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

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