Given a regular polygon of N sides with radius(distance from the center to any vertex) R. The task is to find the area of the polygon.
Examples:
Input : r = 9, N = 6 Output : 210.444 Input : r = 8, N = 7 Output : 232.571
In the figure, we see that the polygon can be divided into N equal triangles.
Looking into one of the triangles, we see that the whole angle at the centre can be divided into = 360/N parts.
So, angle t = 180/N.
Looking into one of the triangles, we see,
h = rcost a = rsint
We know,
area of the triangle = (base * height)/2 = r2sin(t)cos(t) = r2*sin(2t)/2
So, area of the polygon:
A = n * (area of one triangle) = n*r2*sin(2t)/2 = n*r2*sin(360/n)/2
Below is the implementation of the above approach:
C++
// C++ Program to find the area // of a regular polygon with given radius #include <bits/stdc++.h> using namespace std;
// Function to find the area // of a regular polygon float polyarea( float n, float r)
{ // Side and radius cannot be negative
if (r < 0 && n < 0)
return -1;
// Area
// degree converted to radians
float A = ((r * r * n) * sin ((360 / n) * 3.14159 / 180)) / 2;
return A;
} // Driver code int main()
{ float r = 9, n = 6;
cout << polyarea(n, r) << endl;
return 0;
} |
Java
// Java Program to find the area // of a regular polygon with given radius import java.util.*;
class GFG
{ // Function to find the area
// of a regular polygon
static double polyarea( double n, double r)
{
// Side and radius cannot be negative
if (r < 0 && n < 0 )
return - 1 ;
// Area
// degree converted to radians
double A = ((r * r * n) * Math.sin(( 360 / n) * 3.14159 / 180 )) / 2 ;
return A;
}
// Driver code
public static void main(String []args)
{
float r = 9 , n = 6 ;
System.out.println(polyarea(n, r));
}
} // This code is contributed // By ihritik (Hritik Raj) |
Python3
# Python3 Program to find the area # of a regular polygon with given radius # form math lib import sin function from math import sin
# Function to find the area # of a regular polygon def polyarea(n, r) :
# Side and radius cannot be negative
if (r < 0 and n < 0 ) :
return - 1
# Area
# degree converted to radians
A = (((r * r * n) * sin(( 360 / n) *
3.14159 / 180 )) / 2 );
return round (A, 3 )
# Driver code if __name__ = = "__main__" :
r, n = 9 , 6
print (polyarea(n, r))
# This code is contributed by Ryuga |
C#
// C# Program to find the area // of a regular polygon with given radius using System;
class GFG
{ // Function to find the area
// of a regular polygon
static double polyarea( double n, double r)
{
// Side and radius cannot be negative
if (r < 0 && n < 0)
return -1;
// Area
// degree converted to radians
double A = ((r * r * n) * Math.Sin((360 / n) * 3.14159 / 180)) / 2;
return A;
}
// Driver code
public static void Main()
{
float r = 9, n = 6;
Console.WriteLine(polyarea(n, r));
}
} // This code is contributed // By ihritik (Hritik Raj) |
PHP
<?php // PHP Program to find the area of a // regular polygon with given radius // Function to find the area // of a regular polygon function polyarea( $n , $r )
{ // Side and radius cannot be negative
if ( $r < 0 && $n < 0)
return -1;
// Area
// degree converted to radians
$A = (( $r * $r * $n ) * sin((360 / $n ) *
3.14159 / 180)) / 2;
return $A ;
} // Driver code $r = 9;
$n = 6;
echo polyarea( $n , $r ). "\n" ;
// This code is contributed by ita_c ?> |
Javascript
<script> // javascript Program to find the area // of a regular polygon with given radius // Function to find the area // of a regular polygon function polyarea(n , r)
{ // Side and radius cannot be negative
if (r < 0 && n < 0)
return -1;
// Area
// degree converted to radians
var A = ((r * r * n) * Math.sin((360 / n) * 3.14159 / 180)) / 2;
return A;
} // Driver code var r = 9, n = 6;
document.write(polyarea(n, r).toFixed(5)); // This code contributed by Princi Singh </script> |
Output:
210.444
Time Complexity: O(1)
Auxiliary Space: O(1)