Question

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the minor segment.

[ Use $\pi$ = 3.14]

Answer 1

OA = OB = radius = 10 cm

$\theta ={90}^o$

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

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

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 = 28.5 $c{m}^2$