Question

In $△$ ABC , D and E are the mid points of AB and AC respectively. Find the ratio of the areas of $△$ ADE and $△$ ABC.

Answer 1

Given:

In ΔABC, D and E are the midpoints of AB and AC respectively.

Therefore, $DE\parallel BC$ (By Converse of mid-point theorem)

Also, $DE=\frac{1}{2}BC$

In $△ADEand△ABC$

$\angle ADE=\angle B$ (Corresponding angles)

$\angle DAE=\angle BAC$ (common)

$△ADE\sim △ABC$ (By AA Similarity)

We know that the ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.

$\frac{ar\left(△ADE\right)}{ar\left(△ABC\right)}=\frac{A{D}^2}{A{B}^2}$

$\frac{ar\left(△ADE\right)}{ar\left(△ABC\right)}=\frac{A{D}^2}{2A{D}^2}$

[AD = 2AB as D is the mid point]

$\frac{ar\left(△ADE\right)}{ar\left(△ABC\right)}=\frac{1^2}{2^2}$

$\frac{ar\left(△ADE\right)}{ar\left(△ABC\right)}=\frac{1}{4}$

Hence, the ratio of the areas ADE and ABC is $ar\left(△ADE\right):ar\left(△ABC\right)=1:4.$