Question

Two isosceles triangles have equal vertical angles and their areas are in the ratio $9:16$. What is the ratio of their corresponding altitudes?

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

Answer 1

The correct option is C $3:4$

In $△ABC$

$AB=AC$

$\Rightarrow \angle B=\angle C=\frac{180-\angle A}{2}$.....(I)

and in $△PQR$

$PQ=PR$

$\Rightarrow \angle Q=\angle R=\frac{180-\angle P}{2}$.......(II)

$\therefore$vertical angle of the two triangles are equal.

If two triangles are equal then the ratio of their area is equal to the square of their corresponding altitudes

$\frac{arc△ABC}{arc△PQR}=\frac{A{D}^2}{P{S}^2}$

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

$\Rightarrow \frac{AD}{PS}=\frac{3}{4}$