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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 $x=4\sqrt{2}$.
Angle to Chord
Age 14 to 16
ShortChallenge Level 
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.
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