# How do you find the dimensions of a rectangle whose perimeter is 46 cm and whose area is 128cm^2?

Jul 3, 2016

Do some quadratic equation solving to get a dimension of $9.438 \times 13.562$.

#### Explanation:

We are looking for the length and width of this rectangle.

In order to find length and width, we need formulas which include length and width. Since we have perimeter and area, we will use the formulas for perimeter ($P$) and area ($A$):
$P = 2 l + 2 w$
$A = l w$

We can solve for either length or width - I'll start with width. Dividing by $w$ in $A = l w$ gives us a formula for length in terms of area and width:
$l = \frac{A}{w}$

We can substitute this into the equation for perimeter, $P = 2 l + 2 w$:
$P = 2 l + 2 w \to P = 2 \left(\frac{A}{w}\right) + 2 w$

Since we know the perimeter is $46 \text{ cm}$, and the area is $128 {\text{ cm}}^{2}$, we can plug these into the formula:
$46 = 2 \left(\frac{128}{w}\right) + 2 w$

Now divide everything by $2$ to simplify:
$23 = \frac{128}{w} + w$

Multiply by $w$ to cancel the fraction:
$23 w = 128 + {w}^{2}$

Finally, rearrange and subtract $23 w$ from both sides:
${w}^{2} - 23 w + 128 = 0$

This is a quadratic equation whose solutions can be found using the quadratic formula:
$w = \frac{- \left(- 23\right) \pm \sqrt{{\left(- 23\right)}^{2} - 4 \left(1\right) \left(128\right)}}{2 \left(1\right)}$
$w = \frac{23 \pm \sqrt{17}}{2}$
$w \approx 13.562 \text{ cm}$ $\text{ and }$ $w \approx 9.438 \text{ cm}$

We will use $l = \frac{A}{w}$ to find the corresponding lengths:
$l = \frac{128}{13.562} \approx 9.438 \text{ cm " and " "l=128/9.438~~13.562 " cm}$

As you can see, the rectangle seems to have two different possible lengths and widths, but they're actually the same. So the dimensions of the rectangle are $9.438 \times 13.562$.