Question

The length of a rectangle is 5 more than twice the width. The area of the rectangle is207 ft? What is the length and width of the rectangle?

Answer 1

Length: 23 ft

Width: 9 ft

Step-by-step explanation

The area of a rectangle is:

`A=L* W`

Where L and W are length and width. We know that for this rectangle the area is 207 ft²

and the length is 5 more than twice its width:

`L=5+2W`

We have a system of equations:

`\begin{cases}L=5+2W \\ LW=207\end{cases}`

We can replace L from the first equation into the second equation:

`(5+2W)W=207`

And solve for W. To do this we have to rewrite the equation above:

`2W^2+5W-207=0`

We have a quadratic equation, that can be solved with:

`\begin{gathered} ax^2+bx+c=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}`

In this case x = W, a = 2, b = 5 and c = -207:

`\begin{gathered} W=\frac{-5\pm\sqrt[]{5^2-4\cdot2\cdot(-207)}}{2\cdot2} \\ W=\frac{-5\pm\sqrt[]{25+1656}}{4} \\ W=\frac{-5\pm\sqrt[]{1681}}{4} \\ W=(-5\pm41)/(4) \end{gathered}`

Note that we have here two results, one negative and one positive. The only valid result is the positive one, because the width can't be negative:

`W=(-5+41)/(4)=(36)/(4)=9`

The width of the rectangle is 9 feet.

Then, to find the length we just have to replace the width into the first equation:

`L=5+2\cdot W=5+2\cdot9=5+18=23`

The length is 23 feet.