Segment area for a regular hexagon inscribed in a circle

Could you please tell me that:

1. can a regular hexagon with a side length x can be inscribed inside a circle of a radius x?

2. If 1. is true, then I want to find out the area of the segment for a sector in the diagram below. Is the formula given below correct?

$A=\dfrac{\pi\times x^2\times 60^{\circ}}{360^{\circ}}-\dfrac{1}{2}\times x^2\times \sin(60^{\circ})$ • 1. Yes 2. What is the area of a segment? Which segment anyway? – Bernard Aug 15 '18 at 15:40
• Sorry a little confusing isn’t it? The little area with the curved part left over if you took away the area of the equilateral triangle from – Will Kim Aug 15 '18 at 15:41
• The sector. Sorry about shading all 6 of these. I meant to shade only the top one. – Will Kim Aug 15 '18 at 15:42
• Did you mean 360 in the denominator rather than 36? – Deepak Aug 15 '18 at 15:43
• Yes, just changed it and also updated the diagram. Thanks. – Will Kim Aug 15 '18 at 15:45

The answer to both questions is Yes.

1. Yes, you can inscribe a regular hexagon within a circle of radius equal to the side length of the hexagon. Your construction makes this self evident - consider the hexagon as six equilateral triangles.

2. The formula is correct. The area of the segment is found by subtracting the area of the triangle from the area of the sector. Where the angle subtending the segment $\theta$ is given in radian measure, the area $A$ of the segment is $A = \frac 12 r^2\theta - \frac 12 r^2 \sin\theta$.

Converting to degree measure by multiplying by $\frac{\pi}{180}$ and using the angle $60^{\circ}$ and $r = x$, we get:

$A = \frac 12 \pi x^2 \frac{60}{180} - \frac 12 x^2 \sin 60^{\circ}$, which can be rearranged to what you have.

• Thank you for the reassurance. It makes sense now. – Will Kim Aug 15 '18 at 15:55

It's correct, except it's simple to calculate if the angle measures are in radians: the sought area is the difference between the area of the circular sector with central angle $\pi/3$, which is equal to $\frac12x\cdot x\frac\pi3$, and the area of the equilateral triangle with side $x$, which is $\;\frac 12x\cdot x\cos\frac\pi6$ (each of these formulæ, is half the product of the height $x$ by the length of the base), whence $$A=\frac12x^2\Bigl(\frac\pi 3-\cos\frac\pi6\Bigr)=\frac12x^2\Bigl(\frac\pi 3-\sin\frac\pi3\Bigr).$$

• Thank you so much. – Will Kim Aug 15 '18 at 15:58