In the given figure, $ABCD$ is a rectangle. $P$ and $Q$ are the midpoints of sides $CD$ and $BC$ respectively. Then the ratio of area of shaded portion to the area of unshaded portion is:
In the given figure, $ABCD$ is a rectangle. $P$ and $Q$ are the midpoints of sides $CD$ and $BC$ respectively. Then the ratio of area of shaded portion to the area of unshaded portion is:
Show Hint
- $5:4$
- $3:5$
- $5:3$
$5:8$
The Correct Option is C
Solution and Explanation
Place \(A(0,0),\ B(w,0),\ C(w,h),\ D(0,h)\). Then \[ P=\left(\tfrac{w}{2},\,h\right),\qquad Q=\left(w,\,\tfrac{h}{2}\right). \] Unshaded region is \(\triangle APQ\). Its area \[ [APQ]=\tfrac12\left|\det\!\begin{pmatrix}\tfrac{w}{2}& h \\[2pt] w & \tfrac{h}{2}\end{pmatrix}\right| =\tfrac12\left|\tfrac{wh}{4}-wh\right| =\tfrac{3wh}{8}. \] Rectangle area \(=wh\), so shaded area \(=wh-\tfrac{3wh}{8}=\tfrac{5wh}{8}\). Hence \[ \text{Shaded:Unshaded}=\frac{5/8}{3/8}= \boxed{5:3}. \]
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