Limit of area of n-sided inscribed regular polygons

For regular polygons inscribed in a circle, the more the sides, the smaller their length. When the sides infinitely increase, the perimeter of the polygon will be equal to the circumference of the circle. Using the conclusion, we could also prove the similar conclusion that the limit of the area of the inscribed regular polygons will be the approximation of the their circumscribed circle when their sides keep increasing.

Method 1 using the perimeter conclusion

In the figure, a 2n-sides regular polygon and a n-sided regular polygon share the same circumscribed circle.

Join the center of the circle to the vertices of 2n-sided regular polygon, divide the polygon into 2n isosceles triangles.

A_{2n} = 2na_{2n}

(1)

in which A_{2n} is the area of the 2n-sided regular polygon and a_{2n} is the divided single triangle.

Mark two of the triangles as \triangle AOC and \triangle BOC . Point D is the intersection point of OC and AB . We have the area of two triangles,

A_{\triangle AOB} = \dfrac{1}{2}AB\cdotp OD

A_{\triangle ACB} = \dfrac{1}{2}AB\cdotp CD

A_{\triangle AOB}+ A_{\triangle ACB} = \dfrac{1}{2}AB(OD+CD)

\therefore 2a_{2n} = \dfrac{1}{2}AB\cdotp R

\therefore 2na_{2n} = \dfrac{1}{2}nAB\cdotp R

(2)

in which nAB is the perimeter of n-sided regular expression P_n .

Using the equation (1), we get

A_{2n} = \dfrac{1}{2}P_nR

(3)

From the article Limit of perimeter of n-sided inscribed regular polygons, we know,

When n\to \infty , P_n \to C , the perimeter of the circumscribed circle of the polygon. So,

A_{2n} = \dfrac{1}{2}\cdotp 2\pi R\cdotp R= \pi R^2

(4)

Method 2 using the limit of nsinPi/n

Given n-sided inscribed regular polygon and its circumcircle, join the center of the circle with the vertices, we know the regular polygon is made of n isosceles triangles. These isosceles triangles have two equal sides of length R . The included angle of the two sides or the central angle is,

\dfrac{2\pi }{n}

(1)

Construct altitude h from point O to the side, which bisects the central angle. So,

h = R\cos \dfrac{\pi}{n}

(2)

The side of the regular polygon is,

s = 2R \sin \dfrac{\pi}{n}

(3)

So the area of the regular polygon is

A_n =n\cdotp \dfrac{1}{2}hs = nR^2 \sin \dfrac{\pi}{n} \cos \dfrac{\pi}{n}

(4)

According to trigonometric product-to-sum identity or prosthaphaeresis

\sin \alpha\cos \beta = \dfrac{1}{2} [\sin(\alpha+\beta ) +\sin(\alpha - \beta) ]

\sin \dfrac{\pi}{n} \cos \dfrac{\pi}{n}= \dfrac{1}{2} \sin \dfrac{2\pi}{n}

So equation (4) is converted to

A_n = \dfrac{1}{2}nR^2 \sin \dfrac{2\pi}{n}= \dfrac{n}{2\pi}\pi R^2 \sin \dfrac{2\pi}{n}

=\pi R^2\dfrac{\sin \dfrac{2\pi}{n} }{ \dfrac{2\pi}{n}}

(5)

From the article, Limit of perimeter of n-sided inscribed regular polygons

\lim\limits_{ \theta\to 0} \dfrac{\sin \theta}{\theta}=1 ,

Therefore,

\lim\limits_{n\to \infty} \dfrac{\sin \dfrac{2\pi}{n} }{ \dfrac{2\pi}{n}}=1

So when n is infinitely large, equation (5) is simplified as

A_n = \pi R^2

Steven Zheng posted 1 year ago