Question

In the following figure, $AE=EB,BD=2DC$ What is the ratio of the areas of PED and ABC?

• A. $\frac{1}{4}$
• B. $\frac{1}{6}$
• C. $\frac{1}{9}$
• D. $\frac{1}{12}$

Answer 1

The correct option is D $\frac{1}{12}$

Let B is origin and the position vector of A and C are $2\overrightarrow{a}$ and $3\overrightarrow{b}$
Then P.V. of $E=\overrightarrow{a}$ and P.V. of $D=2\overrightarrow{b}$
Now, let P divides AD in $\lambda :1$ ratio
and P divides EC in $\mu :1$
$\therefore \frac{2\overrightarrow{b}\lambda +2\overrightarrow{a}}{\lambda +1}=\frac{3\overrightarrow{b}\mu +\overrightarrow{a}}{\mu +1}$
$2\overrightarrow{b}\lambda \mu +2\overrightarrow{b}\lambda +2\overrightarrow{a}\mu +2\overrightarrow{a}=3\overrightarrow{b}\lambda \mu +\overrightarrow{a}\lambda +3\overrightarrow{b}\mu +\overrightarrow{a}$
$\overrightarrow{a}(2\mu +2-\lambda -1)=\overrightarrow{b}(3\lambda \mu +3\mu -2\lambda \mu -2\lambda )$
But $\overrightarrow{a}$ and $\overrightarrow{b}$ are not collinear.
$2\mu -\lambda +1=0$ and $\lambda \mu +3\mu -2\lambda =0$
We get $\mu =1$
Now, P.V. of P is $=\frac{\overrightarrow{a}+3\overrightarrow{b}}{2}$
Now, $\frac{ar\Delta PED}{ar\Delta ABC}=\frac{\frac{1}{2}|(\overrightarrow{a}-\frac{\overrightarrow{a}+3\overrightarrow{b}}{2})\times (2\overrightarrow{b}-\frac{\overrightarrow{a}+3\overrightarrow{b}}{2})|}{\frac{1}{2}|2\overrightarrow{a}\times 3\overrightarrow{b}|}$
$=\frac{\frac{1}{4}|(\overrightarrow{a}-3\overrightarrow{b})\times (\overrightarrow{b}-\overrightarrow{a})|}{6|\overrightarrow{a}\times \overrightarrow{b}|}=\frac{1}{12}$