Chapter 9.4, Problem 26WE

To determine

To find:The possibility of drawing two chords in the ratio $1:2$ and their distances from the centre are in the ratio $2:1$ . Also find the length of each chords in terms of the radius. If not, then prove that it is impossible.

Answer to Problem 26WE

Length are $\frac{2}{\sqrt{5}}r\text{ and }\frac{4}{\sqrt{5}}r$ .

Explanation of Solution

Given information:

Two chords whose lengths are in the ratio $1:2$ and distances from the centre are in the ratio $2:1$ .

Concept used:

Using the property “Perpendicular to any chord from center of a circle, bisects the chord”. And Pythagoras theorem.

Calculation:

Circle with chords in the given ratio is drawn below.

  

It is given that.

  $\begin{array}{l}\frac{AB}{CD}=\frac{1}{2}\\ \Rightarrow CD=2AB\\ \frac{OE}{OF}=\frac{2}{1}\\ \Rightarrow OF=\frac{OE}{2}\end{array}$

Using Pythagoras theorem in triangle AOE.

  $\begin{array}{l}O{A}^2=A{E}^2+O{E}^2\\ \therefore \left(\text{Perpendicular form center to any chord,bisects the chord}\right)\\ r^2={\left(\frac{AB}{2}\right)}^2+O{E}^2\mathrm{......}\left(1\right)\end{array}$

Using Pythagoras theorem in triangle COF.

  $\begin{array}{l}O{C}^2=C{F}^2+O{F}^2\\ r^2={\left(\frac{CD}{2}\right)}^2+O{F}^2\\ r^2={\left(\frac{2AB}{2}\right)}^2+{\left(\frac{OE}{2}\right)}^2\therefore \left(CD=2AB\right)\\ r^2={\left(AB\right)}^2+{\left(\frac{OE}{2}\right)}^2\\ \text{From }\left(1\right).\\ {\left(AB\right)}^2+{\left(\frac{OE}{2}\right)}^2={\left(\frac{AB}{2}\right)}^2+O{E}^2\\ {\left(\frac{AB}{2}\right)}^2={\left(\frac{OE}{2}\right)}^2\\ AB=OE\\ \text{From }\left(1\right).\\ r^2={\left(\frac{AB}{2}\right)}^2+A{B}^2\\ r^2=\frac{A{B}^2}{4}+A{B}^2\\ r^2=\frac{5}{4}A{B}^2\\ A{B}^2=\frac{4}{5}{r}^2\\ AB=\frac{2}{\sqrt{5}}r\end{array}$

  $\begin{array}{l}CD=2AB\\ CD=2\times \frac{2}{\sqrt{5}}r\\ CD=\frac{4}{\sqrt{5}}r\end{array}$

Therefore, the answer is $\frac{2}{\sqrt{5}}r\text{ and }\frac{4}{\sqrt{5}}r$