Let PQR be a triangle. The points A, B and C are on the sides QR, RP and PQ respectively such that Q A A R = R B B P = P C C Q = 1 2 . Then Area ( P Q R ) Area ( A B C ) is equal to [2023]

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Solution

Q.
Let PQR be a triangle. The points A, B and C are on the sides QR, RP and PQ respectively such that $\frac{Q A}{A R} = \frac{R B}{B P} = \frac{P C}{C Q} = \frac{1}{2} .$ Then $\frac{Area (∆ P Q R)}{Area (∆ A B C)}$ is equal to [2023]

1 4

2 3

3 2

4 5/2

Ans.
(2)

$Let the position vectors of P, Q, R be \vec{0}, \vec{a}, \vec{b} .$

$\Rightarrow P.V. of A = \frac{2 \vec{a} + \vec{b}}{3}, P.V. of B = \frac{2 \vec{b}}{3}$

$and P.V. of C = \frac{\vec{a}}{3}$

$∴$ $\vec{A B} = \frac{2 \vec{b}}{3} - (\frac{2 \vec{a} + \vec{b}}{3}) = \frac{\vec{b}}{3} - \frac{2 \vec{a}}{3} = \frac{\vec{b} - 2 \vec{a}}{3}$

$\vec{C A} = \frac{2 \vec{a} + \vec{b}}{3} - \frac{\vec{a}}{3} = \frac{\vec{a} + \vec{b}}{3}$

$Area of ∆ P Q R = \frac{1}{2}$ $| \vec{P Q} \times \vec{P R} |$ $= \frac{1}{2}$ $| \vec{a} \times \vec{b} |$

$Area of ∆ A B C = \frac{1}{2}$ $| \vec{C A} \times \vec{A B} |$

$= \frac{1}{2}$ $| (\frac{\vec{a} + \vec{b}}{3}) \times (\frac{\vec{b} - 2 \vec{a}}{3}) |$ $= \frac{1}{2}$ $| \frac{\vec{a} \times \vec{b}}{3} |$

$∴$ $\frac{Area (∆ P Q R)}{Area (∆ A B C)} =$ $\frac{\frac{1}{2} | \vec{a} \times \vec{b} |}{\frac{1}{2} | \frac{\vec{a} \times \vec{b}}{3} |}$ $= 3$

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