# 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++

`// 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

`// 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

`# 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#

`// 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

`<?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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