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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?

User Kiloumap
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ANSWER

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.

User Leng
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