Question

If $\alpha$ is the angle (in degrees) of a minor sector of a circle of radius $r$, then the area of the major sector is

• A. $\frac{\pi r^2({360}^{\circ }-\alpha )}{{360}^{\circ }}$
• B. $\frac{\pi r^2\alpha }{{360}^{\circ }}$
• C. $\frac{2\pi r({360}^{\circ }-\alpha )}{{360}^{\circ }}$
• D. $\frac{2\pi r\alpha }{{360}^{\circ }}$

Answer 1

The correct option is A $\frac{\pi r^2({360}^{\circ }-\alpha )}{{360}^{\circ }}$

The angle made by the minor sector at the centre of circle is $\alpha$.
$\therefore$ Angle made by major sector $=({360}^{\circ }–\alpha )$
Since, the area of a sector making an angle $\theta$ at centre of circle of radius $r$ is given as $\left(\frac{{\theta }^{\circ }}{{360}^{\circ }}\right)\times \pi r^2$
$\therefore$ Area of major sector for given angle $({360}^{\circ }–\alpha )=\frac{\pi r^2({360}^{\circ }-\alpha )}{{360}^{\circ }}$
Hence, the correct answer is option a.