In figure below, $DE\ ||\ BC$.
If $DE\ :\ BC\ =\ 3\ :\ 5$. Calculate the ratio of the areas of $ΔADE$ and the trapezium $BCED$.

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Given:

In the given figure $DE\ ||\ BC$ and $DE\ :\ BC\ =\ 3\ :\ 5$.

To do:

We have to find the ratio of the areas of $ΔADE$ and the trapezium $BCED$.

Solution:

In $ΔADE$ and $ΔABC$,

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

$\angle DAE = \angle BAC$  (Common)

Therefore,

$ΔADE ~ ΔABC$ (By AA Similarity)

The ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Therefore,

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

$\frac{Ar(ΔADE)}{Ar(ΔABC)} = (\frac{3}{5})^2$

$\frac{Ar(ΔADE)}{Ar(ΔABC)} = \frac{9}{25}$

Let $\frac{Ar(ΔADE)}{Ar(ΔABC)} = \frac{9}{25}=\frac{9k}{25k}$

This implies,

Area of $ΔADE = 9k\ sq\ units$

Area of $ΔABC = 25k\ sq\ units$

Area of trapezium $BCED =$ Area of $ΔABC –$ Area of $ΔADE$

                                              $= 25k – 9k$

                                              $= 16k\ sq\ units$

Therefore,

$\frac{Ar(ΔADE)}{Ar(trap BCED)} = \frac{9k}{16k}$

$\frac{Ar(ΔADE)}{Ar(trapBCED)} = \frac{9}{16}$

The ratio of the areas of $ΔADE$ and the trapezium $BCED$ is $9:16$.

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Updated on: 10-Oct-2022

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