Question

$\Delta$ DEF$\sim$ $\Delta$ MNK. If DE$=5$ and MN$=6$, then find the value of $\frac{A\left(\Delta DEF\right)}{A\left(\Delta MNK\right)}$.

Answer 1

Given $\Delta DEF\sim \Delta MNK$ and $DE=5,MN=6$

$\therefore \frac{A\left(\Delta DEF\right)}{A\left(\Delta MNK\right)}=\frac{D{E}^2}{M{N}^2}$ (ratio of areas of similar triangles is equal to ratio of squares of the corresponding sides).

$=\frac{5^2}{6^2}$

$=\frac{25}{36}$

$\therefore \frac{A\left(\Delta DEF\right)}{A\left(\Delta MNK\right)}=\frac{25}{36}$