内接菱形的最大可能圆的半径，而菱形内接矩形

给出一个长l宽b的矩形，它内接一个菱形，菱形又内接一个圆。任务是找到这个圆的半径。
例子：

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

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

方法：从图中可以看出，对角线x & y等于矩形的长和宽。
此外，菱形内的圆半径r为 = xy/2√(x^2+y^2)。
因此，根据l & b的圆半径是 = lb/2√(l^2+b^2)。
下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
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;
}

Java

// 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..

Python3

# 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

C#

// 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

Javascript

// javascript 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
    var r = ((l * b) / (2 * Math.sqrt((Math.pow(l, 2) + Math.pow(b, 2)))));
    return r;
}
 
var l = 5, b = 3;
document.write(circleradius(l, b).toFixed(5)) ;
 
// This code is contributed by shikhasingrajput

输出：

1.28624

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