Question

The area of a rectangle is
`\frac{ {x}^(2) - 4 }{2x}in^(2)`, and its length is
`((x + 2)^(2) )/(2)in`. Find the width

Answer 1

Finding the width of a rectangle

We know its area and its length

`\begin{gathered} Area=(x^2-4)/(2x) \\ Length=((x+2)^2)/(2) \end{gathered}`

We know

Width · length = area

Let's call w: width, L: length and a: area

Then

`\begin{gathered} w\cdot l=a \\ w=(a)/(l) \end{gathered}`

Replacing the given information

`\begin{gathered} w=((x^2-4)/(2x))/(((x+2)^2)/(2)) \\ w=(x^2-4)/(2x)\cdot(2)/(\mleft(x+2\mright)^2) \\ =(\mleft(x^2-4\mright)\cdot2)/(2x\cdot\mleft(x+2\mright)^2) \\ =((x^2-4))/(x\mleft(x+2\mright)^2) \end{gathered}`

We know that

`x^2-4=(x-2)(x+2)`

Replacing it in the equation of w

`\begin{gathered} w=((x-2)(x+2))/(x(x+2)^2) \\ =((x-2))/(x(x+2)) \end{gathered}`

Answer width = (x - 2) / [ x ( x + 2 ) ]