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

Accepted Answer

B. Area of $	riangle ABP = \frac{1}{2} 	imes AB 	imes BP$. Area of $	riangle APQ = \frac{1}{2} 	imes AB 	imes PQ$. Let Area($	riangle ABP$) $= A$. Given Area($AQCD$) $= 4A$. Since Area($ABP$), Area($APQ$), and Area($AQCD$) form a geometric progression with first term $A$ and third term $4A$, the second term Area($APQ$) is: $$	ext{Area}(APQ) = \sqrt{A 	imes 4A} = 2A$$ Thus, the areas are in the ratio $1 : 2 : 4$. Since triangles $ABP$, $APQ$, and trapezium $AQCD$ all share the same height perpendicular to $BC$ (or can be seen via bases on $BC$), their areas are directly proportional to the base lengths $BP$, $PQ$, and $QC$ respectively. Therefore, $BP : PQ : QC = 1 : 2 : 4$.

Ratio of segment lengths in rectangle

Explanation

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