WAEC 2011 · Paper 2 · Q12

(a)
The area of trapezium $PQRS$ is $60\text{ cm}^2$ . $PQ \parallel RS$ , $|PQ| = 15\text{ cm}$ , $|RS| = 25\text{ cm}$ and $\angle PSR = 30^\circ$ . Calculate the: (i) perpendicular height of $PQRS$ ; (ii) $|PS|$ .
Separate values with commas, e.g. 3, −2
(b)
Ade received $\frac35$ of a sum of money, Nelly $\frac13$ of the remainder while Austin took the rest. If Austin's share is greater than Nelly's share by ₦3,000, how much did Ade receive?

Worked solution (try it first)

(a)(i)

Area of a trapezium $= \frac12 \times (\text{sum of the parallel sides}) \times \text{height}$ .
So $60 = \frac12 \times (15 + 25) \times h$
$= 20h$ , and $h = 3\text{ cm}$ .

(ii)

Drop a perpendicular from $P$ to $SR$ .
It is the height, 3 cm, and it is opposite the $30^\circ$ angle at $S$ , with $PS$ as the hypotenuse: $\sin 30^\circ = \frac{3}{|PS|}$ .
So $|PS| = \frac{3}{\sin 30^\circ}$
$= \frac{3}{\frac12}$
$= 6\text{ cm}$ .

(b)

Let the sum be ₦ $y$ .
Ade got $\frac35y$ , leaving $\frac25y$ .
Nelly got $\frac13$ of the remainder: $\frac13 \times \frac25y = \frac{2}{15}y$ .
Austin got the rest of the remainder: $\frac25y - \frac{2}{15}y = \frac{4}{15}y$ .
Austin's share is ₦3,000 more than Nelly's: $\frac{4}{15}y - \frac{2}{15}y = 3000$ .
So $\frac{2}{15}y = 3000$ and $y = 22\,500$ .
Ade received $\frac35 \times 22\,500 =$ ₦13,500.

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