Question

Calculate the perimeter of quadrilateral ABCD. (Hint: use distance formula}

Answer 1

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

Answer 2

Check the picture below.

`~\hfill \stackrel{\textit{\large distance between 2 points}}{d = √(( 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=√([ -1- (-5)]^2 + [ 6- 2]^2) \\\\\\ ~\hfill \boxed{√(32)} \\\\\\ B(\stackrel{x_1}{-1}~,~\stackrel{y_1}{6})\qquad C(\stackrel{x_2}{3}~,~\stackrel{y_2}{2}) ~\hfill BC=√([ 3- (-1)]^2 + [ 2- 6]^2) \\\\\\ ~\hfill \boxed{√(32)}`

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

`~\dotfill\\\\ √(32)\implies √(16\cdot 2)\implies √(4^2\cdot 2)\implies 4√(2)\implies \stackrel{\textit{four times that much}}{4(4√(2))\implies \blacksquare~~ 16√(2) ~~\blacksquare}`