WAEC 2015 · Paper 2 · Q5

Plane mensuration
Trigonometric ratios

A trapezium $PQRS$ is such that $PQ \parallel RS$ and the perpendicular from $P$ to $RS$ is $40\text{ cm}$ . If $|PQ| = 20\text{ cm}$ , $|SP| = 50\text{ cm}$ and $|SR| = 60\text{ cm}$ , calculate, correct to 2 significant figures, the:

(a)
area of the trapezium;
(b)
$\angle QRS$ .

Worked solution (try it first)

Draw the trapezium with $SR$ (60 cm) at the bottom and $PQ$ (20 cm) at the top.
Drop perpendiculars from $P$ and $Q$ to $SR$ , meeting it at $N$ and $M$ .
Both are 40 cm long.

(a)

Area $= \frac12(20 + 60) \times 40$
$= 1600\text{ cm}^2$ .

(b)

In the right-angled triangle $PNS$ : $|SN| = \sqrt{50^2 - 40^2} = 30$ cm.
$NM = PQ = 20$ cm, so $|MR| = 60 - 30 - 20 = 10$ cm.
In triangle $QMR$ , the height 40 is opposite $\angle QRS$ and $MR = 10$ is adjacent: $\tan\angle QRS = \frac{40}{10} = 4$ .
So $\angle QRS \approx 76^\circ$ .

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