Question

A rectangular piece of plywood has a diagonal which measures two feet more than the width. The length of the plywood is twice the width. What is the length of the plywood's diagonal? Round to the nearest tenth.

Short answer

The length of the diagonal is$3.6$ft.

Step-by-step solution

Step 1. Given ifnormation.

A rectangular piece of plywood has a diagonal which measures two feet more than the width. The length of the plywood is twice the width.

Step 2. Determine the length of the plywood diagonal.

Let the width is $x$.

From the given information:

A rectangular piece of plywood has a diagonal which measures two feet more than the width. The length of the plywood is twice the width.

So, the required length is $2x$and diagonal is $x+2$.

Step 3. Apply the Pythagorean theorem to frame the equation.

$\begin{array}{rcl}{x}^2+{\left(2x\right)}^2& =& {\left(x+2\right)}^2\\ x^2+4x^2& =& x^2+4x+4\\ 4x^2& =& 4x+4\\ 4x^2-4x-4& =& 0\\ 4\left(x^2-x-1\right)& =& 0\\ x^2-x-1& =& 0\end{array}$

Step 4. Solve the equation.

Apply quadratic formula:

$x^2-x-1=0$

Here, $a=1,b=-1,c=-1$.

$\begin{array}{rcl}x& =& \frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ & =& \frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2-4\times \left(1\right)\times \left(-1\right)}}{2\left(1\right)}\\ & =& \frac{1\pm \sqrt{1+4}}{2}\\ & =& \frac{1\pm \sqrt{5}}{2}\\ & \approx & \frac{1\pm 2.2}{2}\end{array}$

Step 5. Solve for x.

$\begin{array}{rcl}x& =& \frac{1+2.2}{2}\\ & =& \frac{3.2}{2}\\ & =& 1.6\end{array}$ and $\begin{array}{rcl}x& =& \frac{1-2.2}{2}\\ & =& -\frac{1.2}{2}\\ & =& -0.6\end{array}$

The width can't be negative.

So, the required value of width is $x=1.6$ft.

Step 6. Simplified answer.

The length of the diagonal is $x+2$ft.

Substitute $x=1.6$in the above expression.

Hence, the length of the diagonal is$1.6+2=3.6$ft.