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)

3

Answer 2

Let P is 0,Q is q​ and R is r
A is 32q​+r​,B is 32r​ and C is 3q​​
Area of △PQR is =21​∣q​×r∣
Area of △ABC is 21​∣AB×AC∣
AB=3r−2q​​,AC=3−r−q​​
Area of △ABC=61​∣q​×r∣ Area(△ABC)Area(△PQR)​=3