Question

In the figure, the center of the circle is A and ABCDEF is a regular hexagon of side 6 cm. $\begin{array}{cc}\text{Find}& :(\text{i) Area of segment BPF}\\ & \text{(ii) Area of shaded portion.}\end{array}$

Answer 1

The area of segment is given as Area of sector -area of triangle

Area of sector is $\frac{120}{360}\times \pi \times 6^2=12\pi$

Area of triangle is $\frac{1}{2}\times 6\times 6\times \sin 120\frac{1}{2}\times 36\times \frac{\sqrt{3}}{2}=9\sqrt{3}$

The area of segment is $12\pi -9\sqrt{3}$

The area of shaded portion is given as area of hexagon - area of triangle.

Area of hexagon is $3\sqrt{3}{a}^{2}3\sqrt{3}\times 6^{2}108\sqrt{3}$

Area of shaded region is $108\sqrt{3}-9\sqrt{3}=99\sqrt{3}$