Question

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

• A. 1 : 1 : 2
• B. 1 : 2 : 1
• C. 1 : 2 : 4
• D. 2 : 4 : 1

Answer 1

The Correct Option is D

Given:
AB = 9cm, BC = 6cm 
We are also given that the areas of the ABP, APQ and AQCD are in geometric progression.

Therefore, it can be assumed as:
Area of ABP, APQ and AQCD as k, 2k and 4k respectively.

As per the question, the ratio of BP, PQ and QC will be the ratio of the respective triangles.
So, we can draw a line from Point A to C.

Let the area of \( \triangle AQC \) be \( x \). The area of \( \triangle ADC \) is given by:

\( \triangle ADC = ADQC = AQC = 4k - x \)

This is equal to the sum of the areas of triangles \( \triangle APB \), \( \triangle AQP \), and \( \triangle ACQ \), which can be expressed as:

\( 4k - x = 3k + x \)

Simplifying this equation:

\( 4k - x = 3k + x \Rightarrow x = \frac{k}{2} \)

Now, let's calculate the ratio of \( BP : PQ : CQ \):

\( BP : PQ : CQ = k : 2k : \frac{k}{2} \)

By simplifying the ratio:

\( = 2 : 4 : 1 \)

Therefore, the correct option is:

(D): 2 : 4 : 1