Question

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

Answer 1

Let $△$ABC and $△$DEF be the given triangles in which AB = AC, DE = DF, $\angle$ A = $\angle$ D and

Draw AL $\perp$ BC and DM $\perp$ EF

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

$\Rightarrow$ $\frac{AB}{AC}-\frac{DE}{DF}$
$\therefore$ In $△$ABC and $△$DEF, we have

$\frac{AB}{DE}-\frac{AC}{DF}$ and $\angle$A = $\angle$D
$\Rightarrow$ $△$ABC and $△$DEF [By SAS similarity axiom]

But, the ratio of the areas of two similar As is the same as the ratio of the squares of their corresponding heights.

$\frac{Area\left(△ABC\right)}{Area\left(△DEF\right)}-\frac{A{L}^2}{D{M}^2}$ $\Rightarrow$ $\frac{16}{25}$-${\left(\frac{AL}{DM}\right)}^2$ $\Rightarrow$ $\frac{AL}{DM}$ - $\frac{4}{5}$

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