Perimeter of a Sector Calculator

Perimeter of a Sector Example 1

Find the perimeter of the sector below to 1 decimal place.

The sector has a central angle of 115° and a radius of 6 cm.

The perimeter of a sector:

\[ P = 2r + L \] \[ L = {\theta \over 180} \pi r \] where P is the perimeter, L is the arc length, θ is the central angle and r is the radius of the circle.

First, we are going to find the value of the arc length, L:

So if we substitute the values of the angle and radius into this equation, we get: \[ L = {115 \over 180} \pi \cdot 6 = {115 \cdot 6 \over 180} \pi = {690 \over 180} \pi \]

If we simplify this equation, we get: \[ L = {23 \over 6} \pi = 12.042 \; cm \; (to \; 3 \; dp) \]

We can now find the perimeter: \[ P = 2r + L = 2 \cdot 6 + 12.042 = 12 + 12.042 = 24.042 \; cm \]

This gives us a final answer of: \[ P = 24.0 \; cm \; to \; 1 \; decimal \; place \]

Perimeter of a Sector Example 2

Find the perimeter of the sector below to 1 decimal place.

The sector has a central angle of 72° and a radius of 8 ½ in.

The perimeter of a sector:

\[ P = 2r + L \] \[ L = {\theta \over 180} \pi r \] where P is the perimeter, L is the arc length, θ is the central angle and r is the radius of the circle.

First, we are going to find the value of the arc length, L:

So if we substitute the values of the angle and radius into this equation, we get: \[ L = {72 \over 180} \pi \cdot 8 {1 \over 2} = {72 \over 180} \pi \cdot {17 \over 2} \]

Multiplying this out gives us: \[ L = { 72 \cdot 17 \over 180 \cdot 2} \pi = {1224 \over 360} \]

If we simplify this equation, we get: \[ L = {17 \over 5} \pi = 10.681 \; in (to 3 dp) \]

We can now find the perimeter: \[ P = 2r + L = 2 \cdot 8 {1 \over 2} + 10.681 = 17 + 10.681 = 27.681 \; in \]

This gives us a final answer of: \[ P = 27.7 \; in \; to \; 1 \; decimal \; place \]

Perimeter of a Sector Example 3

Find the perimeter of the sector below, giving your answer to 1 decimal places.

The sector has a central angle of 278° and a radius of 9.6 cm.

The perimeter of a sector:

\[ P = 2r + L \] \[ L = {\theta \over 180} \pi r \] where P is the perimeter, L is the arc length, θ is the central angle and r is the radius of the circle.

First, we are going to find the value of the arc length L:

So if we substitute the values of the angle and radius into this equation, we get: \[ L = {278 \over 180} \pi \cdot 9.6 = {278 \cdot 9.6 \over 180} \pi \]

Multiplying this out gives us: \[ L = { 2668 \over 180} \pi = {1112 \over 75} \pi = 46.579 \; cm \]

We can now find the perimeter: \[ P = 2r + L = 2 \cdot 9.6 + 46.579 = 19.2 + 46.579 = 65.779 \; cm \]

This gives us a final answer of: \[ P = 65.8 \; cm \; to \; 1 \; decimal \; place \]

Perimeter of a Sector Example 4

Find the perimeter of the sector below, where the angle is in radians.

The sector has a central angle of 2 ½ radians and a radius of 16 m.

The perimeter of a sector, where the angle is in radians is:

\[ P = 2r + L \] \[ L = \theta r \] where P is the perimeter, L is the arc length, θ is the central angle and r is the radius of the circle.

First, we are going to find the value of the arc length L:

So if we substitute the values of the angle and radius into this equation, we get: \[ L = 2 {1 \over 2} \cdot 16 \]

Multiplying this out gives us: \[ L = { 5 \over 2} \cdot 16 = 40 \; m \]

We can now find the perimeter: \[ P = 2r + L = 2 \times 16 + 40 = 32 + 40 = 72 \; m \]

This gives us a final answer of: \[ P = 72 m \]

Perimeter of a Sector Example 5

Find the perimeter of the sector below, where the angle is in radians.

The sector has a central angle of 4.8 radians and a radius of 17 ½ inches.

The perimeter of a sector, where the angle is in radians is:

\[ P = 2r + L \] \[ L = \theta r \] where P is the perimeter, L is the arc length, θ is the central angle and r is the radius of the circle.

First, we are going to find the value of the arc length L:

So if we substitute the values of the angle and radius into this equation, we get: \[ L = 4.8 \cdot 17 {1 \over 2} \]

Simplifying this expression gives us: \[ L = 4.8 \cdot { 35 \over 2} = 84 \; in. \]

We can now find the perimeter: \[ P = 2r + L = 2 \times 17 {1 \over 2} + 84 = 35 + 84 = 119 \; in. \]

This gives us a final answer of: \[ P = 119 \; inches \]