Question

A chord of a circle of radius $15cm$ subtends an angle of ${60}^o$ at the centre. Find the area of the corresponding minor and major segment of the circle.
(Use $\pi =3.14$ and $\sqrt{3}=1.73$ )

Answer 1

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

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

$=\frac{55\times 15}{7}=117.85c{m}^2$

Area of$△ABC$$=\frac{1}{2}ab\sin \varphi$

Here, $a=b=r$

Area of$△ABC$$=\frac{1}{2}{r}^2\sin \varphi$

$=\frac{1}{2}\times 15\times 15\times \frac{\sqrt{3}}{2}$

$=225\frac{\sqrt{3}}{4}$

Area of$△ABC$$=97.3125c{m}^2$

Minor segment $BDC=AreaofsectorABDC-Areaof△ABC$

$=117.85-97.3125$

Minor segment $BDC=20c{m}^2$

Area of circle $=\pi r^2=20.53c{m}^2$

$=\frac{22}{7}\times 15\times 15=707014{cm}^2$

So, major segment $BDC=$ Area of circle-minor segment

$=707.14-20.5$

Major segment $BDC$$=686.642c{m}^2$