To find the perimeter and the area of the rectangle you must have its length and width
So we will find AB as width and BC as a length
The rule of the distance between two points is
[tex]d=\sqrt[]{(x2-x1)^2+(y2-y1)^2}[/tex]
A = (2, 6) and B = (4, 3)
x1 = 2 and x2 = 4
y1 = 6 and y2 = 3
Substitute them in the rule to find AB
[tex]AB=\sqrt[]{(4-2)^2+(3-6)^2}=\sqrt[]{4+9}=\sqrt[]{13}[/tex]
B = (4, 3) and C = (10, 6)
x1 = 4 and x2 = 10
y1 = 3 and y2 = 6
[tex]BC=\sqrt[]{(10-4)^2+(6-3)^2}=\sqrt[]{36+9}=\sqrt[]{45}=3\sqrt[]{5}[/tex]
The perimeter of the rectangle
[tex]P=2\lbrack\sqrt[]{13}+3\sqrt[]{5}\rbrack=20.6[/tex]
The area of the rectangle
[tex]A=\sqrt[]{13}\times3\sqrt[]{5}=24.2[/tex]
The perimeter = 20.6 units
The area = 24.2 square units
