# The length of a rectangle is twice its width. The perimeter is 60 ft. How do you find its area?

Apr 23, 2018

$A = 200 f {t}^{2}$

#### Explanation:

Let the width be $x$, then the length is $2 x$

$\text{Perimeter} = x + 2 x + x + 2 x = 60$

$6 x = 60$

$x = 10 \text{ } \leftarrow$ this is the width

$2 x = 20 \text{ } \leftarrow$ this is the length

$\text{Area} = l \times b$

$A = 20 \times 10$

$A = 200 f {t}^{2}$

Apr 23, 2018

The area is $200 f t {.}^{2}$

#### Explanation:

Always draw a picture for this type of question to help you visualize what's going on. When you do that, label the shorter side of your rectangle $w$. Label the longer side $l = 2 w$, or just $2 w$, since the length is twice the width.
Add up the sides and set them equal to $60$:
$w + w + 2 w + 2 w = 60$
$6 w = 60$
$\frac{6 w}{6} = \frac{60}{6}$
$w = 10$

So the width is 10 ft, and the length that is two times as long is 20 ft.
The formula for area is length times width, so plug in our new values: $l \cdot w = 20 \cdot 10 = 200 f t {.}^{2}$

Hope that helps!!