Perimeter of a small rectangular park is 560 feet and measure of its
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17 Feb 2025, 02:27
OFFICIAL EXPLANATION
Let the length \& the breadth of the small rectangular park be L \& B respectively.
As the perimeter of the rectangular park is 560 feet, we get $\(2(L+B)=560 \Rightarrow L+B=280\)$
Also as the diagonal of the park is 200 feet, we get $\(L^2+B^2=200^2=40000\)$ where
$$
\(\begin\mathrm{L}^2+\mathrm{B}^2=(\mathrm{L}+\mathrm{B})^2-2 \mathrm{LB}=(280)^2-2 \mathrm{LB}=78400-2 \mathrm{LB} \\
\)
$$
which gives
\(78400-2 \mathrm{LB}=40000 \Rightarrow \mathrm{LB}=\frac{38400}{2}=19200
\)
Hence the answer is (A).