Question

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

• A. $1:2$
• B. $2:3$
• C. $3:4$
• D. $2:1$

Answer 1

The correct option is C $3:4$

Let $ABC,PQR$ be two isosceles triangle with equal vertical angles.

Let in $\Delta ABC,AB=AC$

$\Rightarrow \angle B=\angle C=\frac{180-\angle A}{2}$

And in $\Delta PQR,PQ=PR$

$\Rightarrow \angle Q=\angle R=\frac{180-\angle P}{2}$

Given two vertical angles are equal.

$\therefore \angle A=\angle P$

$\Rightarrow \angle B=\angle C=\angle Q=\angle R$

By $AAA$ postulate, "two triangles are similar if they have three corresponding angles congruent."

$\therefore \Delta ABC\sim \Delta PQR$

We know ratio between the areas of two similar triangles is same as the ration between the squares of their corresponding altitudes and corresponding heights of two given triangles are $AD$ and $PS$.

$\frac{Areaof\Delta ABC}{Areaof\Delta PQR}=\frac{A{D}^2}{P{S}^2}$

$\Rightarrow \frac{9}{16}=\frac{A{D}^2}{P{S}^2}$

$\Rightarrow AD:PS=3:4$