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 $$\triangle ABP, \triangle ABQ$$ and $$\triangle ABC$$ are in arithmetic progression. If the area of ∆ABC is 1.5 times the area of $$\triangle ABP$$, the length of PQ, in cm, is
Correct Answer: 2
Given that ABC is a right-angled triangle with AB = 5 and BC = 12 => Area of the triangle = 0.5 * 5 * 12 = 30.
Let us assume BP = p, BQ = q
=> Area of ABP = 0.5 * 5 * p = 2.5p
=> Area of ABQ = 0.5 * 5 * q = 2.5q
Given the area of ABC is 1.5 times that of ABP
=> 30 = 1.5 * 2.5p => 20 = 2.5p => p = 8.
Given Areas of ABP, ABQ and ABC are in A.P. => 2 * 2.5q = 2.5 * 8 + 30 => 5q = 50 => q = 10.
PQ = BQ - BP = q - p = 10 - 8 = 2.
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