Question

Two isosceles triangles have equal vertical angles and their areas are in the ratio 16 : 25. Find the ratio of their corresponding heights. [4 MARKS]

Answer 1

Concept : 1 Mark
Application : 2 Marks
Calculation : 1 Mark

Let $\Delta ABC$ and $\Delta DEF$ be the given triangles in which

$AB=AC,DE=DF,\angle A=\angle D$

Draw $AL\perp BCandDM\perp EF$

Now, $\frac{AB}{AC}=1$ and $\frac{DE}{DF}=1$ [ ∵ AB = AC and DE = DF]

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

$\Rightarrow \frac{AB}{DE}=\frac{AC}{DF}.....\left(1\right)$

$\therefore In\Delta ABCand\Delta DEF,$

$\frac{AB}{DE}=\frac{AC}{DF}$ [ From (1)]

and $\angle A=\angle D$

$\Rightarrow \Delta ABC\sim \Delta DEF$ [By SAS similarity]

But, the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding heights.

$\frac{Area\left(\Delta ABC\right)}{Area\left(\Delta DEF\right)}=\frac{A{L}^2}{D{M}^2}$

$\Rightarrow \frac{16}{25}={\left(\frac{AL}{DM}\right)}^2$

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

$\therefore$ AL : DM = 4 : 5 , i.e, the ratio of their corresponding height = 4 : 5 .