In the figure triangle PQR, PQ = 8cm, PR = 12cm and S is a point on PR such that ∠PQS = ∠PRQ. ⓘ Write down the other pair of equal angles. ⓘ Use similar triangles to calculate the length of PS. ⓘ Given that the area of triangle PQS is 24cm², calculate the area of triangle PQR.

If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Let the ratio of corresponding sides be $k = \frac{s_1}{s_2}$.
The ratio of their areas is $\frac{\text{Area}_1}{\text{Area}_2} = k^2 = \left(\frac{s_1}{s_2}\right)^2$.

Given the ratio of corresponding sides is $4:7$, so $k = \frac{4}{7}$.

The ratio of their areas is $\boxed{\text{16:49}}$.

We have:

• $\angle QPS = \angle QPR$ (Common angle)
• $\angle PQS = \angle PRQ$ (Given)

Since two pairs of corresponding angles are equal, $\triangle PQS \sim \triangle PQR$ by the Angle-Angle (AA) similarity criterion.
Therefore, the third pair of corresponding angles must also be equal:

The other pair of equal angles is $\boxed{\angle QSP = \angle PQR}$.

Since $\triangle PQS \sim \triangle PQR$, the ratio of their corresponding sides is equal:

We use the ratio $\frac{PS}{PQ} = \frac{PQ}{PR}$.
Given $PQ = 8 \text{ cm}$ and $PR = 12 \text{ cm}$.

The length of PS is $\boxed{\frac{16}{3} \text{ cm}}$.

From part (ii), the ratio of corresponding sides is $\frac{PQ}{PR} = \frac{8}{12} = \frac{2}{3}$.

The ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides.

Given Area($\triangle PQS$) = $24 \text{ cm}^2$.

The area of triangle PQR is $\boxed{54 \text{ cm}^2}$.