Question

What will be the area of the segment, if the arc subtends an angle of ${90}^{\circ }$ at the centre of the circle of radius $6\text{cm}$?

• A. $9(\pi -2){\text{cm}}^2$
• B. $(6\pi -12){\text{cm}}^2$
• C. $3(3\pi -1){\text{cm}}^2$
• D. $12(3\pi -1){\text{cm}}^2$

Answer 1

The correct option is A $9(\pi -2){\text{cm}}^2$

The given information can be depicted as follows,

The formula to calculate the area of segment is given as,

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

The area of the sector whose arc subtends an angle of ${90}^{\circ }$ with radius $6\text{cm}$ is obtained as,

$\begin{array}{cc}\text{Area of sector}& =\frac{\theta }{360}\pi \left(r^2\right)\\ & =\frac{90}{360}\pi \left(6^2\right)\\ & =\frac{1}{4}\pi \left(36\right)\\ & =9\pi \end{array}$

Now, the area of triangle $AOB$ with base and height equal to $6\text{cm}$ is calculated as,

$\begin{array}{cc}\text{Area of triangle}AOB& =\frac{1}{2}\times b\times h\\ & =\frac{1}{2}\times 6\times 6\\ & =18\end{array}$

Thus, the area of the segment is,

$\begin{array}{cc}\text{Area of segment}& =9\pi -18\\ & =9(\pi -2)\end{array}$

Hence, the required area is $9(\pi -2){\text{cm}}^2$.