Question

A chrod of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:
i) Minor segment
ii) Major segment
(use π=3.14).

Answer 1

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

1. r = 10 cm

ϴ = 90°

Area of minor sector

$= \frac{θ}{360°} \times \pi \: r {}^{2}$

$= \frac{90°}{360°} \times 3.14 \times 10 \times 10$

$= 78.5 \: cm {}^{2}$

Area of ∆ AOB

$= \frac{1}{2}r {}^{2} \sin θ$

$\frac{1}{2}(10) {}^{2} \sin90°$

$= \frac{100}{2} \times 1 = 50 \: cm {}^{2}$

1. ∴ Area of the minor segment

= Area of minor sector – Area of ∆ OAB

= 78.5 cm² – 50 cm² = 28.5 cm²

2. Area of major sector

= Area of circle – Area of minor sector

$\pi \: r {}^{2} - 78.5$

$= 3.14 \times 10 \times 10 - 78.5$

$= 314 - 78.5 = 235.5 \: cm {}^{2}$

$\huge \sf \green{Alternative \: Method}$

Area of major sector

$= \frac{(360° - θ)}{360°} \times \pi \: r {}^{2}$

$= \frac{(360° - 90°)}{360°} \times 3.14 \times (10) {}^{2}$

$= 235. \: cm {}^{2}$