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Area of a circle inscribed in a regular hexagon
• Last Updated : 16 Mar, 2021

Given a regular Hexagon with side length a, the task is to find the area of the circle inscribed in it, given that, the circle is tangent to each of the six sides.
Examples:

```Input: a = 4
Output: 37.68

Input: a = 10
Output: 235.5``` Approach
From the figure, it is clear that, we can divide the regular hexagon into 6 identical equilateral triangles.
We take one triangle OAB, with O as the centre of the hexagon or circle, & AB as one side of the hexagon.
Let M be mid-point of AB, OM would be the perpendicular bisector of AB, angle AOM = 30 deg
Then in right angled triangle OAM,

tanx = tan30 = 1/√3
So, a/2r = 1/√3
Therefore, r = a√3/2
Area of circle, A =Πr²=Π3a^2/4

Below is the implementation of the approach

## C++

 `// C++ Program to find the area of the circle``// which can be inscribed within the hexagon``#include ``using` `namespace` `std;` `// Function to find the area``// of the inscribed circle``float` `circlearea(``float` `a)``{` `    ``// the side cannot be negative``    ``if` `(a < 0)``        ``return` `-1;` `    ``// area of the circle``    ``float` `A = (3.14 * 3 * ``pow``(a, 2)) / 4;` `    ``return` `A;``}` `// Driver code``int` `main()``{``    ``float` `a = 4;``    ``cout << circlearea(a) << endl;` `    ``return` `0;``}`

## Java

 `//Java program to find the``//area of the circle``//which can be inscribed within the hexagon` `import` `java.util.*;` `class` `solution``{``static` `double` `circlearea(``double` `a)``{` `// the side cannot be negative``    ``if` `(a < ``0``)``    ``return` `-``1``;` `// area of the circle``    ``double` `A = (``3.14` `* ``3` `* Math.pow(a,``2``) ) / ``4``;` `    ``return` `A;``}``public` `static` `void` `main(String arr[])``{``    ``double` `a = ``4``;``    ``System.out.println(circlearea(a));``}``}`

## Python 3

 `# Python 3 program to find the``# area of the circle``# which can be inscribed within the hexagon` `# Function to find the area``# of the inscribed circle``def` `circlearea(a) :` `    ``# the side cannot be negative``    ``if` `a < ``0` `:``        ``return` `-``1` `    ``#  area of the circle``    ``A ``=` `(``3.14` `*` `3` `*` `pow``(a,``2``)) ``/` `4` `    ``return` `A`  `# Driver code    ``if` `__name__ ``=``=` `"__main__"` `:` `    ``a ``=` `4``    ``print``(circlearea(a))`  `# This code is contributed by ANKITRAI1`

## C#

 `// C# program to find ``// the area of the circle``// which can be inscribed``// within the hexagon``using` `System;` `class` `GFG``{``static` `double` `circlearea(``double` `a)``{` `    ``// the side cannot be negative``    ``if` `(a < 0)``    ``return` `-1;` `    ``// area of the circle``    ``double` `A = (3.14 * 3 *``                ``Math.Pow(a, 2)) / 4;` `    ``return` `A;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``double` `a = 4;``    ``Console.WriteLine(circlearea(a));``}``}` `// This code is contributed``// by inder_verma`

## PHP

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## Javascript

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Output:
`37.68`

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