WAEC 2018 · Paper 2 · Q9

(a)
In the diagram, $T$ is on $PS$ and $Q$ is on $PR$ with $TQ \parallel SR$ . $|PT| = 4\text{ cm}$ , $|TS| = 6\text{ cm}$ , $|PQ| = 6\text{ cm}$ and $\angle SPR = 30^\circ$ . Calculate, correct to the nearest whole number: (i) $|SR|$ ; (ii) the area of $TQRS$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

In triangle $PTQ$ , two sides and the angle between them are known, so use the cosine rule: $|TQ|^2 = 6^2 + 4^2 - 2(6)(4)\cos 30^\circ$
$= 52 - 41.57$
$\approx 10.43$ , so $|TQ| \approx 3.23$ cm.
$TQ \parallel SR$ , so triangle $PTQ$ is similar to triangle $PSR$ , with every length multiplied by $\frac{|PS|}{|PT|} = \frac{10}{4} = 2.5$ .
So $|SR| = 2.5 \times 3.23 \approx 8.07$ , which is 8 cm to the nearest whole number.
(Also $|PR| = 2.5 \times 6 = 15$ cm.)

Area of a triangle $= \frac12ab\sin C$ .
Area of $PSR = \frac12 \times 10 \times 15 \times \sin 30^\circ$
$= 37.5\text{ cm}^2$ .
Area of $PTQ = \frac12 \times 4 \times 6 \times \sin 30^\circ$
$= 6\text{ cm}^2$ .
So area of $TQRS = 37.5 - 6 = 31.5$ , which is 32 cm² to the nearest whole number.