# Area and Perimeter of a Semicircle and Quadrant of a Circle

We will learn how to find the Area and perimeter of a semicircle and Quadrant of a circle.

Area of a semicircle = $$\frac{1}{2}$$πr2

Perimeter of a semicircle = (π + 2)r.

because a semicircle is a sector of sectorial angle 180°.

Area of a quadrant of a circle = $$\frac{1}{4}$$πr2.

Perimeter of a quadrant of a circle = ($$\frac{π}{2}$$ + 2)r.

because a quadrant of a circle is a sector of the circle whose sectorial angle is 90°.

Here r is the radius of the circle.

Solved Examples on Area and perimeter of a semicircle and Quadrant of a circle:

1. The area of a semicircular region is 308 cm^2. Find its perimeter. (Use π = $$\frac{22}{7}$$.)

Solution:

Let r be the radius. Then,

area = $$\frac{1}{2}$$ ∙ πr^2

⟹ 308 cm^2 = $$\frac{1}{2}$$ ∙ $$\frac{22}{7}$$ ∙ r^2

⟹ 308 cm^2 = $$\frac{22}{14}$$ ∙ r^2

⟹ $$\frac{22}{14}$$ ∙ r^2 = 308 cm^2

⟹ r^2 = $$\frac{14}{22}$$ ∙ 308 cm^2

⟹ r^2 = $$\frac{7}{11}$$ ∙ 308 cm^2

⟹ r^2 = 7 × 28 cm^2

⟹ r^2 = 196 cm^2

⟹ r^2 = 14^2 cm^2

⟹ r = 14 cm.

Therefore, the radius of the circle is 14 cm.

Now, perimeter = (π + 2)r

= ($$\frac{22}{7}$$ + 2) ∙ 14 cm

= $$\frac{36}{7}$$  ×  14 cm

= 36 × 2 cm

= 72 cm.

2. The perimeter of a sheet of paper in the shape of a quadrant of a circle is 75 cm. Find its area. (Use π = $$\frac{22}{7}$$.)

Solution:

Then,

perimeter = ($$\frac{π}{2}$$ + 2)r

⟹ 75 cm = ($$\frac{1}{2}$$ ∙ π + 2)r

⟹ 75 cm = ($$\frac{ 1 }{2}$$ ∙ $$\frac{22}{7}$$  + 2)r

⟹ 75 cm = ($$\frac{11}{7}$$  + 2)r

⟹ 75 cm = $$\frac{25}{7}$$r

⟹ $$\frac{25}{7}$$r = 75 cm

⟹ r = 75 × $$\frac{7}{25}$$ cm

⟹ r = 3 × 7 cm

⟹ r = 21 cm.

Therefore, the radius of the circle is 21 cm.

Now, area = $$\frac{1}{4}$$πr^2

= $$\frac{1}{4}$$ ∙  $$\frac{22}{7}$$ ∙ 21^2 cm^2

= $$\frac{1}{4}$$ ∙  $$\frac{22}{7}$$ ∙ 21 ∙ 21 cm^2

= $$\frac{693}{2}$$ cm^2

= 346.5 cm^2.

Therefore, area of the sheet of paper is 346.5 cm^2.

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