PK 2019 Q18

Let \(O\) be the centre of the circle, and \(P\), \(Q\), \(P'\) and \(Q'\) the vertices of the square. The triangles \(OPQ\) and \(OP'Q'\) are congruent since they are right-angled and have two equal sides (\(PQ = P'Q'\) since they are edges of a square, and \(OP = OP'\) because each is a radius). Hence \(OQ = OQ'\), and \(O\) is thus the midpoint of the edge of the square.

Let the side of the square be \(2x\). And \(OP = 1\) since it is a radius. And by Pythagoras’ Theorem on triangle \(OPQ\) we have \((2x)^2 + x^2 = 1^2\), so \(4x^2 + x^2 = 1\), and \(5x^2 = 1\). This gives \(x^2 = \tfrac{1}{5}\). The area of the square is \((2x)^2 = 4x^2 = 4 \times \tfrac{1}{5} = \tfrac{4}{5}\).