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 length 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++

`// 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;` `}` |

## 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` |

## PHP

`<?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` `?>` |

## Javascript

`<script>` `// 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` `</script>` |

**Output:**

1.28624

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