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

• Last Updated : 18 Mar, 2021

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```

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In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students. 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 ``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

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## Javascript

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Output:
`1.28624`

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