Extended Mathematics for IGCSE
Similarities
Author- Kunal Sethi
Produced by- Aga Khan Productions
Similar Polygons
Definition: Similarity
Given two polygons, if corresponding angles are congruent, and
corresponding sides are proportional, then the correspondence is called
similarity, and the polygons are then said to be similar.
In simpler terms, similar polygons are any two polygons with equal
corresponding angles and with the same shape (ratio of corresponding sides are
equal). “~” is the sign for “is similar to”.
If triangles ABC and DEF are similar, then we could also write ABC~DEF. For
two polygons to be similar:
1) corresponding angles must be equal
2) the ratio of corresponding sides must also be equal
In the case above, for ABC to be similar to DEF, the above facts must be
true. Let us see if they are really similar. The following can be said about the
two triangles:
1) L A=L D, LB=LE and LC=L F
2) AB = BC = CA 8 = 6 = 10 = 2
DE EF FD , 4 3 5
1
As all corresponding angles and the ratio of each pair of corresponding sides is
2:1, we can conclude that ∆ABC~∆DEF.
NB: It is possible for a polygon to have one of the facts true and the other fact
may not be true. If so, the two polygons aren’t similar as they do not follow
both the facts.
Example 1: With corresponding sides having same ratio but corresponding
angles not equal
G
8
J
8
K
N
70º
110º
6
6
6
6
70º
H
8
I
L
110º
M
8
Above, we can notice that the ratio of corresponding sides is 1 : 1
GH = HI = IJ = JG 6 = 8 = 6 = 8
KL LM MN NK , 6 8 6 8
But the corresponding angles are not equal (90º ≠ 70º, 90º ≠ 110º). So these
two polygons cannot be similar.
Example 2: Polygons with corresponding angles being equal but unequal ratio
of corresponding sides. They could even have the same shape!
O
7
4
P
7
R
S
4
4
Q
T
12
V
4
12
U
The two polygons above are both rectangles and their corresponding angles
are all 90º and equal. But the ratios of the corresponding sides are not equal.
OP ≠ PQ ≠ QR ≠ RO 4 ≠ 7 ≠ 4 ≠ 7
ST TU UV VS , 4 12 4 12
So these two rectangles aren’t similar as they don’t follow they don’t follow the
definition of similar polygons i.e don’t follow the two facts of similarities.
Congruent Polygons
Definition: Congruent
If two polygons are exactly the same in shape, their corresponding lengths and
angles are equal, then two the two polygons are said to be congruent.
Students often confuse between similar and congruent polygons. In simpler
terms, congruent polygons are exactly same; one can exactly fit over the other
while similar polygons must have proportional sides. The sign for “is congruent
to” is “ “.
Example 1: They could be up-side-down
The corresponding sides are equal (AB=EF, BC=FG, CD=GH and DA=HE).
Also the corresponding angles are equal LA=LF, LB=LE, LC=LH and LD=LG.
Although being up-side-down, as long as they have equal corresponding sides
and angles, they are congruent.
Example 2: One could be a reflection of the other
Although the above two triangles are a reflection of another, they are
congruent as they have equal corresponding sides and angles.
Special Case: Similar Triangles
Triangles are polygons so to prove that they are similar, their corresponding
angles should be equal and corresponding sides should be proportional.
Though, in the case of triangles, these facts can be followed but they are a bit
time consuming. For triangles, if they follow any of the two rules of similarity,
they are similar. Rules and examples below might help you get a better
understanding of the concept.
Rule 1: If two angles of one triangle are congruent to two angles of another
triangle, then the triangles are similar. It is called the AA (angle-angle) rule.
The rule above is because if you have two congruent angles both triangles, then
third one also has to be congruent as the sum of the two and subtracting from
180° to get the third angle will be the same for both triangles.
In ∆C
LA + LB + LC = 180°
40° + LB + 60°= 180°
LB = 180° - 100°
LB = 80°
In DEF
LE=80° as 180°-(60°+40°)= 80°. We can notice that LA=LD, LB=LE and LC=LF so
∆C~∆DEF. Just by having two angles, we can prove whether the triangles are
similar or not (AA rule).
Having known ∆C~∆DEF, we can also say.
AB = BC = CA
DE EF FD
Example 1:
LB LD, which means they are equal. Since angle DCE and ACB are vertical
angles, they are also equal and thus 30˚. Since you have two corresponding
angles, the third angle can be found.
LA + LB +LC = 180˚
LA + 50˚ + 30˚ = 180˚
LA = 100˚
LE can also be found in the same way. If you will find for LE using LD and
LC, you will get it as 100 also. So this can prove that ∆C~∆DEF.
Example 2:
As stated above, lines LM and MN are equal for ∆ LMN which means LL and
LN are equal. In ∆OPQ, lines OP and PQ are equal so LO and LP are equal.
In ∆ LMN
As LL and LN are equal, let’s name them both “x”.
40 + 2x = 180
2x = 140
x = 70
For ∆OPQ, the same method above can be used making LO and LP “x”. You
will get that also LO and LP are 70. This again proves that ∆ LMN~∆OPQ.
Rule 2: If two corresponding sides of two triangles are proportional, then
the triangles are similar. This is called the SS (side-side) rule.
Let’s see if the above theory is true or not using the following examples.
Example 1:
∆RST is a right angled triangle. So we can find side RT using Pythagoras
theorem.
(RS)² + (ST)² = (RT)²
6² + 8² = (RT) ²
36 + 64 = (RT) ²
=
RT = 10
∆UVW is also a right angled triangle. We can also find UW using Pythagoras
theorem.
(UV)² + (VW)² = (UW)²
3² + 4² = (UW) ²
9 + 16 = (UW) ²
=
UW = 5
If they are really similar, the ratio of corresponding sides should be equal.
RS = ST = RT
6 =8 = 10
=2
UV VW UW
3 4 5
1
The corresponding sides of the two triangles are proportional; the ratio of their
corresponding sides is 2:1 and so we can say that ∆RST~∆UVW and this proves
the SS theorem for right angle triangles.
Example 2: It is not a must the two triangles should be right angles. We can
even prove the SS rule by using ordinary triangles and in this case we have to
use either sine or cosine rule. But at least one angle and two sides are needed.
We know that ∆C~∆MNO as two sides are equal (SS rule). But to prove it
really works, then following can be done.
In ∆C, by using the sine rule, we can find side LB.
Sine A = Sin B
a
b
Sin 80 = Sin B (cross multiply)
10
8
Sin B = 0.787…
B = 52
If ∆C~∆MNO, according to the AA rule (rule 1), at least two angles should
be equal. We have all ready found that LB = 52. If LN also equals 52, then
∆C~∆MNO.
Let’s find LN
Sin 80 = Sin N (cross multiply)
5
4
Sin N = 0.787…
N = 52
LN = LB, which means ∆C~∆MNO as all three angles are equal. So this has
proved that using the SS rule (rule 2) also works i.e having only two sides (in
this case also on angle) to find whether triangles are similar or not.
Perimeters for Similar Polygons
The perimeter of a polygon is basically the sum of all sides. The perimeters
of two polygons can also be compared. If the ratio of corresponding sides of
two polygons is 2 : 3, then their perimeters’ ratio will also be 2: 3.
Rule 3: If two polygons are similar, the ratio of their perimeters equals the
ratio of the lengths of any pair of corresponding sides,
This can also be if the sides are multiplied by a positive number, then the
perimeter is also multiplied by the same number.
This can be proved by the following example.
Example:
Lets check whether ABCD~EFGH by finding the ratio of corresponding sides
6 = 10 = 9 = 8
=2
3 5 4.5 4
1
They all come up to a ratio of 2:1 which proves that ABCD~EFGH. If rule 3 is
true, then the perimeters of the two quadrilaterals should also have the ratio
2:1.
Perimeter of ABCD = 6 + 10 + 9 + 8 = 33
Perimeter of EFGH = 3 + 5 + 4.5 + 4 = 16.5
33 = 2
16.5 1
As the ratio of the two perimeters is also 2:1, rule 3 is proved correct. That if the
ratio of corresponding sides is x:y, then the ratio of the perimeters is also x:y.
Let’s say we multiply the sides by 3, will the perimeter also be multiplied by 3?
3 (6 + 10 + 8 + 9) = 99 = 3 (33)
Areas for Similar Polygons
The areas of two similar polygons can also be related. If the ratio of
corresponding sides of two polygons is 2:3, then their areas’ ratio will also be
22: 32.
Rule 4: The ratio of the areas of similar polygons is equal to the ratio of the
squares of the corresponding sides’ ratio.
The above rule can be proved and explained better in the following examples.
Example 1:
Figures
are in cm
Area = length x base
Area of PQRS = 6 x 4
= 24cm2
Area of TUV W = 9 x 6
= 54cm2
The ratio of corresponding sides are 2:3
6=4 =2
9 6
3
According to rule 4, if the corresponding sides’ ratio is 2:3, then the ratio of
the areas should be 22: 32 (4:9). Let’s see if this is so.
24 = 4 = 22
54 9 32
So this proves rule 4 right. That if the sides have a ratio of x:y, the ratio of the
areas will be x2:y2. This even further proves that PQRS~TUVW
Example 2:
Figures
are in cm
Area = ½ x base x height
Area ∆C = ½ x 15 x 5
= 37.5 cm2
Area ∆MNO= ½ x 9 x 3
= 13.5 cm2
Corresponding sides of the triangles are in a 5:3 proportion.
15 = 5 = 5
9 3
3
We found the areas to be 37.5 cm2 and 13.5 cm2. According to rule 4, they
should be in the proportion 52:32 (25:9).
37.5 = 25 = 52
13.5 9
32
We can also conclude that ∆C~∆MNO as it fulfills the rule 4 conditions of
similar polygons.
Special Case: Similar Circles
Circle Perimeters/circumferences:
Circles’ radii and circumference can also be compared.
Rule 5: The ratio of the circumferences of two circles is equal to the ratio of
the lengths of their radii or the ratio of the lengths of their diameters.
This can be simply said as “the circumference of a circle varies directly with
its radius or diameter. If this is so, then all circles are similar as the only variable
factor is their radius or diameter which affects the circumference and the radii
or diameters can be in any ratio or proportion. This means that if the radii or
diameters are multiplied or divided by a positive number, then the perimeter is
also multiplied by the same number.
Example:
Figures
are in cm
Circumference = Pie x radius
Circle A circumference = pie x 15 = 47.12cm
Circle B circumference = pie x 9 = 28.27cm
47.12 = 15 = 5
28.27 9
3
This proves rule 5 right. The ratio of their radii and circumferences is the same.
What if we multiply by the radii 2? Will the circumferences also be multiplied
by two as mentioned previously?
Circle A = pie x 30 = 94.24cm = 2 (47.12)
Circle B = pie x 18 = 56.55cm = 2 (28.27)
Circle Areas :
Rule 6: The ratio of two circles’ areas is equal to the ratio of the squares of
the radii or diameter
This can also simply be said as the area of a circle varies directly with square of
the radius or diameter. This means if the radii are multiplied by a positive
number, then the area is multiplied by the square of that number.
Example :
Figures
are in cm
Area of a circle = pie x radius2
Area of Circle A = pie x 42 = 50.27cm2
Area of Circle B = pie x 32 = 28.27cm2
The ratio of the circles’ radii is 4:3. So the ratio of their areas should be 4 2:32
(16:9) according to rule 6.
50.27 = 16 = 42
28.27 9 32
As mentioned, if the radii are multiplied by a number, then the area is
multiplied by the square of that number. Does this actually happen? Let’s
multiply the radii by 3.
Circle A = pie x 122 = 452.39cm2 = 32 (50.27)
Circle B = pie x 92 = 254.47cm2 = 32 (28.27)
Similar 3D Shapes
3D shapes are basically shapes that have three dimensions i.e width, length and
height. 3D shapes or solid objects can also be compared and related to each
other. But one has to be an accurate enlargement of the other.
Solid Volumes:
Rule 7: The ratio of the volumes of similar solids is equal to the ratio of the
cubes of the corresponding sides’ ratio.
If a solid has sides to the ratio 2:3, then the ratio of their volumes will be 23 :33
(8: 27). Examples below explain this better.
Example 1: Cubes
Figures
are in cm
Volume = length x width x height
Volume of cube 1 = 8 x 4 x 6 = 192cm3
Volume of cube 2 = 4 x 2 x 3 = 24 cm3
The ratio of the cubes sides is 2:1. According to rule 7, the ratio of their
volumes should be 23:13 (8:1)
192 = 8 = 23
24 1 13
We can simplify and say, if a cube has the sides’ ratio as x3:y3, then the ratio of
their volumes is x:y. This proves Rule 7.
Example 2: Spheres
Figures
are in cm
Volume = 4/3 x pie x radius3
Volume of sphere 1 = 4/3 x pie x 43 = 268.08cm3
Volume of sphere 2 = 4/3 x pie x 33 = 113.1cm3
According to rule 7, the ratios should be 43:33 (64:27).
268.08 = 64 = 43
113.1
27 33
What if the radii were multiplied by 2? What would happen?
Sphere 1 = 4/3 x pie x 83 = 2144.66cm3 = 23 (268.08)
Sphere 2 = 4/3 x pie x 63 = 904.78cm3 = 23 (113.1)
So we can conclude that “if the sides or lengths are multiplied by a positive
number, then the volume is multiplied by the cube of that number.
Solid Areas:
Rule 8: The ratio of the areas of similar solids is equal to the ratio of the
squares of the corresponding sides’ ratio.
Example 1: Cylinders
Figures
are in cm
Area = 2 x pie x radius x height
Area of cylinder 1 = 2 x pie x 6 x 21 = 791.68cm2
Area of cylinder 2 = 2 x pie x 2 x 7 = 87.96cm2
791.68 = 9 = 32
87.96 1 12
This has also proved rule 8 correct. As in previous cases, does the area also be
multiplied by the number you multiplied the sides or lengths with? Let’s try
multiplying by 3.
Cylinder 1 = 2 x pie x 18 x 63 = 7125.13cm2 = 32 (791.68)
Cylinder 2 = 2 x pie x 6 x 21 = 798.68cm2 = 32 (87.96)
So the same rule applies everywhere u multiply the sides of a shape with any
number.
Revision Exercise:
1. ∆C~∆XYZ. Find the sides and angles marked by a letter.
2. Prove that ABCD~EFGH
3. Two similar polygons vary in the ratio 3:4. Find
i) The side of the larger polygon if the corresponding side of the smaller
polygon is 9 cm. Find
ii) The perimeter ratio of the two
iii) The area ratio between the two
iv) The new perimeter when the old perimeter is 42cm and the sides are
multiplied by 3
v) The new area if the sides of the two polygons are multiplied by 3 and the old
area was 108cm2
4. If two triangles are similar in a 2:3 proportion, and the lengths of the smaller
rectangle are 6cm and 9cm, find the area of the larger rectangle and then prove
rule 4.
5. The ratio of the circumferences of two circles is 3:2 and the smaller circle has
a radius of 8. Find the length of the larger circle.
6. The ratio of the area of two circles is 9:4. Find the ratio between their radii
and state how many times is the length of the radius of the larger circle than
that of the smaller circle.
7. Two similar solid sphere have the volume of 269.08 cm3 and 113.1 cm3. If
their radii are 4cm and 3 cm respectively, then prove rule 7.
8. Two similar cylindrical tins have the base radii 6cm and 8 cm respectively. If
the area of the larger tin is 1508cm2, find
i) The height of the larger tin
ii) The area of the smaller tin without using its height.