Chapter 12.3, Problem 45E

(a)

To determine

To calculate:

The perimeter of the new square.

Answer to Problem 45E

The perimeter of the new square $ABCD$ is $10\sqrt{2}feets$ .

Explanation of Solution

Given information:

Calculation:

Consider that $A\text{'}B\text{'}C\text{'}D\text{'}$ be a square.

Perimeter of the square $A\text{'}B\text{'}C\text{'}D\text{'}$ is $20feet$ .

Formula for the perimeter of a square is $4a$ .

Therefore,

  $\begin{array}{l}4a=20\\ a=5\end{array}$

Side of the square is $5feets$ .

Therefore,

  $\begin{array}{l}AA\text{'}=2.5feets.\\ AB\text{'}=2.5feets.\end{array}$

Here $AA\text{'}D$ is a right-angled triangle.

By using Pythagorean Theorem, we can find the length of $AD$ .

From the right-angled triangle $AA\text{'}D$ ,

  $\left(A{D}^2\right)={\left(AA\text{'}\right)}^2+{\left(A\text{'}D\right)}^2$

Substitute the values,

  $\begin{array}{l}\left(A{D}^2\right)={\left(2.5\right)}^2+{\left(2.5\right)}^2\\ \left(A{D}^2\right)=6.25+6.25\\ \left(A{D}^2\right)=12.5\\ AD=\sqrt{12.5}\\ AD=\sqrt{\frac{125}{10}}\\ AD=\sqrt{\frac{25}{2}}\\ AD=\frac{5}{\sqrt{2}}\end{array}$

Therefore, length of each side of the square $ABCD$ is $\frac{5}{\sqrt{2}}$ .

Therefore, perimeter of the $ABCD$ is $\frac{5}{\sqrt{2}}\left(4\right)=10\sqrt{2}feets$ .

(b)

To determine

To calculate:

The sum of perimeters of all the squares.

Answer to Problem 45E

The sum of perimeters of all the squares is $69feets$ .

Explanation of Solution

Given information:

Calculation:

The following is the calculation of the sum of the series:

Perimeter of the square $A\text{'}B\text{'}C\text{'}D\text{'}$ is $20feet$ .

Perimeter of the square $ABCD$ is $10\sqrt{2}feets$ .

Similarly, the perimeter of the next square is $10$ .

The sum of the perimeters of all squares is

  $20+10\sqrt{2}+10+\mathrm{...}$

This is a geometric series.

Let,

  $\begin{array}{l}a=20\\ r=\frac{10\sqrt{2}}{20}\\ r=\frac{1}{\sqrt{2}}\end{array}$

The formula for the sum of the infinite geometric series is given by:

  $\begin{array}{l}S=\frac{a}{1-r}\\ S=\frac{20}{1-\frac{1}{\sqrt{2}}}\\ S=\frac{20\left(\sqrt{2}\right)}{\sqrt{2}-1}\\ S=\frac{28.27}{0.41}\\ S=68.97feets\\ S\approx 69feets\end{array}$

Therefore, the sum of the perimeters of the all the squares is $69feets$ .