The areas of two similar triangles are 121 cm² and 25 cm² respectively. If the median of the bigger triangle is 10 cm. Find the corresponding median of the smaller triangle.

Let the two triangles be ABC and PQR with medians 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 medians.

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

$\Rightarrow \frac{121}{25}=\frac{{10}^2}{{\mathrm{PS}}^2}$

$\Rightarrow \frac{10}{\mathrm{PS}}=\sqrt{\frac{121}{25}}$

$\Rightarrow \frac{10}{\mathrm{PS}}=\frac{11}{5}$

$\Rightarrow \mathrm{PS}=\frac{10\times 5}{11}$

$\Rightarrow \mathrm{PS}=\frac{50}{11} \mathrm{cm}$

Hence, the median of the smaller triangle is $\frac{\mathbf{50}}{\mathbf{11}}\mathbf{ }\mathbf{cm}$.