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

Answer: (D)

2 : 4 : 1

Answer 2

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.

Now, suppose the area of $△AQC$ be x, which implies the area of $△ ADC = ADQC = AQC = 4k - x$
which is equal to the sum of the area of triangles APB, AQP and ACQ.

So, $4k - x = 3k + x$
$⇒ x =$ $\frac{k}{2}$

Therefore, the ratio of BP: PQ: CQ is:
$= k : 2k :$ $\frac{k}{2}$
$= 2 : 4 : 1.$
Therefore, the correct option is $(D): 2: 4: 1.$