WAEC 2019 · Paper 2 · Q4

(a)
In the diagram, $PQRS$ is a quadrilateral, $\angle PQR = \angle PRS = 90^\circ$ , $|PQ| = 3\text{ cm}$ , $|QR| = 4\text{ cm}$ and $|PS| = 13\text{ cm}$ . Find the area of the quadrilateral.

Worked solution (try it first)

Split the quadrilateral along the diagonal $PR$ into two right-angled triangles.
In triangle $PQR$ (right angle at $Q$ ): $|PR| = \sqrt{3^2 + 4^2} = 5$ cm.
In triangle $PRS$ (right angle at $R$ ): $|RS| = \sqrt{13^2 - 5^2}$
$= \sqrt{144}$
$= 12$ cm.
Area $= \frac12 \times 3 \times 4 + \frac12 \times 5 \times 12$
$= 6 + 30$
$= 36\text{ cm}^2$ .