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

We need to find the ratio BP : PQ : QC in rectangle ABCD where P and Q are points on side BC.

Given information:

• Rectangle ABCD with AB = 9 cm, BC = 6 cm
• P and Q are points on BC
• Areas of ABP, APQ, and AQCD are in geometric progression (GP)
• Area of AQCD = 4 × Area of ABP

Let us define: Area of triangle ABP = k

Since the area of AQCD is four times the area of ABP:

Area of AQCD = 4k

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Since the areas ABP, APQ, and AQCD are in GP, we have:

k, Area(APQ), 4k are in GP

For three terms a, b, c to be in geometric progression, the middle term satisfies: $b^2 = ac$

[Area(APQ)]² = k × 4k = 4k²

Therefore: Area(APQ) = 2k

So our GP is: k, 2k, 4k with common ratio = 2

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Since ABCD is a rectangle, the diagonal AC divides it into two equal triangles, each with area = 27 cm².

Area(ADC) = 27 (half the rectangle)

Area(ABC) = 27 (the other half)

The quadrilateral AQCD can be split as: Area(AQCD) = Area(ADC) + Area(AQC)

If Area(AQC) = m, then:

Area(AQCD) = 27 + m = 4k

Also, Area(ABC) = Area(ABP) + Area(APQ) + Area(AQC) = k + 2k + m = 3k + m = 27

From these equations:

• 27 + m = 4k
• 3k + m = 27

27 - 3k = 4k - 27

54 = 7k

k = $\frac{54}{7}$

And: m = 27 - 3k = 27 - $\frac{162}{7}$ = $\frac{27}{7}$

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Now we will use the fact that all these figures share the same height (AB = 9 cm) as their base.

For triangles with the same height, their areas are proportional to their bases:

• Area(ABP) ∝ BP
• Area(APQ) ∝ PQ
• Area(AQC) ∝ QC

From our calculations:

• Area(ABP) = k = $\frac{54}{7}$
• Area(APQ) = 2k = $\frac{108}{7}$
• Area(AQC) = m = $\frac{27}{7}$

The ratio of areas = $\frac{54}{7}$ : $\frac{108}{7}$ : $\frac{27}{7}$ = 54 : 108 : 27 = 2 : 4 : 1

Since the areas are proportional to the bases:

BP : PQ : QC = 2 : 4 : 1

Therefore, BP : PQ : QC = 2 : 4 : 1