Question

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

Answer 1

Answer:

Perimeter of rectangle= $2 (\sqrt{10} + \sqrt{29})$

= $2\sqrt{10} + 2\sqrt{29}$.

Step-by-step explanation:

Given: $E(x_{1}, y_{1}) = (1. -1), F(x_{2}, y_{2}) = (-4, 1), G(x_{3}, y_{3}) = (-3, 4), H(x_{4}, y_{4}) = (2, 2)$

Using distance formula:

Length of EF = $\sqrt{(x_{2} - x_{1} )^2 + (y_{2} - y_{1})^2}$

= $\sqrt{(-4 - 1)^2 + (1 - (-1))^2}$

= $\sqrt{(-4 - 1)^2 + (1 + 1))^2}$

=  $\sqrt{(-5)^2 + (2))^2}$

=  $\sqrt{25 + 4}$

= $\sqrt{29}$

Length of FG = $\sqrt{(x_{3} - x_{2} )^2 + (y_{3} - y_{2})^2}$

= $\sqrt{(-3 - (-4)^2 + (4 - 1)^2}$

= $\sqrt{(1)^2 + (3)^2}$

= $\sqrt{1 + 9}$

= $\sqrt{10}$

Length of GH =  $\sqrt{(x_{4} - x_{3} )^2 + (y_{4} - y_{3})^2}$

=  $\sqrt{(2 - (-3))^2 + (2 - 4)^2}$

= $\sqrt{(5)^2 + (-2))^2}$

=  $\sqrt{25 + 4}$

= $\sqrt{29}$

Length of HE =  $\sqrt{(x_{4} - x_{1} )^2 + (y_{4} - y_{1})^2}$

= $\sqrt{(2 - 1)^2 + (2 - (-1))^2}$

= $\sqrt{(1)^2 + (3)^2}$

= $\sqrt{1 + 9}$

= $\sqrt{10}$

∵ EFGH is a rectangle ∴ EH = FG and EF = HG

Perimeter of rectangle = 2 ( EF + FG + GH + HE)

= 2 (EF + FG)

= $2 (\sqrt{10} + \sqrt{29})$

= $2\sqrt{10} + 2\sqrt{29}$

Therefore option (b) is the correct answer.