Let O be the centre of the circle.
Then ÐPOR=90^o 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, x²=4²+4²=2 ×4² and so x=4Ö2.