Perimeters of Similar Shapes Formulas | List of Perimeters of Similar Shapes Formulas You Should Know - BYJUS

Perimeter of Similar Shapes Formulas

The total distance around a shape is referred to as its perimeter. It is the length of any two-dimensional geometric shape’s outline or boundary. Depending on the dimensions, the perimeter of different shapes can be equal....Read MoreRead Less

What is Geometric Similarity?

Similarity in geometry is defined as the similarity in shape of two geometric shapes. The lengths of the corresponding sides of two geometric shapes will be proportional when they are similar. To put it in an alternative way, two geometric shapes are said to be similar if they have many of the same features, but are not truly identical. We can start by comparing the corresponding sides of the shapes to see if they are similar.

In geometry, perimeter refers to the path or boundary that surrounds a shape. Another way to describe a shape is by the length of its outline. When we talk about two similar shapes then their corresponding sides are in a particular ratio, which is called the scale factor. When the scale factor of two similar shapes is a : b, then the ratio of their perimeters is also a : b.

List of Formulas

In the following section we will look at the relationship between the perimeter of similar geometric shapes. There will be a variety of similar shapes in the worked examples, but we can start with similar triangles and the ratio between side lengths and their perimeters.

The total distance around the boundary of a geometric shape is known as its perimeter. The ratio of two similar perimeters of two similar shapes is equal to the   ratio of the lengths of their corresponding sides. Let’s assume that the   triangles  in the image are similar. P1 and P2 are the perimeters of the triangles    PQR and XYZ. Let s1 and s2 be the corresponding side lengths of the two triangles.

Following this, we have:

P1 : P2 = s1 : s2

The expression  can also be written in the fraction form as:

$$\frac{p1}{p2}=\frac{s1}{s2}$$

Perimeters of Similar Shapes

The ratio of the perimeters of similar geometric shapes equals the ratio of their corresponding side lengths. Since we already considered two similar triangles, PQR and XYZ, then this implies that,

$$\frac{\text{Perimeter of triangle PQR}}{\text{Perimeter of triangle XYZ}}=\frac{\text{PQ}}{\text{XY}}=\frac{\text{QR}}{\text{YZ}}=\frac{\text{PR}}{\text{XZ}}$$.

Where, PQ, QR, PR are the side lengths of the triangle PQR.

XY, YZ, XZ are the side lengths of the triangle XYZ.

Solved Examples

Example 1: Determine the perimeters of the shapes given below and  determine the ratio of perimeters of the below shapes.

Solution:

As we know, the perimeter of a geometric shape is the total distance around the boundary of the shape.

Hence, the perimeter of the first shape = 12 + 20 + 16 = 48 inches

The perimeter of the second shape = 2 + 5 + 4 = 12 inches

We also know that,

The ratio of the perimeters of two similar shapes is equal to the ratio of their corresponding side’s lengths.

Therefore, the ratio of perimeters of shapes A and B

$$=\frac{48}{12}$$

$$=\frac{4}{1}$$

= 4 : 1

Therefore, ratio of perimeters = 4:1 = scale factor.

Example 2: This pair of triangles are s similar. Find the missing perimeter. (P denotes the perimeter).The scale factor of A to B is 2.6 : 9. Find the perimeter of triangle A.

Solution:

Given that, scale factor = 2.6 : 9 = ratio of perimeters

Therefore,

$$\frac{\text{Perimeter of triangle A}}{\text{Perimeter of triangle B}}=\frac{P_A}{P_B}=\frac{2.6}{9}$$

$$\frac{P_A}{36}=\frac{2.6}{9}$$  (since the perimeter of triangle B is 36 feet)

$$P_A=\frac{2.6}{9}\times 36$$

$$P_A=10.4$$ feet

Hence, The perimeter of the shape A is 10.4 feet.

Example 3 : This pair of shapes are  similar. Find the missing perimeter. (P denotes the perimeter). The scale factor of X to Y is 7.5 : 1. Find the perimeter of Y.

Solution:

Given that, scale factor = 7.5 : 1 = ratio of perimeters.

Therefore,

$$\frac{\text{Perimeter of shape X}}{\text{Perimeter of shape X}}=\frac{P_X}{P_Y}=\frac{1}{7.5}$$

$$\frac{15}{P_Y}=\frac{1}{7.5}$$             (since the perimeter of shape X is 15 inches)

$$P_Y=7.5 \times 15$$     (Cross multiplied)

$$P_Y=112.5$$ inches.

Hence, the perimeter of the shape Y is 112.5 inches.

Example 4: If the scale factor between two regular geometric shapes G and F is 3:0.8 and the perimeter of shape G is 60 yards, calculate the perimeter of shape F.

Solution:

Given that, scale factor = 3 : 0.8 = ratio of perimeters.

Therefore,

$$\frac{\text{Perimeter of shape G}}{\text{Perimeter of shape F}}=\frac{P_G}{P_F}=\frac{3}{0.8}$$  (since the perimeter of shape G is 30 Yards)

$$0.8 \times 60 = P_F \times 3$$                   (Cross multiplied)

$$P_F=16$$ yards.

Hence, the perimeter of the shape F is 16 yards.

Example 5: The given pair of shapes is similar. Find the missing perimeter. (P denotes the perimeter).The scale factor of S to R is 6 : 7. Find the perimeter of R.

Solution:

Given that, the scale factor of S to R = 6 : 7

So, the scale factor of R to S would be 7 : 6, which is also equal to the ratio of the perimeters of the shapes R and S.

Therefore, $$\frac{\text{Perimeter of shape R}}{\text{Perimeter of shape S}}=\frac{P_R}{P_S}=\frac{7}{6}$$

$$\frac{P_R}{24}=\frac{7}{6}$$            (since the perimeter of shape S is 24 feet)

$$P_R=\frac{7}{6} \times 24$$    (Cross multiplied)

$$P_R=28$$ feets

Hence, the perimeter of the shape R is 28 feets.