Question

In a right-angled triangle Δ 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, ΔABQ , and Δ ABC are in arithmetic progression. If the area of Δ ABC is 1.5 times the area of Δ ABP , the length of PQ , in cm , is

Answer 1

The correct answer is 2.

Answer 2

Given that ABC is a right-angled triangle with \(AB = 5\) and \(BC = 12\)
 => Area of the triangle = \(0.5 \times 5 \times 12 = 30\). Let us assume \(BP = p, BQ = q\)
=> Area of ABP =\(0.5 \times 5\times p = 2.5p\)
 => Area of ABQ = \(0.5\times5\times q = 2.5q\) Given the area of \(ABC\)  is \( 1.5\) times that of \( ABP\)
 =>\(30 = 1.5 \times2.5p\)
=> \(20 = 2.5p\)
=>\(p = 8.\) Given Areas of \(ABP, ABQ \)and ABC are in A.P.
 => \(2\times 2.5q = 2.5\times8+ 30\)
=> \(5q = 50\)
=> \(q = 10.\)
\(PQ=BQ - BP = q-p=10-8=2.\)

Correct  Answer is  \(2\)