Question

Let $PQR$ be a triangle. The points $A,B$ and $C$ are on the sides $QR,RP$ and $PQ$ respectively such that $\frac{QA}{AR}=\frac{RB}{BP}=\frac{PC}{CQ}=\frac{1}{2}$. Then $\frac{A\text{ Area }(△PQR)}{\text{ Area }(△ABC)}$ 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 $△PQR$ is $=\frac{1}{2}|\overrightarrow{q}\times \overrightarrow{r}|$
Area of $△ABC$ 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 $△ABC=\frac{1}{6}|\overrightarrow{q}\times r|$ $\frac{\text{Area}(△PQR)}{\text{Area}(△ABC)}=3$