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The areas of two similar triangles are $81\ cm^2$ and $49\ cm^2$ respectively. Find the ration of their corresponding heights. What is the ratio of their corresponding medians?
Given:
The areas of two similar triangles are $81\ cm^2$ and $49\ cm^2$ respectively.
To do:
We have to find the ratio of their corresponding heights and their corresponding medians.
Solution:
Consider two similar triangles, $ΔABC$ and $ΔPQR$, $AD$ and $PS$ be the altitudes of $ΔABC$ and $ΔPQR$ respectively.
By area of similar triangles theorem,
$\frac{ar(ΔABC)}{ar(ΔPQR)} = \frac{AB^2}{PQ^2}$
$\frac{81}{49} = \frac{AB^2}{PQ^2}$
$ \begin{array}{l}
\frac{AB}{PQ} =\sqrt{\frac{81}{49}}\\
\\
\frac{AB}{PQ} =\frac{9}{7}
\end{array}$
In $ΔABD$ and $ΔPQS$,
$\angle B = \angle Q$
$\angle ABD = \angle PSQ = 90^o$
Therefore,
$ΔABD ∼ ΔPQS$ (By AA similarity)
$\frac{AB}{PQ} = \frac{AD}{PS}$ (corresponding parts of similar triangles are proportional)
$\frac{AD}{PS} = \frac{9}{7}$
Similarly,
The ratio of two similar triangles is equal to the ratio of the squares of their corresponding medians.
Therefore,
Ratio of altitudes $=$ Ratio of medians $= \frac{9}{7}$.
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