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# Apothem of a n-sided regular polygon

• Last Updated : 22 Jun, 2022

Given here the side length a of a regular n-sided polygon, the task is to find the length of its Apothem.
Apothem is the line drawn from the center of the polygon that is perpendicular to one of its sides.
Examples:

```Input a = 9, n = 6
Output: 7.79424

Input: a = 8, n = 7
Output: 8.30609``` Approach

In the figure, we see the polygon can be divided into n equal triangles.
Looking into one of the triangles, we see the whole angle at the centre can be divided into = 360/n
So, angle t = 180/n
now, tan t = a/2h
So, h = a/(2*tan t)
here, h is the apothem,
so, apothem = a/(2*tan(180/n))

Below is the implementation of the above approach.

## C++

 `// C++ Program to find the apothem``// of a regular polygon with given side length``#include ``using` `namespace` `std;` `// Function to find the apothem``// of a regular polygon``float` `polyapothem(``float` `n, ``float` `a)``{` `    ``// Side and side length cannot be negative``    ``if` `(a < 0 && n < 0)``        ``return` `-1;` `    ``// Degree converted to radians``    ``return` `a / (2 * ``tan``((180 / n) * 3.14159 / 180));``}` `// Driver code``int` `main()``{``    ``float` `a = 9, n = 6;``    ``cout << polyapothem(n, a) << endl;` `    ``return` `0;``}`

## Java

 `// Java Program to find the apothem of a``// regular polygon with given side length``import` `java.util.*;` `class` `GFG``{` `    ``// Function to find the apothem``    ``// of a regular polygon``    ``double` `polyapothem(``double` `n, ``double` `a)``    ``{` `        ``// Side and side length cannot be negative``        ``if` `(a < ``0` `&& n < ``0``)``            ``return` `-``1``;` `        ``// Degree converted to radians``        ``return` `(a / (``2` `* java.lang.Math.tan((``180` `/ n)``                ``* ``3.14159` `/ ``180``)));``    ``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``double` `a = ``9``, n = ``6``;``    ``GFG g=``new` `GFG();``    ``System.out.println(g.polyapothem(n, a));``}` `}``//This code is contributed by Shivi_Aggarwal`

## Python3

 `# Python 3 Program to find the apothem``# of a regular polygon with given side``# length``from` `math ``import` `tan` `# Function to find the apothem``# of a regular polygon``def` `polyapothem(n, a):``    ` `    ``# Side and side length cannot be negative``    ``if` `(a < ``0` `and` `n < ``0``):``        ``return` `-``1` `    ``# Degree converted to radians``    ``return` `a ``/` `(``2` `*` `tan((``180` `/` `n) ``*``                   ``3.14159` `/` `180``))` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``a ``=` `9``    ``n ``=` `6``    ``print``(``'{0:.6}'``.``format``(polyapothem(n, a)))``    ` `# This code is contributed by``# Sahil_Shelangia`

## C#

 `// C# Program to find the apothem of a``// regular polygon with given side length``using` `System;` `class` `GFG``{` `// Function to find the apothem``// of a regular polygon``static` `double` `polyapothem(``double` `n,``                          ``double` `a)``{` `    ``// Side and side length cannot``    ``// be negative``    ``if` `(a < 0 && n < 0)``        ``return` `-1;` `    ``// Degree converted to radians``    ``return` `(a / (2 * Math.Tan((180 / n) *``                       ``3.14159 / 180)));``}` `// Driver code``public` `static` `void` `Main()``{``    ``double` `a = 9, n = 6;``    ``Console.WriteLine(Math.Round(polyapothem(n, a), 4));``}``}` `// This code is contributed by Ryuga`

## PHP

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## Javascript

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Output:

`7.79424`

Time Complexity: O(1)

Auxiliary Space: O(1)

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