# Program to find the side of the Octagon inscribed within the square

• Last Updated : 18 Mar, 2021

Given a square of side length ‘a’, the task is to find the side length of the biggest octagon that can be inscribed within it.
Examples:

```Input: a = 4
Output: 1.65685

Input: a = 5
Output: 2.07107```

Approach:

=> From the figure, it can be seen that, side length of the Octagon = b
=> Also since the polygons are regular, therefore 2*x + b = a
=> From the right angled triangle, x^2 + x^2 = b^2
=> Hence, x = b/âˆš2
=> So, âˆš2b + b = a
=> Therefore, b = a/(âˆš2 +1)

Below is the implementation of the above approach:

## C++

 `// C++ Program to find the side of the octagon``// which can be inscribed within the square` `#include ``using` `namespace` `std;` `// Function to find the side``// of the octagon``float` `octaside(``float` `a)``{` `    ``// side cannot be negative``    ``if` `(a < 0)``        ``return` `-1;` `    ``// side of the octagon``    ``float` `s = a / (``sqrt``(2) + 1);``    ``return` `s;``}` `// Driver code``int` `main()``{` `    ``// Get he square side``    ``float` `a = 4;` `    ``// Find the side length of the square``    ``cout << octaside(a) << endl;` `    ``return` `0;``}`

## Java

 `// Java Program to find the side of the octagon``// which can be inscribed within the square` `import` `java.io.*;` `class` `GFG {``    ` `// Function to find the side``// of the octagon``static` `double` `octaside(``double` `a)``{` `    ``// side cannot be negative``    ``if` `(a < ``0``)``        ``return` `-``1``;` `    ``// side of the octagon``    ``double` `s = a / (Math.sqrt(``2``) + ``1``);``    ``return` `s;``}` `// Driver code``    ` `    ``public` `static` `void` `main (String[] args) {``        ` `    ``// Get he square side``    ``double` `a = ``4``;` `    ``// Find the side length of the square``    ``System.out.println( octaside(a));` `        ` `        ` `    ``}``}``//This Code  is contributed by ajit`

## Python3

 `# Python 3 Program to find the side``# of the octagon which can be``# inscribed within the square``from` `math ``import` `sqrt` `# Function to find the side``# of the octagon``def` `octaside(a):``    ` `    ``# side cannot be negative``    ``if` `a < ``0``:``        ``return` `-``1` `    ``# side of the octagon``    ``s ``=` `a ``/` `(sqrt(``2``) ``+` `1``)``    ``return` `s` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# Get he square side``    ``a ``=` `4` `    ``# Find the side length of the square``    ``print``(``"{0:.6}"``.``format``(octaside(a)))``    ` `# This code is contributed``# by Surendra_Gangwar`

## C#

 `// C# Program to find the side``// of the octagon which can be``// inscribed within the square``using` `System;` `class` `GFG``{``    ` `// Function to find the side``// of the octagon``static` `double` `octaside(``double` `a)``{` `    ``// side cannot be negative``    ``if` `(a < 0)``        ``return` `-1;` `    ``// side of the octagon``    ``double` `s = a / (Math.Sqrt(2) + 1);``    ``return` `s;``}` `// Driver code``static` `public` `void` `Main ()``{``    ``// Get he square side``    ``double` `a = 4;``    ` `    ``// Find the side length``    ``// of the square``    ``Console.WriteLine( octaside(a));``}``}` `// This code is contributed``// by akt_mit`

## PHP

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## Javascript

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Output:

`1.65685`

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