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In $ΔABC$, $D$ and $E$ are the mid- points of $AB$ and $AC$ respectively. Find the ratio of the areas $ΔADE$ and $ΔABC$.
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Given:
In $ΔABC$, $D$ and $E$ are the mid-points of $AB$ and $AC$ respectively.
To do:
Here, we have to find the ratio of the areas $ΔADE$ and $ΔABC$.
Solution:
$D$ and $E$ are the mid-points of $AB$ and $AC$ respectively.
This implies,
$DE||BC$ (By converse of mid-point theorem)
$\therefore DE = (\frac{1}{2}) BC$
In $ΔADE$ and $ΔABC$,
$\angle ADE = \angle ABC$ (Corresponding angles)
$\angle DAE = \angle BAC$ (Common)
Therefore,
$ΔADE ~ ΔABC$ (By AA Similarity)
We know that,
The ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
$\frac{Ar(ΔADE)}{Ar(ΔABC)} = (\frac{AD}{AB})^2$
$\frac{Ar(ΔADE)}{Ar(ΔABC)} = (\frac{AD}{2AD})^2$ ($D$ is the mid-point of$AB$)
$\frac{Ar(ΔADE)}{Ar(ΔABC)} =(\frac{1}{2})^2$
$\frac{Ar(ΔADE)}{Ar(ΔABC)} =\frac{1}{4}$
The ratio of the areas $ΔADE$ and $ΔABC$ is $1:4$.
Updated on: 10-Oct-2022
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