Question

a chord of a circle of radius 14cm subtends an angle 120 degrees at centre.find the area of corresponding major segment of circle?π=3.14

Answer 1

Answer:

Area of major segment= 410.66cm²

Step-by-step explanation:

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

                                    = π×r²- π×r²÷3

                                    = 2×π×r²÷3

                                     =2×22×14²÷21

                                      =410.66

Answer 2

The area of corresponding major segment of circle is 120.54 sq. cm.

Explanation:

Area of segment : $\dfrac{1}{2}(\theta -\sin\theta)\times r^2$

, where $\theta$ = Central angle.

r= radius

As per given , we have

$\theta=120^{\circ}$

In radian ,  $\theta=\dfrac{120}{180}\pi=\dfrac{2\pi}{3}$

r= 14 cm

Then , the area of corresponding major segment of circle

=  $\dfrac{1}{2}(\dfrac{2\pi}{3}-\sin120^{\circ})\times (14)^2$

=  $\dfrac{1}{2}(\dfrac{2\pi}{3}-\sin2(60^{\circ}))\times (196)$

=  $(\dfrac{2\pi}{3}-(2\sin60^{\circ}\cos60^{\circ} ))\times (98)\ \ [\because\ \sin2x=2\sin x\cos x]$

=  $(\dfrac{2\pi}{3}-(2(\dfrac{\sqrt{3}}{2})(\dfrac{1}{2})) )\times (98) [\because\ \sin 60^{\circ}=\dfrac{\sqrt{3}}{2},\ \sin60^{\circ}=\dfrac{1}{2}]$

=$98(\dfrac{2\pi}{3}-\dfrac{\sqrt{3}}{2})$

=$98(\dfrac{2(3.14)}{3}-\dfrac{1.73}{2})$

=$98(1.23)=120.54$

Hence, the area of corresponding major segment of circle is 120.54 sq. cm.

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