 # How Do You Find the Perimeter and Area of a Rectangle?

Now, let’s learn how to find the perimeter and area of rectangles. A rectangle is a quadrilateral where all the angles are $90$°, and where parallel sides have the same length as each other.

Rule

### Rectangle

A rectangle is a quadrilateral where all the angles are $90$°.

In a rectangle, parallel sides have the same length. ## Perimeter of a Rectangle

You can find the perimeter of a rectangle by adding the lengths of all four sides together. Because each set of parallel sides has the same length as each other, you can use the formula below. We call length $l$ and width $w$.

Formula

### PerimeterofaRectangle

$\begin{array}{llll}\hfill P& =2\cdot \text{length}+2\cdot \text{width}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2\cdot l+2\cdot w\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 1

Find the perimeter of the rectangle You plug the numbers into the formula above and get $\begin{array}{llll}\hfill P& =2\cdot 4\phantom{\rule{0.17em}{0ex}}\text{cm}+2\cdot 2\phantom{\rule{0.17em}{0ex}}\text{cm}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =8\phantom{\rule{0.17em}{0ex}}\text{cm}+4\phantom{\rule{0.17em}{0ex}}\text{cm}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =12\phantom{\rule{0.17em}{0ex}}\text{cm}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The perimeter of the rectangle is $12$ cm.

## Area of a Rectangle

You can find the area of a rectangle by multiplying the width and the length of the rectangle. Like before, we call the width $w$ and the length $l$.

Formula

### AreaofaRectangle

$\begin{array}{llll}\hfill A& =\text{length}\cdot \text{width}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =l\cdot w\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 2

Find the area of the rectangle By putting the numbers into the formula above, you get: $\begin{array}{llll}\hfill A& =4\phantom{\rule{0.17em}{0ex}}\text{cm}\cdot 3\phantom{\rule{0.17em}{0ex}}\text{cm}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =12\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The area of the figure is $12$ cm2.