WAEC 2021 · Paper 2 · Q9

In the diagram, $PQRS$ is a trapezium with $\overline{QR} \parallel \overline{PS}$ . $U$ and $T$ are points on $\overline{PS}$ such that $|PU| = 5\text{ cm}$ , $|QU| = 12\text{ cm}$ and $\angle PUQ = \angle STR = 90^\circ$ , and $\angle RST = 50^\circ$ . If the area of $\triangle PQR = 20\text{ cm}^2$ , calculate, correct to the nearest whole number, the:

(a)
perimeter;
(b)
area, of the trapezium.

Worked solution (try it first)

$QU$ is perpendicular to $PS$ , so it is the height of the trapezium: 12 cm.
Triangle $PQR$ has base $QR$ and the same height: $\frac12 \times |QR| \times 12 = 20$ , so $|QR| = \frac{10}{3} \approx 3.33$ cm, and $|UT| = |QR| = 3.33$ cm.
$|PQ| = \sqrt{12^2 + 5^2} = 13$ cm.
In right-angled triangle $RTS$ , $RT = 12$ is opposite the $50^\circ$ angle: $|RS| = \frac{12}{\sin 50^\circ}$
$\approx 15.665$ cm and $|TS| = \frac{12}{\tan 50^\circ}$
$\approx 10.069$ cm.
So $|PS| = 5 + 3.33 + 10.069 \approx 18.40$ cm.

(a)

Perimeter $\approx 13 + 3.33 + 15.665 + 18.40$
$\approx 50.4$ , which is 50 cm to the nearest whole number.

(b)

Area $= \frac12 \times (3.33 + 18.40) \times 12$
$\approx 130.4$ , which is 130 cm² to the nearest whole number.

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