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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 $ΔABC$.
"
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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