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 $B:L = 1:4.$
$\text{Option A}$