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
⌨️ Shortcuts:[1-4 / A-D] Select[Enter] Check[B] Star[S] Solution
Official Correct Answer: B. We have a rectangle ABCD where AB is 9 \text{ cm} and BC is 6 \text{ cm}. Points P and Q are on side BC. The areas of the figures ABP, APQ, and AQCD are in geometric progression. Also, the area of AQCD is four times the area of triangle ABP. We need to find the ratio BP : PQ : QC.
Consider. ABCD is a rectangle, so AB is the length, 9 \text{ cm}, and BC is the width, 6 \text{ cm}. So, if I imagine the rectangle, AB is the base, BC is the height, CD is the top, and AD is the other vertical side.
Points P and Q are on BC. So, BC is 6 \text{ cm}, and P and Q divide BC into three segments: BP, PQ, and QC. We need to find the ratio of these three segments.
Given that the areas of ABP, APQ, and AQCD are in geometric progression. Also, the area of AQCD is four times the area of triangle ABP.
Let BP = x, PQ as y, and QC as z. So, x + y + z = BC = 6 \text{ cm}.
We need to find the ratio x : y : z.
First, let's find the areas of the figures.
1. Area of triangle ABP: Since AB is 9 \text{ cm} and BP is x, the area is (1/2)*AB*BP = (1/2)*9*x = (9x)/2.
2. Area of figure APQ: Hmm, APQ is a quadrilateral. Since P and Q are on BC, APQ is a triangle? Wait, no. APQ is a quadrilateral because A is a corner, P is on BC, and Q is also on BC. So, APQ is a triangle? Wait, no, because A is connected to P and Q, but P and Q are on BC. So, APQ is a triangle with base PQ and height from A to BC.
Wait, no. Let me think again. APQ is a figure with vertices A, P, Q. Since A is at the corner, and P and Q are on BC, which is the adjacent side. So, APQ is a triangle with base PQ on BC and vertex at A. So, the area would be (1/2)*PQ*AB, because the height from A to BC is AB, which is 9 \text{ cm}. Wait, no, the height is actually the same as the height of the rectangle, which is BC, but that's 6 \text{ cm}. Wait, no, the height from A to BC is the same as the length of AD, which is 6 \text{ cm}.
Wait, maybe I'm confusing the sides. Let me clarify.
In rectangle ABCD, AB is 9 \text{ cm} (length), BC is 6 \text{ cm} (width). So, the sides are AB = CD = 9 \text{ cm}, and BC = AD = 6 \text{ cm}.
So, point A is at the bottom-left corner, B is at the bottom-right, C is at the top-right, and D is at the top-left.
Points P and Q are on BC. So, BC is the vertical side from B (bottom) to C (top). So, P and Q are somewhere along this vertical side.
So, triangle ABP: this is a right triangle with base AB = 9 \text{ cm} and height BP = x \text{ cm}. So, area is (1/2)*9*x = (9x)/2.
Similarly, figure APQ: since P and Q are on BC, APQ is a quadrilateral. Wait, but A is connected to P and Q, which are on BC. So, APQ is a triangle? Because A is connected to P and Q, but P and Q are colinear on BC. So, APQ is a triangle with base PQ on BC and vertex at A. So, the area would be (1/2)*PQ*AB? Wait, no, because the height from A to BC is the same as the width of the rectangle, which is 6 \text{ cm}. Wait, no, the height from A to BC is actually the same as the length of AD, which is 6 \text{ cm}. So, the area of triangle APQ would be (1/2)*PQ*AD = (1/2)*y*6 = 3y.
Wait, but APQ is a triangle with base PQ on BC and height from A to BC, which is 6 \text{ cm}. So, yes, area is (1/2)*PQ*6 = 3y.
Similarly, the area of AQCD: AQCD is a quadrilateral. . A is connected to Q, Q is on BC, then C, then D, and back to A. So, AQCD is a trapezoid. The area of a trapezoid is (1/2)*(sum of the two parallel sides)*height. The two parallel sides are AQ and CD. Wait, no, AQ is not a side; AQ is a line from A to Q. Hmm, maybe I should think differently.
Alternatively, AQCD can be considered as the area of the rectangle minus the area of triangle ABP and the area of triangle APQ. Wait, but that might complicate things. Alternatively, since AQCD is a quadrilateral, maybe it's a trapezoid with bases AD and QC, but I'm not sure.
Wait, let me think again. The figure AQCD is bounded by points A, Q, C, D. So, from A to Q is a line, then Q to C, then C to D, then D back to A. So, AQCD is a quadrilateral. To find its area, perhaps it's easier to subtract the areas of ABP and APQ from the total area of the rectangle.
The total area of the rectangle is AB*BC = 9*6 = 54 \text{ cm}².
So, area of AQCD = total area - area of ABP - area of APQ = 54 - (9x/2) - 3y.
But we also know that the area of AQCD is four times the area of ABP. So, area of AQCD = 4*(9x/2) = 18x.
So, 54 - (9x/2) - 3y = 18x.
Let me write that equation:
54 - (9x/2) - 3y = 18x.
Simplify:
54 = 18x + (9x/2) + 3y.
Convert 18x to (36x/2) to have a common denominator:
54 = (36x/2 + 9x/2) + 3y = (45x/2) + 3y.
So, 54 = (4