Question

Two isosceles triangles have equal vertical angles and their area are in the ratio of the $16:25$. Find the ratio of their corresponding heights.

Answer 1

$AB=AC$ (given)

$\therefore \frac{AB}{AC}=1-\left(1\right)$

$DE=DF$ (given)

$\frac{DE}{DF}=1-\left(2\right)$

comparing$\left(1\right)\&\left(2\right)$

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

$In△ABC\&△DEF$

$\frac{AB}{DE}=\frac{AC}{DF}⟹\angle A=\angle D$

$\therefore$ By using SAS similar condition

$△ABC\sim △DEF$

$\frac{ar\left(△ABC\right)}{ar\left(△DEF\right)}=\frac{{AD}^2}{{DQ}^2}$

$\therefore$ The area of two similar $△les$ are in the ratio of the square of the corresponding altitude.

$\frac{16}{25}=\frac{{AP}^2}{{DQ}^2}$

$⟹\frac{AP}{DQ}=\frac{4}{5}$