Chapter 4.4, Problem 56PPSE

Solution

To Find: The sides of one rectangle whose area is equal to the area of the remaining part.

The sides of one rectangle whose area is equal to the area of the remaining part is $y=x+x\sqrt{2}$ .

Given:

Suppose cutting the small square from a square sheet of cardboard. Find the sides of one rectangle whose area is equal to the area of the remaining part.

  

Explanation:

From the diagram,

Let the side of the smaller square be $y$ units.

And its area is $y^2$ units.

Side of cardboard is $x+y$ units

And its area be ${\left(x+y\right)}^2$ square units

The difference of the area of the square piece of cardboard $\left({\left(x+y\right)}^2\right)$ and the area of the smaller square to be cut $\left(y^2\right)$ , is the area of the remaining part:

  ${\left(x+y\right)}^2-y^2$

Can be the equation factorable:

Since the equation is a difference of two squares, it can be factored.

  $\begin{array}{c}{\left(x+y\right)}^2-y^2=\left(x+y-y\right)\left(x+y+y\right)\\ =x\left(x+2y\right)\end{array}$

By setting the area of the small square to be cut out equal to the area of the remaining part, find the $y$ interns of $x$ :

  $\begin{array}{c}{y}^2={\left(x+y\right)}^2-y^2\\ y^2=x^2+2xy+y^2-y^2\\ y^2=x^2+2xy\end{array}$

Thus, $y^2-2xy-x^2=0$

Use quadratic formula to solve for $y$

Substitute $a=1,b=-2x$ and $c=-x^2$ :

  $\begin{array}{c}y=\frac{2x\pm \sqrt{{\left(-2x\right)}^2-4\left(1\right)\left(-x^2\right)}}{2\left(1\right)}\\ =\frac{2x\pm \sqrt{4x^2+4x^2}}{2}\\ =\frac{2x\pm 2x\sqrt{2}}{2}\\ =x\pm x\sqrt{2}\end{array}$

Eliminate the negative values.

So, $y=x+x\sqrt{2}$