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Problem 547

Geometry Difficulty 2.5 Find the answer CEMC Cayley · Canada · 2020

In the diagram, the circle has centre OO and square OPQROPQR has vertex QQ on the circle.

Figure 0

If the area of the circle is 72π\pi, the area of the square is

3838
4848
2525
1212
3636

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

Let ss be the side length of the square. Therefore, OR=RQ=sOR=RQ=s.

Let rr be the radius of the circle. Therefore, OQ=rOQ=r since OO is the centre of the circle and QQ is on the circumference of the circle.

Since the square has a right-angle at each of its vertices, then △ORQ\triangle ORQ is right-angled at RR.

Figure 1

By the Pythagorean Theorem, OR2+RQ2=OQ2OR^2 + RQ^2 = OQ^2 and so s2+s2=r2s^2 + s^2 = r^2 or 2s2=r22s^2 = r^2.

In terms of rr, the area of the circle is πr2\pi r^2.

Since we are given that the area of the circle is 72π72\pi, then πr2=72π\pi r^2 = 72\pi or r2=72r^2 = 72.

Since 2s2=r2=722s^2 = r^2 = 72, then s2=36s^2 = 36.

In terms of ss, the area of the square is s2s^2, so the area of the square is 36.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.