What is the diameter of a circle inscribed in a regular polygon of 12 sides, each of length 1 cm ?

Concept:

Sine Rule:

In a triangle Δ ABC, where a is the side opposite to A, b is the side opposite to B, c is the side opposite to C and where R is the circum-radius:

\(\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R\)

The right-angled triangle follow the Pythagoras theorem,

Hypotenuse2 = Perpendicular2 + Base2

The hypotenuse is the greatest side of the right-angled triangle

Solution:

Given, Length of sides of polygon a = 1

Number of sides of polygon n = 12

The right-angled triangle follow the Pythagoras theorem,

\(\frac{sin 15}{0.5}=\frac{sin 75}{R}\)

\(\Rightarrow \frac{\sqrt3-1}{(2\sqrt2)0.5}=\frac{\sqrt6+\sqrt2}{4R}\)

\(\Rightarrow \frac{2(\sqrt3-1)}{(2\sqrt2)}=\frac{\sqrt6+\sqrt2}{4R}\)

\(\Rightarrow \frac{(\sqrt3-1)}{(\sqrt2)}=\frac{\sqrt6+\sqrt2}{4R}\)

\(\Rightarrow R=\frac{\sqrt6+\sqrt2}{4}\times \frac{\sqrt2}{\sqrt3-1}\)

\(\Rightarrow D = 2R=\frac{\sqrt6+\sqrt2}{4}\times \frac{2\sqrt2}{\sqrt3-1}\) ...... (Diameter D = 2R)

\(\Rightarrow D =\frac{\sqrt6+\sqrt2}{2}\times \frac{\sqrt2}{\sqrt3-1}\)

\(\Rightarrow D =\frac{\sqrt6+\sqrt2}{\sqrt2 \sqrt2}\times \frac{\sqrt2}{\sqrt3-1}\)

\(\Rightarrow D =\frac{\sqrt6+\sqrt2}{\sqrt2}\times \frac{1}{\sqrt3-1}\)

\(\Rightarrow D =\frac{\sqrt6+\sqrt2}{\sqrt6-\sqrt2}\)

\(\Rightarrow D =\frac{\sqrt6+\sqrt2}{\sqrt6-\sqrt2} \times \frac{\sqrt6+\sqrt2}{\sqrt6+\sqrt2}\)

\(\Rightarrow D =\frac{6+2\sqrt{12}+2}{4}\)

\(\Rightarrow D =\frac{8+2\sqrt{12}}{4}\)

\(\Rightarrow D =\frac{8+2\sqrt{4 \times3}}{4}\)

\(\Rightarrow D =2+\sqrt3\)