In the diagram, $A B$ and $B D$ are radii of a circle with centre $B$. The area of sector $A B D$ is $2 \pi$, which is oneeighth of the area of the circle. The area of the shaded region is
(A) $2 \pi-4$
(B) $\pi$
(C) $2 \pi-2$
(D) $2 \pi-4.5$
(E) $2 \pi-8$

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-136.jpg?height=325&width=352&top_left_y=561&top_left_x=1361)
In the diagram, $A B$ and $B D$ are radii of a circle with centre $B$. The area of sector $A B D$ is $2 \pi$, which is one-eighth of the area of the circle. The area of the shaded region is
(A) $2 \pi-4$
(B) $\pi$
(C) $2 \pi-2$
(D) $2 \pi-4.5$
(E) $2 \pi-8$

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-223.jpg?height=355&width=393&top_left_y=1362&top_left_x=1473)

Solution

Since the sector $A B D$ represents one-eighth a circle that has an area of $2 \pi$, the area of the whole circle is $8 \times 2 \pi$ or $16 \pi$. Also, since the whole circle has $360^{\circ}$ around the centre $\angle A B D=\frac{360^{\circ}}{8}=45^{\circ}$.

We find the radius of the circle by solving, $\pi r^{2}=16 \pi$.

Therefore, $r=4 \quad(r>0)$.

Since $r=A B=4$ and $\angle A B C=45^{\circ}$, then $\angle B A C=45^{\circ}$ and $A C=B C=2 \sqrt{2}$.

Thus the area of $\triangle A B C$ is $\frac{1}{2}(2 \sqrt{2})(2 \sqrt{2})=4$.

We observe that the area of the shaded region equals the area of the sector minus the area of $\triangle A B C$. Thus, the required shaded area is $2 \pi-4$.