In the above square ABCD, PR is parallel to CD and the length of CD
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08 Jan 2025, 02:10
ABCD is a square with side 10 units each. As PR is parallel with CD , the length of $\(\mathrm{PR}=10\)$ only.
The area of quadrilateral $\(P Q R S\)$ is same as the sum of the areas of triangle $\(P R S\)$ \& triangle $\(P Q R=$ $\frac{1}{2} \times P R \times h_1+\frac{1}{2} \times P R \times h_2\)$ where $\(h_1 \& h_2\)$ are the perpendicular distances of $\(S\)$ to $\(R P \& Q\)$ to $\(R P\)$ respectively.
So, the area of quadrilateral PQRS is $\(\frac{1}{2} \times \mathrm{PR} \times\left(\mathrm{h}_1+\mathrm{h}_2\right)=\frac{1}{2} \times \mathrm{PR} \times \mathrm{SQ}=\frac{1}{2} \times 10 \times 10=50\)$ which means the area of the shaded portion i.e. the area of square $\(A B C D\)$ excluding the area of quadrilateral PQRS is (area of square ABCD ) - (area of quadrilateral PQRS ) $=$ $\(10 \times 10-50=100-50=50\)$
Hence column A has same quantity as column B, so the answer is (C).