Question

It says," use the rectangle. Round to the nearest tenth."

Answer 1

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

`d=\sqrt[]{(x2-x1)^2+(y2-y1)^2}`

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

`AB=\sqrt[]{(4-2)^2+(3-6)^2}=\sqrt[]{4+9}=\sqrt[]{13}`

B = (4, 3) and C = (10, 6)

x1 = 4 and x2 = 10

y1 = 3 and y2 = 6

`BC=\sqrt[]{(10-4)^2+(6-3)^2}=\sqrt[]{36+9}=\sqrt[]{45}=3\sqrt[]{5}`

The perimeter of the rectangle

`P=2\lbrack\sqrt[]{13}+3\sqrt[]{5}\rbrack=20.6`

The area of the rectangle

`A=\sqrt[]{13}*3\sqrt[]{5}=24.2`

The perimeter = 20.6 units

The area = 24.2 square units