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

Given: Triangle \( ABC \) is right-angled with \( AB = 5 \), \( BC = 12 \). 

Area of triangle \( ABC \):
\( \text{Area} = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 5 \times 12 = 30 \)

Let \( BP = p \), \( BQ = q \)

Area of triangle \( ABP = \frac{1}{2} \times AB \times p = 2.5p \)

Area of triangle \( ABQ = \frac{1}{2} \times AB \times q = 2.5q \)

Given: Area of \( \triangle ABC = 1.5 \times \text{Area of } \triangle ABP \)

\( 30 = 1.5 \times 2.5p \Rightarrow 30 = 3.75p \Rightarrow p = \frac{30}{3.75} = 8 \)

Also given: Areas of \( \triangle ABP, \triangle ABQ, \triangle ABC \) are in A.P.

So, \( 2 \times \text{Area of } ABQ = \text{Area of } ABP + \text{Area of } ABC \)

\( 2 \times 2.5q = 2.5 \times 8 + 30 \Rightarrow 5q = 20 + 30 = 50 \Rightarrow q = 10 \)

Therefore: \( PQ = BQ - BP = q - p = 10 - 8 = \boxed{2} \)