Question

The area of the circle and the area of the regular polygon of n sides of perimeter equal to that of circle are in the ratio of $a\tan \left( {\dfrac{\pi }{n}} \right)$: \[\dfrac{{b\pi }}{n}\], then a + b =

Answer 1

Hint – In this question use the concept of area of circle and area of n sided regular polygon which is given as $\pi {r^2}$ and $\dfrac{{n{L^2}}}{4}\cot \left( {\dfrac{\pi }{n}} \right)$ respectively, and perimeter of any shape is the sum of all the sides of the shape so use these concepts to reach the solution of the question.

Complete step-by-step answer:
As we know that the area of the circle is given as $\pi {r^2}$ where r is the radius of the circle.
Now the area of the polygon with n sides is given as half multiplied by the perimeter of the polygon and multiplied by the apothem.
Where apothem is the length of the segment joining the center of the polygon and the midpoint of any side of n sided polygon.
And the perimeter of any shape is the sum of all the side lengths.
So the area of the n sided polygon = $\dfrac{1}{2} \times P \times A$
Where, P = perimeter and A = apothem.
Now let the length of each n sided polygon = L
So the perimeter of the n sided polygon P = $n \times L$
Therefore area of the n sided polygon = $\dfrac{1}{2} \times \left( {n \times L} \right) \times A$
And the area of the polygon is also written as $\dfrac{{n{L^2}}}{4}\cot \left( {\dfrac{\pi }{n}} \right)$
And the perimeter of the circle is nothing but the circumference of the circle = $2\pi r$
Now it is given that the perimeter of the n sided polygon is equal to the perimeter of the circle.
Therefore, $n \times L = 2\pi r$....................... (1)
Now it is also given that the area of the circle to the n sided polygon is $\dfrac{{a\tan \left( {\dfrac{\pi }{n}} \right)}}{{\dfrac{{b\pi }}{n}}}$................. (2)
And in original the ratio of the area of the circle to the n sided polygon is
$ \Rightarrow \dfrac{{\pi {r^2}}}{{\dfrac{{n{L^2}}}{4}\cot \left( {\dfrac{\pi }{n}} \right)}}$................... (3)
Therefore, both equation (2) and (3) are same so equate them we have,
$ \Rightarrow \dfrac{{\pi {r^2}}}{{\dfrac{{n{L^2}}}{4}\cot \left( {\dfrac{\pi }{n}} \right)}} = \dfrac{{a\tan \left( {\dfrac{\pi }{n}} \right)}}{{\dfrac{{b\pi }}{n}}}$
Now as we know that $\cot x = \dfrac{1}{{\tan x}}$ so use this property in above equation we have,
$ \Rightarrow \dfrac{{\pi {r^2}\tan \left( {\dfrac{\pi }{n}} \right)}}{{\dfrac{{n{L^2}}}{4}}} = \dfrac{{a\tan \left( {\dfrac{\pi }{n}} \right)}}{{\dfrac{{b\pi }}{n}}}$
Now cancel out the common terms we have,
$ \Rightarrow \dfrac{{4\pi {r^2}}}{{n{L^2}}} = \dfrac{{an}}{{b\pi }}$...................... (4)
Now from equation (1) we have,
$ \Rightarrow L = \dfrac{{2\pi r}}{n}$
Now substitute this value in equation (4) we have,
$ \Rightarrow \dfrac{{4\pi {r^2}}}{{n{{\left( {\dfrac{{2\pi r}}{n}} \right)}^2}}} = \dfrac{{an}}{{b\pi }}$
Now simplify this we have,
$ \Rightarrow \dfrac{{4\pi {r^2}}}{{n\left( {\dfrac{{4{\pi ^2}{r^2}}}{{{n^2}}}} \right)}} = \dfrac{{an}}{{b\pi }}$
$ \Rightarrow \dfrac{n}{\pi } = \dfrac{{an}}{{b\pi }}$
So on comparing a = b = 1
So the value of a + b = 1 + 1 = 2
So this is the required answer.

Note – Whenever we face such types of questions the key concept we have to remember is the formula of the area of the circle and the n sided regular polygon which is stated above, then according to the given condition (i.e. Perimeter of the circle is equal to the perimeter of the n sided polygon) compare the ratio of the area of the circle to the area of the n sided polygon to the given ratio and simplify we will get the required value of a + b.