Question

A chord of a circle of radius 15 cm subtends an angle 60 degree at the centre find the area (in $c{m}^2$) of the corresponding major segments of the circle (nearest integer).

Answer 1

From given, we have,

Area of sector $=\pi r^2\frac{\theta }{360}$

$=\frac{22}{7}\times {15}^2\frac{60}{360}$

$=117.85$

Area of triangle $=\frac{1}{2}{r}^2\sin \theta$

$=\frac{1}{2}\times {15}^2\sin 60$

$=97.312$

Minor segment = Area of sector - Area of triangle

$=117.85-97.312=20.53$

Area of circle

$=\pi r^2$

$=\frac{22}{7}\times {15}^2=707.14$

Now,

Major segment = area of circle - minor segment

$=707.14-20.5$

$\approx 686$