Question

Two isosceles triangles are having equal vertical angles and their areas are in the ratio 9 : 16. Find the ratio of their corresponding altitudes.

Answer 1

As shown in the above figure, let $△ABC$ and $△DEF$ are the isosceles triangles with altitudes $AM$ and $DN$ and with $\angle A=\angle D$, therefore,

$\angle B=\angle C$ and $\angle E=\angle F$ that is

$\angle B=\angle C=\angle E=\angle F$

Thus, $△ABC\sim △DEF$

We know that the arc of similar triangles are proportional to squares of their corresponding altitude, therefore with the given ratio of area of triangles $9:16$, we have,

$\frac{Ar(△ABC)}{Ar(△DEF)}=\frac{A{M}^2}{D{N}^2}\Rightarrow \frac{9}{16}=\frac{A{M}^2}{D{N}^2}\Rightarrow \sqrt{\frac{A{M}^2}{D{N}^2}}=\sqrt{\frac{9}{16}}\Rightarrow \frac{AM}{DN}=\frac{3}{4}\Rightarrow AM:DN=3:4$

Hence the ratio of their altitudes is $3:4$.