Question

Show that the ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Answer 1

$\Delta ABC$ and $\Delta DEF$ be similar triangles.

AM and DN are the altitudes of the $\Delta ABC$ and $\Delta DEF$ respectively

$\frac{Areaof\Delta ABC}{Areaof\Delta DEF}=\frac{\left(\frac{1}{2}\right)\times \left(AM\right)\times \left(BC\right)}{\left(\frac{1}{2}\right)\times \left(DN\right)\times \left(EF\right)}--------\left(1\right)$

$\Delta AMB$ and $\Delta DNE$ are similar and $\Delta ABC$ and $\Delta DEF$ are similar.

$\frac{AM}{DN}=\frac{AB}{DE}$

$\frac{AB}{DE}=\frac{BC}{EF}$

$\Rightarrow \frac{AM}{DN}=\frac{BC}{EF}--------\left(2\right)$

Substituting (2) in equation (1), we can get the desired result.

$\frac{Areaof\Delta ABC}{Areaof\Delta DEF}=\frac{(BC{)}^2}{(EF{)}^2}$