Check the picture below, now the distance from 2,0 to 4,0 there's no need to do much calculation since that's just 2 units, as you see there.
[tex]~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{2}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{4}) ~\hfill d1=\sqrt{[ 1- 2]^2 + [ 4- 0]^2} \\\\\\ d1=\sqrt{(-1)^2+4^2}\implies \boxed{d1=\sqrt{17}} \\\\[-0.35em] ~\dotfill[/tex]
[tex](\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{-2}) ~\hfill d2=\sqrt{[ -1- 1]^2 + [ -2- 4]^2} \\\\\\ d2=\sqrt{(-2)^2+(-6)^2}\implies \boxed{d2=\sqrt{40}} \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{-2})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{0}) ~\hfill d3=\sqrt{[ 4- (-1)]^2 + [ 0- (-2)]^2} \\\\\\ d3=\sqrt{(4+1)^2+(0+2)^2}\implies \boxed{d3=\sqrt{29}}[/tex]
[tex]~\dotfill\\\\ \stackrel{\textit{\Large Perimeter}}{2~~ + ~~\sqrt{17}~~ + ~~\sqrt{40}~~ + ~~\sqrt{29}~~\approx~~17.83}[/tex]
