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

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/?2,
=> So, ?2b + b = a
=> Therefore, b = a/(?2 +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

 ``

## Javascript

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Output:

`1.65685`

Time Complexity: O(1)  since no loop is used the algorithm takes up constant time to perform the operations

Space Complexity: O(1) since no extra array is used so the space taken by the algorithm is constant

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