Given a rhombus with diagonals **a** and **b**, which contains an inscribed circle. The task is to find the area of that circle in terms of a and b.

**Examples:**

Input: l = 5, b = 6 Output: 11.582 Input: l = 8, b = 10 Output: 30.6341

**Approach:** From the figure, we see, the radius of inscribed circle is also a height **h=OH** of the right triangle **AOB**. To find it, we use equations for triangle’s area :

Area AOB = 1/2 * (a/2) * (b/2) = ab/8 = 12ch

where **c = AB** i.e. a hypotenuse. So,

r = h = ab/4c = ab/4√(a^2/4 + b^2/4) = ab/2√(a^2+b^2)

and therefore area of the circle is

A = Π * r^2 = Π a^2 b^2 /4(a2 + b2)

**Below is the implementation of above approach:**

`// C++ Program to find the area of the circle ` `// which can be inscribed within the rhombus ` `#include <bits/stdc++.h> ` `using` `namespace` `std; `
` ` `// Function to find the area ` `// of the inscribed circle ` `float` `circlearea(` `float` `a, ` `float` `b) `
`{ ` ` ` ` ` `// the diagonals cannot be negative `
` ` `if` `(a < 0 || b < 0) `
` ` `return` `-1; `
` ` ` ` `// area of the circle `
` ` `float` `A = (3.14 * ` `pow` `(a, 2) * ` `pow` `(b, 2)) `
` ` `/ (4 * (` `pow` `(a, 2) + ` `pow` `(b, 2))); `
` ` `return` `A; `
`} ` ` ` `// Driver code ` `int` `main() `
`{ ` ` ` `float` `a = 8, b = 10; `
` ` `cout << circlearea(a, b) << endl; `
` ` ` ` `return` `0; `
`} ` |

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`// Java Program to find the area of the circle ` `// which can be inscribed within the rhombus ` ` ` `public` `class` `GFG { `
` ` ` ` `// Function to find the area `
` ` `// of the inscribed circle `
` ` `public` `static` `float` `circlearea(` `double` `a, ` `double` `b) `
` ` `{ `
` ` `// the diagonals cannot be negative `
` ` `if` `(a < ` `0` `|| b < ` `0` `) `
` ` `return` `-` `1` `; `
` ` ` ` `//area of the circle `
` ` `float` `A = (` `float` `) ((` `3.14` `* Math.pow(a, ` `2` `) * Math.pow(b, ` `2` `)) `
` ` `/ (` `4` `* (Math.pow(a, ` `2` `) + Math.pow(b, ` `2` `)))) ; `
` ` ` ` `return` `A ; `
` ` `} `
` ` ` ` `// Driver code `
` ` `public` `static` `void` `main(String[] args) { `
` ` `float` `a = ` `8` `, b = ` `10` `; `
` ` ` ` `System.out.println(circlearea(a, b)); `
` ` ` ` `} `
`// This code is contributed by ANKITRAI1 ` `} ` |

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`# Python 3 Program to find the area of the circle ` `# which can be inscribed within the rhombus ` ` ` ` ` `# Function to find the area ` `# of the inscribed circle ` `def` `circlearea(a, b): `
` ` ` ` `# the diagonals cannot be negative `
` ` `if` `(a < ` `0` `or` `b < ` `0` `): `
` ` `return` `-` `1`
` ` ` ` `# area of the circle `
` ` `A ` `=` `((` `3.14` `*` `pow` `(a, ` `2` `) ` `*` `pow` `(b, ` `2` `))` `/`
` ` `(` `4` `*` `(` `pow` `(a, ` `2` `) ` `+` `pow` `(b, ` `2` `)))) `
` ` `return` `A `
` ` `# Driver code ` `if` `__name__ ` `=` `=` `"__main__"` `: `
` ` `a ` `=` `8`
` ` `b ` `=` `10`
` ` `print` `( circlearea(a, b)) `
` ` `# This code is contributed by ChitraNayal ` |

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`// C# Program to find the area of the circle ` `// which can be inscribed within the rhombus ` `using` `System; `
` ` `public` `class` `GFG { `
` ` ` ` `// Function to find the area `
` ` `// of the inscribed circle `
` ` `public` `static` `float` `circlearea(` `double` `a, ` `double` `b) `
` ` `{ `
` ` `// the diagonals cannot be negative `
` ` `if` `(a < 0 || b < 0) `
` ` `return` `-1 ; `
` ` ` ` `//area of the circle `
` ` `float` `A = (` `float` `) ((3.14 * Math.Pow(a, 2) * Math.Pow(b, 2)) `
` ` `/ (4 * (Math.Pow(a, 2) + Math.Pow(b, 2)))) ; `
` ` ` ` `return` `A ; `
` ` `} `
` ` ` ` `// Driver code `
` ` `public` `static` `void` `Main() { `
` ` `float` `a = 8, b = 10 ; `
` ` ` ` `Console.WriteLine(circlearea(a, b)); `
` ` ` ` `} `
`// This code is contributed by inder_verma.. ` `} ` |

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`<?php ` `// PHP Program to find the area ` `// of the circle which can be ` `// inscribed within the rhombus ` ` ` `// Function to find the area ` `// of the inscribed circle ` `function` `circlearea(` `$a` `, ` `$b` `) `
`{ ` ` ` ` ` `// the diagonals cannot be negative `
` ` `if` `(` `$a` `< 0 || ` `$b` `< 0) `
` ` `return` `-1; `
` ` ` ` `// area of the circle `
` ` `$A` `= (3.14 * pow(` `$a` `, 2) * pow(` `$b` `, 2)) / `
` ` `(4 * (pow(` `$a` `, 2) + pow(` `$b` `, 2))); `
` ` `return` `$A` `; `
`} ` ` ` `// Driver code ` `$a` `= 8; ` `$b` `= 10; `
`echo` `circlearea(` `$a` `, ` `$b` `); `
` ` `// This code is contributed by anuj_67 ` `?> ` |

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**Output:**

30.6341

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