The perimeter of a certain sector of a circle of radius 5.6 m is 20.0

Q) The perimeter of a certain sector of a circle of radius 5.6 m is 20.0 m. Find the area of the sector.

Ans:

We know that the perimeter of a circle’s sector making θ angle, is given by: P = 2 R + 2 π R $\frac{\theta}{360}$

∴ π R $\frac{\theta}{360}$ = $\frac{P - 2 R}{2}$ ……..(i)

And the area of the circle’s sector making θ angle, is given by: A = π R ² $\frac{\theta}{360}$

by substituting values from equation (i), we get:

A = π R ² $\frac{\theta}{360}$

A = R x π R $\frac{\theta}{360}$ = R x $\frac{P - 2 R}{2}$

Here we are given, radius R = 5.6 m, Perimeter P = 20 m

By substituting these values in above formula, we get:

A = R x $\frac{P - 2 R}{2}$

∴ A = 5.6 x $\frac{20 - 2 \times 5.6}{2}$

∴ A = 5.6 x $\frac{20 - 11.2}{2}$

∴ A = 5.6 x $\frac{8.8}{2}$ = 5.6 x 4.4 = 24.64

Therefore, the area of the sector is 24.64 m²

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