Question

A regular hexagon & a regular dodecagon are inscribed in the same circle. If the side of the dodecagon is $(\sqrt{3}-1)$, then the side of the hexagon is:

• A. $\sqrt{2}+1$
• B. $\frac{\sqrt{3}+1}{2}$
• C. $2$
• D. $\sqrt{2}$

Answer 1

The correct option is D $\sqrt{2}$

We know that

$\sin \theta /2=\frac{\left(side/2\right)}{r}$

Where $\theta =$ Angle taken by one side of inscribed polygon

$r=$ circum radius

$\therefore$ For do-decagon $=\theta =\frac{{360}^o}{12}={30}^o$

$\therefore \sin {15}^o=\frac{\sqrt{3}-1}{\left(2\right)\left(r\right)}$

$\therefore r=\left(\frac{\sqrt{3-1}}{2}\right)\csc {15}^o$

$\therefore$ For hexagon $\theta =\frac{{360}^o}{6}={60}^o$

$\therefore \sin {30}^o=\frac{side}{\left(2\right)\frac{\sqrt{3}-1}{2}\csc {15}^o}$

$\therefore side=\frac{\sqrt{3}-1}{\left(2\right)}\times \frac{1}{\sin {15}^o}$

$=\frac{\sqrt{3}-1\left(2\sqrt{2}\right)}{\left(2\right)(\sqrt{3}-1)}$

$\therefore side=\sqrt{2}cm$