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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 $=$ $15$ cm

Arc $AXB$ 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 have
$AC\,=15\,\sin30$        [$sin\theta=\frac PH$]
$OC\,=15\,\cos30$       [$cos\theta=\frac BH$]
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

                                       $=(3.14\times15\times15)-20.4375$

                                        $=\,686.0625\,\,cm^2$



Mathematics
NCERT
Standard X

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