circlesgeometry
Could you please tell me that:
can a regular hexagon with a side length x can be inscribed inside a circle of a radius x?
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?
$A=\dfrac{\pi\times x^2\times 60^{\circ}}{360^{\circ}}-\dfrac{1}{2}\times x^2\times \sin(60^{\circ})$
Here are some equal angle inscribed angles (they all subtend half the angle of the red arc). It is clear that the area of the last sector is properly contained in the areas of the others.
$\hspace{2.5cm}$
The area of the sector is the sum of the area between the arc and its chord (a constant area) and the area of the triangle, which is $\frac12$ the product of the length of the chord (a constant length) and the altitude of the triangle, which varies as the vertex moves around the circle.
The length of a side is $2r sin(\frac{\pi}{n})$ by trigonometry.
The distance is $r(1-cos\frac{\pi}{n})$, which is the difference between the radius and the length of the line from the center of the circle to the center of the side.
Best Answer
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 $$A=\frac12x^2\Bigl(\frac\pi 3-\cos\frac\pi6\Bigr)=\frac12x^2\Bigl(\frac\pi 3-\sin\frac\pi3\Bigr).$$