Triangle Area in Arithmetic Progression

CAT 2023 Slot 1 · QA · Medium · Geometry

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In a right-angled triangle ABC, the altitude AB is 5 cm, and the base BC is 12 cm. P and Q are two points on BC such that the areas of ΔABP\Delta ABP, ΔABQ\Delta ABQ and ΔABC\Delta ABC are in arithmetic progression. If the area of ΔABC\Delta ABC is 1.5 times the area of ΔABP\Delta ABP, the length of PQ, in cm, is

Answer

2

Explanation

Area of ΔABC=12×12×5=30 cm2\Delta ABC = \frac{1}{2} \times 12 \times 5 = 30\text{ cm}^2.

Given Area(ΔABC)=1.5×Area(ΔABP)\text{Area}(\Delta ABC) = 1.5 \times \text{Area}(\Delta ABP): Area(ΔABP)=301.5=20\text{Area}(\Delta ABP) = \frac{30}{1.5} = 20 Since Area(ΔABP)=12×BP×AB  ⟹  20=12×BP×5  ⟹  BP=8 cm\text{Area}(\Delta ABP) = \frac{1}{2} \times BP \times AB \implies 20 = \frac{1}{2} \times BP \times 5 \implies BP = 8\text{ cm}.

The areas of ΔABP\Delta ABP, ΔABQ\Delta ABQ, and ΔABC\Delta ABC are in AP: 20,Area(ΔABQ),3020, \text{Area}(\Delta ABQ), 30 Area(ΔABQ)=20+302=25\text{Area}(\Delta ABQ) = \frac{20 + 30}{2} = 25 Since Area(ΔABQ)=12×BQ×5=25  ⟹  BQ=10 cm\text{Area}(\Delta ABQ) = \frac{1}{2} \times BQ \times 5 = 25 \implies BQ = 10\text{ cm}.

Thus, PQ=BQ−BP=10−8=2 cmPQ = BQ - BP = 10 - 8 = 2\text{ cm}.

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