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Question:

In triangle ABC, P and R are points on side BA while Q and S are points on side BC. What is the minimum possible area (in sq. units) of triangle ABC if the area of triangle BRS and ABC are integers, and AP : PB = CQ : QB = PR : RB = QS : SB = 2 : 3?

Explanation:

In ∆BAC and ∆BPQ

$\frac{B P}{B A} = \frac{B Q}{B C} = \frac{3}{5}$ and ∠B is common

⇒ ∆BAC ~ ∆BPQ (S-A-S Rule)

⇒ $\frac{A r e a o f ∆ B P Q}{A r e a o f ∆ B A C} = {(\frac{3}{5})}^{2} = \frac{9}{25}$ …(1)

In ∆BRS and ∆BPQ

$\frac{B R}{B P} = \frac{B S}{B Q} = \frac{3}{5}$ and ∠B is common

⇒ ∆BAC ~ ∆BRQ (S-A-S Rule)

⇒ $\frac{A r e a o f ∆ B R S}{A r e a o f ∆ B P Q} = {(\frac{3}{5})}^{2} = \frac{9}{25}$ …(2)

From (1) and (2) we get,

Area of ∆BAC : Area of ∆BPQ : Area of ∆BRS = 625 : 225 : 81

Since area of ∆BAC and ∆BRS are integers, hence the minimum possible area of ∆BAC = 625 sq. units.

Hence, 625.

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