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 $\sqrt{3}=1.73$) [3 MARKS]

Answer 1

Concept: 1 Mark
Application: 2 Marks

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

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
$=r\times cos{60}^{\circ }=12\times 0.5=6cm$.

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

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

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