JAMB 1985 · UME · Q42

Plane mensuration

$PQRS$ is a trapezium of area $14\text{ cm}^2$ in which $PQ \parallel RS$ . If $PQ = 4$ cm and $SR = 3$ cm, find the area of $\triangle SQR$ in $\text{cm}^2$ .

Worked solution (try it first)

Area of a trapezium $= \frac12(a + b)h$ : $\frac12(4 + 3)h = 14$ , so $3.5h = 14$ and $h = 4$ cm.
$\triangle SQR$ has base $SR = 3$ cm, and its height is the distance between the parallel sides, 4 cm.
Area $= \frac12 \times 3 \times 4$
$= 6.0\text{ cm}^2$ , option B.

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