Question

A chord of a circle of radius 12 cm subtends an angle of ${120}^{\circ }$ at the centre. Find the area in $c{m}^2$ of the corresponding segment of the circle. (Use $\pi$ = 3.14 and $\sqrt{3}$ = 1.73)

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Answer 1

The area of the sector = $\frac{120}{360}\times \pi \times {12}^2$ = 150.8 $c{m}^2$

Perpendicular from chord is drawn to the centre bisects the chord. The angle subtended by each triangle at the centre is ${60}^{\circ }$.

Height of perpendicular = r x cos 60${}^{\circ }$ = 12 x 0.5 = 6 cm.

Length of chord = 2 × r × sin 60${}^{\circ }$ = 24 $\times \frac{\sqrt{3}}{2}$= 20.6 cm

The area of triangle = 0.5 × 20.6 × 6 = 61.8 $c{m}^2$

The area of segment = 150.8 - 61.8 = 89 $c{m}^2$