Question

A chord PQ of a circle of radius 10 cm subtends an angle of ${60}^{\circ }$ at the centre of the circle. Find the area of major and minor segments of the circle.

Answer 1

Area of sector formed by the arc $=\frac{\theta }{360}\times \pi r^2=\frac{60}{360}\times \frac{22}{7}\times {10}^2=\frac{1100}{21}=52.38c{m}^2$

In $△$OAB, let OM be $\perp$ bisector of AB,

$\angle AOM=\angle BOM$

Let OM be x cm

In $△$OMA,

$\frac{OM}{OA}=cos{30}^o\frac{x}{10}=\frac{\sqrt{3}}{2}x=5\sqrt{3}=8.66cmOM=8.66cm$

In $△$OMA,

$\frac{AM}{OA}=sin{30}^o\frac{AM}{10}=\frac{1}{2}AM=5cmAB=2\times AM=2\times 5=10cm$

Area of $△$OAB $=\frac{1}{2}\times OM\times AB=\frac{1}{2}\times 8.66\times 10=43.3c{m}^2$

Area of minor segment = Area of sector formed by the arc - Area of $△OAB=52.38-43.3=9.08c{m}^2$

Area of major segment = Area of circle - Area of minor segment

$=\pi r^2-9.08=3.14\times 10\times 10-9.08=314-9.08=304.92c{m}^2$