Question

In $\triangle ABC$, if $P, Q, R$ divides sides $BC, AC$ and $AB$ respectively in the ratio $k:1$ and $\frac{\text{area of } \triangle PQR}{\text{area of } \triangle ABC}=\frac{1}{3}$, then $k$ is equal to

• A. 1/3
• B. 2
• C. 3
• D. None of these

Answer 1

Answer: (B)

2

Answer 2

Let $\Delta$ be the area of $\triangle ABC$. If points $P, Q, R$ divide sides $AB, BC, CA$ respectively in the ratio $k:1$, the ratio of the area of $\triangle PQR$ to the area of $\triangle ABC$ is given by the formula $\frac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle ABC)} = \frac{k^2 - k + 1}{(k+1)^2}$.

Assuming the question intended this standard configuration (or that the formula holds regardless of the cyclic permutation of sides), we set this ratio equal to the given value $1/3$:

$\frac{k^2 - k + 1}{(k+1)^2} = \frac{1}{3}$

$3(k^2 - k + 1) = (k+1)^2$

$3k^2 - 3k + 3 = k^2 + 2k + 1$

$2k^2 - 5k + 2 = 0$

Factoring the quadratic equation: $(2k - 1)(k - 2) = 0$

The solutions are $2k - 1 = 0 \implies k = 1/2$ or $k - 2 = 0 \implies k = 2$.

Since the ratio is $k:1$, $k$ must be positive. Both $1/2$ and $2$ are positive.

Comparing with the given options, $k=2$ is option b.