Question

A chord of a circle of radius $10cm$ subtends a right angle at the centre. Find the area of corresponding minor segment $(\pi =3.14)$

Answer 1

Area of sector OAPB

$=\frac{\theta }{360}\times \pi \times r^2$

$=\frac{90}{360}\times \pi r^2$

$=\frac{1}{4}\times 3.14\times 10\times 10$

$=78.5c{m}^2$

$Areaof△OAB=\frac{1}{2}{r}^{2}sin\theta =\frac{1}{2}\times 10\times 10$

$=50c{m}^2$

Area of minor segment = Area of sector $OAPQ$ - area of $△\left(AOB\right)=78.5c{m}^2-50c{m}^2$

$=28.5c{m}^2$