ZEBAR SCHOOL FOR CHILDREN
CLASS IX – MATHEMATICS (2023 – 2024)
Ch.6: LINES AND ANGLES
Worksheet No. 2
ANSWER-KEY
• Choose the correct option:
1. If AOB is a straight line, then x is
(a) 15°
(b) 30°
(c) 45°
(d) 20°
2. If two interior angles on the same side of the transversal intersecting two parallel lines are in ratio
5:4, then the measure of smaller angle is:
(a) 20°
(b) 100°
(c) 80°
(d) 60°
3. In the figure, which of the following statement is true?
(i) a+b=d+c
(ii) a+c+e= 180°
(iii) b+f=c+e
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (ii) and (iii) only
4. AB is the mirror; PQ is the incident ray and QR is the reflected ray. If ∠PQR=108°, then ∠AQP is:
(a) 36°
(b) 30°
(c) 90°
(d) 72°
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5. If two complementary angles are in the ratio 2: 2, then two angles are:
(a) 15° and 75°
(c) 30° and 150°
(b) 23° and 67°
(d) 10° and 170°
In the following questions, a statement of assertion (A) is followed by a statement of reason (R).
Mark the correct choice as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation
of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of
assertion (A).
(c) Assertion (A) is true, but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
6. Assertion: In the figure, if PQ∥RS, then ∠ABC=90°.
Reason: If two parallel lines are intersected by a transversal, then
each pair of alternate interior angles are equal.
Ans: (d) Assertion (A) is false but reason (R) is true.
7. Assertion: The supplement of the angle (120+a−2b)° is (60−a+2b)°
Reason: If the sum of two angles is 180° then the angles are supplementary.
Ans: (a)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation
of assertion (A).
Class IX / Mathematics / Ch.6 / Worksheet-2 / AK /2023-2024
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8. Assertion: The bisectors of the angles of a linear pair form a right angle.
Reason: If the sum of two adjacent angles is 180°, then the non-common arms of the
angles are in a straight line.
Ans: (b) Both assertion and reason are true, but reason is not the correct explanation of
assertion.
9. One of the Math teachers draws a straight-line AB shown on the blackboard.
D
A
a. Now he told Riya to draw another line CD as in the figure.
b. The teacher told Ajay to mark ∠AOD as 2z.
O
c. Suraj was told to mark ∠AOC as 4y.
d. Krina made angle ∠COE = 60°.
e. Peter marked ∠BOE and ∠BOD as y and x respectively.
C
B
E
i) Find the value of y.
ii) Find the value of x.
iii) Find the sum of supplement of ∠AOD and reflex ∠COE.
Solution:
(i)∠AOC + ∠COE + ∠BOE = 180°
( linear pair as Straight line AB)
4y + 60° + y = 180°
=> 5y = 120°
=> y = 24°
(ii)∠COE + ∠BOE + ∠DOB = 180°
( linear pair as Straight line CD)
60° + y +x = 180°
=> 60° + 24° + x= 180°
=>x = 96°
or we could use x = 4y ( vertically opposite angle)
Hence x = 4(24) = 96°
(iii)x + 2z = 180° ( linear pair )
96° + 2z = 180°
=> 2z =84°
=> z = 42°
x = 96
y = 24
z = 42
(iii) supplement of ∠AOD= 180°- 84°= 96°
complement of ∠COE = 90°- 60°= 30°
sum of supplement of ∠AOD and complement of ∠COE= 96°+30°=126°
10. There are two parallel streets PQ and RS in a city as shown in
figure. There are two internal roads AC and AB that connect to both parallel roads. If ∠PAB=70°
and ∠ACS=138°, then
(i) Find ∠ABC.
(ii) Find ∠BAC.
(iii) Find reflex ∠CAQ.
Solution: (i) ∠ABC= 70° (alternate interior angles)
(ii) ∠BAC= 138°-70°= 68°
(iii) ∠CAQ = 180°- 138° = 42°
reflex∠CAQ =360°-42°= 318°
Class IX / Mathematics / Ch.6 / Worksheet-2 / AK /2023-2024
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11. If the complement of an angle is equal to the supplement of the thrice of it. Find its measure.
Solution: Let the required angle be x°
Then its complement angle = 90°−x
and supplement of thrice of the angle= 180°−3x
∴(90°−x) =180°−3x
⇒3x−x=180°−90°
⇒2x=90°⇒2x=90°
⇒x=45°
∴ Required angle = 45°
12. In the figure, OD is the bisector of ∠AOC, OE is the bisector of ∠BOC and OD ⊥ OE. Show that
points A, O and B are collinear.
Solution: Given: OD ⊥ OE, OD and OE are the bisectors of ∠AOC and ∠BOC.
To Prove: A, O and B are collinear, i.e., AOB is a straight line.
Since OD and OE bisect angles ∠AOC and ∠BOC respectively, . . . . . . . .(i)
∴
∠AOC=2∠DOC.......(i)
and
∠COB=2∠COE..........(ii)
On adding equations (i) and (ii), we get,
∠AOC+∠COB=2∠DOC+2∠COE
⇒
∠AOC+∠COB=2(∠DOC+∠COE)
⇒
∠AOC+∠COB=2∠DOE
⇒
∠AOC+∠COB=2×90°
O
⇒
∠AOC+∠COB=180°
∴
∠AOB=180°
∠AOC and ∠COB are forming linear pair.
Therefore, AOB is a straight line.
Hence, points A, O and B are collinear.
13. AB, CD and EF are three concurrent lines passing through the point O such that OF bisects
∠BOD. If ∠BOF = 35°, find ∠BOC and ∠AOD.
Solution: Three concurrent lines are: AB, CD and EF
Also, OF is the bisector of ∠BOD and it is given that ∠BOF = 35°.
Therefore,
Also,
Since, ∠BOD and ∠AOC are vertically opposite angles. Therefore,
From (i) equation:
Class IX / Mathematics / Ch.6 / Worksheet-2 / AK /2023-2024
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We know that ∠AOC and ∠BOC form a linear pair.
Thus,
Similarly, ∠AOC and ∠AOD form a linear pair.
Thus,
14. In the figure, if AB II HF and DE II FG, then find the measure of ∠FDE.
Solution: AB ‖ HF and CD cuts them ∠HFC = ∠FDA
(Corresponding angle)
∴∠FDA = 28°
Now, ∠FDA + ∠FDE + ∠EDB = 180°(Angles on straight line)
∴28° + ∠FDE + 72° = 180°
∴ ∠FDE = 80°
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15. In the figure, ∠1=60° and ∠2=( )rd of a right angle. Show that l∥m.
3
Solution: In the figure, a transversal n intersects two lines l and m
∠1=60° and
𝟐
∠2=(𝟑)rd of right angle
𝟐
∠2=(𝟑)rd ×90°=60°
∴∠1=∠2
But there are corresponding angles
∴l||m
16. In the given figure, PO⊥AB. If x: y: z = 1:3:5, then find the degree measure of x, y and z.
Solution: As given in the question that PO is perpendicular to AB.
so, ∠ AOP = 90°
also given that X : Y : Z = 1 : 3 : 5
⇒ ∠ POQ + ∠ QOR + ∠ ROA = 90° = 1x + 3x + 5x = 90°
⇒ 9x = 90
⇒ x = 90/9
⇒ x = 10
Now putting the value of x.
we get, angle X = 10°
angle Y = 3 × 10 = 30°
and angle Z = 5 × 10 = 50°
Class IX / Mathematics / Ch.6 / Worksheet-2 / AK /2023-2024
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17. In the figure, show that AB∥EF.
Solution: Given in figure, AB ||EF
∠BAC=57°,∠ACE=22°
∠ECD=35° and ∠CEF=145°
To prove: AB||EF,
Proof : ∠ECD+∠CEF=35°+145°=180°
But these are co-interior angles
∴EF||CD
Since, 57°=22°+35°
Hence, ∠BAC=∠ACE+∠ECD
∴AB||CD
∴AB||EF (Lines parallel to the same line are parallel to each other)
18. In the figure, AB∥CD. Find the value of x.
Solution: Draw EF ∥ AB ∥ CD.
F
EF∥CD and CE is the transversal.
Then,
∠ECD + ∠CEF = 180° (Angles on the same side of a
transversal line are supplementary)
⇒ 130° + ∠CEF = 180°
⇒ ∠CEF = 50°
Again, EF∥AB and AE is the transversal.
E
Then,
∠BAE + ∠AEF = 180° Angles on the same side of a transversal line are supplementary
⇒ x° + 20° + 50° = 180°
∠AEF = ∠AEC + ∠CEF
⇒ x° + 70° = 180°
G
⇒ x° = 110°
19. In the given figure, if AB∥CD and ∠AFE= 30°, then find ∠FCD.
Solution: Draw FG∥AB ∠GFE + 30° = 90° (co-interior angles)
∠GFE=60°
∠FEB= 180°- 60° = 120° (co-interior angles)
∠FCD = 120° (corresponding angles)
20. In the figure, AB∥CD∥EF and GH ∥ LK, find ∠HKL.
Solution: From figure, CD || GF, so, alternate angles are equal.
∠CHG =∠HGP = 60°
∠HGP =∠KPF = 60° [Corresponding angles of parallel lines are
equal]
Hence, ∠KPG =180° – 60° = 120°
∠GPK = ∠AKL= 120° [Corresponding angles of parallel lines are
equal]
∠AKH = ∠KHD = 25° [alternate angles of parallel lines]
Therefore, ∠HKL = ∠AKH + ∠AKL = 25° + 120° = 145°
Class IX / Mathematics / Ch.6 / Worksheet-2 / AK /2023-2024
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