JAMB 1989 · UME · Q36

$PQRS$ is a rhombus. If $PR^2 + QS^2 = kPQ^2$ , determine $k$ .

Worked solution (try it first)

The diagonals of a rhombus bisect each other at right angles, so each side is the hypotenuse of a right-angled triangle with legs $\frac{PR}{2}$ and $\frac{QS}{2}$ .
By Pythagoras, $PQ^2 = \frac{PR^2}{4} + \frac{QS^2}{4}$ .
Multiply by 4: $PR^2 + QS^2 = 4PQ^2$ , so $k = 4$ , option D.