
If the perimeter of a circle is equal to the perimeter of a regular polygon of ‘n’ sides, then their areas are in the ratio
$
{\text{A}}{\text{. tan}}\left( {\dfrac{\pi }{n}} \right):\;\left( {\dfrac{\pi }{n}} \right) \\
{\text{B}}{\text{. cos}}\left( {\dfrac{\pi }{n}} \right):\;\left( {\dfrac{\pi }{n}} \right) \\
{\text{C}}{\text{. sin}}\left( {\dfrac{\pi }{n}} \right):\;\left( {\dfrac{\pi }{n}} \right) \\
{\text{D}}{\text{. cot}}\left( {\dfrac{\pi }{n}} \right):\;\left( {\dfrac{\pi }{n}} \right) \\
$
Answer
591.6k+ views
Hint: To find the ratio of the areas of the circle and polygon, first find the relation between the radius of the circle and the length of the side of the polygon. Then put them in the formula for their respective areas, and then get the ratio.
Complete step-by-step answer:
Let the radius of the circle be r and the side length of the polygon be a.
We know that perimeter of circle = $2\pi r$
Perimeter of an n sided polygon = $n.a$
As the two perimeters are given to be equal-
$
2\pi r = n.a \\
\dfrac{{2\pi r}}{n} = a \\
$ …..relation 1
This way we found the side length a in terms of r.
Now, in order to calculate the ratio of the areas of the two surfaces, put the value of r and a in their respective formulas.
Area of circle = $\pi {r^2}$
Area of an n-sided polygon = $\dfrac{1}{4}n{a^2}\cot \left( {\dfrac{\pi }{n}} \right)$
Substituting the value of ‘a’ in the above formula from relation 1-
$
\dfrac{1}{4}n{a^2}\cot \left( {\dfrac{\pi }{n}} \right) \\
\dfrac{1}{4}n{\left( {\dfrac{{2\pi r}}{n}} \right)^2}\cot \left( {\dfrac{\pi }{n}} \right) \\
\dfrac{1}{4}n\left( {\dfrac{{4{\pi ^2}{r^2}}}{{{n^2}}}} \right)\cot \left( {\dfrac{\pi }{n}} \right) \\
\left( {\dfrac{{{\pi ^2}{r^2}}}{n}} \right)\cot \left( {\dfrac{\pi }{n}} \right) \\
$
Now the ratio of the circle and the polygon becomes-
$
\;\pi {r^2}:\left( {\dfrac{{{\pi ^2}{r^2}}}{n}} \right)\cot \left( {\dfrac{\pi }{n}} \right)\; \\
1:\left( {\dfrac{\pi }{n}} \right)\cot \left( {\dfrac{\pi }{n}} \right) \\
\tan \left( {\dfrac{\pi }{n}} \right):\left( {\dfrac{\pi }{n}} \right) \\
$
So the correct option is A.
Note: The formula for the area of a polygon can be determined by inscribing a circle inside the polygon. For direct calculation purposes, we use the final formula directly. Rest the question is based on simple calculations.
Complete step-by-step answer:
Let the radius of the circle be r and the side length of the polygon be a.
We know that perimeter of circle = $2\pi r$
Perimeter of an n sided polygon = $n.a$
As the two perimeters are given to be equal-
$
2\pi r = n.a \\
\dfrac{{2\pi r}}{n} = a \\
$ …..relation 1
This way we found the side length a in terms of r.
Now, in order to calculate the ratio of the areas of the two surfaces, put the value of r and a in their respective formulas.
Area of circle = $\pi {r^2}$
Area of an n-sided polygon = $\dfrac{1}{4}n{a^2}\cot \left( {\dfrac{\pi }{n}} \right)$
Substituting the value of ‘a’ in the above formula from relation 1-
$
\dfrac{1}{4}n{a^2}\cot \left( {\dfrac{\pi }{n}} \right) \\
\dfrac{1}{4}n{\left( {\dfrac{{2\pi r}}{n}} \right)^2}\cot \left( {\dfrac{\pi }{n}} \right) \\
\dfrac{1}{4}n\left( {\dfrac{{4{\pi ^2}{r^2}}}{{{n^2}}}} \right)\cot \left( {\dfrac{\pi }{n}} \right) \\
\left( {\dfrac{{{\pi ^2}{r^2}}}{n}} \right)\cot \left( {\dfrac{\pi }{n}} \right) \\
$
Now the ratio of the circle and the polygon becomes-
$
\;\pi {r^2}:\left( {\dfrac{{{\pi ^2}{r^2}}}{n}} \right)\cot \left( {\dfrac{\pi }{n}} \right)\; \\
1:\left( {\dfrac{\pi }{n}} \right)\cot \left( {\dfrac{\pi }{n}} \right) \\
\tan \left( {\dfrac{\pi }{n}} \right):\left( {\dfrac{\pi }{n}} \right) \\
$
So the correct option is A.
Note: The formula for the area of a polygon can be determined by inscribing a circle inside the polygon. For direct calculation purposes, we use the final formula directly. Rest the question is based on simple calculations.
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