Let $$\triangle ABC \sim \triangle QPR  and  \frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \frac{4}{25}$$. If AB = 12 cm, BC = 8 cm and AC = 10 cm, then QP is equal to:

As Given,  $$\triangle ABC \sim \triangle QPR $$ So,

$$\frac {AB}{QP} = \frac{BC}{PR} = \frac{AC}{QR}$$

and also we know,

$$ \frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \frac {AB^2}{QP^2} = \frac{BC^2}{PR^2} = \frac{AC^2}{QR^2}$$, 

So here we Get,  $$\frac {AB}{QP} = \frac{BC}{PR} = \frac{AC}{QR} =  \frac{\sqrt{ar(\triangle ABC)}}{\sqrt{ar(\triangle PQR)}}=\frac{\sqrt4}{\sqrt25}$$

$$\therefore \frac {AB}{QP} = \frac{BC}{PR} = \frac{AC}{QR} = \frac{2}{5}$$

$$\Rightarrow \frac {AB}{QP} = \frac{2}{5}$$

$$\Rightarrow  QP = \frac{5AB}{2} =\frac{5\times12}{2} = \frac{60}{2} = 30 cm$$