Question

what is the perimeter of a rectangle with vertices at (3,2) (-2,2) (-2,-2) and (3,-2

Answer 1

To find the perimeter

Let ABCD be the vertices of (3,2) (-2,2) (-2,-2) and (3,-2​) respectively

A(3,2) B (-2,2) C(-2,-2) and D (3,-2)

​

We will find the distance AB, BC, CD and DA

Using the distance formula;

$|d|=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}$

Find the distance AB

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

x₁ = 3 y₁=2 x₂=-2 y₂=2

substituting the value into the formula:

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

Evaluate

$|AB|=\sqrt[]{(-5)^2+0}$$=\sqrt[]{25}$

$=5$

Find the distance BC

B (-2,2) C(-2,-2)

x₁ = -2 y₁=2 x₂=-2 y₂=-2

Substitute into the formula

$|BC|=\sqrt[]{(-2+2)^2+(-2-2)^2}$

Evaluate

$=\sqrt[]{0+(-4)^2}$

$=\sqrt[]{16}$$=4$

Find the distance CD

C(-2,-2) D (3,-2)

x₁ = -2 y₁=-2 x₂=-3 y₂=-2

Substitute into the formula

$|CD|=\sqrt[]{(-3+2)^2+(-2+2)^2}$

Evaluate

$|CD|=\sqrt[]{(-5)^2+0}$$=\sqrt[]{25}$$=5$

Find the distance DA

D(3 -2) A(3, 2)

x₁ = 3 y₁=-2 x₂=3 y₂=2

Substitute into the formula

$|DA|=\sqrt[]{(3-3)^2+(2+2)^2}$

Evaluate

$=\sqrt[]{0+4^2}$$=\sqrt[]{16}$$=4$

To get the perimeter, we will add all the distance

That is:

Perimeter = AB + BC + CD + DA

=5 + 4 + 5 + 4

=18

Perimeter = 18