The areas of two similar triangles are 100 cm² and 64 cm² respectively. If the altitude of the smaller triangle is 5.5 cm. Find the corresponding altitude of the bigger triangle.

Let the two triangles be ABC and PQR with altitudes AD and PS, respectively.

We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding altitudes.

$\therefore \frac{\text{ Area of }∆\mathrm{ABC}}{\text{ Area of }∆\mathrm{PQR}}=\frac{{\mathrm{AD}}^2}{{\mathrm{PS}}^2}$

$\Rightarrow \frac{100}{64}=\frac{{\mathrm{AD}}^2}{5.5^2}$

$\Rightarrow \frac{\mathrm{AD}}{5.5}=\sqrt{\frac{100}{64}}$

$\Rightarrow \frac{\mathrm{AD}}{5.5}=\frac{10}{8}$

$\Rightarrow \mathrm{AD}=\frac{10}{8}\times 5.5$

$\Rightarrow \mathrm{AD}=6.875 \mathrm{cm}$

Hence, the altitude of the bigger triangle is 6.875 cm.