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In two isosceles triangles, the angles opposite to their bases are equal and the ratio of their areas is $ 36: 25 $. Find the ratio of the corresponding altitudes of those triangles.
Given:
In two isosceles triangles, the angles opposite to their bases are equal and the ratio of their areas is \( 36: 25 \).
To do:
We have to find the ratio of their corresponding altitudes.
Solution:
Let the two triangles be as shown below:
$AB=AC$, $PQ=PR$ and $\angle A=\angle P$
$AD$ and $PS$ are altitudes.
$\frac{ar(\triangle ABC)}{ar(\triangle PQR)}=\frac{36}{25}$....(i)
In $\triangle ABC$ and $\triangle PQR$,
$\angle A=\angle P$
$\frac{AB}{PQ}=\frac{AC}{PR}$ (Since $\frac{AB}{AC}=\frac{PQ}{PR}$)
Therefore,
$\triangle ABC \sim\ \triangle PQR$ (By SAS similarity)
We know that,
If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.
$\frac{ar(\triangle ABC)}{ar(\triangle PQR)}=\frac{AB^2}{PQ^2}$
This implies,
$\frac{AB^2}{PQ^2}=\frac{36}{25}$
$\frac{AB}{PQ}=\sqrt{\frac{36}{25}}$
$\frac{AB}{PQ}=\frac{6}{5}$
In $\triangle ABD$ and $\triangle PQS$,
$\angle B=\angle Q$ (Since $\triangle ABC \sim\ \triangle PQR$)
$\angle ADB=\angle PSQ=90^o$
Therefore,
$\triangle ADB \sim\ \triangle PSQ$ (By AA similarity)
This implies,
$\frac{AB}{PQ}=\frac{AD}{PS}$
$\frac{AD}{PS}=\frac{6}{5}$
The ratio of their corresponding altitudes is $6:5$.
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