# The least possible perimeter of a rectangle of area $100m^{2}$ is

$\begin{array}{1 1}(1)10&(2)20\\(3)40&(4)60\end{array}$

Area $ab=100$
Perimeters $s= 2(a+b)$
$s = 2 \bigg( a+ \large\frac{100}{a} \bigg)$
$\large\frac{ds}{dt} $$=2 \bigg( 1-\large\frac{100}{a^2} \bigg) \frac{da}{dt} 1- \large\frac{100}{a^2}$$=0=> a= 10$
When $a=10, b= \large\frac{100}{10}$$=10$
$\qquad= 2(10+10) =40$
Hence 3 is the correct answer.