We will learn how to find the Area and perimeter of a semicircle and Quadrant of a circle.
Area of a semicircle = \(\frac{1}{2}\)πr^{2}
Perimeter of a semicircle = (π + 2)r.
because a semicircle is a sector of sectorial angle 180°.
Area of a quadrant of a circle = \(\frac{1}{4}\)πr^{2}.
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:
Let the radius be r.
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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