Given a n-sided regular polygon of side length a.The task is to find the length of it’s diagonal.
Examples:
Input: a = 9, n = 10 Output: 17.119 Input: a = 4, n = 5 Output: 6.47213
Approach:
We know that the sum of interior angles of a polygon = (n – 2) * 180 where, n is the no. of sides in the polygon.
So, each interior angle = (n – 2) * 180/n
Now, we have to find BC = 2 * x. If we draw a perpendicular AO on BC, we will see that the perpendicular bisects BC in BO and OC, as triangles AOB and AOC are congruent to each other.
Now, t = (n – 2) * 180/2n
So, sint = x/a
Therefore, x = asint
Hence, diagonal=2x = 2asint = 2asin((n – 2) * 180/2n)
// C++ Program to find the diagonal // of a regular polygon with given side length #include <bits/stdc++.h> using namespace std;
// Function to find the diagonal // of a regular polygon float polydiagonal( float n, float a)
{ // Side and side length cannot be negative
if (a < 0 && n < 0)
return -1;
// diagonal
// degree converted to radians
return 2 * a * sin ((((n - 2) * 180) / (2 * n)) * 3.14159 / 180);
} // Driver code int main()
{ float a = 9, n = 10;
cout << polydiagonal(n, a) << endl;
return 0;
} |
// Java Program to find the diagonal // of a regular polygon with given side length class GFG {
// Function to find the diagonal // of a regular polygon static float polydiagonal( float n, float a) {
// Side and side length cannot be negative
if (a < 0 && n < 0 ) {
return - 1 ;
}
// diagonal
// degree converted to radians
return ( float ) ( 2 * a * Math.sin((((n - 2 ) * 180 ) / ( 2 * n)) * 3.14159 / 180 ));
}
// Driver code public static void main(String[] args) {
float a = 9 , n = 10 ;
System.out.printf( "%.3f" ,polydiagonal(n, a));
}
} // This code is contributed by 29AjayKumar |
# Python3 Program to find the diagonal # of a regular polygon with given side length import math as mt
# Function to find the diagonal # of a regular polygon def polydiagonal(n, a):
# Side and side length cannot
# be negative
if (a < 0 and n < 0 ):
return - 1
# diagonal degree converted to radians
return ( 2 * a * mt.sin((((n - 2 ) * 180 ) /
( 2 * n)) * 3.14159 / 180 ))
# Driver code a, n = 9 , 10
print (polydiagonal(n, a))
# This code is contributed # by Mohit kumar 29 |
// C# Program to find the diagonal // of a regular polygon with given side length using System;
public class GFG{
// Function to find the diagonal // of a regular polygon static float polydiagonal( float n, float a) {
// Side and side length cannot be negative
if (a < 0 && n < 0) {
return -1;
}
// diagonal
// degree converted to radians
return ( float ) (2 * a * Math.Sin((((n - 2) * 180) / (2 * n)) * 3.14159 / 180));
}
// Driver code static public void Main (){
float a = 9, n = 10;
Console.WriteLine(polydiagonal(n, a));
}
} // This code is contributed by @Sachin... |
<?php // PHP Program to find the diagonal of a // regular polygon with given side length // Function to find the diagonal // of a regular polygon function polydiagonal ( $n , $a )
{ // Side and side length cannot
// be negative
if ( $a < 0 && $n < 0)
return -1;
// diagonal
// degree converted to radians
return 2 * $a * sin(((( $n - 2) * 180) /
(2 * $n )) * 3.14159 / 180);
} // Driver code $a = 9;
$n = 10;
echo polydiagonal( $n , $a );
// This code is contributed // by Sach_Code ?> |
<script> // javascript Program to find the diagonal // of a regular polygon with given side length // Function to find the diagonal // of a regular polygon function polydiagonal(n , a) {
// Side and side length cannot be negative
if (a < 0 && n < 0) {
return -1;
}
// diagonal
// degree converted to radians
return (2 * a * Math.sin((((n - 2) * 180)
/ (2 * n)) * 3.14159 / 180));
} // Driver code var a = 9, n = 10;
document.write(polydiagonal(n, a).toFixed(3)); // This code contributed by Princi Singh </script> |
17.119
Time Complexity: O(1)
Auxiliary Space: O(1)