Question

A hexagon is inscribed in a circle. If the difference between the area of the circle and the area of the hexagon is 36 m2, use the formula for the area of a sector to approximate the radius r of the circle. (Round your answer to three decimal places.)

Answer 1

A buddy is in the same class and we both had trig in high school, so we are wondering how everyone else is doing since we're having so much trouble. I managed to finally figure out how to solve using sectors only it requires knowing the relationship between the sides of a 30-60-90 triangle, which is a couple chapters down the road.

This is my work:
\[A=\frac{1}{2}r^{2}θ\]
\[A=12rh\]
\[\frac{1}{2}r^{2}θ−\frac{1}{2}rh=36\]
\[\theta = \frac{\pi}{3}\]
\[\frac{1}{2}r^{2}(\frac{\Pi}{3})−\frac{1}{2}rh=36\]
\[h=\frac{\sqrt{3}}{2}r\]
\[\frac{1}{2}r^{2}(\frac {\Pi}{3})−(\frac{1}{2})r(\frac{\sqrt{3}}{2})r=36\]
\[\frac{1}{2}r^{2}(\frac{\Pi}{3})−\frac{1}{2}(\frac{\sqrt{3}}{2})r^{2}=36\]
\[\frac{1}{2}r^{2}[(\frac{\Pi}{3})−(\frac{\sqrt{3}}{2})]=36\]
\[6(\frac{1}{2}r^{2}[(\frac{\Pi}{3})−(\frac{\sqrt{3}}{2})])=36\]
\[3r^{2}[(\frac{\Pi}{3})−(\frac{\sqrt{3}}{2})])=36\]
\[r^{2}[(\frac{\Pi}{3})−(\frac{\sqrt{3}}{2})])=12\]
\[r=\sqrt{12 \div [(\frac{\Pi}{3})−(\frac{\sqrt{3}}{2})])}\]
\[r \approx 8.139\]
While this works, is there a way to do it with trigonometric functions, but still apply sector theory?

Answer 2

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Answer 3

It is a regular hexagon right?

Answer 4

Yeah, that's assumed from the problem given, so each of the six triangles formed from the hexagon would be equilateral triangles with 60 degree angle measures.

Answer 5

Yes excatly! :D

Answer 6

And then the area of the circle would be
pi*r*r - area of hexagon
pi*r*r - [3 sqrt 3*r^2/2] = 36

Just find this and you'll get radius!! :D

Answer 7

That's what I did with my work shown above. I was wondering if there was a way to do it without involving any roots and by using trig functions because our book technically hasn't covered 30-60-90 triangles yet.

Answer 8

Yes you can do it but this method is the EASIEST and the fastest.
And where would you require 30-60-90 triangles? :D

Answer 9

I did it when I split the equilateral triangle in half to find h in terms of r and got h = (sqrt3/2)r

Answer 10

Hahah...But why?
:')
You already have a direct formula for equilateral triangle = sqrt 3*a^2/4
Why do you want to slpit it? XD

Answer 11

. . . I like right triangles :D
I usually end up breaking things down too far and over-analyze them. It was making a lot of sense to break it that way even if its the long way.

Answer 12

Hahahaha!! LOL
But you understood this now right? :D

Answer 13

Not really, I remember getting that equation before and I was wondering how the pi*r*^2 part worked because the formula for the area of a sector has a measure of theta involved.