Question

What is the perimeter of the rectangle whose vertices are at (6, 3), (−6, 3), (−6, −1), and (6, −1)?

Answer 1

Given:

The vertices of a rectangle are (6, 3), (−6, 3), (−6, −1), and (6, −1).

To find:

The perimeter of the rectangle.

Solution:

Distance formula:

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

Let the vertices of a rectangle are A(6, 3), B(−6, 3), C(−6, −1), and D(6, −1).

Using the distance formula, we get

$AB=\sqrt{(-6-6)^2-(3-3)^2}$

$AB=\sqrt{(-12)^2-(0)^2}$

$AB=\sqrt{144}$

$AB=12$

Similarly,

$BC=\sqrt{\left(-6-\left(-6\right)\right)^2+\left(-1-3\right)^2}=4$

$CD=\sqrt{\left(6-\left(-6\right)\right)^2+\left(-1-\left(-1\right)\right)^2}=12$

$AD\sqrt{\left(6-6\right)^2+\left(-1-3\right)^2}=4$

Now, the perimeter of the rectangle is:

$Perimeter=AB+BC+CD+AD$

$Perimeter=12+4+12+4$

$Perimeter=32$

Therefore, the perimeter of the rectangle is 32 units.