JAMB 2015 · UTME · Q16

Plane mensuration

In the figure, $PQ \parallel SR$ , $ST \parallel RQ$ , $PS = 7\text{ cm}$ , $PT = 7\text{ cm}$ and $SR = 4\text{ cm}$ . Find the ratio of the area of $QRST$ to the area of $PQRS$ .

Worked solution (try it first)

$TQ \parallel SR$ and $ST \parallel RQ$ , so $QRST$ is a parallelogram and $TQ = SR = 4$ cm.
So $PQ = 7 + 4 = 11$ cm.
Area of $QRST$ = base × height $= 4 \times 7 = 28\text{ cm}^2$ .
Area of trapezium $PQRS$ $= \frac12(11 + 4) \times 7$
$= 52.5\text{ cm}^2$ .
The ratio is $28 : 52.5$ .
Double both to clear the decimal: $56 : 105$ , option B.

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