Question

Let $P Q R$ be a triangle. The points $A, B$ and $C$ are on the sides $Q R, R P$ and $P Q$ respectively such that $\frac{Q A}{A R}=\frac{R B}{B P}=\frac{P C}{C Q}=\frac{1}{2}$. Then $\frac{A \text { Area }(\triangle P Q R)}{\text { Area }(\triangle A B C)}$ is equal to

• A. 4
• B. 3
• C. 2
• D. $\frac{5}{2}$

Answer 1

Answer: (B)

Let $P$ is $\overrightarrow{0}, Q$ is $\vec{q}$ and $R$ is $\vec{r}$
$A$ is $\frac{2 \overrightarrow{ q }+\overrightarrow{ r }}{3}, B$ is $\frac{2 \overrightarrow{ r }}{3}$ and $C$ is $\frac{\overrightarrow{ q }}{3}$
Area of $\triangle PQR$ is $=\frac{1}{2}|\overrightarrow{ q } \times \overrightarrow{ r }|$
Area of $\triangle ABC$ is $\frac{1}{2}|\overrightarrow{ AB } \times \overrightarrow{ AC }|$
$\overrightarrow{ AB }=\frac{\overrightarrow{ r }-2 \overrightarrow{ q }}{3}, \overrightarrow{ AC }=\frac{-\overrightarrow{ r }-\overrightarrow{ q }}{3}$
Area of $\triangle ABC =\frac{1}{6}|\overrightarrow{ q } \times r |$ $\frac{\text{Area}(\triangle PQR )}{\text{Area}(\triangle ABC )}=3$