Question

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

Answer 1

Concept: 1 Mark
Application: 2 Marks

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

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
$=r\times cos{30}^{\circ }=15\times \frac{\sqrt{3}}{2}=12.75cm$.

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

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

The area of minor segment
$=117.83-95.63=22.21c{m}^2$

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

$=\pi r^2-22.21$

$=3.14\times {15}^2-22.21$

$=706.5-22.21$

$=684.29c{m}^2$