Question #5329

(a) Let A sub n be the area of a polygon with n equal sides inscribed in a circle with radius r. By dividing the polygon into n congruent triangles with central angle 2*(pi)/n, show that A sub n=1/2*(n/r^2)*sin*(2pi/n)

Expert's answer

Let S be the area of each of n triangles, then A_n = n*S.

So we have to find

the area of S.

Notice that S is an isosceles trianlge with two sides of

length r.

The angle betwenn these sides is 2*pi/n.

It is known that the

area of a triangle with sides a, b, and angle phi between them is

a*b*sin(phi)/2.

In our case

a=b=r

phi=2*pi/n.

Hence

S = 1/2

* r^2 * sin(2*pi/n)

Therefore

A_n = n*S = 1/2 * n * r^2 *

sin(2*pi/n)

So we have to find

the area of S.

Notice that S is an isosceles trianlge with two sides of

length r.

The angle betwenn these sides is 2*pi/n.

It is known that the

area of a triangle with sides a, b, and angle phi between them is

a*b*sin(phi)/2.

In our case

a=b=r

phi=2*pi/n.

Hence

S = 1/2

* r^2 * sin(2*pi/n)

Therefore

A_n = n*S = 1/2 * n * r^2 *

sin(2*pi/n)

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