JAMB 1988 · UME · Q33

In the figure, $PS = 7$ cm and $RY = 9$ cm. If the area of parallelogram $PQRS$ is $56\text{ cm}^2$ , find the area of trapezium $PQTS$ .

Worked solution (try it first)

The area of parallelogram $PQRS$ is base × height, so the height is $56 \div 7 = 8$ cm.
$QR = PS = 7$ cm, and $PYTS$ is also a parallelogram, so $YT = PS = 7$ cm.
So $QT = 7 + 9 + 7 = 23$ cm.
Trapezium $PQTS$ has parallel sides 7 and 23 cm: area $= \frac12(7 + 23) \times 8$
$= 120\text{ cm}^2$ , option C.