if a regular polygon of 100 sides be inscribed in a circle prove that its area differs from thatof the circleby lessthan 1/1500of the latter?
- PhilipLv 61 month ago
Let us first consider a regular n-gon, n(=/>) 3, inscribed in a circle of radius r.;
Our n-gon consists of n equal isosceles triangles each subtending angle x = 2pi/n at
the circle center. If we choose one of these triangles and draw a perpendicular to the middle of it's base we create 2 equal right angled triangles each subtending angle x/2 = pi/n at the circle center. The height, h, of each triangle is rcos(pi/n). The base,
b, of each right angled triangle is rsin(pi/n). The base, B, of the containing isosceles triangle = 2b = 2rsin(pi/n). The area, A, of each isosceles triangle is (1/2)hB =;
(1/2)rcos(pi/n)2rsin(pi/n) = (r^2)cos(pi/n)*sin(pi/n) = (1/2)(r^2)sin(2pi/n). The total area of the n-gon is (n/2)(r^2)sin(2pi/n). In our particular case, n = 100. Our n-gon's
area = 50r^2sin(pi/50) = 50r^2sin(3.6°) = 50r^2(0.06279051953) = 3.139525976r^2.
Circle area - 100-gon area = (pi- 3.1395256)r^2 = 0.0026667713r^2.;
[circle area - 100-gon area]/(circle area) = 0.000657843762 < 0.0006666666667;
- VamanLv 71 month ago
100 sides. Each triangle makes an angle 3.6 degrees. Area of the triangle will be 1/2 a(radius) a i
sin1.8*2*100. put a=1. we get area=3.1411. area= pi=3.1416
- DixonLv 71 month ago
Consider the triangle formed by one side of the polygon and the center of the circle. In fact consider the right angled triangle of the bisected side.
The angle at the center is θ = 360/200
Opposite = r sinθ
Adjacent = r cosθ
Area = 1/2 r² cosθ sinθ
Area of polygon = 100 r² cosθ sinθ
Area of circle = πr²
Fractional difference = (100 r² cosθ sinθ - πr²)/πr²
Fractional difference = (100 cosθ sinθ - π)/π
Fractional difference = -0.0006584
1/Fractional difference= - 1/1520
- PopeLv 71 month ago
Maybe you should lay down some ground rules. I began work on a geometric proof, much as Archimedes did. I soon discovered that the actual ratio is quite close to 1/1500, so my inequalities and iterated dissections would fall short.
Did you want a simple calculator comparison? Here is how that would work. Let the circle have radius r.
area of circle = πr²
area of inscribed 100-gon = 50r²sin(3.6°)
difference of areas = πr² - 50r²sin(3.6°)
[πr² - 50r²sin(3.6°)] : πr²
= 1 : π / [π - 50sin(3.6°)]
≈ 1 : 1520
Taking that as proof is a bit of a stretch. It is only a calculation.
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- 1 month ago
Wow this is an interesting question. I don't know the answer, but I can help you make a start:
Whenever you're attempting a really difficult problem like this, start by picking an easier version of the same problem, and look for some sort of pattern.
So let's start with a triangle inside a circle (since that's the "simplest" possible polygon):
Since it's a "regular triangle" (as opposed to an irregular triangle), this means that it must be an equilateral triangle (all sides and angles must be the same):
If the circle has a radius of 1, then the area of the circle is pi.
What is the area of the triangle? Well, start with a point in the centre of the circle, branch out to each vertex of the triangle and use some trigonometry to work it out.
A 100-sided polygon actually can be split up into a bunch of triangles. Like think about it: each triangle is formed by making a line from the centre of the polygon to 2 adjacent vertices. The interior angles of each triangle will add up to 180 degrees. You can use trigonometry to work out the area of each individual triangle, multiply that number by 100, then you've got the area to the 100-sided polygon.
Then just compare that area to the area of the circle and get a ratio. Show that the area of the polygon is more than (1499*pi/1500) and you've solved the problem.