# Biggest Reuleaux Triangle inscribed within a square which is inscribed within an ellipse

Given an ellipse with major axis length and minor axis 2a & 2b respectively which inscribes a square which in turn inscribes a reuleaux triangle. The task is to find the maximum possible area of this reuleaux triangle.

Examples:

```Input: a = 5, b = 4
Output: 0.0722389

Input: a = 7, b = 11
Output: 0.0202076
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution. Approach: As, the side of the square inscribed within an ellipse is, x = √(a^2 + b^2)/ab. Please refer Area of the Largest square that can be inscribed in an ellipse.
Also, in the reuleaux triangle, h = x = √(a^2 + b^2)/ab.
So, Area of the reuleaux triangle, A = 0.70477*h^2 = 0.70477*((a^2 + b^2)/a^2b^2).

Below is the implementation of the above approach:

## C++

 `// C++ Program to find the biggest Reuleaux triangle ` `// inscribed within in a square which in turn ` `// is inscribed within an ellipse ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the biggest reuleaux triangle ` `float` `Area(``float` `a, ``float` `b) ` `{ ` ` `  `    ``// length of the axes cannot be negative ` `    ``if` `(a < 0 && b < 0) ` `        ``return` `-1; ` ` `  `    ``// height of the reuleaux triangle ` `    ``float` `h = ``sqrt``(((``pow``(a, 2) + ``pow``(b, 2)) ` `                    ``/ (``pow``(a, 2) * ``pow``(b, 2)))); ` ` `  `    ``// area of the reuleaux triangle ` `    ``float` `A = 0.70477 * ``pow``(h, 2); ` ` `  `    ``return` `A; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``float` `a = 5, b = 4; ` `    ``cout << Area(a, b) << endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java Program to find the biggest Reuleaux triangle ` `// inscribed within in a square which in turn ` `// is inscribed within an ellipse ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` `     `  `// Function to find the biggest reuleaux triangle ` `static` `float` `Area(``float` `a, ``float` `b) ` `{ ` ` `  `    ``// length of the axes cannot be negative ` `    ``if` `(a < ``0` `&& b < ``0``) ` `        ``return` `-``1``; ` ` `  `    ``// height of the reuleaux triangle ` `    ``float` `h = (``float``)Math.sqrt(((Math.pow(a, ``2``) + Math.pow(b, ``2``)) ` `                ``/ (Math.pow(a, ``2``) * Math.pow(b, ``2``)))); ` ` `  `    ``// area of the reuleaux triangle ` `    ``float` `A = (``float``)(``0.70477` `* Math.pow(h, ``2``)); ` ` `  `    ``return` `A; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main (String[] args) ` `{ ` `    ``float` `a = ``5``, b = ``4``; ` `    ``System.out.println(Area(a, b)); ` `} ` `} ` ` `  `// This code is contributed by anuj_67.. `

## Python3

 `# Python3 Program to find the biggest Reuleaux  ` `# triangle inscribed within in a square ` `# which in turn is inscribed within an ellipse  ` `import` `math; ` ` `  `# Function to find the biggest  ` `# reuleaux triangle  ` `def` `Area(a, b): ` ` `  `    ``# length of the axes cannot  ` `    ``# be negative  ` `    ``if` `(a < ``0` `and` `b < ``0``):  ` `        ``return` `-``1``;  ` ` `  `    ``# height of the reuleaux triangle  ` `    ``h ``=` `math.sqrt(((``pow``(a, ``2``) ``+` `pow``(b, ``2``)) ``/` `                   ``(``pow``(a, ``2``) ``*` `pow``(b, ``2``))));  ` ` `  `    ``# area of the reuleaux triangle  ` `    ``A ``=` `0.70477` `*` `pow``(h, ``2``);  ` ` `  `    ``return` `A;  ` ` `  `# Driver code  ` `a ``=` `5``; ` `b ``=` `4``;  ` `print``(``round``(Area(a, b), ``7``)); ` ` `  `# This code is contributed by chandan_jnu `

## C#

 `// C# Program to find the biggest Reuleaux triangle ` `// inscribed within in a square which in turn ` `// is inscribed within an ellipse ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `     `  `// Function to find the biggest reuleaux triangle ` `static` `double` `Area(``double` `a, ``double` `b) ` `{ ` ` `  `    ``// length of the axes cannot be negative ` `    ``if` `(a < 0 && b < 0) ` `        ``return` `-1; ` ` `  `    ``// height of the reuleaux triangle ` `    ``double` `h = (``double``)Math.Sqrt(((Math.Pow(a, 2) +  ` `                                    ``Math.Pow(b, 2)) / ` `                                   ``(Math.Pow(a, 2) *  ` `                                   ``Math.Pow(b, 2)))); ` ` `  `    ``// area of the reuleaux triangle ` `    ``double` `A = (``double``)(0.70477 * Math.Pow(h, 2)); ` ` `  `    ``return` `A; ` `} ` ` `  `// Driver code ` `static` `void` `Main() ` `{ ` `    ``double` `a = 5, b = 4; ` `    ``Console.WriteLine(Math.Round(Area(a, b),7)); ` `} ` `} ` ` `  `// This code is contributed by chandan_jnu `

## PHP

 ` `

Output:

```0.0722389
```

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