Question

In a DABC, if P,Q,R divide sides AB, BC, CA respectively in k : 1. If $\frac { \text { area of } \triangle \mathrm { PQR } } { \text { area of } \triangle \mathrm { ABC } } = \frac { 1 } { 3 }$ , then k equals

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

Answer 1

Answer: (B)

2

Answer 2

Area of (D ARQ) = $\frac { 1 } { 2 } \frac { \mathrm { kc } } { \mathrm { k } + 1 } \frac { \mathrm {~b} } { \mathrm { k } + 1 }$sinA

= $\frac { 1 } { 2 } \frac { \mathrm { k } } { ( \mathrm { k } + 1 ) ^ { 2 } }$ bc sinA

= $\frac { \mathrm { k } } { ( \mathrm { k } + 1 ) ^ { 2 } }$ D

similarly area of (DBRP)

= area (D PQC) =

̃ area (D PQR) = D – = $\left( \frac { \mathrm { k } ^ { 2 } - \mathrm { k } + 1 } { \mathrm { k } ^ { 2 } + 2 \mathrm { k } + 1 } \right)$ D

Now

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