In figure below, $DE\ ||\ BC$.
If $DE\ =\ 4\ m$, $BC\ =\ 6\ cm$ and $Area\ (ΔADE)\ =\ 16\ cm^2$, find the $Area\ of\ ΔABC$.
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Given:

In the given figure $DE\ ||\ BC$.

$DE\ =\ 4\ m$, $BC\ =\ 6\ cm$ and $Area\ (ΔADE)\ =\ 16\ cm^2$.

To do:

We have to find the $Area\ of\ ΔABC$.

Solution:

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.

Therefore,

$\frac{Ar(ΔADE)}{Ar(ΔABC)} = (\frac{DE}{BC})^2$

$\frac{16}{Ar(ΔABC)} = (\frac{4}{6})^2$

$Ar(ΔABC) = \frac{36 \times 16}{16}$

$Ar(ΔABC) = 36\ cm^2$

The $Area\ of\ ΔABC$ is $36\ cm^2$.

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Updated on: 10-Oct-2022

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