Question

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

Answer 1

Finding the width of a rectangle

We know its area and its length

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

We know

Width · length = area

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

Then

$\begin{gathered} w\cdot l=a \\ w=\frac{a}{l} \end{gathered}$

Replacing the given information

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

We know that

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

Replacing it in the equation of w

$\begin{gathered} w=\frac{(x-2)(x+2)}{x(x+2)^2} \\ =\frac{(x-2)}{x(x+2)} \end{gathered}$

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