Let PQR be a triangle. The points $\mathrm{A}, \mathrm{B}$ and C are on the sides $\mathrm{QR}, \mathrm{RP}$ and PQ respectively such that $\frac{\mathrm{QA}}{\mathrm{AR}}=\frac{\mathrm{RB}}{\mathrm{BP}}=\frac{\mathrm{PC}}{\mathrm{CQ}}=\frac{1}{2}$. Then $\frac{\operatorname{Area}(\triangle \mathrm{PQR})}{\operatorname{Area}(\triangle \mathrm{ABC})}$ is equal to

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Vector Algebra

Let PQR be a triangle. The points

A, B

and C are on the sides

QR, RP

and PQ respectively such that

\frac{QA}{AR} = \frac{RB}{BP} = \frac{PC}{CQ} = \frac{1}{2}

. Then

\frac{Area ( △ PQR )}{Area ( △ ABC )}

is equal to

Answer: (B)

Take

P

as origin with

Q = q

and

R = r

. The given ratios place

A = \frac{2 q + r}{3}

B = \frac{2 r}{3}

and

C = \frac{q}{3}

. Then

A B = \frac{r - 2 q}{3}

and

A C = \frac{- q - r}{3}

, so

A B \times A C = \frac{1}{3} (q \times r)

, giving

[△ A B C] = \frac{1}{6} ∣ q \times r ∣

while

[△ P QR] = \frac{1}{2} ∣ q \times r ∣

. Their ratio is

3.

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Single correct option