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

Answer:

Step-by-step explanation:

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

A(3,-2) ;  B(6,-2)

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

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