(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)
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Expert's answer
2011-12-01T08:38:26-0500
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)
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