Question 19.
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, ΔABQ and ΔABC are in arithmetic progression. If the area of ΔABC is 1.5 times the area of ΔABP, the length of PQ, in cm, is
A
B
C
D
Question Explanation
Text ExplanationVideo Explanation
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.
