In the diagram, $A B C$ is a semi-circle with diameter $A C$, centre $O$ and radius 1. Also, $O B$ is perpendicular to $A C$. Using $A B$ as a diameter, a second semi-circle $A E B$ is drawn. The region inside this second semi-circle that lies outside the original semi-circle is shaded, as shown. The area of this shaded region is
(A) $\frac{\pi}{4}$
(D) $\frac{3}{4}$
(B) $\frac{1}{2}$
(C) $\frac{3 \pi}{4}+\frac{1}{2}$
(A) $\frac{\pi}{4}$
(D) $\frac{3}{4}$
(E) $\frac{\pi}{2}-\frac{1}{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-131.jpg?height=35&width=14&top_left_y=610&top_left_x=829)
The easiest way to calculate the area of the shaded region is to take the area of the semi-circular region $A E B$ and subtract the area of the region labelled (1). But the area of this region (1) is the area of the quarter circle $A B O$ minus the area of the triangle $A B O$. We now calculate these areas.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-212.jpg?height=290&width=444&top_left_y=1617&top_left_x=1361)
Since $A O=B O=1$, then the area of $\triangle A B O$ is $\frac{1}{2}(1)(1)=\frac{1}{2}$. Since $A O=B O=1$, then radius of the quarter circle $A B O$ is 1 , and so the area of the quarter circle is $\frac{1}{4} \pi(1)^{2}=\frac{\pi}{4}$.

Lastly, since $A O=B O=1$, then $A B=\sqrt{2}$, and so the radius of the semi-circle $A E B$ is $\frac{\sqrt{2}}{2}$, which means that the area of the semi-circle $A E B$ is $\frac{1}{2} \pi\left(\frac{\sqrt{2}}{2}\right)^{2}=\frac{\pi}{4}$.

Thus, the area of the shaded region is

(Area of semicircle $A E B$ ) - (Area of region (1))

$$
\begin{aligned}
& =(\text { Area of semicircle } A E B)-[(\text { Area of quarter circle } A O B)-(\text { Area of } \triangle A O B)] \\
& =\frac{\pi}{4}-\left[\frac{\pi}{4}-\frac{1}{2}\right] \\
& =\frac{1}{2}
\end{aligned}
$$

ANSWER: (B)