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In figure below, $DE\ ||\ BC$.
If $DE\ =\ 4\ m$, $BC\ =\ 8\ cm$ and $Area\ (ΔADE)\ =\ 25\ cm^2$, find the $Area\ of\ ΔABC$.
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Given: In the given figure $DE\ ||\ BC$. $DE\ =\ 4\ m$, $BC\ =\ 8\ cm$ and $Area\ (ΔADE)\ =\ 25\ cm^2$. 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, $\frac{Ar(ΔADE)}{Ar(ΔABC)} = (\frac{DE}{BC})^2$ $\frac{25}{Ar(ΔABC)} = (\frac{4}{8})^2$ $Ar(ΔABC) = \frac{64 \times 25}{16}$ $Ar(ΔABC) = 100\ cm^2$ The $Area\ of\ ΔABC$ is $100\ cm^2$.
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