Question

Find the ratio of areas of a circle and regular polygon of \[n\] sides of the equal perimeter.

Answer 1

Hint: First, we will use the formula to calculate the area of circle, \[\pi {r^2}\], where \[r\] is the radius of the circle and the formula to calculate of area of polygon of \[n\] sides, \[\dfrac{1}{4}n{a^2}\cot \left( {\dfrac{\pi }{n}} \right)\], where \[a\] is the length of one side, and use the fact that \[2\pi r = na\] to find the required ratio.

Complete step by step answer:

We are given the areas of a circle and regular polygon of \[n\] sides of the equal perimeter.
Let us assume that \[r\] be the radius of the circle and \[{A_1}\] be the area of this circle.
We know that the formula to calculate the area of a circle, \[\pi {r^2}\], where \[r\] is the radius of the circle.
Thus, we have \[{A_1} = \pi {r^2}\].
Let us also assume that \[a\] be the length of one side of the regular polygon of \[n\] sides and \[{A_2}\] be the area of this polygon.
We also know that the formula to calculate of area of polygon of \[n\] sides, \[\dfrac{1}{4}n{a^2}\cot \left( {\dfrac{\pi }{n}} \right)\], where \[a\] is the length of one side.
Thus, we have \[{A_2} = \dfrac{1}{4}n{a^2}\cot \left( {\dfrac{\pi }{n}} \right)\].
We will now use the condition that perimeter of circle and polygon to find a.
\[ \Rightarrow 2\pi r = na\]
Dividing the above equation by \[n\] on both sides, we get
\[
   \Rightarrow \dfrac{{2\pi r}}{n} = \dfrac{{na}}{n} \\
   \Rightarrow a = \dfrac{{2\pi r}}{n} \\
 \]
Substituting the above value of \[a\] in the area of polygon \[{A_2}\], we get
\[
  {A_2} = \dfrac{1}{4}n{\left( {\dfrac{{2\pi r}}{n}} \right)^2}\cot \left( {\dfrac{\pi }{n}} \right) \\
  {A_2} = \dfrac{1}{4}\dfrac{{{\pi ^2}{r^2}}}{n}\cot \left( {\dfrac{\pi }{n}} \right) \\
 \]
Dividing \[{A_1}\] by the above value of \[{A_2}\] to find the required ratio, we get
\[
   \Rightarrow \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{\pi {r^2}}}{{\dfrac{{{\pi ^2}{r^2}}}{n}\cot \left( {\dfrac{\pi }{n}} \right)}} \\
   \Rightarrow \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{1}{{\dfrac{\pi }{n}\cot \left( {\dfrac{\pi }{n}} \right)}} \\
   \Rightarrow \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{1}{{\dfrac{\pi }{n}\cot \left( {\dfrac{\pi }{n}} \right)}} \\
   \Rightarrow \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{1}{{\dfrac{\pi }{n}}} \cdot \dfrac{1}{{\cot \left( {\dfrac{\pi }{n}} \right)}} \\
 \]
Using the trigonometric value, \[\dfrac{1}{{\cot \left( a \right)}} = \tan \left( a \right)\] in the above equation, we get
\[
   \Rightarrow \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{1}{{\dfrac{\pi }{n}}} \cdot \tan \left( {\dfrac{\pi }{n}} \right) \\
   \Rightarrow \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{\tan \left( {\dfrac{\pi }{n}} \right)}}{{\dfrac{\pi }{n}}} \\
 \]
Hence, the required ratio is \[\tan \left( {\dfrac{\pi }{n}} \right):\dfrac{\pi }{n}\].

Note: We need to know the formula of areas of circle and polygons with \[n\] sides. Do not get confused while writing the labels and their values to avoid calculation mistakes. We will use the formula to calculate of area of polygon of \[n\] sides, \[\dfrac{1}{4}n{a^2}\cot \left( {\dfrac{\pi }{n}} \right)\], where \[a\] is the length of one side. We need to know the formula to calculate the perimeter of the circle, \[2\pi r\], where \[r\] is the radius of the circle. The perimeter of the polygon of \[n\] sides is \[na\], which is the sum of all sides of the polygon.