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$ABC$ is a triangle in which $∠\ A\ =\ 90^o$, $AN\ ⊥\ BC$, $BC\ =\ 12\ cm$ and $AC\ =\ 5\ cm$. Find the ratio of the areas of $ΔANC$ and $ΔABC$.
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Given:


$ABC$ is a triangle in which $∠\ A\ =\ 90^o$, $AN\ ⊥\ BC$, $BC\ =\ 12\ cm$ and $AC\ =\ 5\ cm$. 


To do:


We have to find the ratio of the areas of $ΔANC$ and $ΔABC$.

Solution:


In $ΔANC$ and $ΔABC$,

$\angle ACN = \angle ACB$  (Common)

$\angle ANC = \angle BAC=90^o$

Therefore,

$ΔANC ∼ ΔABC$  (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.

This implies,

$\frac{ar(ΔANC)}{ar(ΔABC)} = (\frac{AC}{BC})^2$

                                                 $= (\frac{5}{12})^2 = \frac{25}{144}$

Therefore, $ar(ΔANC):ar(ΔABC) = 25:144$.

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Updated on: 10-Oct-2022

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