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

Answer: Given in solution

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About mvtrinh

Retired high school math teacher.
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