Similar Triangles
Properties of Similar Triangles
1. In similar triangles, corresponding angles are equal
Given ABC is similar to EFG
then <ABC = <EFG
<ACB = <EGF
<BAC = <FEG
a
a
2. In similar triangles, the ratios of corresponding sides are equal.
Given ABC is similar to EFG
then
AB AC BC


EF
EG FG
3. In similar triangles, the ratio of the areas is equal to the square of the ratio of the
corresponding sides
Given ABC is similar to EFG
AC I
F
IJ  F
FBC I
G J
G
G
J
H K HEG K HFG K
Area of ABC
AB

Area of EFG
EF
2
2
2
Example:
Solve for the unknowns
y
x
9
4
6
8
Because of AAA, there are two congruent triangles in this question.
This means that the ratios of the corresponding sides are equivalent.
y+9
9
x
4
4
14
6
Solve for x by using the ratio
6
4

14
x
thus 6x = 56 (through cross multiplication)
x 
28
3
Solve for y by using the ratio
6
9

14
y 9
Thus 126 = 54 + 6y (through cross multiplication)
y = 12
Example:
Show that ACD is ~ to  ABE
D
E
C
A
B
<AEB = <ADC (F theorem of parallel lines)
<ABE = <ACD (F theorem of parallel lines)
<EAB = <DAC since they are the same angle
Through AAA, ACD is ~ to  ABE
Example:
Two triangles are given below are similar. If the area of the smaller triangle is 6 units
squared, find the area of eth larger triangle.
5
3
6
4
We know that congruency implies the ratios of the side lengths squared equals the ratio of
the areas.
3
 
6
2


6

 
Area
of
L
arg
er



9
6

36
Area
Cross multiply to get
9A = 216
and A = 24 units squared
Example:
This time I will use the same triangles but give you different information. The triangles are
similar and the ratios of their areas is 1/9. Solve for x.
25
 25 


 x 
x
2
1
 
9
If we take the square of each side
 25   1 

 
 x  3
Thus x = 75
Example:
Two right triangles are similar. The base of the larger is 4 times the base of the smaller.
The area of the smaller triangle is 6cm2. Find the area of the larger triangle.
Draw a diagram
A =6
A= ?
x
 x 


 4x 
2
4x
6
 
 A
 1  6

  
 16   A 
By reducing the left hand side ratio, this becomes
Through cross multiplication, A = 96cm2
1
 
4
2
6
 
 A
Example:
The ratios of the areas of 2 similar triangles is 5/7. If the length of one side of the smaller
triangle is 8cm, find the exact corresponding side length of the larger triangle.
8
 
x
2
5
 
7
Since we want to get closer to x (inside the bracket) we will square root
everything. This is not necessary, it just makes things easier.
 8   5 
 
 x   7 
Cross multiply and we get
To solve for x multiply and divide
8 7  5x
x 
8 7
5
We do not like to have radicals in eth denominator, but you will learn how to do
the process of ‘rationalizing the denominator’ next year
Example:
Two right triangles are similar. One side length of the smaller triangle is ¼ the length of the
corresponding side length of the larger triangle. If the area of the smaller triangle is y, find
the area of the larger triangle.
Let x be a side length of the larger triangle and let A be the area of the larger triangle.
1 
 x
4 
 x 




1
 
4
2
2
y
 
 A
y
 
 A
The left hand ratio reduces to be
By squaring we get
Cross multiplication gives us A = 16y
 1  y

  
 16   A 
1
 
 4    1 
 1  4
 
 
Example:
The triangles are similar. One side of the smaller triangle is 1/3 the length of the
corresponding side of the larger triangle. If another side on the smaller triangle is 8, find
the corresponding side length of the larger triangle.
1 
 x
 3    8 
 x   y 




1  8 
    
3  y 
We need to reduce and cross multiply to solve for y
So y = 24