Problem 539

Let $s$ be the side length of the square. Therefore, $OR=RQ=s$. Let $r$ be the radius of the circle. Therefore, $OQ=r$ 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 $\triangle ORQ$ is right-angled at $R$. [[IMAGE0]] By the Pythagorean Theorem, $OR^2 + RQ^2 = OQ^2$ and so $s^2 + s^2 = r^2$ or $2s^2 = r^2$. 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 $2s^2 = r^2 = 72$, then $s^2 = 36$. In terms of $s$, the area of the square is $s^2$, so the area of the square is 36.