Respuesta :

Check the picture below.

[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\\\\ A(\stackrel{x_1}{-5}~,~\stackrel{y_1}{2})\qquad B(\stackrel{x_2}{-1}~,~\stackrel{y_2}{6}) ~\hfill AB=\sqrt{[ -1- (-5)]^2 + [ 6- 2]^2} \\\\\\ ~\hfill \boxed{\sqrt{32}} \\\\\\ B(\stackrel{x_1}{-1}~,~\stackrel{y_1}{6})\qquad C(\stackrel{x_2}{3}~,~\stackrel{y_2}{2}) ~\hfill BC=\sqrt{[ 3- (-1)]^2 + [ 2- 6]^2} \\\\\\ ~\hfill \boxed{\sqrt{32}}[/tex]

[tex]C(\stackrel{x_1}{3}~,~\stackrel{y_1}{2})\qquad D(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-2}) ~\hfill CD=\sqrt{[ -1- 3]^2 + [ -2- 2]^2} \\\\\\ ~\hfill \boxed{\sqrt{32}} \\\\\\ D(\stackrel{x_1}{-1}~,~\stackrel{y_1}{-2})\qquad A(\stackrel{x_2}{-5}~,~\stackrel{y_2}{2}) ~\hfill DA=\sqrt{[ -5- (-1)]^2 + [ 2- (-2)]^2} \\\\\\ ~\hfill \boxed{\sqrt{32}}[/tex]

[tex]~\dotfill\\\\ \sqrt{32}\implies \sqrt{16\cdot 2}\implies \sqrt{4^2\cdot 2}\implies 4\sqrt{2}\implies \stackrel{\textit{four times that much}}{4(4\sqrt{2})\implies \blacksquare~~ 16\sqrt{2} ~~\blacksquare}[/tex]

Ver imagen jdoe0001

It's a rhoumbus as AC and BD both diagonals are perpendicular to each other .

Find one side

  • A(-5,2)
  • D(-1,-2)

AD

  • √(-1+5)²+(2+2)²
  • √4²+4²
  • 4√2

So

Perimeter

  • 4a
  • 4(4√2)
  • 16√2units
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