Question

In given figure two isosceles triangles have equal vertical angles and their areas are in the ratio 16:25. Find the ratio of their corresponding heights.

Answer 1

Let $△ABC$ and $△DEF$ be the given triangles such that AB=AC and DE=DF, $\angle A=\angle D$

and $\frac{Area\left(△ABC\right)}{Area\left(△DEF\right)}=\frac{16}{25}$.......(i)

Draw $AL\perp BC$ and $DM\perp EF$.

Now, $AB=AC,DE=DF$

$\Rightarrow$ $\frac{AB}{AC}=1$ and $\frac{DE}{DF}=1$

$\Rightarrow$ $\frac{AB}{AC}=\frac{DE}{DF}$

$\Rightarrow$ $\frac{AB}{DE}=\frac{AC}{DF}$

Thus, in triangles ABC and DEF, we have

$\frac{AB}{DE}=\frac{AC}{DF}$ and $\angle A=\angle D$ [Given]

So, by SAS-similarity criterion, we have

$△ABC\sim △DEF$

$\Rightarrow$ $\frac{Area\left(△ABC\right)}{Area\left(△DEF\right)}=\frac{{AL}^2}{{DM}^2}$

$\Rightarrow$ $\frac{16}{25}=\frac{{AL}^2}{{DM}^2}$ [Using (i)]

$\Rightarrow$ $\frac{AL}{DM}=\frac{4}{5}$

Hence, $AL:DM=4:5$