SIMILAR POLYGONS.
Two polygons with the same shape are called similar polygons. If the two polygons are similar, the
symbol used to show it is ~.
Facts to be considered when two polygons are similar.
Corresponding angles are equal.
The ratios of the corresponding sides have to be equal, meaning that they should be in the same
proportion.
(IF NONE OF THESE APPLY, THEN THE POLYGONS ARE NOT SIMILAR)
P
A
Q
B
Rectangle 1
Rectangle 2
D
C
Rectangle ABCD ~ Rectangle PQRS.
AB = 6 = 1
CD = 6 = 1
BC = 6 = 1
PQ 12 2
RS 12 2
QR 12 2
S
R
AD = 6 = 1
PS 12 2
SIMILAR TRIANGLES.
For proving that two polygons are similar, you have to show that all the angles and the ratios of the
corresponding sides are equal. But for proving that two triangles are similar is that the pairs of the two
corresponding angles of the two triangles should be equal. Notes: if triangle ABC is similar to triangle
QRT then we write ABC ~ QRT
If ABC ~ QRT, then angle B = angle R, angle C= angle T and angle A= angle Q.
BC
.
QR QT RT
P
C
A
B
Q
R
Also AB = AC=
TRIANGLE ABC= TRIANGLE PRQ
THEREFORE= AB/PR = BC/RQ = AC/PQ
P
C
B
y
A
x
R
Given: AC = 9cm, BC=4cm, BP=18cm, RP=15cm. Show that ABC ~ PBR. Hence find x and y.
Solution; from triangles ABC and PBR;
<C = <R (Alternate Angles), <A=<P (Alternate Angles) and <ABC=<PBR (Vertically Opposite Angles)
Therefore triangle ABC ~ triangle PBR (A.A.A)
For similar triangles:
AB = BC= AC
PB BR PR
AC= AC = 9 = y
PR PB 15 18
y= 18 x 9
15
y= 10.8 cm
AC= BC = 9 = 4
PR BR 15 x
x= 15 x 4
9
x= 6.7 cm
Example 2 ;
W
5
w
4
X
3
Z
y
Y
Find w and v.
w= 4
5 3
w= 4/3 x 5
w=6.67 cm
v= 4
4 3
.
v= 4/3 x 4
v=5.3 cm
C
6
B
x
y
(y-1)
D
y = 6
y+6
10
10y=6(y+6)
4
E
6
x
A
= 10
y(9)-1
x= 10/6 x 6
6
10y= 6y+36
x= 13.5 cm
4y=36
Y=9cm
CONGRUENCY.
Two figures are congruent if they fit exactly to each other. They must be of the same size and the same
shape. Two triangles are congruent if they satisfy either one of the following:
(S=SIDE AND A=ANGLE)
1) Two pairs of sides are congruent and the included angles are congruent. (S.A.S)
B
Q
.
C
A
P
R
2) Two pairs of angles and pair of corresponding sides are congruent (A.A.S)
.
A
P
B
Q
C
ABC=PQR
3) Three pairs of corresponding sides are congruent (S.S.S)
.
Q
P
.
R
C
C
B
A
R
4) If two triangles have congruent hypotenuses and one pair of sides congruent, then the
triangles are congruent. (R.H.S.)
.
Example: Triangle LMN is isosceles with LM=LN; X and Y are points on LM, LN respectively
such that LX=LY. Prove that triangles LMY and LNX are congruent.
L
X
M
L
L
Y
N
X
N
Y
.
M
<MLY= <NLX
LX= LY (GIVEN)
LM= LN (GIVEN
So, the two pairs of sides are congruent and the included angles are congruent (S.A.S). This
proves that triangle LMY and triangle LNX are congruent.
Perimeters of similar shapes.
The ratio of perimeters of two similar polygons is equal to the ratio of the corresponding
sides
Similar triangles whose scale factor is 2 : 1.
The ratios of corresponding sides are 4/2, 10/5, 6/3. These all reduce to 2/1. It is
then said that the scale factor of these two similar triangles is 2 : 1. The perimeter of
∆ABC is 10 cm, and the perimeter of ∆DEF is 20cm. When you compare the ratios of
the perimeters of these similar triangles, you also get 2 : 1.
If two similar triangles have a scale factor of a: b, then the ratio of their perimeters is
a :b.
AREAS OF SIMILAR SHAPES.
Consider the two similar rectangles below;
a
.
.
1
.
ka
2
b
kb
Ratio of corresponding sides:
ka = kb= k
a
b
Ratio of Areas: Area of rectangle 2 = ka x kb
Area of rectangle 1
axb
=k²ab =k²
ab
Given two similar shapes with the ratio of their corresponding side k, the ratio of
the areas will be k².
Example 2:
Two triangles are similar with areas 18cm² and 32cm² respectively. If the base of a smaller
triangle is 6cm, find the base of the larger triangle
Solution:
.
18cm²
32cm²
6cm
?
Ratio of sides, k = b
6
Ratio of areas , k²= 32
18
Square root
b
²
6
=
32
18
b =4
6 3
3b= 24
b=8cm
The base of the larger triangle is 8cm.
Example 3:
Two rectangles are similar and the ratio of their areas is 16:9. If the length of the smaller
rectangle is 8cm, find the length of the larger.
.
8 cm
L2=?
A1:A2 = 16:9
Ratios of sides, k=l2
8
Ratios of areas, k²= 16
9
l2
8
²
=
16
9
l2 = 4
8 3
3 l2= 10.7
The length of the larger rectangle is 10.7cm.
Example 4:
Given XY=2cm, BC=3cm and area of XYCB= 10cm², find the area of triangle AXY.
A
A
.
A
X
Y
X
C
B
Area of AXY
Ratio of sides, k= (3/2)²
Ratio of area, k²= 10+x
x
9=10+x
4
x
(10+x) 4=9x
40+4x=9x
40=9x-4x
40=5x
5 5
X=8cm²
EXAMPLE 5:
The triangles ABC and EBD are similar (AC and DE are not parallel). If AB=8cm, BE=4cm and
the area of triangle DBE=6cm², find the area of triangle ABC.
A
A
D
B
E
Ratio of sides, k = (4/8) ²
Ratio of areas, k² = 6
x
16 = 6
64 x
16x = 384
16
16
X = 24cm²
C
Ratio of surface areas of similar figures; if the ratio of corresponding sides of
figure X and figure Y is x : y, then the ratio of their corresponding surface is x2 : y2.
VOLUMES OF SIMILAR FIGURES.
If the ratio of corresponding sides of similar figures is k. the ratio of their
areas will be k² and the ratio of their volumes will be k³.
Consider the two similar cuboids below:
h
kh
b
a
kb
ka
Ratio of sides: ka = kb = kh = k
a b h
Volume: ka x kb x kh = k³
axbxh
Example:
The radii of two spheres are 4cm and 5cm. find the ratio of their volumes.
5cm
4cm
R1
³=
R2
V1
V2
(4/3)³ = V1
V2
64 = V1
125 V2
64:125 = V1:V2
Example 2:
Two cylinders have radii 6 cm and 10 cm respectively. If the volume of a smaller cylinder is
400cm³, find the volume of the larger one.
6cm
V1=400cm³
(R1/R2) ³=V1
V2
(6/10) ³= V1
V2
27 = 400
125
V2
V2= 125 X 400
27
V2= 1852cm³
Example 3:
10cm
A cone of height 16cm is cut from a cone of radius 7cm and height 24cm. Find the volume of
the trustrum (truncated cone). (use
as 22 )
7
24cm
Volume= 1 X 22 X 7 X 7 X 24
3 7
V= 3696 =
1232cm³
3
16
24
³=x
1232
4096 X 1232 = 13824 X x
5046272 = 13824
13824
13824
X = 365
1232cm³ - 365cm³ = 867cm³
Example :
16
6
m
³ = 54 = 27
6
16
m³ = 27
216
8
8m³= 5832
m³= 729
m= 9cm
8
54
m