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The area of Rectangle $ = L \ast B$

Perimeter of Rectangle $ = 2 \ast (L+B),$ where $L=$ length $,B=$ width

Given that,  the area of a rectangle and the square of its perimeter is in the ratio of $1: 25.$

$\frac{L\ast B}{[2\ast (L+B)]^{2}} = \frac{1}{25}$

$\Rightarrow  25\ast L\ast B = [2\ast (L+B)]^{2}$

$\Rightarrow 25LB = 4\ast (L^2+B^2+2LB)$

$\Rightarrow 25LB = 4L^2+4B^2+8LB$

$\Rightarrow 4L^2-17LB+4B^2=0$

$\Rightarrow 4L^2-16LB-1LB+4B^2=0$

$\Rightarrow 4L(L-4B)-B(L-4B)=0$

$\Rightarrow (L-4B)(4L-B)=0$

$\Rightarrow L-4B = 0\;\text{(or)}\;4 L – B = 0$

$\Rightarrow L = 4B \;\text{(or)}\; 4L  = B$

$\Rightarrow \frac{L}{B} = \frac{4}{1} \;\text{(or)}\; \frac{L}{B} = \frac{1}{4}$

As we know the shorter length in the rectangle is the width & the longer side is the length.

So, the required ratio  is $B:L = 1:4.$

$\text{Option A}$
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