Chapter 8.3, Problem 105E

To determine

To find: the total area of the shaded region.

Answer to Problem 105E

The area of shaded region is,

  $124{\text{in}}^{\text{2}}$ .

Explanation of Solution

Given information:

The sides of a square are 16 inches in length.

Given figure

  

New squares are formed by joining the mild points of sides of the squares.

Process repeats 5 times.

Calculation:

The square has 16 inches length side.

New squares are formed by joining the mild points of sides of the squares.

This will results 4 isosceles right angled triangle with sides 8, 8, $\sqrt{128}$ inches respectively.

Area of triangle can be found as shown:

  $\begin{array}{c}{A}_1=\frac{1}{2}\times 8\times 8\\ =32{\text{in}}^{\text{2}}\end{array}$

Area of two triangles

  $\begin{array}{c}{B}_1=2A_1\\ =64{\text{in}}^{\text{2}}\end{array}$

The square has $\sqrt{128}$ inches length side.

New squares are formed by joining the mild points of sides of the squares.

This will results 4 isosceles right angled triangle with sides $\frac{\sqrt{128}}{2}$ , $\frac{\sqrt{128}}{2}$ , $8$ inches respectively.

Area of triangle can be found as shown:

  $\begin{array}{c}{A}_2=\frac{1}{2}\times \frac{\sqrt{128}}{2}\times \frac{\sqrt{128}}{2}\\ =16{\text{in}}^{\text{2}}\end{array}$

Area of two triangles

  $\begin{array}{c}{B}_2=2A_1\\ =16{\text{in}}^{\text{2}}\end{array}$

Repeating the above process the geometric series will be

  $64,32,16,\mathrm{...}=64{\left(\frac{1}{2}\right)}^0,64{\left(\frac{1}{2}\right)}^1,64{\left(\frac{1}{2}\right)}^2,\mathrm{...}$

The sum of this series is,

  $S_n=\frac{a_1\left(1-r^n\right)}{1-r}$

We know that,

  $\begin{array}{c}{a}_1=64\\ r=\frac{1}{2}\\ n=5\end{array}$

Substituting,

  $\begin{array}{c}{S}_5=\frac{64\left(1-{\left(\frac{1}{2}\right)}^5\right)}{1-\left(\frac{1}{2}\right)}\\ =124{\text{in}}^{\text{2}}\end{array}$

Therefore, the area of shaded region is,

  $124{\text{in}}^{\text{2}}$