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;` `}` |

## 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

`<?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)` |

## Javascript

`<script>` `// javascript program to find the` `//area of the circle` `//which can be inscribed within the hexagon` `function` `circlearea(a) {` ` ` `// the side cannot be negative` ` ` `if` `(a < 0)` ` ` `return` `-1;` ` ` `// area of the circle` ` ` `var` `A = (3.14 * 3 * Math.pow(a, 2)) / 4;` ` ` `return` `A;` `}` `var` `a = 4;` `document.write(circlearea(a));` `// This code is contributed by 29AjayKumar` `</script>` |

**Output:**

37.68

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