WAEC 2024 · Paper 1 · Q21

Angles, triangles & polygons

In the diagram, $\overline{MN} \parallel \overline{KL}$ , $\overline{ML}$ and $\overline{KN}$ intersect at $X$ . $|MN| = 12\text{ cm}$ , $|MX| = 10\text{ cm}$ and $|KL| = 9\text{ cm}$ . If the area of $\triangle MXN$ is $16\text{ cm}^2$ , calculate the area of $\triangle LXK$ .

The paper marks this diagram “not drawn to scale”; the redraw uses the true measurements.

Worked solution (try it first)

$MN \parallel KL$ , so the alternate angles at $K$ and $N$ are equal, and so are those at $L$ and $M$ .
Triangles $LXK$ and $MXN$ are similar.
The matching sides $KL$ and $MN$ give the length scale factor $\frac{9}{12} = \frac34$ .
Areas scale by the square of that: $\left(\frac34\right)^2 = \frac{9}{16}$ .
So the area of $\triangle LXK$ is $16 \times \frac{9}{16} = 9\text{ cm}^2$ , option B.

Report a problem with this question