Question

Rectangle ABCD is graphed in the coordinate plane. The following are the vertices of the rectangle: A(3,−2), B(6, -2), C(6, 5) D(3,5). What is the area of this rectangle

Answer 1

The answer is 72,14 step by step answer and equation

Answer 2

Explanation:

`Distance=\sqrt{(x_2-x_1)^(2)+(y_2-y_1)^(2)} \\`

A(3,-2) ; B(6,-2)

`AB=\sqrt{(6-3)^(2)+(-2-[-2])^(2)}\\\\ =\sqrt{3^(2)+(-2+2)^(2)}\\\\=√(9+0)=√(9)\\\\`

AB = 3 units

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

`BC=\sqrt{(6-6)^(2)+(5-[-2])^(2)} \\\\=\sqrt{0+(5+2)^(2)}\\\\=\sqrt{(7)^(2)}\\`

BC = 7 units

Area of rectangle = length * width

= AB * BC

= 3 * 7

= 21 square units