Chapter 11.5, Problem 13CE

To determine

Prove $\frac{C}{d}=\frac{C\text{'}}{d\text{'}}$

Explanation of Solution

Given:

  $\odot O\text{ and }\odot \text{O'}$ with circumferences C and C’ and diameters d and d’ respectively.

Calculation:

Proof:

Inscribe in each circle a regular polygon of n sides. Let p and p’ be the perimeters.

  

  $p=\underset{n-times}{\underbrace{s+s+s+\mathrm{...}+s}}=ns\mathrm{.............}[perimeter]$

  $p\text{'}=\underset{n-times}{\underbrace{s\text{'}+s\text{'}+s\text{'}+\mathrm{...}+s\text{'}}}=ns\text{'}\mathrm{.............}[perimeter]$

So,

  $\frac{p}{p\text{'}}=\frac{ns}{ns\text{'}}=\frac{s}{s\text{'}}\mathrm{...................}(1)$

The $m\angle AOB=\frac{360°}{n}=m\angle A\text{'}O\text{'}B\text{'}$ , since the angle is divided into equal n angles by regular n -gon.

Also, in the isosceles triangles AOB and A’O’B’ ,

  $m\angle OAB=m\angle OBA=m\angle O\text{'}A\text{'}B\text{'}=m\angle O\text{'}B\text{'}A\text{'}$ ,since the third angle is equal in both the triangles.

So, by AAA Similarity Postulate , $\Delta AOB\sim \Delta A\text{'}O\text{'}B\text{'}$

So,

  $\begin{array}{l}\frac{AB}{A\text{'}B\text{'}}=\frac{OB}{O\text{'}B\text{'}}=\frac{2OB}{2O\text{'}B\text{'}}\\ \Rightarrow \frac{s}{s\text{'}}=\frac{r}{r\text{'}}=\frac{d}{d\text{'}}\mathrm{.................}(2)\end{array}$

where d and d’ are diameters of $\odot O\text{ and }\odot O\text{'}$ respectively.

So, from equations (1) and (2),we have:

  $\frac{p}{p\text{'}}=\frac{d}{d\text{'}}$

As $n\to \infty ,p\to C\text{ and }p\text{'}\to C\text{'}$ where $C\text{ and }C\text{'}$ are circumference of $\odot O\text{ and }\odot \text{O'}$ .

So, as n becomes very large,

  $\begin{array}{l}\frac{C}{C\text{'}}=\frac{d}{d\text{'}}\\ \Rightarrow \frac{C}{d}=\frac{C\text{'}}{d\text{'}}\mathrm{................}[\text{multiply each side by }\frac{C\text{'}}{d}]\end{array}$

Hence proved.