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 <bits/stdc++.h> ` `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; ` `} ` |

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## 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)); ` `} ` `} ` |

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## 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 ` |

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## 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 ` |

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## PHP

`<?php ` `// PHP Program to find the area of ` `// the circle which can be inscribed ` `// within the hexagon ` ` ` `// Function to find the area ` `// of the inscribed circle ` `function` `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` `$A` `; ` `} ` ` ` `// Driver code ` `$a` `= 4; ` `echo` `circlearea(` `$a` `) . ` `"\n"` `; ` ` ` `// This code is contributed ` `// by Akanksha Rai(Abby_akku) ` |

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**Output:**

37.68

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