# $AD$ is an altitude of an equilateral triangle $ABC$. On $AD$ as base, another equilateral triangle $ADE$ is constructed. Prove that Area($\triangle ADE$):Area($\triangle ABC$)$=3:4$.

Given:

$AD$ is an altitude of an equilateral triangle $ABC$. On $AD$ as base, another equilateral triangle $ADE$ is constructed.

To do:

We have to prove that Area($\triangle ADE$):Area($\triangle ABC$)$=3:4$.
Solution:

Let $AB=BC=AC=2x$

In an equilateral triangle altitude is also the perpendicular bisector.

This implies,

$BD=DC=x$

$\triangle ADB$ is a right triangle. Therefore, by using Pythagoras theorem,

$AB^2=AD^2+BD^2$

$(2x)^2=AD^2+x^2$

$AD^2=4x^2-x^2$

$AD^2=3x^2$

$AD=\sqrt{3x^2}$

$AD=\sqrt3x$

$\triangle ABC$ and $\triangle ADE$ are equilateral triangles and hence equiangular.

Therefore,

$\triangle ABC \sim\ \triangle ADE$

We know that,

If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

This implies,

$\frac{ar(\triangle ADE)}{ar(\triangle ABC)}=\frac{AD^2}{BC^2}$

$=\frac{3x^2}{(2x)^2}$

$=\frac{3x^2}{4x^2}$

$=\frac{3}{4}$

Hence proved.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

56 Views