# The perimeter of a rectangular driveway is 68 feet. The area is 280 square feet. What are the dimensions of the driveway?

Nov 21, 2016

1)w=20ft, l=14ft
2)w=14ft, l=20ft

#### Explanation:

Let's define the variables:

$P :$perimeter
$A :$ area
$l :$length
$w :$ width

$P = 2 l + 2 w = 68$

Simplify (divide by $2$)

$l + w = 34$

Solve for $l$

$l = 34 - w$

$A = l \cdot w = 280$

Substitute $34 - w$ instead of $l$

$A = \left(34 - w\right) w = 280$

$- {w}^{2} + 34 w = 280$
$- {w}^{2} + 34 w - 280 = 0$

Multiply by $- 1$

${w}^{2} - 34 w + 280 = 0$

Factorize

$\left(w - 20\right) \left(w - 14\right) = 0$

Set each expression equal to zero

1)w-20=0
$w = 20$

2)w-14=0
$w = 14$

Option $1$) substitute $20$ instead of $w$

$l + w = 34$
$l + 20 = 34$
$l = 14$

Option$2$) substitute $14$ instead of $w$

$l + w = 34$
$l + 14 = 34$
$l = 20$

1)w=20ft, l=14ft
2)w=14ft, l=20ft

Nov 21, 2016

The dimensions are $20$ and $14$ feet. See explanation.

#### Explanation:

We are looking for the dimensions of a rectangle, so we are looking for 2 numbers $a$ and $b$ which satisfy the set of equations:

$\left\{\begin{matrix}2 a + 2 b = 68 \\ a \cdot b = 280\end{matrix}\right.$

To solve this set we calculate $b$ from the first equation:

$a + b = 34 \implies b = 34 - a$

Now we substitute $b$ in the second equation:

$a \cdot \left(34 - a\right) = 280$

$34 a - {a}^{2} = 280$

$- {a}^{2} + 34 a - 280 = 0$

$\Delta = 1156 - 1120 = 36$

$\sqrt{\Delta} = 6$

${a}_{1} = \frac{- 34 - 6}{- 2} = 20$

${a}_{2} = \frac{- 34 + 6}{- 2} = 14$

Now we have to calculate $b$ for each calculated value of $a$

${b}_{1} = 34 - {a}_{1} = 34 - 20 = 14$

${b}_{2} = 34 - {a}_{2} = 34 - 14 = 20$

So we see that the dimensions are $20$ and $14$ feet.