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 = $\frac{120}{360}$ × π × ${12}^2$ = 150.8 $c{m}^2$

Perpendicular from chord is drawn to the centre. By RHS congruence, the two triangles will be congruent. The perpendicular divides the chord into equal halves. The angle subtended by each triangle at the centre is ${60}^{\circ }$.

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

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

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

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