Chapter 8.4, Problem 28E

To determine

Find the area of the region using Archimedes geometric series.

Answer to Problem 28E

Thearea of the region is $\frac{32}{3}$ .

Explanation of Solution

Given:

The given diagram is

  

The area of the region is $\frac{4}{3}$ the area of the inscribed triangle.

Calculation:

Archimedes proves that the area of each green triangles is $\frac{1}{8}$ th of the area of the blue triangle.

From the diagram,

Area of each yellow triangle $=\frac{1}{8}\times$ area of green triangle

Area of each red triangle $=\frac{1}{8}\times$ area of yellow triangle

and so on

In the given diagram, one large blue triangle, two green triangles, four yellow triangles, eight red triangles and so on.

Area $=A+2\cdot \left(\frac{A}{8}\right)+4\left(\frac{A}{8^2}\right)+8\left(\frac{A}{8^3}\right)+\mathrm{...}$

  $A$ represents the area of the large blue triangles.

From the above statement, the series is

  $A\left(1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\mathrm{...}\right)$

It is a infinite geometric series.

The sum of an infinite geometric series with first term $a_1$ and common ratio $r$ is

  $S=\frac{a_1}{1-r}$ , where $r<1$ .

If $r\ge 1$ , then the series has no sum.

In the given series $a_1=1$ and $r=\frac{\frac{1}{4}}{1}=\frac{1}{4}$ .

Total profit $S=\frac{a_1}{1-r}$

  $\begin{array}{l}=\frac{1}{1-\frac{1}{4}}\\ =\frac{4}{3}\end{array}$

Area $=\frac{4}{3}\cdot A=\frac{4}{3}\cdot 8=\frac{32}{3}$

Hence the area of the regionis $\frac{32}{3}$ .