Question

A chrod of circle of radius 7 cm subtends a right angle at centre at the centre. Find the area of the major sector of the circle.

Answer 1

$\huge \sf \purple{Solution.}$

r = 7 cm and θ = 90°

Area of the minor sector

$= \pi \: r {}^{2} \frac{θ}{360}$

$= \frac{22}{7}(7) {}^{2} \frac{90}{360}$

$= \frac{77}{2} = 38.5 \: cm {}^{2}$

$= \frac{22}{7} \times 7 \times 7 \times \frac{1}{4} = \frac{11 \times 7}{2}$

Area of the circle

$\pi \: r {}^{2}$

$= \frac{22}{7}(7) {}^{2} = 154 \: cm {}^{2}$

∴ Area of the major sector

= Area of the circle – Area of the minor sector

= 154 – 38.5

= 115.5 cm².