Question

The area of the sector whose arc subtends an angle of ${90}^{\circ }$ is given as $88{\text{cm}}^2$. Find the area of the segment formed.

• A. $32{\text{cm}}^2$
• B. $88{\text{cm}}^2$
• C. $33{\text{cm}}^2$
• D. $56{\text{cm}}^2$

Answer 1

The correct option is A $32{\text{cm}}^2$

The given information can be represented as,


The highlighted part is the area of the segment, which is obtained as

$\text{Area of segment}=\text{Area of sector}-\text{Area of triangle}AOB$

Given the area of sector is $88{\text{cm}}^2$ and the angle subtented by an arc is ${90}^{\circ }$. Therefore, the radius of the circle can be calculated using the formula of area of sector, i.e.,

$\begin{array}{cc}\text{Area of sector}& =\frac{\theta }{{360}^{\circ }}\pi r^2\\ 88& =\frac{{90}^{\circ }}{{360}^{\circ }}\times \frac{22}{7}\times r^2\\ r^2& =88\times \frac{{360}^{\circ }}{{90}^{\circ }}\times \frac{7}{22}\\ r^2& =112\\ r& =4\sqrt{7}c{m}^2\end{array}$

Now the area of triangle is calculated as, $\text{Area of triangle}=\frac{1}{2}bh$, where $b$ is the base of triangle and $h$ is its height.

Since, the given triangle is right-angled triangle, therefore the base and height is same as the radius of the circle.

Thus, the area of triangle is calculated as,

$\begin{array}{cc}\text{Area of triangle}& =\frac{1}{2}bh\\ & =\frac{1}{2}{r}^2\\ & =\frac{1}{2}(4\sqrt{7}{)}^2\\ & =\frac{1}{2}\left(112\right)\\ & =56c{m}^2\end{array}$

Thus, the area of the segment is

$\begin{array}{cc}\text{Area of segment}& =88-56\\ & =32c{m}^2\end{array}$.

Hence, the required area is $32{\text{cm}}^2$.