of concurrence of Perpendicular bisectors of sides is circum centre.
2)
PT is median, PS is altitude
3
i)
Centroid
ii) ortho centre iii) incentre
iv)
ircumcentre
4)
i)
2:1
ii) 2:3
iii) 3:1
iv)
1:2
5)
i)
4cm.
ii) 8cm.
iii) 10 cm.
iv)
5cm.
6)
Incentre
4)
5)
(i), (iv) eT]jáTT (v) \T
n\οÃD çÜuó„TC²\T
(iii) eT]jáTT (vi) \T n~ó¿ÃD çÜuó„TC²\T
(ii) \+‹¿ÃD çÜuóT„ C²\T
|Ÿ³+ (i) n~ó¿£ ¿ÃD çÜuóT„ È+ eT]jáTT $wŸeTu²VŸQ çÜuóT„ È+
|Ÿ³+ (ii) \+‹¿ÃD çÜuóT„ È+ eT]jáTT $wŸeTu²VŸQ çÜuóT„ È+
|Ÿ³+ (iii) n~ó¿£ ¿ÃD çÜuó„TÈ+ eT]jáTT dŸeTu²VŸQ çÜuó„TÈ+
nuó²«dŸ+ - 5.2
1)
b, d eT]jáTT e ¿=\Ôá\
çÜuóT„ C²“• @sÁÎsÁTkÍïsTT.
4)
S = 500, N = 200, P = 550
5)
6)
|Ÿ³+(i) x = 680 eT]jáTT y = 520
7)
530 8) 800, 300, 700
9)
11) 400, 800, 600
2) 750
3) 650
CUT x = 1100 in NET x = 510
|Ÿ³eTT (ii)ýË, x = 360 eT]jáTT y = 1440
500 , 500 10) i) Ôá|Ÿð ii) ÿ|Ÿð iii) Ôá|Ÿð
12) 10800
nuó²«dŸ+ - 5.3
1)
YXP, ZYQ and XZR
2) ACD = 1330 and DAR = 1200
3)
350, 170
5)
|Ÿ³eTT (i)ýË, x = 1000, y = 1450; |Ÿ³eTT (ii)ýË x = 1300, y = 500
4) 500
nuó²«dŸ+ - 5.4
2)
i)
ii)
3)
3, 5, 6, eT]jáTT 5, 6, 9 dŸeTÖVŸä\ çÜuóT„ C²\“ @sÁÎsÁTkÍïsTT. m+<ŠT¿£q>± s
 +&ƒT uóT„ C²\ yîTTÔá+ï
eTÖ&ƒe uóT„ È+ ¿£+fÉ mÅ£”Øe.
3 , 6 , 9 çÜuóT„ C²“• @sÁÎsÁ#<á TŠ .m+<ŠT¿£q>± s +&ƒT uóT„ C²\T 3™d+.MT, 6™d+.MT\ yîTTÔá+ï 9¿ì dŸeÖq+
nsTTq~.
650, 4.3 cm.
4) 500, 500
5)
740 , 550
6)
(i), (iii) is true
nuó²«dŸ+ - 5.5
3
ÿ|ŸÚÎ
ii) Ôá|ڟ Î, \+‹¿ÃD çÜuóT„ È+ jîTT¿£Ø \+‹ ¿
¹ +ç<Š+ \+‹¿ÃD+ –q• osÁü+ e<ŠÝ –+³T+~
iii) ÿ|ŸÚÎ
iv) ÿ|ŸÚÎ
v) Ôá|ڟ Î, çÜuóT„ È+ýË \+‹ dŸeT~ÇK+&ƒq s
¹ K\ $T[Ôá_+<ŠTeÚ |Ÿ]eÔáï ¿¹ +ç<Š+ neÚÔáT+~.
PT eT<óŠ«>·Ôá ¹sK, PS –q•Ü
i) >·TsÁTÔáÇ¿¹ +ç<Š+ ii) \+‹ ¿
¹ +ç<Š+ iii) n+ÔáseY Ôáï ¿¹ +ç<ŠeTT iv) |Ÿ]eÔáï ¿¹ +ç<ŠeTT
4)
i)
2:1
ii) 2:3
iii) 3:1
iv)
1:2
5)
i)
4cm.
ii) 8cm.
iii) 10 cm.
iv)
5cm.
6)
n+ÔáseÁ Ôáï ¿¹ +ç<ŠeTT
1)
2)
i)
Exercise - 5.6
1)
i)
AB and DE, BC and EF, CA and FD
ii) A and D, B and E, C and F
2)
i)
figures (iv) and (vi) are correct;
ii) figures iii) and v) are not correct
3)
i)
AB  FD = 2cm. ; BC  DE = 3cm. ; AC  FE = 4cm.; ABC  FDE
ii)
PQ  YZ = 4.5 cm.; QP  YX = 3.2 cm.; PR  ZX = 4.8 cm. PQR  YZX
iii) AB  DC = 3 cm. ; AD  CB = 5 cm. BD  DB (Common side) ABD  CDB
iv) PQ=TR ; PR=TQ ; QR = RQ ; PQR  TRQ
4
i)
XZ=KL = 3.5cm ; ZY=LM = 5cm. ; KLM = XZY = 45º ; KLM  XZY
ii)
ED=FD = 4.8cm. ; DG = GD (Common side) ; EDG = FDG = 42º; EDG  FDG
iii) QR = SP = 5cm. ; RP = PR (Common side) ; QRP = SPR ; = SPR  QRP
iv) i)
5
MN = JI = 3cm. ; NO = IH = 3.8cm. ; MNO = JIH = 120º ; MNO  JIH
i)
X = T = 95º ; Z = S = 40º ; XZ = TS = 4cm. ; MNO  JIH
ii)
N = P = 50º ; NOM = POQ = 75º ; NO = PO = 4cm. ; NOM  POQ
iii) F = T = 118º ; G = R = 26º ; FG = TR = 5cm. ; FGE  TRS
iv) SPR = QPR = 50º ; PRS = PRQ = 45º ; PR = PR ; SPR  QPR
6
i)
SR = GE = 4cm. ; SQ = GF = 3cm. ; Q = F = 90º ; SQR  GFE
ii)
BC = DF = 4.5cm. ; AB = ED = 2.8cm. ; A = E = 90º ; ABC  EDF
iii) XY = ZX = 4.5cm. ; XP = XP = 2.8cm. ; XPY = XPZ = 90º ; XPY  XPZ
iv) CA = AC ; DC = BA = 2.8cm. ; ADC = ABC ; ADC  CBA
Unit Exercise
1)
Only one right angle
2)
3)
No, it is possible only in acute angle triangle
4)
i)
3cm, 7cm, 5cm
5)
i)
Right angle triangle : 900, 200, 700 and 900, 350, 550
ii)
Obtuse angle triangle : 1200, 200, 400 and 1000, 300, 500
ii)
XY
3 cm, 9 cm, 11 cm
iii) 3cm, 7cm, 9cm
iii) Acute angle triangle : 550, 650, 600 and 850, 550, 400
6)
A line which passes through centroid, incentre, circumcentre and orthocentre
7)
2:1 from the vertex
8)
orthocentre, circumcentre
AB eT]jáTT DE, BC eT]jáTT EF, CA
nuó²«dŸ+ - 5.6
eT]jáTT FD ii) A eT]jáTT D, B eT]jáTT E, C eT]jáTT F
(iii) ¿±<ŠT (iv) neÚqT (v) ¿±<ŠT
(vi) neÚqT
1)
i)
2)
(i)
neÚqT
3)
i)
AB  FD = 2™d+.MT. ; BC  DE = 3™d+.MT.; AC  FE = 4™d+.MT.; ABC  FDE
ii)
PQ  YZ = 4.5™d+.MT.; QP  YX = 3.2™d+.MT.; PR  ZX = 4.8™d+.MT., PQR  YZX
(ii) ¿±<ŠT
iii) AB  DC = 3 ™d+.MT.; AD  CB = 5™d+.MT. BD  DB (–eTˆ& uóT„ ÈeTT) ABD  CDB
iv) PQ=TR ; PR=TQ ; QR = RQ ; PQR  TRQ
4
i)
XZ=KL = 3.5™d+.MT. ; ZY=LM = 5™d+.MT. ; KLM = XZY = 45º ; KLM  XZY
ii)
ED=FD = 4.8™d+.MT.; DG = GD (–eTˆ& uóT„ ÈeTT) ; EDG = FDG = 42º; EDG  FDG
iii) QR = SP = 5™d+.MT. ; RP = PR (–eTˆ&
iv) i)
5
MN = JI = 3™d+.MT. ; NO = IH
uó„TÈeTT) ; QRP = SPR ; = SPR  QRP
= 3.8™d+.MT.; MNO = JIH = 120º ; MNO  JIH
i)
X = T = 95º ; Z = S = 40º ; XZ = TS = 4™d+.MT. ; MNO  JIH
ii)
N = P = 50º ; NOM = POQ = 75º ; NO = PO = 4™d+.MT. ; NOM  POQ
iii) F = T = 118º ; G = R = 26º ; FG = TR = 5™d+.MT. ; FGE  TRS
iv) SPR = QPR = 50º ; PRS = PRQ = 45º ; PR = PR ; SPR  QPR
6
i)
SR = GE = 4™d+.MT. ; SQ = GF = 3™d+.MT. ; Q = F = 90º ; SQR  GFE
ii)
BC = DF = ™d+.MT. ; AB = ED = 2.8™d+.MT. ; A = E = 90º ; ABC  EDF
iii) XY = ZX = 4.5™d+.MT. ; XP = XP = 2.8™d+.MT. ; XPY = XPZ = 90º ; XPY  XPZ
iv) CA = AC ; DC = BA = 2.8™d+.MT. ; ADC = ABC ; ADC  CBA
jáT֓{Ù nuó²«dŸ+
1)
3)
4)
5)
6)
7)
ÿ¿£ \+‹ ¿ÃD+
2) XY
¿±<ŠT, n\οÃD çÜuóT„ È+ýË eÖçÔáyTû @sÁÎ&ƒTÔáT+~
i) 3™d+.MT., 7™d+.MT., 5™d+.MT. ii) 3 ™d+.MT., 9 ™d+.MT., 11 ™d+.MT. iii) 3 ™d+.MT., 7™d+.MT., 9™d+.MT.
i) \+‹¿ÃD çÜuóT„ È+ : 900, 200, 700 eT]jáTT 900, 350, 550
ii) n~ó¿£ ¿ÃD çÜuóT„ È+ : 1200, 200, 400 eT]jáTT 1000, 300, 500
iii) n\οÃD çÜuóT„ È+ : 550, 650, 600 eT]jáTT 850, 550, 400
>·TsÁTÔáÇ¿¹ +ç<Š+, \+‹ ¿¹ +ç<Š+, n+Ôás¿¹Á +ç<Š+, |Ÿ]eÔáï ¿¹ +ç<Š+\ <‘Çs bþjûT s¹ K
osÁü+ qT+& 2:1
8) \+‹ ¿
¹ +ç<Š+, |Ÿ]eÔáï ¿¹ +ç<Š+
9)
orthocentre, circumcentre 10) 620 , 1180
11) 720 , 180 , 900
12) 200
13)
Scalene triangle
Isosceles triangle
Equilateral triangle
Acute triangle
400, 600, 800
500, 500, 800
600, 600, 600
Right triangle
900, 400, 500
900, 450, 450
—
Obtuse triangle
1100, 400, 300
1000, 400, 400
—
14) i)
ii)
SAS Congruency LNM  PQO
ASA Congruency PAS  RTC
iii) RHS Congruency FGH  IJK
iv) SSS Congruency ABC  MKL
CHAPTER - 6
Exercise - 6.1
1)
i)
3)
7
iii)
17
20
ii)
15
38
4)
x+1
1)
3)
i) 2
No mode
ii)
4)
1.
4)
i) 7
0.25
ii)
5)
1)
2)
3)
4)
i) 2016-17
Double Bar Graph
i) Education
Pie chart
ii)
7.5
5)
Exercise - 6.2
B and K
iii)
T , Uni modal data
Exercise - 6.3
19
2)
7
Exercise - 6.4
2015-16
iii)
ii)
food
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1)
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6)
Unit Exercise
73, 64 Kohli has better average than Rohit
2)
x = 2 and the observations are 14, 10, 6, 4, 2
5)
Pie chart
iii)
2)
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No mode
2)
2
146
3)
7
3)
7, 5.5
120
+
38 and 43
Double Bar Graph
9)
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10) 620, 1180
11) 720 , 180 , 900
12) 200
13)
$wŸeT u²VŸQ çÜuóT„ È+
14)
dŸeT~Çu²VŸQ çÜuóT„ È+ dŸeTu²VŸQ çÜuóT„ È+
n\οÃD çÜuóT„ È+
40 , 60 , 80
\+‹¿ÃD çÜuóT„ È+
900, 400, 500
n~ó¿£ ¿ÃD çÜuóT„ È+
1100, 400, 300
i) uóT„ .¿Ã.uóT„ . dŸsÇÁ dŸeÖqÔáÇeTT LNM  PQO
ii) ¿Ã.uóT„ .¿Ã dŸsÇÁ dŸeÖqÔáÇeTT PAS  RTC
iii) \+.¿£.uóT„ . dŸsÇÁ dŸeÖqÔáÇeTT FGH  IJK
iv) uó„T.uó„T.uó„T. dŸsÁÇdŸeÖqÔáÇeTT ABC  MKL
0
0
0
500, 500, 800
600, 600, 600
900, 450, 450
—
1000, 400, 400
—
n<ó‘«jáT+ - 6
nuó²«dŸ+ - 6.1
ii)
15
iii)
17
20
4)
7.5
5)
x+1
‹VŸQÞø¿+£ ýñ<TŠ
ii)
4)
B eT]jáTT K
T , @¿£ ‹VŸQÞø¿+
£
iii)
1)
4)
i) 7
0.25
ii)
5)
19
7
1)
2)
3)
4)
i)
s +&ƒT ¿£MTˆ\ s¹ U ºçÔá+
i) #á<TŠ eÚ
™|Õ ºçÔá+
ii)
ii)
1)
4)
6)
73, 64 sÃV¾²ÔY ¿£+fñ ¿ÃV¾²¢ dŸsdŸ] mÅ£”Øe
x = 2 eT]jáTT |Ÿ]o\q\T 14, 10, 6, 4, 2
™|Õ ºçÔá+
1)
i)
3)
38
1)
3)
i)
7
2
2016-17
2)
`625
‹VŸQÞø¿+£ ýñ<TŠ
2)
2
2)
146
3)
7
2015-16
iii)
120
€VŸäsÁ+
iii)
`3000
3)
7, 5.5
nuó²«dŸ+ - 6.2
<ŠÔï+Xø+
nuó²«dŸ+ - 6.3
nuó²«dŸ+ - 6.4
jáT֓{Ù nuó²«dŸ+
2)
5)
38 eT]jáTT 43
+
s +&ƒT ¿£MTˆ\ s¹ U ºçÔá+