# The length of a rectangle is 5ft more than twice its width, and the area of the rectangle is 88ft. How do you find the dimensions of the rectangle?

Jun 23, 2016

Length$= 16$ feet, Width$= \frac{11}{2}$ feet.

#### Explanation:

Let the length & width be $l$ feet, & $w$ feet, rep.

By what is given, $l = 2 w + 5. \ldots \ldots \ldots \ldots \ldots \left(1\right) .$

Next, using the formula : Area of rectangle = length $\times$ width, we get another eqn.,
$l \cdot w = 88 ,$ or, by $\left(1\right)$, $\left(2 w + 5\right) \cdot w = 88 ,$ i.e., $2 {w}^{2} + 5 w - 88 = 0.$

To factorise this, we observe that $2 \cdot 88 = 2 \cdot 8 \cdot 11 = 16 \cdot 11$, & $16 - 11 = 5$.

So we replace, $5 w$ by $16 w - 11 w$, to get,

$2 {w}^{2} + 16 w - 11 w - 88 = 0.$

$\therefore 2 w \left(w + 8\right) - 11 \left(w + 8\right) = 0.$

$\therefore \left(w + 8\right) \left(2 w - 11\right) = 0.$

$\therefore w$ = width$= - 8 ,$ which is not permissible, $w = \frac{11}{2.}$

Then $\left(1\right)$ gives, $l = 16.$

It is easy to verify that the pair $\left(l , w\right)$ satisfies the given conds.

Hence the dimensions of the rectangle are length$= 16$ feet, width$= \frac{11}{2}$ feet.

Jun 23, 2016

Length of rectangle is $16 f t$ and Width is $5.5$ft

#### Explanation:

The area of the rectangle should be $88 s q . f t$ in place of $88$ ft mentioned in the question.
Let the width of the rectangle be $x \therefore$ the length will be $2 x + 5 \therefore$Area of rectangle is $\left(2 x + 5\right) \cdot x = 88 \mathmr{and} 2 {x}^{2} + 5 x - 88 = 0 \mathmr{and} 2 {x}^{2} + 16 x - 11 x - 88 = 0 \mathmr{and} 2 x \left(x + 8\right) - 11 \left(x + 8\right) = 0 \mathmr{and} \left(2 x - 11\right) \left(x + 8\right) = 0 \therefore x = 5.5 \mathmr{and} x = - 8$ Width cannot be negative So x=5.5 ; 2x+5=16 Hence length is $16 f t$ and Width is $5.5$ft[Ans]