Question

Rectangle ABCD is graphed in the coordinate plane. The following are the vertices of the rectangle: A(2, -6), B(5, -6), C(5, -2) and D(2, -2). What is the perimeter of rectangle ABCD?

Answer 1

Answer:

Step-by-step explanation:

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

A(2, -6), B(5, -6),

       $AB =\sqrt{(5-2)^{2}+(-6-[-6])^{2}}\\\\=\sqrt{(5-2)^{2}+(-6+6)^{2}}\\\\=\sqrt{3^{2}+0}\\\\=\sqrt{3^{2}}\\\\=3 units$

B(5,-6) ; C(5,-2)

$BC = \sqrt{(5-5)^{2}+(-2-[-6])^{2}}\\\\ = \sqrt{0+(-2+6)^{2}}\\\\ = \sqrt{4^{2}}$

BC = 4 units

C(5, -2) ; D (2,-2)

$CD = \sqrt{(2-5)^{2}+(-2-[-2])^{2}}\\\\ = \sqrt{(-3)^{2}+(-2+2)^{2}}\\\\ = \sqrt{(-3)^{2}}$

CD = 3 units

A(2,-6) ; D(2,-2)

$AD = \sqrt{(2-2)^{2}+(-2+6)^{2}}\\\\ = \sqrt{0 +(4)^{2}}\\\\ = \sqrt{(4)^{2}}$

= 4 units

Perimeter = AB + BC + CD + AD

                = 3 + 4 + 3 + 4

                 = 14 units