Question

A chord of a circle of radius 10 cm subtends a right angle at the centre. The area of the minor segments (given, $\pi$ = 3.14) is

(a) $32.5c{m}^2$ (b) $34.5c{m}^2$ (c) $28.5c{m}^2$ (d) $30.5c{m}^2$

Answer 1

Given that

OA = Ob = radius = 10 cm

$\theta ={90}^o$

Area of segment APB = Area of sector OAPB - Area of $△$ AOB

Area of sector OAPB $=\frac{\theta }{{360}^o}\times \pi r^2=\frac{90}{{360}^o}\times 3.14\times {10}^2=\frac{1}{4}\times 3.14\times 100=78.5c{m}^2$

Area of $△AOB=\frac{1}{2}\times base\times height==\frac{1}{2}\times OA\times OB=\frac{1}{2}\times 10\times 10=50c{m}^2$

Area of segment APB = Area of sector OAPB - Area of $△$ AOB

$=78.5-50=18.5c{m}^2$