# Area of a square inscribed in a circle which is inscribed in a hexagon

Given a regular hexagon with side **A**, which inscribes a circle of radius **r**, which in turn inscribes a square of side **a**.The task is to find the area of this square.**Examples**:

Input : A = 5 Output : 37.5 Input : A = 8 Output : 96

**Approach**:

We know the radius of the circle inscribed within the hexagon is, **r=A√3/2**(Please refer here)

Also, side length of circle within the circle is, **a=√r=√3A/√2**

So, Area of the Square, **Area=(√3A/√2)^2**

## C++

`// C++ Program to find the area of the square` `// inscribed within the circle which in turn` `// is inscribed in a hexagon` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the area of the square` `float` `area(` `float` `a)` `{` ` ` `// side of hexagon cannot be negative` ` ` `if` `(a < 0)` ` ` `return` `-1;` ` ` `// area of the square` ` ` `float` `area = ` `pow` `((a * ` `sqrt` `(3)) / (` `sqrt` `(2)), 2);` ` ` `return` `area;` `}` `// Driver code` `int` `main()` `{` ` ` `float` `a = 5;` ` ` `cout << area(a) << endl;` ` ` `return` `0;` `}` |

## Java

`// Java Program to find the area of the square` `// inscribed within the circle which in turn` `// is inscribed in a hexagon` `import` `java.io.*;` `class` `GFG {` `// Function to find the area of the square` `static` `float` `area(` `float` `a)` `{` ` ` `// side of hexagon cannot be negative` ` ` `if` `(a < ` `0` `)` ` ` `return` `-` `1` `;` ` ` `// area of the square` ` ` `float` `area = (` `float` `)Math.pow((a * Math.sqrt(` `3` `)) / (Math.sqrt(` `2` `)), ` `2` `);` ` ` `return` `area;` `}` `// Driver code` ` ` `public` `static` `void` `main (String[] args) {` ` ` `float` `a = ` `5` `;` ` ` `System.out.println( area(a));` ` ` `}` `}` `// This code is contributed by ajit` |

## Python3

`# Python 3 Program to find the area` `# of the square inscribed within the` `# circle which in turn is inscribed` `# in a hexagon` `from` `math ` `import` `pow` `, sqrt` `# Function to find the area` `# of the square` `def` `area(a):` ` ` ` ` `# side of hexagon cannot` ` ` `# be negative` ` ` `if` `(a < ` `0` `):` ` ` `return` `-` `1` ` ` `# area of the square` ` ` `area ` `=` `pow` `((a ` `*` `sqrt(` `3` `)) ` `/` ` ` `(sqrt(` `2` `)), ` `2` `)` ` ` `return` `area` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `a ` `=` `5` ` ` `print` `(` `"{0:.3}"` `.` `format` `(area(a)))` `# This code is contributed` `# by SURENDRA_GANGWAR` |

## C#

`// C# Program to find the area of` `// the square inscribed within the` `// circle which in turn is inscribed` `// in a hexagon` `using` `System;` `class` `GFG` `{` `// Function to find the area` `// of the square` `static` `float` `area(` `float` `a)` `{` ` ` `// side of hexagon cannot be negative` ` ` `if` `(a < 0)` ` ` `return` `-1;` ` ` `// area of the square` ` ` `float` `area = (` `float` `)Math.Pow((a * Math.Sqrt(3)) /` ` ` `(Math.Sqrt(2)), 2);` ` ` `return` `area;` `}` `// Driver code` `public` `static` `void` `Main ()` `{` ` ` `float` `a = 5;` ` ` `Console.WriteLine( area(a));` `}` `}` `// This code is contributed by inder_verma..` |

## PHP

`<?php` `// PHP Program to find the area` `// of the square inscribed within` `// the circle which in turn is` `// inscribed in a hexagon` `// Function to find the area` `// of the square` `function` `area(` `$a` `)` `{` ` ` `// side of hexagon cannot` ` ` `// be negative` ` ` `if` `(` `$a` `< 0)` ` ` `return` `-1;` ` ` `// area of the square` ` ` `$area` `= pow((` `$a` `* sqrt(3)) /` ` ` `(sqrt(2)), 2);` ` ` `return` `$area` `;` `}` `// Driver code` `$a` `= 5;` `echo` `area(` `$a` `) . ` `"\n"` `;` `// This code is contributed` `// by Akanksha Rai(Abby_akku)` `?>` |

## Javascript

`<script>` ` ` `// javascript Program to find the area of the square` `// inscribed within the circle which in turn` `// is inscribed in a hexagon` `// Function to find the area of the square` `function` `area(a)` `{` ` ` `// side of hexagon cannot be negative` ` ` `if` `(a < 0)` ` ` `return` `-1;` ` ` `// area of the square` ` ` `var` `area = Math.pow((a * Math.sqrt(3)) / (Math.sqrt(2)), 2);` ` ` `return` `area;` `}` `// Driver code` `var` `a = 5;` `document.write( area(a).toFixed(5));` `// This code contributed by shikhasingrajput` `</script>` |

**Output:**

37.5

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