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

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.