Q:

On a coordinate plane, rectangle E F G H is shown. Point E is at (1, negative 1), point F is at (negative 4, 1), point G is at (negative 3, 4), and point H is at (2, 2). What is the perimeter of rectangle EFGH? StartRoot 10 EndRoot + StartRoot 29 EndRoot units 2 StartRoot 10 EndRoot + 2 StartRoot 29 EndRoot units 22 units 39 units

Accepted Solution

A:
Answer:Perimeter of rectangle= [tex]2 (\sqrt{10} + \sqrt{29})[/tex]= [tex]2\sqrt{10} + 2\sqrt{29}[/tex].Step-by-step explanation:Given: [tex]E(x_{1}, y_{1}) = (1. -1), F(x_{2}, y_{2}) = (-4, 1), G(x_{3}, y_{3}) = (-3, 4), H(x_{4}, y_{4}) = (2, 2)[/tex]Using distance formula:Length of EF = [tex]\sqrt{(x_{2} - x_{1} )^2 + (y_{2} - y_{1})^2}[/tex]= [tex]\sqrt{(-4 - 1)^2 + (1 - (-1))^2}[/tex]= [tex]\sqrt{(-4 - 1)^2 + (1 + 1))^2}[/tex]=  [tex]\sqrt{(-5)^2 + (2))^2}[/tex]=  [tex]\sqrt{25 + 4}[/tex]= [tex]\sqrt{29}[/tex]Length of FG = [tex]\sqrt{(x_{3} - x_{2} )^2 + (y_{3} - y_{2})^2}[/tex]= [tex]\sqrt{(-3 - (-4)^2 + (4 - 1)^2}[/tex]= [tex]\sqrt{(1)^2 + (3)^2}[/tex]= [tex]\sqrt{1 + 9}[/tex]= [tex]\sqrt{10}[/tex]Length of GH =  [tex]\sqrt{(x_{4} - x_{3} )^2 + (y_{4} - y_{3})^2}[/tex]=  [tex]\sqrt{(2 - (-3))^2 + (2 - 4)^2}[/tex]= [tex]\sqrt{(5)^2 + (-2))^2}[/tex]=  [tex]\sqrt{25 + 4}[/tex]= [tex]\sqrt{29}[/tex]Length of HE =  [tex]\sqrt{(x_{4} - x_{1} )^2 + (y_{4} - y_{1})^2}[/tex]= [tex]\sqrt{(2 - 1)^2 + (2 - (-1))^2}[/tex]= [tex]\sqrt{(1)^2 + (3)^2}[/tex]= [tex]\sqrt{1 + 9}[/tex]= [tex]\sqrt{10}[/tex]∵ EFGH is a rectangle ∴ EH = FG and EF = HGPerimeter of rectangle = 2 ( EF + FG + GH + HE) = 2 (EF + FG)= [tex]2 (\sqrt{10} + \sqrt{29})[/tex]= [tex]2\sqrt{10} + 2\sqrt{29}[/tex]Therefore option (b) is the correct answer.