# The perimeter of a rectangle is 36ft, and the area of the rectangle is 72ft^2 . How do you find the dimensions?

Apr 18, 2016

You must write a system of equations to represent the problem.

#### Explanation:

The formula for perimeter of a rectangle is $p = 2 L + 2 W$. The formula for area is $A = L \times W$

Thus, $L \times W = 72 , 2 L + 2 W = 36$

$W = \frac{72}{L} \to 2 L + 2 \left(\frac{72}{L}\right) = 36$

$2 L + \frac{144}{L} = 36$

$\frac{2 {L}^{2}}{L} + \frac{144}{L} = \frac{36 L}{L}$

We can now eliminate the denominators since all fractions are equal.

$2 {L}^{2} + 144 = 36 L$

$2 {L}^{2} - 36 L + 144 = 0$

This is a trinomial of the form $y = a {x}^{2} + b x + c , a \ne 1$ Therefore, this can be factored by finding two numbers that multiply to $a \times c$ and that add to b, and following the process shown below. These two numbers are $- 12$ and $- 24$

$2 {L}^{2} - 12 L - 24 L + 144 = 0$

$2 L \left(L - 6\right) - 24 \left(L - 6\right) = 0$

$\left(2 L - 24\right) \left(L - 6\right) = 0$

$L = 12 \mathmr{and} 6$

Since the length can be the width and vice-versa, the sides of the rectangle measure 12 and 6.

Hopefully this helps!