5. The areas of two similar triangles are in the ratio 16 : 9. The length of one side of the smaller triangle is $12 \mathrm{~cm}$, and the length of the height to that side is $15 \mathrm{~cm}$. Calculate the area and the length of the corresponding side of the larger triangle.
5. Let $P_{1}$ and $P_{2}$ be the areas, and $a_{1}$ and $a_{2}$ the corresponding sides of similar triangles.

It is given that $P_{2}=\frac{a_{2} \cdot v_{a_{2}}}{2}=\frac{12 \cdot 15}{2}=90 \mathrm{~cm}^{2}$.

Since $P_{1}: P_{2}=16: 9$, it follows that $P_{1}=160 \mathrm{~cm}^{2}$.

1 POINT

Given that $P_{1}: P_{2}=16: 9=k^{2}$, then $k=\frac{4}{3}$.

2 POINTS

Since $a_{1}: a_{2}=k$, it follows that $a_{1}=16 \mathrm{~cm}$.

The area of the larger triangle is $160 \mathrm{~cm}^{2}$, and the length of the corresponding side is $16 \mathrm{~cm}$.