Radius of the biggest possible circle inscribed in rhombus which in turn is inscribed in a rectangle

Give a rectangle with length l & breadth b, which inscribes a rhombus, which in turn inscribes a circle. The task is to find the radius of this circle.

Examples:

Input: l = 5, b = 3
Output: 1.28624

Input: l = 6, b = 4
Output: 1.6641

Approach: From the figure, it is clear that diagonals, x & y, are equal to the lenght and breadth of the rectangle.
Also radius of the circle, r, inside a rhombus is = xy/2√(x^2+y^2).
So, radius of the circle in terms of l & b is = lb/2√(l^2+b^2).

Below is the implementation of the above approach:

C++

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// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the radius
// of the inscribed circle
float circleradius(float l, float b)
{
  
    // the sides cannot be negative
    if (l < 0 || b < 0)
        return -1;
  
    // radius of the circle
    float r = (l * b) / (2 * sqrt((pow(l, 2) + pow(b, 2))));
    return r;
}
  
// Driver code
int main()
{
    float l = 5, b = 3;
    cout << circleradius(l, b) << endl;
  
    return 0;
}

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Java














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// Java implementation of above approach 


  


import java.io.*; 


  


class GFG { 


      


// Function to find the radius 


// of the inscribed circle 


static float circleradius(float l, float b) 


{ 


  


    // the sides cannot be negative 


    if (l < 0 || b < 0) 


        return -1; 


  


    // radius of the circle 


    float r = (float)((l * b) / (2 * Math.sqrt((Math.pow(l, 2) + Math.pow(b, 2))))); 


    return r; 


} 


  


    // Driver code 


    public static void main (String[] args) { 


        float l = 5, b = 3; 


    System.out.print (circleradius(l, b)) ; 


    } 


} 


// This code is contributed by inder_verma.. 



















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Python3














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# Python 3 implementation of  


# above approach 


from math import sqrt 


  


# Function to find the radius 


# of the inscribed circle 


def circleradius(l, b): 


      


    # the sides cannot be negative 


    if (l < 0 or b < 0): 


        return -1


  


    # radius of the circle 


    r = (l * b) / (2 * sqrt((pow(l, 2) + 


                             pow(b, 2)))); 


    return r 


  


# Driver code 


if __name__ == '__main__': 


    l = 5


    b = 3


    print("{0:.5}" . format(circleradius(l, b))) 


  


# This code is contribute  


# by Surendra_Gagwar 



















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C#














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// C# implementation of above approach 


using System; 


  


class GFG  


{ 


      


// Function to find the radius 


// of the inscribed circle 


static float circleradius(float l, 


                          float b) 


{ 


  


    // the sides cannot be negative 


    if (l < 0 || b < 0) 


        return -1; 


  


    // radius of the circle 


    float r = (float)((l * b) / 


              (2 * Math.Sqrt((Math.Pow(l, 2) + 


                   Math.Pow(b, 2))))); 


    return r; 


} 


  


// Driver code 


public static void Main ()  


{ 


    float l = 5, b = 3; 


    Console.WriteLine(circleradius(l, b)); 


} 


} 


  


// This code is contributed 


// by inder_verma 



















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PHP














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<?php 


// PHP implementation of above approach  


  


// Function to find the radius  


// of the inscribed circle  


function circleradius($l, $b)  


{  


  


    // the sides cannot be negative  


    if ($l < 0 || $b < 0)  


        return -1;  


  


    // radius of the circle  


    $r = ($l * $b) / (2 * sqrt((pow($l, 2) + 


                                pow($b, 2))));  


    return $r;  


}  


  


// Driver code  


$l = 5; 


$b = 3;  


echo circleradius($l, $b), "\n";  


  


// This code is contributed by ajit  


?> 



















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Output:


1.28624






          
          
          
          

            

    

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