Let $O$ be the centre of the circle.

Then $\angle POR=90^{\circ}$ as the angle subtended by an arc at the centre of a circle is twice the angle subtended by that arc at a point on the circumference of the circle.

So triangle $POR$ is an isosceles right-angled triangle with $PO=RO=4cm$. Let the length of $PR$ be $x$ cm.

Then, by Pythagoras' Theorem, $x^2=4^2+4^2=2 \times 4^2$ and so $x=4\sqrt{2}$.

This problem is taken from the UKMT Mathematical Challenges.

You can find more short problems, arranged by curriculum topic, in our short problems collection.

Visualising. Angle properties of polygons. Angles - points, lines and parallel lines. Similarity and congruence. Pythagoras' theorem. Circle properties and circle theorems. Regular polygons and circles. Creating and manipulating expressions and formulae. Sine, cosine, tangent. 2D shapes and their properties.