Question

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides, Prove it.

Answer 1

Given & $ABC\sim \Delta EFG$

Draw $AD\perp BC$ & $EH\perp FG$

In ${\Delta }^{\prime \prime }ABD$ & ${\Delta }^{\prime \prime }EFH$

$\measuredangle B=\measuredangle F$ (angle of similar ${\Delta }^{\prime \prime }$)

$\measuredangle D=\measuredangle H$ (both ${90}^0$)

$\therefore$ ${\Delta }^{\prime \prime }ABD\sim {\Delta }^{\prime \prime }EFH$ (AA criterion)

$\therefore$ $\frac{AB}{EF}=\frac{AD}{EH}$

Now, $\frac{ar\left(ABC\right)}{ar\left(EFG\right)}=\frac{\frac{1}{2}AD\times BC}{\frac{1}{2}EH\times FG}=\frac{BC}{FG}\times \frac{AD}{EH}=\frac{BC}{FG}\times \frac{AB}{EF}={\left(\frac{BC}{FG}\right)}^2$ proved.