**Introduction**

Similarity is defined as a property of being similar. It is denoted as ||| ^{lr}. We are learning polygons such that the angles of one
are equal to the corresponding angles of the other but the corresponding sides are proportional.

In similarity learning ,for two polygons (any) to be **similar**, they must have the same shape but not need the same
size.

**Example 1: **

**The areas of two similarity triangles are 20 cm ^{2} and 45 cm^{2}.The sums of their perimeters are 25 cm. Find the
perimeter of each triangle.**

we are learning representation of triangles and call the two triangles t_{1} and t_{2} and let the scale factor of
the two similarity triangles be *a* : *b.*

area of triangle formula as follow:

* *area t_{1}/ area t_{2}=(a/b)^{2 }(From theorem)

20/45= (a/b) ^{2}

Reduce the fraction,

4/9= (a/b)^{2}

Take square roots of both sides.

2/3=a/b

*a:* *b* is the reduced form of the scale factor. 2: 3 is then the reduced form

of the comparison of the perimeters.

Let 2*x* =
perimeter of t_{1}

and 3*x* =
perimeter of t_{2}

Then 2*x* + 3*x* = 25 (The sum of the perimeters is 25 cm)

5x=25

x=5

Therefore Perimeter t_{1}=2(5) =10 cm

Perimeter t_{2}=3(5) =15 cm

**Answer: P t _{1}=10 cm**

** P t _{2} =15 cm**

**Example 2:**

In below ΔPQR˜ΔMNW. Find the perimeter of ΔMNW.

Perimeter of ΔPQR/Perimeter of ΔMNW=PQ/MN

24/Perimeter of ΔMNW=6/9

6(Perimeter of ΔMNW) =216 Using cross product property.

Perimeter of ΔMNW = 36 inches.

** Answer: 36 inches**

**Example problem in similarity learning :**

**Finding the of similarity right angled triangles whose scale factor is 3 : 4.**

Area ΔABC=1/2(9) (12)

=54

Area ΔPQR=1/2(16) (12)

=96

the ratio of the areas as folllow:

Now we can compare the ratio of the areas of the above similarity triangles.

Area ΔABC/area ΔPQR=54/96

= 9/16

= (3/4)^{2}

**
Answer : (3/4) ^{2}**