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

The Correct Option is 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 $△ABC=\frac{1}{6}|\overrightarrow{q}\times r|$ $\frac{\text{Area}(△PQR)}{\text{Area}(△ABC)}=3$

Answer 2

Let \( P = \vec{0}, Q = \vec{q}, R = \vec{r} \). The coordinates of \( A, B, C \) can be written as: \[ A = \frac{2\vec{q} + \vec{r}}{3}, \quad B = \frac{2\vec{r} + \vec{p}}{3}, \quad C = \frac{2\vec{p} + \vec{q}}{3}. \] The area of \( \triangle PQR \) is: \[ \text{Area of } \triangle PQR = \frac{1}{2} \left| \vec{q} \times \vec{r} \right|. \] The area of \( \triangle ABC \) is: \[ \text{Area of } \triangle ABC = \frac{1}{2} \left| \frac{\vec{r} - 2\vec{q}}{3} \times \frac{\vec{q} - 2\vec{r}}{3} \right|. \] Simplifying: \[ \text{Area of } \triangle ABC = \frac{1}{6} \left| \vec{q} \times \vec{r} \right|. \] Thus, the ratio of areas is: \[ \frac{\text{Area of } \triangle PQR}{\text{Area of } \triangle ABC} = \frac{\frac{1}{2} \left| \vec{q} \times \vec{r} \right|}{\frac{1}{6} \left| \vec{q} \times \vec{r} \right|} = 3. \]