In a rectangle ABCD, AB = 9 cm and BC = 6 cm. P and Q are two points on BC such that the areas of the figures ABP, APQ, and AQCD are in geometric progression. If the area of the figure AQCD is four times the area of triangle ABP, then BP : PQ : QC is:
Explanation:
Area of △ABP, △APQ and AQCD are in GP. Let their areas be A, Ar and Ar2 Now, Area of AQCD = 4 × Area of △ABP ⇒ Ar2 = 4 × A ⇒ r = 2
∴ Area of △ABP : Area of △APQ : Area of AQCD = 1 : 2 : 4
From the figure given above Area of △ABP = 1/2 × BP × AB = 1/2 × BP × 9 = 4.5 × BP Area of △APQ = 1/2 × PQ × AB = 1/2 × PQ × 9 = 4.5 × PQ Area of AQCD = 1/2 × (QC + AD) × AB = 1/2 × (QC + 6) × 9 = 4.5 × (QC + 6)
Now, Area of △APQ = 2 × Area of △ABP ⇒ 4.5 × PQ = 2 × 4.5 × BP ⇒ PQ = 2BP
∴ QC = 6 - BP - PQ = 6 - 3BP
Now, Area of AQCD = 4 × Area of △ABP ⇒ 4.5 × (QC + 6) = 4 × 4.5 × BP ⇒ 4.5 × (6 - 3BP + 6) = 4 × 4.5 × BP ⇒ 6 - 3BP + 6 = 4BP ⇒ BP = 12/7
∴ PQ = 24/7 and
∴ QC = 6 - 3 × 12/7 = 6/7
⇒ BP : PQ : QC = 12/7 : 24/7 : 6/7 = 2 : 4 : 1
Hence, option (a).
» Your doubt will be displayed only after approval.
Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing us your valuable feedback about Apti4All and how it can be improved.