Question

A chord of a circle of radius 15cm subtends an angle of 60° at the centre . find the area of the corresponding minor and major segments of the circle .
(Use pie = 3.14 and √3 = 1.73 )​

Answer 1

$\underline{ \sf \large{Solution - }}$

Area of minor segment =

$\sf: \implies \rm r^{2} \bigg[ \dfrac{ \pi \theta}{360} - \dfrac{ \sin\theta }{2} \bigg]$

$\sf: \implies \rm 15^{2} \bigg[ \dfrac{ 3.14 \times 60}{360} - \dfrac{ \sin60 }{2} \bigg]$

$\sf: \implies \rm 225 \bigg[ \dfrac{ 1.57}{3} - \dfrac{ \sqrt{3} }{2 \times 2} \bigg]$

$\sf: \implies \rm 225 \bigg[ \dfrac{ 1.57}{3} - \dfrac{ 1.73 }{4} \bigg]$

$\sf: \implies \rm 225 \bigg[ \dfrac{ 6.28 - 5.19}{12} \bigg]$

$\sf: \implies \rm 225 \bigg[ \dfrac{ 1.09}{12} \bigg]$

$\sf: \implies \rm 75 \bigg[ \dfrac{ 1.09}{4} \bigg]$

$\sf: \implies \rm 75 \times 0.2725$

$\sf: \implies \rm 20.4375 \: cm ^{2}$

Thus,

➢ Area of minor segment = 20.4375 cm²

Now,

$\sf: \implies Area \: of \: circle = \pi r ^{2}$

$\sf: \implies \rm 3.14 \times 15 ^{2}$

$\sf: \implies \rm 3.14 \times 225$

$\sf: \implies \rm 706.5 \: cm ^{2}$

Thus,  

➢ Area of circle = 706.5 cm²

Now,  

➢ Area of major segment = Area of circle - Area of minor segment.

$\sf: \implies 706.5 - 20.4375$

$\sf: \implies \rm 686.0625 \: cm ^{2}$

Thus,

➢ Area of major segment = 686.0625 cm²