# Area and Perimeter of a Sector of a Circle

We will discuss the Area and perimeter of a sector of a circle

We know that

Therefore,

Area of a sector of a circle = $$\frac{ \theta^{\circ}}{360^{\circ}}$$ × Area of the circle = $$\frac{ θ}{360}$$ ∙ πr2

where r is the radius of the circle and $$\theta^{\circ}$$ is the sectorial angle.

Also, we know that

Therefore,

Arc MN = $$\frac{ \theta^{\circ}}{360^{\circ}}$$ × Circumference of the circle = $$\frac{ θ}{360}$$ ∙ 2πr = $$\frac{πθr}{180}$$

where r is the radius of the circle and $$\theta^{\circ}$$ is the sectorial angle.

Thus,

perimeter of a sector of a circle = ($$\frac{πθ}{180}$$ ∙ r + 2r) = ($$\frac{πθ}{180}$$ + 2)r

where r is the radius of the circle and θ° is the sectorial angle.

Problems on Area and Perimeter of a Sector of a Circle:

1. A plot of land is in the shape of a sector of a circle of radius 28 m. If the sectorial angle (central angle) is 60°, find the area and the perimeter of the plot. (Use π = $$\frac{22}{7}$$.)

Solution:

Area of the plot = $$\frac{60^{\circ}}{360^{\circ}}$$ ×  πr2 [Since θ = 60]

= $$\frac{1}{6}$$ ×  πr2

= $$\frac{1}{6}$$ × $$\frac{22}{7}$$ × 282 m2.

= $$\frac{1}{6}$$ × $$\frac{22}{7}$$ × 784 m2.

= $$\frac{17248}{42}$$ m2.

= $$\frac{1232}{3}$$ m2.

= 410$$\frac{2}{3}$$ m2.

Perimeter of the plot = ($$\frac{πθ}{180}$$ + 2)r

= ($$\frac{22}{7}$$ ∙ $$\frac{60}{180}$$ + 2) 28 m

= ($$\frac{22}{21}$$ + 2) 28 m

= $$\frac{64}{21}$$ ∙ 28 m

= $$\frac{1792}{21}$$ m

= $$\frac{256}{3}$$ m

= 85$$\frac{1}{3}$$ m.