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The ratio of the area of a regular polygon of n sides inscribed in a circle to that of the polygon of n sides and of perimeter equal to that of the circle are in the ratio of ans:tan(pi/n):(pi/n)

The ratio of the area of a regular polygon of n sides inscribed in a circle to that of the polygon of n sides and of perimeter equal to that of the circle are in the ratio of

ans:tan(pi/n):(pi/n)
 

Grade:11

1 Answers

Arun
25763 Points
3 years ago

The Area of the regular polygon inscribed in a Circle of radius r would be n sin(pie/n)cos(pie/n) r2

Now the second polygone has perimeter = 2pie.r and its Area would be r2. n sin(pi)/n cos(pi)/n 

and Area of the polygon with perimeter = Circle would be a2 n tan(pi)/n. where a is apothem. 

a = s/2tan(pi)/n = 2pi. r /2ntan(pi)/n= pi. r/2n tan(pi)/n 

So Area would be (pi. r/2n tan(pi)/n)2. n. tan(pi)/n. = [((pi)2.r2)/4 n2 tan2(pi)/n] *[n. tan(pi)/n]

So the ratio would be = [n sin(pie/n)cos(pie/n) r2]/[ ((pi)2.r2)/4 n2tan2(pi)/n] *[n. tan(pi)/n] = [sin2(pi)/n]/(pi/n)2

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