Question

A chord in a circle of radius 15 cm subtends an angle of 60${}^{\circ }$ at the centre. Find the areas in $c{m}^2$ of the corresponding minor and major segments of the circle.(Use $\pi$ = 3.14 and $\sqrt{3}$ = 1.73)

Answer 1

The area of the sector = $\frac{60}{360}$ ×$\pi$ × ${15}^2$ =117.83 $c{m}^2$

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 ${30}^{\circ }$.

Height of perpendicular = r x cos ${30}^{\circ }$ = 15 x $\frac{\sqrt{3}}{2}$ = 12.75cm.

Length of chord = 2 x r x sin ${30}^{\circ }$ = 30 x $\frac{1}{2}$ = 15cm.

The area of triangle = 0.5 x 12.75 x 15 = 95.63 $c{m}^2$

The area of segment = 135.72 - 61.8 = 22.21 $c{m}^2$