In the diagram, the circle and the square have the same centre $O$ and equal areas. The circle has radius 1 and intersects one side of the square at $P$ and $Q$. What is the length of $P Q$ ?
(A) $\sqrt{4-\pi}$
(B) 1
(C) $\sqrt{2}$
(D) $2-\sqrt{\pi}$
(E) $4-\sqrt{\pi}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-088.jpg?height=331&width=317&top_left_y=1466&top_left_x=1294)

Part C: Each correct answer is worth 8.
Since the circle has radius 1 , then its area is $\pi\left(1^{2}\right)=\pi$.

Since the square and the circle have the same area, then the side length of the square is $\sqrt{\pi}$.

Let $M$ be the midpoint of line segment $P Q$.

Since $P Q$ is a chord of the circle, then $O M$ is perpendicular to line segment $P Q$.

Since $O M$ is perpendicular to $P Q$ and $O$ is the centre of the square, then $O M$ is half of the length of one of the sides of the square, so $O M=\frac{1}{2} \sqrt{\pi}$.

By the Pythagorean Theorem in $\triangle O P M$, we have $P M^{2}=O P^{2}-O M^{2}=1^{2}-\left(\frac{1}{2} \sqrt{\pi}\right)^{2}=1-\frac{1}{4} \pi$.

Therefore, $P Q=2 P M=2 \sqrt{1-\frac{1}{4} \pi}=\sqrt{4\left(1-\frac{1}{4} \pi\right)}=\sqrt{4-\pi}$.

ANswer: (A)

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