Given △ABC ~ △PQR, area of △ABC = 54 cm² and area of △PQR = 24 cm². If AD and PM are medians of △'s ABC and PQR respectively, and length of PM is 10 cm, then length of AD is

1. $\dfrac{49}{9}$ cm

2. $\dfrac{20}{3}$ cm

3. 15 cm

4. 22.5 cm

Given, △ABC ~ △PQR.

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

$\therefore \dfrac{\text{Area of △ABC}}{\text{Area of △PQR}} = \dfrac{AD^2}{PM^2} \\[1em] \Rightarrow \dfrac{54}{24} = \dfrac{x^2}{10^2} \\[1em] \Rightarrow x^2 = \dfrac{54 \times 100}{24} \\[1em] \Rightarrow x^2 = \dfrac{5400}{24} \\[1em] \Rightarrow x^2 = 225 \\[1em] \Rightarrow x = 15.$

Hence, length of AD = 15 cm.

Hence, Option 3 is the correct option.