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Two isosceles triangles have equal vertical angles and their areas are in the ratio $36:25$. Find the ratio of their corresponding heights.
Given:
Two isosceles triangles have equal vertical angles and their areas are in the ratio $36:25$.
To do:
We have to find the ratio of their corresponding heights.
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 heights is $6:5$.
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