we know that
if the figure is a rectangle
then
FG=EH -------> the height of the rectangle
FE=GH --------> the base of a rectangle
Let
x-------> the base of a rectangle
y-------> the height of a rectangle
The perimeter of the rectangle is equal to
[tex]Perimeter=2x+2y[/tex]
Step[tex]1[/tex]
Find the length of the base FE
the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
[tex]E(1,-1)\\F(-4,1)[/tex]
substitute in the formula
[tex]dFE=\sqrt{(1+1)^{2}+(-4-1)^{2}}[/tex]
[tex]dFE=\sqrt{29}[/tex] units
[tex]x=\sqrt{29}[/tex] units
Step[tex]2[/tex]
Find the length of the height FG
the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
[tex]F(-4,1)\\G(-3,4)[/tex]
substitute in the formula
[tex]dFG=\sqrt{(4-1)^{2}+(-3+4)^{2}}[/tex]
[tex]dFG=\sqrt{10}[/tex] units
[tex]y=\sqrt{10}[/tex] units
Step[tex]3[/tex]
Find the perimeter of the rectangle
[tex]Perimeter=2x+2y[/tex]
[tex]Perimeter=(2\sqrt{29} +2\sqrt{10})\ units[/tex]
therefore
the answer is
the perimeter of the rectangle is [tex](2\sqrt{29} +2\sqrt{10})\ units[/tex]
