Question

In a circle of radius 21 cm, an arc subtends an angle of ${60}^{\circ }$ at the centre. The area of the segment formed by the corresponding chord of the arc is :

• A. 42 $c{m}^2$
• B. 421.73 $c{m}^2$
• C. 429.43 $c{m}^2$
• D. 40 $c{m}^2$

Answer 1

The correct option is D 40 $c{m}^2$

Radius $=21cm$
Angle subtended by the arc = ${60}^{\circ }$
Area of the sector = $\frac{\theta }{{360}^{\circ }}\times \pi r^2$

Area of the sector = $\frac{60}{360}\pi (21{)}^2$
Area of the sector = $231$ $c{m}^2$
Now, area of the triangle formed between the chord and the radius. Since, the angle subtended is ${60}^{\circ }$ and the other two angles are equal, (ngles opposite to he radius), which will be gain ${60}^{\circ }$. The triangle thus formed is an equilateral triangle.
Hence, area = $\frac{\sqrt{3}}{4}{s}^2$
Area of triangle = $\frac{\sqrt{3}}{4}(21{)}^2$ = $190.95$ $c{m}^2$
Thus, area of the segment = area of the sector - area of triangle
Area of segment = $231-190.95=$ $40c{m}^2$