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'Angle to Chord' printed from https://nrich.maths.org/
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
$x=4\sqrt{2}$.