The corresponding altitudes of two similar triangles are 10 cm and 5.5 cm respectively. Find the ratio between their areas.

Let the two similar 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.

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

$\Rightarrow \frac{\text{ Area of }∆\mathrm{ABC}}{\text{ Area of }∆\mathrm{PQR}}=\frac{{10}^2}{5.5^2}$

$=\frac{100}{2.25}$

$=\frac{400}{9}$

Hence, the ratio between the areas is 400:9.