Question

In a triangle $ABC,$, $P,Q,R$ divide the sides $BC,CA$ and $AB$ respectively in the ratio $k:1$ (in order).

If the ratio $\left(\frac{areaofPQR}{areaofABC}\right),$ then $k$ is equal to?

• A. $1$
• B. $2$
• C. $\frac{1}{2}$
• D. None of these

Answer 1

Answer: (B), (C)

$2$
$\frac{1}{2}$

Area of triangle $ARQ=\frac{b}{(k+1)}.\frac{kc}{(k+1)}.\frac{1}{2}\sin A=\frac{k\Delta }{{(k+1)}^2}$

Similarly, area of $△BPR=$ area of $△PCQ=\frac{k\Delta }{{(k+1)}^2}$

$\Rightarrow$ area of $\Delta PQR=\Delta -\frac{3k\Delta }{{(k+1)}^2}=\left(\frac{k^2-k+1}{k^2+2k+1}\right)\Delta$

Ratio $=\left(\frac{k^2-k+1}{k^2+2k+1}\right)=\frac{1}{3}\Rightarrow k=2$ or $\frac{1}{2}$