Problem 629

Let $O$ be the centre of the circle. Join $OS$ and $OR$.

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Since the diameter of the semicircle is 20, then its radius is half of this, or 10.

Since $OS$ and $OR$ are radii, then $OS=OR=10$.

Consider $\triangle OPS$ and $\triangle OQR$.

Since $PQRS$ is a rectangle, both triangles are right-angled (at $P$ and $Q$).

Also, $PS=QR$ (equal sides of the rectangle) and $OS=OR$ (since they are radii of the circle).

Therefore, $\triangle OPS$ is congruent to $\triangle OQR$. (Right-angled triangles with equal hypotenuses and one other pair of equal corresponding sides are congruent.)

Since $\triangle OPS$ and $\triangle OQR$ are congruent, then $OP=OQ$.

Since $PQ=16$, then $OP = \frac{1}{2}PQ = 8$.

Finally, since $\triangle OPS$ is right-angled at $P$, then we can apply the Pythagorean Theorem to conclude that $PS = \sqrt{OS^2 - OP^2} = \sqrt{10^2-8^2} = \sqrt{100-64}=6$.