## Rational Approximation of Pi

A calculator gives us rational approximations of irrational numbers, so we can “see” that $2\sqrt{3}>\pi$. Explain how a circle inscribed in a regular hexagon allows us to “see” that $2\sqrt{3}>\pi$ without any decimal approximations.
Source: NCTM Mathematics Teacher 2008

SOLUTION

Consider the circle of center $A$ inscribed in a regular hexagon of side length $a$. Triangle $ABC$ is an equilateral triangle. Triangle $AHC$ a 30-60-90 degrees triangle with hypotenuse of length $a$. $AH$ is also the radius of the circle
$r=\sqrt{3}(a/2)$
$a=\dfrac{2r}{\sqrt{3}}$
$=\dfrac{2r\sqrt{3}}{3}$

Method 1: Area of triangle $AHC$ > area of sector $AHC$
$\dfrac{1}{2}\left (r\dfrac{a}{2}\right )>\pi r^2\dfrac{30^\circ}{360^\circ}$
Simplify
$\dfrac{ar}{4}>\dfrac{\pi r^2}{12}$
Divide both sides of inequality by $r\neq 0$ and multiply by $12$
$3a>\pi r$
Substitute the value of $a$ into the above inequality
$3\dfrac{2r\sqrt{3}}{3}>\pi r$
Simplify
$2\sqrt{3}>\pi$

Method 2: Area of hexagon > area of circle
$6\left (\dfrac{1}{2}ar\right )>\pi r^2$
$3ar>\pi r^2$
Substitute the value of $a$ into the above inequality
$3\left (\dfrac{2r\sqrt{3}}{3}\right )r>\pi r^2$
Simplify
$2\sqrt{3}>\pi$