Chapter 10.1, Problem 51E

To determine

To find: Sum of areas of all semicircles.

Answer to Problem 51E

  $A=\frac{\pi }{2}$ square units.

Explanation of Solution

Given:

The given figure is,

  

Number of semicircle in $n^{\text{th}}$ row is $2^n$

Radius of each semicircle is $1/2^n$

Formula used:

A geometric progression with initial term $a$ and common ratio $r$ , Then the sum of this geometric progression with $r<1$ is,

  $S=\frac{a}{1-r}$

Calculation:

Here, total area of semicircle in the 1^st row is,

  $A_1=2\times \frac{1}{2}\pi {\left(\frac{1}{2}\right)}^2$

Total area of semicircle in the 2^nd row is,

  $A_2=2^2\times \frac{1}{2}\pi {\left(\frac{1}{2^2}\right)}^2$

Similarly, total area of semicircle in the $n^{\text{th}}$ row is,

  $A_n=2^n\times \frac{1}{2}\pi {\left(\frac{1}{2^n}\right)}^2$

Hence area of all semicircles is,

  $\begin{array}{l}A={\displaystyle \sum _{n=1}^{\infty }{2}^n\times \frac{1}{2}\pi {\left(\frac{1}{2^n}\right)}^2}\\ ={\displaystyle \sum _{n=1}^{\infty }\frac{\pi }{2}{\left(\frac{1}{2}\right)}^n}\end{array}$

Here, A is geometric progression with initial term $a=\pi /4$ and common ratio $d=1/2$

So,

  $A=\frac{\pi /4}{1-1/2}=\frac{\pi }{2}$

Therefore, areas of all semicircles are $A=\frac{\pi }{2}$ square units.