Question

# A chord of a circle of radius $15 cm$ 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 $\sqrt {3} = 1.73$)

Solution

## In the mentioned figure,$O$ is the centre of circle,$AB$ is a chord$AXB$ is a minor arc,$OA =OB =$ radius $=$ $15$ cmArc $AXB$ subtends an angle $60^o$ at $O.$Area of sector $=\dfrac{\theta}{360}\times \pi\times r^2$where, $\theta= central angle$, $r = radius$Area of sector $AOB=\dfrac{60}{360}\times \pi\times r^2$                                    $=\dfrac{60}{360}\times3.14\times(15)^2$                                    $=117.75\,cm^2$$OC will bisect \angle AOB, you can get this, as \triangle AOC & \triangle BOC are congruent by RHS congruence Rule.By trigonometry, we haveAC\,=15\,\sin30 [sin\theta=\frac PH]OC\,=15\,\cos30 [cos\theta=\frac BH]And, AB=2AC$$\therefore$ $AB=2\times15\sin30 = 15$ cm$\therefore$ $OC=15\cos30 = 15\dfrac{\sqrt3}2 = 15\times\dfrac{1.73}2 = 12.975$ cm$\therefore$ Area of $\triangle AOB = 0.5\times15\times12.975=97.3125 \,cm^2$[ Area of $\triangle$ $=$ $\frac12\times Base\times height$]Area of the minor segment (Area of Shaded region) $=$ Area of sector $AOB -$Area of $\triangle$AOB$\therefore$ Area of minor segment (Area of Shaded region) $=117.75-97.3125=20.4375$ $cm^2$Area of major segment $=$ Area of circle $-$ Area of the minor segment                                       $=(3.14\times15\times15)-20.4375$                                        $=\,686.0625\,\,cm^2$MathematicsNCERTStandard X

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