SIMILARITIES (
)
When we talk about similarities, we can easily relate it to our daily lives.
Lets take an example of cars of the same brand, Nissan Patrol and Nissan
safari. According the photo below, they look similar.
Similarities in triangles.
P
A
B
Q
C
R
We now may say that triangle ABC and the triangle PQR are similar because they both
have the same shape, although they are not congruent( discussed later).
Triangle ABC and triangle PQR would appear similar if we looked at it through a
magnifying glass.
Thus we may state that two similar triangles have the three angles of one triangle
respectively equal to the three angles of the other triangle, but the corresponding sides are
NOT equal.
It follows that similar trangles are equiangular and hve theri corresponding sides
propotional. Equiangular means the angles of the two triangles are equal.
Example :
A
P
find
-PQ
-PR
X
y
3cm
B
𝑃𝑄
4cm
C
Q
8cm
𝑄𝑅
𝑃𝑅
𝐴𝐵 𝐵𝐶
𝐴𝐶
𝑋
3
=
=
=
𝑄𝑅
𝐵𝐶
8
𝑌
4
5
X=8×3÷ 4
X=6CM
R
=
8
4
Y= 8× 5 ÷ 4
Y=10CM
Triangle ABC is similar to triangle PQR.
Why ?
This is because the trinagles ABC and PQR are equiangular hence their corresponding
sides are proportional.
TYPES OF SIMILARITIES OF TRIANGLES.(THEORUMS)
1. SIDE ANGLE SIDE TRIANGLES. (SAS)
In, this type of similarities, the triangles are similar when two sides are corresponding with
the angle included, so ‘if one angle of a triangle is congruent to one angle of another triangle
and the sides that include those angles are proportonal, then two triangles are similar.
For the triangles below, <YXZ and <PQR are equal and that is why these triangles are
called to be similar.
Q
Y
c
d
a
P
X
b
R
Z
Example :
A
X
6CM
B
10CM
4CM
C
Y
FIND YZ
𝐴𝐶 𝐵𝐶
=
𝑋𝑍 𝑌𝑍
6 10
=
4
𝑋
X=4× 10 ÷ 6
X=6.7CM
x
Z
2.ANGLE ANGLE SIDE TRIANGLE (AAS)
In this type of similarities, the triangles are similar when they have two corresponding
angles which are equal and has one side.
B
a
Examples no.1
B
Q
x
16CM
12CM
A
C
P
Find the side labelled x.
∆ ABC is similar to
𝐴𝐵 𝐵𝐶
=
𝑄𝑅 𝑅𝑃
𝑋
12
=
16
6
X=16× 12 ÷ 6
X=32CM
∆ QRP
R
EXAMPLE NO.2
A
Q
y
6CM
16CM
x
R
P
3CM
B
6CM
C
FIND X AND Y
𝐴𝐵
𝑄𝑅
𝐵𝐶
𝐶𝐴
= 𝑃𝑅
𝑃𝑄
16
𝑋
𝑌
6
=3
X=6× 6 ÷ 3
𝐵𝐶
= 𝑃𝑅
6
=3
Y=16× 3 ÷ 6
Y=8CM
6
X=12CM
3. SIDE SIDE SIDE TRIANGLES (SSS)
In this type of similarities, the triangles are similar when three sides of the triangle are
corespnding. So, ‘if all pairs of corresponding sides of two triangles are proportional, then triangles are similar.
NOTE : In side side side triangles, their corresponding sides are manified by a certain
factor ,K.
Q
Y
a
b
X
ka
Z
c
kb
P
R
kc
Thus these triangles above are similar because of their corresponding sides.
EXTENSION OF THIS CONCEPT
D
X
4CM
E
3.8CM
5CM
6CM
F
Y
𝐷𝐸
𝑋𝑌
𝑋𝐹
4
2
6
3
= =
5
= =
𝑌𝑍 7.5
𝐷𝐹
𝑋𝑍
=
2
3
3.8
2
5.7
3
=
5.7CM
THEREFORE :
HENCE, <D=<X, <E=<Y, <F=<Z
𝐷𝐸
𝑋𝑌
=
𝑋𝐹
𝑌𝑍
=
7.5CM
𝐷𝐹
𝑋𝑍
Z
2
=
3
< MEANS ANGLE
THUS, BOTH THE TRIANGLES ABOVE ARE SIMILAR.
EXAMPLE :
Y
2CM
X
P
3.8CM
4CM
Z
9CM
Q
13CM
x
R
𝑋𝑌
𝑋𝑍
= 𝑄𝑅
𝑄𝑃
2
9
=
4
X=4× 9 ÷ 2
𝑄𝑅
X=12CM
EXERCISE (1) ON SIMILARITY OF TRIANGLES
FIND THE UNKNOWN VALUE.
1.
A
m
6CM
M
N
B
18CM
C
9CM
2.
A
x
6CM
4CM
B 6CM
C
3.
9cm
x
12cm
9CM
15cm
4.
3cm
y
5cm
10cm
5.
6cm
4cm
4cm
a
6.
6cm
ff
2cm
6cm
7. A tree of height 4m casts a shadow of lenght 6.5m. find the height of the house casting
a shadow of 26m long.
8.A diagram shows theside view ofa swimming pool being filled with water. Calculate
the length x.
15m
3.7m
1.2m
x
1.8m
9. A tree casts a 23cm shadow. At the same time, a 6cm peson casts a shadow 10cm long.
What is the height of the tree ?
SIMILAR POLYGONS
Suppose à triangular figure made of cardboard is held infront of a light bulb. If the
triangle is held parallel to a screen, a shadow of the triangle apperas on the screen.
(This concept Works with all polygons). The shadow on the screen will be larger
than the cardboard triangle, but i twill be of the same shape.
Thus the cardboard triangle and its shadow on the screen are said to be similar.
Note that similar figures have the same shape, but not necessarily the same size.
Definition : similar polygons are polygons for which all the corresponding angles
are congruent and all the corresponding sides are proportional.
To say that corresponding sides are similar polygons are proportional means that
the ratios of their measures are equal.
Lets look at these pairs of polygons that are not similar.
2cm
3cm
3cm
2cm
3cm
3cm
The rhombus and the square are NOT similar. All the corresponding sides are
propotional, but it is not true that all corresponding angles are congruent.
8cm
5cm
4cm
4cm
8cm
4cm
4cm
The two rectangles are not similar. All the
corresponding angles are congruent, but it is not true
that all the corresponding sides are proporonal.
5cm
NOTE : For polygons to be similar, it requires that all corresponding sides are
proportional and all the corresponding angle must be conguent.
Example :1.
Find the value of x and y and measure <p.
Q
T
7
4
6
S
y
V
x
P
R
9
86°
∆𝑆𝑇𝑉 ~∆𝑃𝑄𝑅
4 𝑋
=
6 9
4
6
=
7
Y
Y=6× 7 ÷ 4
X= 4× 9 ÷ 6
X=6
To find angle p, note that <s and <p are corresponding angles thus <p=<s
<p=60°
2.
Z
X
R
Y
P
Q
1. <x=< ?
2.<p=< ?
𝑥𝑧
𝑧𝑦
3. =
𝑞𝑟
Solutions : 1) Q
2) Y
3) RP
?
EXERCISE(2) ON SIMILAR POLYGONS.
1. An airplane has a length of 24m and a wingspan of 32m. A scale model is
made with a wingspan of 12cm. Calculate the model’s length.
2. A tree casts a shadow 24m long on horizontal ground. A vertical post 3m
high casts a shadow 4 m long at the same time of the day. Calculate the
height of the tree.
3. A building 25m high and 20m wide is shown on a tv program. If its image
on the screen is 12cm wide, what is the height of the image on the screen ?
4. A swimming pool and the concrete walk around it form similar rectangles.
The swimming pool is 12 feet wide by 30 feet long. The width of the
rectangle formed by the outer edge of the walk is 18 feet. What is the length
of the outer rectangle ?
5. A photographic slide is 34mm wide and 22mm high. Projected on a screen,
the image of the slide is 85cm wide. How high is the image ?
EXAMPLE 2 :
A tree casts a 23m shadow. At the same time, a 6m person casts a shadow 10m
long. What is the height of the tree ?
T
M
6M
N
A
10M
𝑀𝐴 𝐴𝑁
=
𝑅𝑇 𝐸𝑅
6
10
= 23
X=13.8m
𝑋
E
R
23M
AREAS AND PERIMETER OF
SIMILAR POLYGONS(TRIANGLE)
PERIMETER OF SIMILAR TRIANGLES
b
a
kb
ka
perimeter1=(a+b)2
perimeter2=(ka+kb)2
𝑃2 2𝐾(𝐴 + 𝐵)
=
𝑃1
2(𝐴 + 𝐵)
=K
𝑃2
𝑃1
=K
A
X
r
kr
kq
q
B
C
Y
Z
P
P1=r+q+p
P2=kr+kq+kp
kp
𝑝2
𝑝1
=
𝑘(𝑝+𝑞+𝑟)
𝑝+𝑄+𝑟
=k
The two triangles above are similar, the ratio of the coresponding sides being k.
Perimeter of triangle ABC= p+q+r
Perimeter of triangle XYZ=kp+kq+kr
Perimeter of ∆𝑋𝑌𝑍
𝑘(𝑝 + 𝑞 + 𝑟)
=
𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟𝑜𝑓 ∆𝐴𝐵𝐶
(𝑝 + 𝑞 + 𝑟)
=k
This illustrates to us an important general rule for all similar shapes.
If two figures are similar and the ratio of their corresponding sides is k, then the
ratio of the perimeter is k.
Example :
The ratio of the perimeters of triangles ABC to triangle PQR is 1 :3. Find the side
of ∆𝐴𝐵𝐶 is two of its sides are 9cm and 15cm and the perimeter of triangle PQR is
90cm given that ∆𝐴𝐵𝐶 ~ ∆𝑃𝑄𝑅 ?
IF YOU DOUBLE THE SIDES,
YOU ALSO DOUBLE THE
PERIMETER OF THE SIMILAR
TRIANGLES.
𝑃1 1
=
90
3
90÷ 3 = 30𝐶𝑀
9+15+X=30
X=6CM.
P
6CM
A
9CM
Q
15CM
PERIMETER= 30CM
R
18CM
B
45CM
27CM
C
PERIMETER=90CM
AREAS OF SIMILAR TRIANGLES
A
B
W
X
kb
b
D
a
C
Z
ka
Y
The two rectangles are similar, the ratio of the coresponding sides being k.
AREA OF ABCD= ab
AREA OF WXYZ= ka× 𝑘𝑏 = 𝑘 2 𝑎𝑏
𝐴𝑅𝐸𝐴 𝑊𝑋𝑌𝑍 𝑘 2 𝑎𝑏
=
𝐴𝑅𝐸𝐴 𝐴𝐵𝐶𝐷
𝑎𝑏
=𝑘 2
This illustrates an important general rule for all similar shapes :
If two figures are similar and the ratio of the corresponding sides is k, then the ratio of
their areas is k2.
NOTE : k is sometimes called the linear scale factor.
Examples :
1. Two similar trinagles have areas of 18cm2 and 32cm2 respectively. If the base of
the smaller triangle is 6cm, find the base of the larger triangle.
18
cm2
32cm2
6cm
X cm
2
32
Ratios of areas (k )=18 =
16
9
16
Ratio of corresponding side= √ 9 =4
3
4
base=3 × 6 = 8𝑐𝑚.
2.
A
3
2
x
y
C
B
XY is parallel to BC.
𝐴𝐵 3
=
𝐴𝑋 2
If the area of ∆𝐴𝑋𝑌 is 4cm2, find the area of triangle ABC.
The triangles ABC and AXY are similar.
3
Ratio of corresponding sides (k)= 2
9
Ratio of areas (k2)= 4
9
Area of ABC= 4 × 4 =9cm2
EXERCISE(3) ON AREA AND PERIMETER OF SIMILAR
POLYGONS.
Find the unknown area.
1.
4cm2
3cm
A
6cm
2.
5CM
3CM
A
58CM2
20CM
10CM
3.
A
27CM2
8CM
12CM
FIND THE LENGTHS OF THE SIMILAR SHAPES.
4.
5
CM2
4CM
20
CM2
x
5.
36CM2
4
CM2
6CM
Y
6.
A
D
E
B
C
Given that AD= 3cm , AB=5cm and the area of triangle ADE=6cm2.
Find :
a) The area of∆𝐴𝐵𝐶
b) The area of DECB
6. What is the linear scale factor of perimeter of similar triangles ?
7. What happens to the perimeter, if the sides have been doubled ?
CONGRUENCE (≅)
Two figures are congruent if they have the same size and shape. When two triangles are
congruent, you can fit one on top of the other so that the two figures match exactly. The
sides and angles that match are corresponding parts. The vertices that match are
corresponding vertices.
A
B
E
C
D
F
Corresponding vertices
A corresponds to D
B corresponds to E
C corresponds to F
corresponding parts
< 𝐴 corresponds to <D
AB corresponds to DE
<B corresponds to <E
BC corresponds to EF
<C corresponds to <F
CA corresponds to FD
When you write a congruence statement, list the corresponding vertices in the same
order. Here are all your ways this can be done for the triangles above. Note that all these
congruence statements imply the same correspondence.
∆𝐴𝐵𝐶 ≅ ∆𝐷𝐸𝐹
∆𝐴𝐶𝐵 ≅ ∆𝐷𝐹𝐸
∆𝐵𝐴𝐶 ≅ ∆𝐸𝐷𝐹
∆𝐵𝐶𝐴 ≅ ∆𝐸𝐹𝐷
∆𝐶𝐴𝐵 ≅ ∆𝐹𝐷𝐸
∆𝐶𝐵𝐴 ≅ ∆𝐹𝐸𝐷
Definition : Congruent
triangles are triangles whose vertices can be
made to correspond in such a way that the corresponding parts of
the triangle are congruent.
Every triangle has got six parts- three sides and three angles. By the definition, if two
triangles are congruent, then the six pairs of corresponding parts are congruent.
Also, if the vertices can be matched so that all six pairs of corresponding parts are
congruent, then the triangles are also congruent.
CONGRUENCE THEORUMS
1. SIDE ANGLE SIDE (SAS)
Suppose 2 sticks are attached so that you can vary the angle formed at A. set an angle
of 60°. There is now only one way to complete the triangle. The distance from B to C
has been fixed.
Triangle ABC so formed is congruent to triangle JKL. Both the triangles have 4 and
6cm sides that form a 60°
L
B
4CM
4CM
A
C
60°
J
6CM
6CM
K
60°
The angle formed by the two sides of a triangle is said to be included by those two sides.
In both triangles, the angle is included by the sides measured 4 and 6.
So, ‘If two sides and the included angle of one triangle are congruent to the two sides and
the included angle of a second triangle, then the two triangles are congruent.’
In thes situations, we cannot use SAS.
1.
4
70°
4
5
5
This cannot use SAS because the two pairs of sides are congruent but the included angle
are not congruent.
2.
5
3
30°
5
3
30°
This cannot use SAS because two sides of one triangle are congruent to two sides of the
other triangle. However, the congruent angles are not included angles for those congruent
sides.
Example :
P
S
4
4
R
5
Q
V
5
T
100°
100°
IS ∆𝑃𝑄𝑅 ≅ ∆𝑆𝑇𝑉 𝐵𝑌 𝑆𝐴𝑆? EXPLAIN
PQ≅ 𝑆𝑇
<𝑄 <𝑇
=
𝑄𝑅
𝑇𝑉
∆𝑃𝑄𝑅 ≅ ∆𝑆𝑇𝑉 by SAS.
2. SIDE SIDE SIDE (SSS)
Suppose you form a quadrilateral from four stiks. You will able to change the measures
of the angle without disconnecting the sticks. The figure forme dis not rigid. Howeer,
triangles are rigid figures.
So this theorum states ‘ If three sides of one triangle are congruent to the three sides of
another triangle, then the two triangles are congruent.
Example :
N
Y
Q
P
SHOW THAT ∆𝑄𝑌𝑁 ≅ ∆𝑄𝑌𝑃
QN≅ 𝑄𝑃
YN≅ 𝑌𝑃
YQ≅YQ
THUS ∆𝑄𝑌𝑁 ≅ ∆𝑄𝑌𝑃
BY SSS.
3. ANGLE SIDE ANGLE (ASA)
The side that is a common side of two angles of a triangle is said to be included by those
angles. In the figure below, the 5cm side is included by the 30° and 70° angles.
This figure suggests another way to
prove triangles congruent. If you fix the
measures of two angles and the included
side, there is only one way to complete
the triangle.
70°
30°
5cm
Thus it states ‘If two angles and
the included side of one triangle
are congruent to the two angles
and the included side of another
triangle, then the two triangles
are congruent.
Example:
B
P
A
5
D
5
C
SHOW THAT ∆𝐵𝐴𝑃 ≅ ∆𝐶𝐷𝑃
<𝐴 <𝐷
=
𝐴𝑃
𝐷𝑃
<BPA≅<CDP
∆BPA≅∆CDP
BY ASA
4. ANGLE ANGLE SIDE (AAS)
Consider these two triangles. The parts with the given measures form a combination
AAS- that is two angles and a side not included by those angles. Are they congruent ?
AAS is a theorum and not a postulate.
Thus this theorum states ‘ If two angles and a nonincluded side of one triangle are
congruent to two angles and the corresponding nonincluded side of another riangle, then
the two triangles are congruent.
Example :
For each of the congruent triangles below, write ASA or AAS as the reason for the
congruence.
1.
A
F
D
B
ANS :
C
ASA(ANGLE SIDE ANGLE)
E
2.
I
L
G
K
H
ANS :
J
AAS(ANGLE ANGLE SIDE)
SAS, SSS, ASA AND AAS are the only combinations of the three pairs of
congruent parts that allow you to prove that two triangls are congruent.
AAA AND SSA are not valid congruence combinations.
EXERCISE(4) OF CONGRUENCE(NO
ANSWERS FOR THIS EXERCISE)
WRITE SSS, SAS, ASA OR AAS SHOW WHY THE TRIANGLES ARE
CONGRUENT
1. T
V
W
V
2.
S
K
R
T
M
3.
Z
Y
W
X
4.
P
V
S
WHY IS <PSV CONGRUENT TO <TSV ?
WHY IS ∆𝑃𝑆𝑉 CONGRUENT TO ∆TSV ?
5.
T
G
E
O
D
F
GIVEN THAT DE≅GF AND <E≅<F
PROVE ∆EDO≅∆FGO.
6.
D
C
A
GIVEN THIS ABOVE :
PROVE THAT ∆DAB≅∆BCD
7.
B
H
G
I
J
LIST THE SIX PAIRS OF CONGRUENT CORRESPONDING PARTS IF ∆GHI ≅∆JHI
8. Points L and M on the side YZ of a triangle XYZ are drawn so that L is between
Y and M. Given that XY= XZ and <YXL =<MXZ, prove that YL=MZ
9. PQRS is a parallelogram in which the bisectors of the angles P and Q meet at X.
prove that the angle PXQ is a right angle.
SIMILAR 3D-SHAPES
3D shapes are the shapes which heve three dimensions. These 3 dimensions are the
length, width and depth.
All 3D shapes are solids.
These 3D shapes can be : cube, cuboid, sphere, cylinder, tetrahedron, prism and much
much more.
STEPS TO DERIVE.
1 Similar solids have same shape and all their corresponding dimensions are proportional.
2 So similar solids do not possess the same dimensions.
3 Ratio of surface areas of similar solids is the square of the similarity ratio.
4 Ratio of volumes of similar solids is the cube of its similarity ratio.
VOLUME
when solid objects are similar, one is an accurate enlargement of the other.
If two objects are similar and the ratio of the corresponding ides is k, then the ratio of
their volumes is k3
A line has one dimension, and the scale factor is used once only.
An area has two dimensions, and the scale factor is used twice.
A volume has three dimensions , and the scale factor is used three times.
Example 1 :
3cm
30cm3
6cm
Two similar cylinders have heights of 3cm and 6cm respectively. If the volume of the
smaller cylinder is 30cm3 ,find the volume of the larger cylinder.
6
Ans : ratio of heights(k)=3 = 2
Ratio of volumes k3= 23 =8
Volume of large cylinder= 8×30
=240cm3
EXAMPLE 2 :
Two similar spheres made of the same material have weights of 32kg and 108kg
respectively. If the radius of the larger sphere is 9cm, find the radius of the smaller
sphere.
Consider the ratio of weights as the ratio of volume.
Ans :
32
Ratio of volume (k3)= 108
8
∛27
2
=3
2
= 3 ×9=6cm
SURFACE AREA
If two figures are similar and the ratio of corresponding sides is k, then the ratio of their
surface area is k2
NOTE : This concept of surface area of
similar shapes is the same like the area of
similar shapes. But, the only difference is
that surface area is for 3d shapes while
area is for 2dshapes.
Example :
60cm2
a
5
10
K=
𝟏𝟎
𝟓
𝟏𝟎𝟎
K2= 𝟐𝟓
𝟏𝟎𝟎
𝟐𝟓
×60= 240cm2
EXERCISE(5) OF 3D
SHAPES.(volume/surface area)
FIND THE UNKNOWN VOLUME V.
1.
5
15
20
2. RADIUS=1.2 CM
4.5
V
RADIUS=12CM
V
3.
V
88
3.1
6.2
4.
54
V
8
12
FIND THE LENGTHS MARKED X
5.
7
x
10
270
6.
3
24
2
x
7.
54
16
6
x
8. Two solid metal spheres have masses of 5kg and 135kg respectively. If the
radius of the smaller one is 4cm, find the radius of the larger one.
9. Two solid spheres have surface area of 5cm2 and 45cm2 respectively and the
mass of the smaller sphere is 2kg. Find the mass of the larger sphere.
10. The masses of two similar objects are 24kg and 81 kg respectively. If the surface
area of the larger object is 540cm2, find the surface area of the smaller object.
11. A cylindrical can has a circumference of 40cm and a capacity of 4.8 litres. Find
the capacity of a similar cylinder of circumference 50cm.
12. A container has a surface area of 5ooocm2 and a capacity of 12.8 litres. Find
the surface area of à similar container which has a capacity of 5.4 litres.
ALL THE ANSWERS FOR THE
EXERCISES IS IN THE END.
I HOPE YOU ENJOYED
THE CHAPTER AND WAS
CLEARLY ENGRAVED IN
YOUR MINDS.
ANSWERS
EXERCISE 1
1.12CM
2.9CM
3.6.75CM
4.6CM
5.6CM
6.4.5CM
7.16M
8.10.8M
9.13.8CM
EXERCISE 2
1.9CM
2.18M
3.15CM
4.45FT
5.550MM
EXERCISE 3
1.16CM2
2.14.5CM2
3.12CM2
4.8CM
5.18CM
6.A) 16.7CM2
B) 10.7CM2
EXERCISE 5
1.540CM3
2.4500CM3
3.11CM3
4.16CM3
5.21CM
6.4CM
7.9CM
8.12CM
9.54KG
10.240CM2
11.9.4LITRES
12.2812.5CM2
DONE BY SUHAIL LALJI