Question

Areas and ratio of angle bisectors of two similar triangles are given. Match the correct ratio in each case.

• A. 9 : 7
• B. 7 : 9
• C. 1 : 2
• D. 2 : 1

Answer 1

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.


Consider $△ABC$ and $△DEF$ with AP DQ as angular bisectors of $\angle A$ and $\angle D$ resp.

$\frac{ar\left(△ABC\right)}{ar(△DEF})=\frac{A{B}^2}{D{E}^2}....(i)$

$\angle A=\angle D$ $(\because △ABC\sim △DEF)$

$\frac{1}{2}\angle A=\frac{1}{2}\angle D$
$\angle BAP=\angle EDQ$

In $△ABP$ and $△DEQ$,
$\angle ABP=\angle DEQ$
$\angle BAP=\angle EDQ$
$\Rightarrow$ $△ABP\sim △DEQ$
(By AA similarity)

$\therefore$ $\frac{AB}{DE}=\frac{AP}{DQ}$
$\Rightarrow$ $\frac{A{B}^2}{D{E}^2}=\frac{A{P}^2}{D{Q}^2}$

So we can say that
$\frac{ar\left(△ABC\right)}{ar(△DEF})=\frac{A{P}^2}{D{Q}^2}$ [From (i)]

$\therefore$ The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding angle bisectors.

Now $\frac{A{P}^2}{D{Q}^2}=\frac{ar\left(△ABC\right)}{ar\left(△DEF\right)}$

1. $\frac{ar\left(△ABC\right)}{ar\left(△DEF\right)}$= $\frac{81c{m}^2}{49c{m}^2}$
$\Rightarrow$ $\frac{A{P}^2}{D{Q}^2}=\sqrt{\frac{81}{40}}=\frac{9}{7}$

2. $\frac{ar\left(△ABC\right)}{ar\left(△DEF\right)}$= $\frac{49c{m}^2}{81c{m}^2}$
$\Rightarrow$ $\frac{A{P}^2}{D{Q}^2}=\sqrt{\frac{49}{81}}=\frac{7}{9}$

3. $\frac{ar\left(△ABC\right)}{ar\left(△DEF\right)}$= $\frac{xc{m}^2}{2xc{m}^2}$
$\Rightarrow$ $\frac{A{P}^2}{D{Q}^2}=\sqrt{\frac{x}{2x}}=\frac{1}{\sqrt{2}}$

4. $\frac{ar\left(△ABC\right)}{ar\left(△DEF\right)}$= $\frac{2xc{m}^2}{xc{m}^2}$
$\Rightarrow$ $\frac{A{P}^2}{D{Q}^2}=\sqrt{\frac{2x}{x}}=\frac{\sqrt{2}}{1}$