Question

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

Answer 1

In the mentioned figure,

$O$ is the centre of circle,

$AB$ is a chord

$AXB$ is a minor arc,

$OA=OB=$ radius $=$ $15$ cm

Arc $AXB$ subtends an angle ${60}^o$ at $O.$

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

where, $\theta =centralangle$, $r=radius$

Area of sector $AOB=\frac{60}{360}\times \pi \times r^2$

$=\frac{60}{360}\times 3.14\times (15{)}^2$

$=117.75c{m}^2$

$OC$ will bisect $\angle AOB$, you can get this, as $△$ AOC & $△BOC$ are congruent by RHS congruence Rule.

By trigonometry, we have

$AC=15\sin 30$ [$\sin \theta =\frac{P}{H}$]
$OC=15\cos 30$ [$\cos \theta =\frac{B}{H}$]
And, $AB=2AC$

$\therefore$ $AB=2\times 15\sin 30=15$ cm

$\therefore$ $OC=15\cos 30=15\frac{\sqrt{3}}{2}=15\times \frac{1.73}{2}=12.975$ cm

$\therefore$ Area of $△AOB=0.5\times 15\times 12.975=97.3125c{m}^2$

[ Area of $△$ $=$ $\frac{1}{2}\times Base\times height$]
Area of the minor segment (Area of Shaded region) $=$ Area of sector $AOB-$Area of $△$AOB

$\therefore$ Area of minor segment (Area of Shaded region) $=117.75-97.3125=20.4375$ $c{m}^2$
Area of major segment $=$ Area of circle $-$ Area of the minor segment

$=(3.14\times 15\times 15)-20.4375$

$=686.0625c{m}^2$