Question

find the dimensions of a rectangle that has a perimeter of 208 feet if its length is twice its width

Answer 1

The perimeter of a rectangle is given by the following formula:

$P=2\cdot(L+W)$

Where L is the length and W is the width.

This rectangle's length is twice is width, thus:

$L=2W$

By replacing L in the perimeter formula, we can solve for W as follows:

$\begin{gathered} P=2\cdot(2W+W) \\ P=2\cdot(3W) \\ P=6W \\ 208ft=6W \\ W=\frac{208ft}{6} \\ W=34.67ft \\ \text{And the Length is then} \\ L=2W=2\cdot34.67ft \\ L=69.33ft \end{gathered}$

The length of the rectangle is 69.33 ft and its width is 34.67ft