Question

A chord of a circle of radius 14cm subtends an angle of ${{30}^{\circ }}$ at the centre. Find the areas of both, minor sector and major sector of the circle. $\left( \text{Take }\pi =\dfrac{22}{7} \right)$.

Answer 1

Hint: Use the formula, area of a sector $=\pi {{r}^{2}}\times \left( \dfrac{\theta }{{{360}^{\circ }}} \right)$ to determine the value of area of minor sector of the circle, where $\theta $ is the angle subtended by the minor arc of the circle. Subtract the area of the minor sector of the circle from the total area of the circle to determine the area of the major sector of the circle.

Complete step-by-step answer:
A circular sector or circle sector, is the portion of a disk enclosed by two radii and an arc, where the smaller area is known as the minor sector and the larger area is known as the major sector.
Now, let us come to the question. We have been given that, a chord of a circle of radius 14cm subtends an angle of ${{30}^{\circ }}$ at the centre. Therefore, $\theta ={{30}^{\circ }}$.

Area of minor sector
$\begin{align}
  & =\pi {{r}^{2}}\times \dfrac{30}{360} \\
 & =\dfrac{\pi {{r}^{2}}}{12} \\
\end{align}$
Substituting the value of $\pi $ and $r$, we get,
Area of minor sector
$=\dfrac {22}7\times 14\times 14\times \dfrac 1{12}$
$=\dfrac {154}3\text{cm}^2$
Now, area of major sector = area of circle - area of minor sector
Therefore, area of major sector
$=\pi r^2 - \dfrac {\pi r^2}{12}$
$=\dfrac {11\pi r^2}{12}$
$=\dfrac {11}{12}\times \dfrac {22}{7}\times 14 \times 14$
$=\dfrac {1694}3\text{cm}^2$

Note: We have used $\pi=\dfrac {22}7$ because it is already provided to us in the question. We can also measure $\theta$ in radian. Therefore, the area of the sector can be written as $\pi r^2 \times \left(\dfrac{\theta}{2\pi}\right)$, here $\theta$ is in radian. It is not necessary to find the area of the minor sector first. We can also determine the area of the major sector first and then subtract it from the total area of the circle to find the area of the minor sector.