Question

$ABCD$ is a quadrilateral such that $AD=9\ \text{cm}$, $BC=13\ \text{cm}$ and $\angle DAB=\angle BCD=90^\circ$. $P$ and $Q$ are two points on $AB$ and $CD$ respectively, such that $DQ:BP=1:2$ and $DQ$ is an integer. How many values can $DQ$ take, for which the maximum possible area of the quadrilateral $PBQD$ is $150\ \text{cm}^2$?

• A. 14
• B. 12
• C. 10
• D. 9
• E. 8

Hint:

When a side lies on a line perpendicular to a fixed segment, the triangle area with that side as base equals $\frac12\times(\text{base})\times(\text{fixed perpendicular distance})$. Split complex polygons into such triangles.

Answer 1

The Correct Option is D

Step 1: Express the area of \(PBQD\) in terms of \(BP\) and \(DQ\).
Given \(DQ:BP = 1:2\), let: \[ BP = 2x, \quad DQ = x, \quad x \in \mathbb{Z}_{>0}. \] Since \(AB \perp AD\) at \(A\), the perpendicular distance from \(D\) to line \(AB\) is: \[ AD = 9. \] Therefore: \[ [\triangle PBD] = \tfrac12 \cdot BP \cdot AD = \tfrac12 \cdot (2x) \cdot 9 = 9x. \] Similarly, since \(BC \perp CD\) at \(C\), the perpendicular distance from \(B\) to line \(CD\) is: \[ BC = 13. \] Hence: \[ [\triangle BQD] = \tfrac12 \cdot DQ \cdot BC = \tfrac12 \cdot x \cdot 13 = \tfrac{13}{2}x. \] Thus the quadrilateral \(PBQD\), being the union of the two triangles, has area: \[ [PBQD] = 9x + \tfrac{13}{2}x = \tfrac{31}{2}x. \]

Step 2: Apply the maximum possible area condition.
We are told the maximum possible area is \(150 \ \text{cm}^2\). Since \(\tfrac{31}{2}x\) increases with \(x\), we solve: \[ \frac{31}{2}x \leq 150 \;\;\Rightarrow\;\; x \leq \frac{300}{31} = 9.677\ldots \;\;\Rightarrow\;\; x_{\max} = 9. \] Hence admissible positive integer values of \(x(=DQ)\) are: \[ 1, 2, 3, \dots, 9. \]

Therefore, the number of admissible values of \(x\) is: \(\boxed{9}\).