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WAEC Mathematics 2024 Paper

Question : 21

Total: 50

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

9cm $^2$
8cm $^2$
10cm $^2$
12cm $^2$

Validate

Solution:

Concept:

Since

\overline{MN} \parallel \overline{KL}

\triangle MXN

and

\triangle LXK

are similar triangles.

For similar triangles, the ratio of their areas equals the square of the ratio of their corresponding sides.

Explanation:

The corresponding sides are

\overline{MN}

and

\overline{KL}

. Their lengths are

12\,\text{cm}

and

9\,\text{cm}

Thus, the ratio of the areas is:

\frac{\text{Area}(\triangle MXN)}{\text{Area}(\triangle LXK)} = \left(\frac{12}{9}\right)^2 = \frac{144}{81}.

Given that the area of

\triangle MXN

16\,\text{cm}^2

, we substitute:

\frac{16}{\text{Area}(\triangle LXK)} = \frac{144}{81}.

Cross-multiply and solve:

\text{Area}(\triangle LXK) = \frac{16 \times 81}{144}.

Simplify:

\text{Area}(\triangle LXK) = \frac{1296}{144} = 9.

Hence, the area of

\triangle LXK

9\,\text{cm}^2

Answer:

9\,\text{cm}^2

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