JAMB 2016 · UTME · Q30

Plane mensuration

In the figure, $PQST$ is a parallelogram and $TSR$ is a straight line. If the area of $\triangle QRS$ is $20\text{ cm}^2$ , find the area of the trapezium $PQRT$ .

Worked solution (try it first)

$\triangle QRS$ has base $SR = 8$ cm: $\frac12 \times 8 \times h = 20$ , so $h = 5$ cm.
This is also the height of the trapezium.
$PQST$ is a parallelogram, so $PQ = TS = 10$ cm.
The other parallel side is $TR = 10 + 8 = 18$ cm.
Area of the trapezium $= \frac12(10 + 18) \times 5$
$= 70\text{ cm}^2$ , option C.

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