In the diagram, two larger circles with radius 1 have centres $P$ and $Q$. Also, the smaller circle has diameter $P Q$. The region inside the two larger circles and outside the smaller circle is shaded.

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-035.jpg?height=360&width=526&top_left_y=991&top_left_x=759)

The area of the shaded region is closest to
(A) 0.36
(B) 0.38
(C) 0.40
(D) 0.42
(E) 0.44
The circle with centre $P$ has radius 1 and passes through $Q$.

This means that $P Q=1$.

Therefore, the circle with diameter $P Q$ has radius $\frac{1}{2}$ and so has area $\pi\left(\frac{1}{2}\right)^{2}=\frac{1}{4} \pi$.

To find the area of the shaded region, we calculate the area of the region common to both circles and subtract the area of the circle with diameter $P Q$.

Suppose that the two circles intersect at $X$ and $Y$.

Join $X$ to $Y, P$ to $Q, P$ to $X, P$ to $Y, Q$ to $X$, and $Q$ to $Y$ (Figure 1).

By symmetry, the area of the shaded region on each side of $X Y$ will be the same.

The area of the shaded region on the right side of $X Y$ equals the area of sector $P X Q Y$ of the left circle minus the area of $\triangle P X Y$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-061.jpg?height=393&width=526&top_left_y=714&top_left_x=645)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-061.jpg?height=390&width=393&top_left_y=716&top_left_x=1186)

Figure 2

Since each of the large circles has radius 1, then $P Q=P X=P Y=Q X=Q Y=1$.

This means that each of $\triangle X P Q$ and $\triangle Y P Q$ is equilateral, and so $\angle X P Q=\angle Y P Q=60^{\circ}$.

Therefore, $\angle X P Y=120^{\circ}$, which means that sector $P X Q Y$ is $\frac{120^{\circ}}{360^{\circ}}=\frac{1}{3}$ of the full circle, and so has area $\frac{1}{3} \pi 1^{2}=\frac{1}{3} \pi$.

Lastly, consider $\triangle P X Y$.

Note that $P X=P Y=1$ and that $\angle X P Q=\angle Y P Q=60^{\circ}$.

Since $\triangle P X Y$ is isosceles and $P Q$ bisects $\angle X P Y$, then $P Q$ is perpendicular to $X Y$ at $T$ and $X T=T Y$.

By symmetry, $P T=T Q$. Since $P Q=1$, then $P T=\frac{1}{2}$.

By the Pythagorean Theorem in $\triangle P T X$ (which is right-angled at $T$ ),

$$
X T=\sqrt{P X^{2}-P T^{2}}=\sqrt{1^{2}-\left(\frac{1}{2}\right)^{2}}=\sqrt{1-\frac{1}{4}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}
$$

since $X T>0$.

Therefore, $X Y=2 X T=\sqrt{3}$.

The area of $\triangle P X Y$ equals $\frac{1}{2}(X Y)(P T)=\frac{1}{2}(\sqrt{3})\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{4}$.

Now, we can calculate the area of the shaded region to the right of $X Y$ to be $\frac{1}{3} \pi-\frac{\sqrt{3}}{4}$, the difference between the area of sector $P X Q Y$ and the area of $\triangle P X T$.

Therefore, the area of the shaded region with the circle with diameter $P Q$ removed is

$$
2\left(\frac{1}{3} \pi-\frac{\sqrt{3}}{4}\right)-\frac{1}{4} \pi=\frac{2}{3} \pi-\frac{\sqrt{3}}{2}-\frac{1}{4} \pi=\frac{5}{12} \pi-\frac{\sqrt{3}}{2} \approx 0.443
$$

Of the given choices, this is closest to 0.44 .

ANSWER: (E)