Given a rectangle of length **l** & breadth **b**, we have to find the largest cricle that can be inscribed in the rectangle. **Examples:**

Input : l = 4, b = 8 Output : 12.56 Input : l = 16 b = 6 Output : 28.26

From the figure, we can see, the biggest circle that could be inscribed in the rectangle will have radius always equal to the half of the shorter side of the rectangle. So from the figure,

radius,

r = b/2&

Area,A = Ď€ * (r^2)

## C++

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

## Java

`// Java Program to find the` `// biggest circle which can be` `// inscribed within the rectangle` `class` `GFG` `{` `// Function to find the area` `// of the biggest circle` `static` `float` `circlearea(` `float` `l,` ` ` `float` `b)` `{` `// the length and breadth` `// cannot be negative` `if` `(l < ` `0` `|| b < ` `0` `)` ` ` `return` `-` `1` `;` `// area of the circle` `if` `(l < b)` ` ` `return` `(` `float` `)(` `3.14` `* Math.pow(l / ` `2` `, ` `2` `));` `else` ` ` `return` `(` `float` `)(` `3.14` `* Math.pow(b / ` `2` `, ` `2` `));` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `float` `l = ` `4` `, b = ` `8` `;` ` ` `System.out.println(circlearea(l, b));` `}` `}` `// This code is contributed` `// by ChitraNayal` |

## Python 3

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

## C#

`// C# Program to find the` `// biggest circle which can be` `// inscribed within the rectangle` `using` `System;` `class` `GFG` `{` `// Function to find the area` `// of the biggest circle` `static` `float` `circlearea(` `float` `l,` ` ` `float` `b)` `{` `// the length and breadth` `// cannot be negative` `if` `(l < 0 || b < 0)` ` ` `return` `-1;` `// area of the circle` `if` `(l < b)` ` ` `return` `(` `float` `)(3.14 * Math.Pow(l / 2, 2));` `else` ` ` `return` `(` `float` `)(3.14 * Math.Pow(b / 2, 2));` `}` `// Driver code` `public` `static` `void` `Main()` `{` ` ` `float` `l = 4, b = 8;` ` ` `Console.Write(circlearea(l, b));` `}` `}` `// This code is contributed` `// by ChitraNayal` |

## PHP

`<?php` `// PHP Program to find the` `// biggest circle which can be` `// inscribed within the rectangle` `// Function to find the area` `// of the biggest circle` `function` `circlearea(` `$l` `, ` `$b` `)` `{` ` ` `// the length and breadth` ` ` `// cannot be negative` ` ` `if` `(` `$l` `< 0 || ` `$b` `< 0)` ` ` `return` `-1;` ` ` `// area of the circle` ` ` `if` `(` `$l` `< ` `$b` `)` ` ` `return` `3.14 * pow(` `$l` `/ 2, 2);` ` ` `else` ` ` `return` `3.14 * pow(` `$b` `/ 2, 2);` `}` `// Driver code` `$l` `= 4;` `$b` `= 8;` `echo` `circlearea(` `$l` `, ` `$b` `).` `"\n"` `;` `// This code is contributed` `// by ChitraNayal` `?>` |

## Javascript

`<script>` `// javascript Program to find the` `// biggest circle which can be` `// inscribed within the rectangle` `// Function to find the area` `// of the biggest circle` `function` `circlearea(l, b)` `{` ` ` `// the length and breadth` ` ` `// cannot be negative` ` ` `if` `(l < 0 || b < 0)` ` ` `return` `-1;` ` ` ` ` `// area of the circle` ` ` `if` `(l < b)` ` ` `return` `(3.14 * Math.pow(l / 2, 2));` ` ` `else` ` ` `return` `(3.14 * Math.pow(b / 2, 2));` `}` `// Driver code` `var` `l = 4, b = 8;` `document.write(circlearea(l, b));` `// This code is contributed by Amit Katiyar` `</script>` |

**Output:**

12.56

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend live classes with industry experts, please refer **Geeks Classes Live**