In the diagram, the circle has centre $O$ and square $O P Q R$ has vertex $Q$ on the circle. If the area of the circle is $72 \pi$, the area of the square is
(A) 38
(B) 48
(C) 25
(D) 12
(E) 36

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-022.jpg?height=322&width=317&top_left_y=495&top_left_x=1316)
Let $s$ be the side length of the square. Therefore, $O R=R Q=s$.

Let $r$ be the radius of the circle. Therefore, $O Q=r$ since $O$ is the centre of the circle and $Q$ is on the circumference of the circle.

Since the square has a right-angle at each of its vertices, then $\triangle O R Q$ is right-angled at $R$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-041.jpg?height=334&width=328&top_left_y=2305&top_left_x=953)

By the Pythagorean Theorem, $O R^{2}+R Q^{2}=O Q^{2}$ and so $s^{2}+s^{2}=r^{2}$ or $2 s^{2}=r^{2}$. In terms of $r$, the area of the circle is $\pi r^{2}$.

Since we are given that the area of the circle is $72 \pi$, then $\pi r^{2}=72 \pi$ or $r^{2}=72$.

Since $2 s^{2}=r^{2}=72$, then $s^{2}=36$.

In terms of $s$, the area of the square is $s^{2}$, so the area of the square is 36 .

ANSWER: (E)