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Similarity
What is Similarity?
Similarity means “like something but not exactly the same”. In geometry, similarity
means the same in shape but not in size. Two polygons which have the same shape are
called Similar Polygons. When objects are similar, one is an accurate enlargement of the
other. The symbol for similarity is ~
Let’s see if the following pairs of shapes are similar or not
Pair 1 –
The lengths have doubled
and the corresponding
angles are the same. This
makes these polygons
similar
W
Pair 2 – S
T
X
Z
V
Y
U
In the above pairs the lengths of the sides may be different but if you compare them, their
ratio is the same.
Pair 2 - ST = TU = UV = VS
WX XY YZ ZW
In the above pairs of polygons, the corresponding angles are also equal
Pair 2 – S=W, T=X, Y = U, Z = V
STUV~WXYZ
This leads to the following conclusion –
Two polygons can only be similar if
- The ratios of pairs of corresponding sides must be equal
- Corresponding angles must be equal.
Sometimes it is possible that 2 polygons do not follow the above conditions. Then we say
that the polygons are not similar
Examples
18cm
6cm
27cm
75o
80o
21 cm
Not similar because corresponding
angles are different
9cm
120o
7cm
120o
32 cm
8cm
not similar because ratio of corresponding
sides is different
Questions
Are the following pairs of polygons similar or not?
100 cm
10 cm
50cm
Yes, the above pair of polygons is similar because they obey the conditions
5 cm
102o
70o
6 cm
21cm
130o
90o
88o
100o
60o
16 cm
80o
4 cm
The above shapes are not similar as the corresponding angles and the ratio of
corresponding sides is not the same
Which of the following must be similar to each other?
a) 2 equilateral triangles - similar
b) 2 squares - similar
c) 2 rectangles – must not be similar but could be
d) 2 circles? – must be similar
Now that we have established the two conditions required for polygons to be similar, we
can easily use them to find the missing angle or a missing length of a side of a polygon.
Examples
Below are two similar polygons. Find angle T and length of side QR
P
6cm
5 cm
T
86o
12 cm
10cm
Q
x
S
8 cm
U
W
R
16 cm
14 cm
V
Angle T is 86 because corresponding angles of similar polygons are equal i.e. angle P =
angle T
Side QR can be found in the following way:
We know that ratios of the corresponding sides are the same so
PQ = QR
TU UV
o
5 = x
10 14
Cross multiply: 10x = 70
10 10
So x = 7 cm
Some more questions….
1. Find angle I
C
3 cm
60o
D
G
9 cm
H
80o
7cm
x
21 cm
18 cm
E
100o
F
4cm
E
J
12 cm
I
Answer –
Angle I = angle E
Angle E = 360o-(100o + 60o + 80o) (all internal angles of a quadrilateral add up to 360o)
Angle E = I = 120o
From the above polygon, find the length of side DE
DE (x) =
CD/GH = x/HI
3/9 = x/18
Cross multiply,
54 cm = 9x
6 cm = x
2. In the given pair of similar polygons, the ratio of the corresponding sides is 2:3.
a) Find angle R
b) Find all the sides of polygon QRST
R
6 cm
N
Q
Cross
M
44o
multiplication
can also be used;
5 cm
9 cm
your answer will
be the same.
P
O
T
S
7 cm
Answer – angle N =angle R
Angle N = 44o then angle R = 44o
b) For this you have to multiply each of the lengths of the sides of MNOP by 3/2. We use
3/2 (reciprocal of the scale factor) because we are increasing the value of the sides of
MNOP by this ratio.
Your answers will be
QR = 9cm, RS = 13.5 cm, ST = 10.5cm, TQ =7.5cm
Similar Triangles
We have seen that for polygons to be similar all corresponding angles must be equal and
the ratio of the corresponding sides must be the same. In triangles, this is not necessary.
Let’s see how:
Look at the following triangles,
35o
55o
To find C , you have to add all three angles in the triangle,
A +B+C =180
90+55+C = 180
C = 180 – 145
C= 35
If you look at the second triangle, angle E is 35
We can now say that C = E and we already know that A = D
These triangles can be called similar because two angles of triangle 1 are equal to two
angles of triangle 2.
AA similarity Postulate – If two angles of one triangle are equal to two angles of
another triangle, then the triangles are similar.
If this happens, then by default the third angle is also equal
Also the ratio of the corresponding sides is equal
BA= AC = CB
FD DE EF
Are the following triangles similar? If Yes, show how?
1)
5
2
4
2.5
3
6
The above triangles are similar because all three angles are equal, even though only two
are indicated; the third one is equal by default.
All sides are in the same proportion i.e. (2.5/5) = (2/4) = (3/6)
All these reduce to ½ = 1:2
2)
B
D
4
21
16
9
E
A
7
F
C
6
The above triangles are not equal because the corresponding angles are not equal and the
corresponding sides are not in the same ratio
Areas and perimeters of similar triangles
Perimeter - Sum of All sides Of a Polygon
Relationship between the corresponding sides of similar triangles and their scale factor
6
J
G
2
9
L
3
I
12
4
K
H
All the corresponding angles are equal in the above diagram
6 = 9 = 12
2 3 4
All the above ratios simplify to 3:1
3:1 is also known as the scale factor of the similar triangles
The reduced ratio of corresponding sides of similar triangles is called the scale
factor
Relationship between the scale factor and the perimeter of similar triangles
Using the above example;
Perimeter of Triangle 1 = 27 cm
Perimeter of triangle 2 = 9cm
27 cm = 3 = 3:1 which is the scale factor of the similar triangle
9 cm 1
Ratio of the perimeter is equal to the reduced ratio of corresponding sides
This leads us to the following theorem:
Theorem 60: if the scale factor of two similar triangles is a:b, then the ratio of their
perimeters will also be a:b
Answers a) scale factor =
reduced ratio of
corresponding sides
5/20 = 7/28 = 9/36
All these reduce to ¼
Scale factor = 1:4
Questions….
1) For the following pair of similar triangles
a) find the scale factor of the corresponding sides
b) find the ratio of their perimeters
X
5cm
U
7cm
Y
28cm
20cm
Z
9cm
V
W
36cm
b) Application of
theorem 60. Perimeter
of XYZ = 21cm and of
UVW = 84cm.
Ratio = 21/84 = ¼ =
1:4
2) The scale factor of 2 similar triangles is 3:4. The perimeter of triangle 1 is 24 cm. what
is the perimeter of triangle 2?
Answer: scale factor = ratio of perimeters (theorem 60)
3:4 = 21: x
Cross multiply = 4(21) = 3x
84/3 = 3x/3
28cm = x
Perimeter of triangle 2 = 28cm
Answer –
a) Perimeter of triangle FGH (z) =
Scale factor = 16/4 = 4/1 = 4:1
Triangle CDE has a perimeter of
76cm
4:1 = 76cm: x (cross multiply)
z = 19cm
3) For the following pair of similar triangles,
a) Find the perimeter of triangle FGH
b) Find the remaining sides of triangle FGH
C
F
16cm
4cm
y
28cm
G
D
H
E
x
b) Use scale factor to find
remaining sides
e.g for x, you can do
4/1 = 32/x and cross multiply
x = 8cm
y = 7cm
32cm
4) If triangle ABC has a perimeter of 27cm and triangle EFG has a perimeter of 13.5 cm,
what is the ratio of the perimeter? The sides of triangle EFG are 6 cm, 5cm and 2.5 cm.
find the sides of triangle ABC?
-Answer
Ratio of perimeter = perimeter of ABC/perimeter of EFG
= 27cm/13.5 cm
= 2/1 = 2:1
Sides of triangle ABC = multiplication of sides by the scale factor
2/1 x 6 = 12cm
2/1 x 5 = 10cm
2/1 x 2.5 = 5cm
Area
Look at the below pair of similar triangles;
3cm
24cm
28cm
14 cm
Scale factor = 1:2 (see before on how to get scale factor)
Area of triangle 1 = 42 cm2
Area of triangle 2 = 168 cm2
What is the ratio of the areas? 42 = 1 = 1:4
168 4
Relationship between scale factor and area ratio?
Scale factor = 1:2
1= 1
2 4
area ratio: 1:4
(1/2) = (1/2)2
As you can see, the ratio of the areas is the square of the ratio of the scale factor proving
that:
Theorem 61: if two similar triangles have a scale factor a:b, then the ratios of their
areas is a2:b2
Questions….
1) Two similar triangles have an area of 18cm2 and 32 cm2 respectively; if the base of
the smaller triangle is 6cm, find the base of the larger triangle?
Answer –
Scale factor = (ratio of areas) 2 – theorem 61
32 = (a/b) 2
18
Simplify the fraction = 16/9
Now reduce it by finding the square root.
Your answer is 4/3 and that is your scale factor.
So to find the base, all you have to do is cross multiply;
4:3 = x: 6
x = 8cm
Answer –
x/4 = (20/5)2
x/4 = (4/1)2
find square root of (4/1)2
x/4 = 2/1
x=4x2
x = 8cm
2) Below are two similar triangles. Find x
20cm2
5cm2
4cm
x
3) The perimeters of two similar triangles is in the ratio 5:6. The sum of their areas is 122
cm2
Find the area of each triangle.
Answer –
Let 5x be a side in triangle 1 and 6x be the corresponding side in triangle 2
Perimeter ratio = scale factor
Scale factor = (ratio of areas)2
Therefore the ratio of perimeter = (ratio of areas)2
5/6 = (5x/6x)2 = 25x2/36x2
Area of each triangle =
25x2 + 36x2 = 122 cm2
61x2 = 122cm2
x2 = 2 cm
Area of
Area of
1 = 25x2
= 25 x 2
= 50cm2
2 = 36x2
= 36 x 2
= 72 cm 2
Try This!
Two similar rectangles have been drawn below. Find A (Area) using the concept of
theorem 61
A
8cm
18cm2
6cm
16 cm
3cm
Answer – the process is
the same.
Find the scale factor
which in this case is 8:3
Then square it, you’ll get
64:9.
Then cross multiply
64:9 = A: 18
A = 128cm2
Similarity in 3-D Shapes
We already know that if two shapes are similar their corresponding sides are in the same
ratio, and their corresponding angles are equal.3D shapes can also be similar provided
they follow the same rules
Examples of similar 3D shapes:
a) Below are two cuboids
6cm
3cm
1cm
2cm
2cm
4cm
The length of cuboid 1 is twice the length of cuboid 2
The width of cuboid 1 is twice the width of cuboid 2
The height of cuboid 1 twice the height of cuboid 2
Therefore the scale factor is 2:1, all angles are same and so the cuboids are similar.
b) Two cones can also be similar.
3x
y
slant height
x
3y
radius – 3z
radius - z
The radius of cone 2 is thrice that of cone 1
The height of cone 2 is thrice that of cone1
The slanted height of cone 2 is thrice that of cone 1
All angles are also equal
Therefore, the cones are also similar.
Are the following shapes similar or not?
8cm
24cm
Radius – 6 cm
Radius – 30cm
Below are 2 cubes – are they similar?
a
aa
a
a
b
b
b
No, these shapes are
not similar because the
ratio of the
corresponding sides is
not equal
6/30 is not equal to
8/24.
b
All cubes are similar without any exception. This is because the length, width and height
of a cube is the same. So the ratio of corresponding sides will also be the same i.e
Length = width = height
a= a = a
b b b
Scale factor = a:b
When dealing with 3D shapes, we only talk about 2 operations namely
a) Volume
b) Surface Area
Volume of similar 3D figures
What is the ratio of their volumes?
Cube1 has a volume of a3
Cube 2 has a volume of b3
The ratio of their volumes is a3:b3
What is the ratio of the corresponding sides (scale factor)?
a: b
This means that for a cube of scale factor a: b the ratio of their volume is a3:b3
This leads to the following conclusion: for any similar 3D shapes with a scale factor
of a: b, the ratio of the volumes will be a3:b3
Example:
Look at the shapes below
60cm3
480cm3
5 cm
10cm
The ratio of corresponding sides is 5:10 which when reduced is 1:2
The ratio of the volume is 60:480 which when reduced you get 1:8
Therefore:
1 = 1 which is the same as ½ = (1/2)3
2 8
This proves our conclusion.
The theorem can also be expressed as follows:
Volume 1 = (a/b)3.
Volume 2
Questions
1) Two similar jugs have heights of 4cm and 6cm respectively. If the capacity of the
smaller jug is 50cm3, find the capacity of the larger jug.
Answer –
Scale factor = 4/6 = 2/3 = 2:3
Volume 1 =
a 3
Volume 2
b
50 cm3 = (2/3)3
x
50cm3/x = 8/27
Cross multiply; x = 50 x 27 = 1350
x= 1350/8
x= 168.75 cm3
2) Below are 2 cones. All the dimensions in each are given. Show that they are similar.
(Find volume of each as well)
50cm
25cm
48cm
24cm
Radius = 7cm
radius = 14cm
Answer - the above cones are similar,
Scale factor of the cones – 7 = 24 = 25 = 1:2 (all sides are corresponding to each other)
14 48 50
Volume of cone 1 = 1/3
1/3 x
x 73
1/3 x 343
r3
Volume of cone 2 = 1/3
1/3 x
x 143
1/3 x 2744
r3
V1 = 1x 343
x3
= 343
V2 3
2744
2744
Answer is 1/8 which is the same as (1/2)3, proving that the cones are similar
2) By referring to the similar cylinders below, find V
5 cm
20 cm3
15cm
V
Answer –
Scale factor – 5/15 = 1:3
In volume, scale factor is cubed =
(1/3)3 = 1/27
1/27 = 20/V
V = 20 x 27
V = 540cm3
3) Find the missing length
Answer – Volume 1 = (a/b)3
Volume 2
3
3cm = (2/x)3
24cm3
(3/24) = 8/x3, cross multiply
24 x 8 = 3x3
192/3 = 3x3/3
64 = x3
4 cm = x
24cm3
3c 3
3cm
2cm
x
4) The area of the bases of two similar glasses is in the ratio 4:25. Find the ratio of
their volumes
Answer Your first step is to find the reduced ratio (scale factor)
4:25 is the area ratio that means that it is in the form of (a:b)2
Reduce the ratio to get the scale factor. Reduce it by finding the square root
Your answer will be 2/5
Now use the normal process to find volume ratio
2/5 = (2/5)3
Final ans – 8/125 = 8:125 (ratio of volumes)
Surface Area
This is the same as the process which you use when you look for areas of 2D shapes
Let’s look at an example:
Below are 2 similar cones
50cm
25cm
48cm
24cm
Radius = 7cm
radius = 14cm
The scale factor/dimensions are in the ratio 1:2.
Let’s find the surface area of the cones
S.A of cone = rl+ r2
Cone 1 = 224
Cone 2 = 896
Cone 2/cone 1 = 224 /896
- Pi
= 1:4
(Scale factor) 2 = area ratio
(1/2)2 = 1/4
So if scale factor is 1:2, then the area ratio is (1/2)2 – Theorem 61
Questions….
1) Two similar pyramids have volumes of 64cm3 and 343cm3. What is the ratio of their
surface areas?
Answer
First step is to find the reduced ratio (scale factor)
64/343 is the volume ratio and so it is in the form of (a: b)3
Reduce it by finding the cube root of the numbers.
Your answer will be 4/7
Now use the normal process to find area
a:b = (a:b)2
4:7 = (4:7)2
Final answer = 16:49
2) Two similar cylinders have a scale factor of 3:5. The surface area of the smaller one is
9cm2. Find the surface area of the larger one?
Answer:
Use theorem 61
(3:5) = (3:5)2
Answer = 9/25 or 9:25
So the surface area of the larger one will be 25cm2
3) Two similar cones have surface areas in the ratio 4:9. Find the ratio of;
a) Their lengths
b) Their volumes
Answer –
Same process as used before. Find square root of 4:9 as it is an area ratio
Your answer is 2:3…. (a)
Now cube this ratio because you have to find the ratio of the volumes
Your answer is 8:27….. (b)
4) The ratio of the surface area of two spheres is 4:1. If the surface area of the larger
sphere is 5542cm2. Find the surface area of the smaller one and hence find radius of
each sphere.
5542cm2
xcm2
Answer Find ratio of smaller sphere;
Cross multiplication required
5542/x = 4/1
x = 5542/4
x = 1385.5 cm2
Find radius of each sphere;
Here you will have to use the formula,
Sphere 1 = 4
r2
5542 cm2 = 4
r2
4
4
r2 = 441
r = 21 cm
Sphere 2 = 4
r2
1385.5 cm2 = 4
r2
4
4
110.25 = r2
r = 10.5 cm
Surface Area of
Sphere =
4
r2
5) Two similar open cylinders are painted together. The larger one has a surface area of
256cm2 and the scale factor of the two cylinders is 4:3. Find the surface area of the
second cylinder.
ii) If the radius of the second cylinder is 12cm, what would be the radius of the first?
Answer –
i) surface area ratio = (4:3)2 = 16:9
Cross multiply = 9 x 256 = 2304/16
S. a of second cylinder = 144cm2
ii) Scale factor = 4:3
Radius of small cylinder = 12cm
Radius of large cylinder (x) =?
Cross multiply = 12 cm x 4 = 48/3
x = 16cm
Try This!!
A few revision questions
1)
Are triangles ABC and DBE similar?
ABC and DBE are similar, because all of the corresponding angles are equal and
And all corresponding angles are equal i.e. angle BDE = angle BAC and angle BED =
angle BCA
2) The Easypac Soup Company produces small cylindrical tins of radius 3cm, and similar
large tins of radius 5cm. If a small tin contains 243ml of soup, how much does a large tin
contain?
You need to work out the ratio of the volume;
Volume ratio for a:b = (a:b)3
(3/5) = (3/5)3
= 27/125
Now you can cross multiply
27/125 = 243/x
125 x 243 = x
27
x = 1125 ml
3)
A
33o
55o
40.3 cm
This question also uses a
concept from the circle
theorem. The theorem
states that angles formed
by a chord in the same
sector are equal. In this
case it is angle A = angle
D and angle B = angle C
B
26.8 cm
92o
X
20.1cm
55o
33o
D
a) Find X
b) Calculate XD
Answer –
X = 920, vertically opposite angle
XD = cross multiplication of corresponding sides
AX/XD = BX/XC
40.3/XD = 26.8/20.1
XD = 30.2cm
C
4) Two cones are similar,
8cm
4cm
11 cm
l cm
a) find l
b) When full the larger cone contains 172cm3 of water. How much water does the
smaller one contain?
Answer –
l=
8cm/4cm = 11cm/l cm
L cm = 5.5 cm
For b) let x be the amount of water in the smaller cone
Find scale factor which is 2:1 (8/4 and 11/5.5)
Now cube it to find volume ratio = (2:1)3 = 8:1
Cross multiply
8/1 = 172/x
x = 21.5 cm3