Given the side of a square **a** which is kept inside a circle. It keeps expanding until all four of its vertices touch the circumference of the circle. Another smaller circle is kept inside the square now and it keeps expanding until its circumference touches all the four sides of the square. The outer and the inner circle form a ring. Find the area of this shaded part as shown in the image below.

**Examples:**

Input:a = 3

Output:7.06858

Input:a = 4

Output:12.566371

**Approach:**

From the above figure, **R = a / sqrt(2)** can be derived where **a** is the side length of the square. The area of the outer circle is **(pi * R * R)**.

Let **s1** be the area of the outer circle **(pi * R * R)** and **s2** be the area of the inner circle **(pi * r * r)**. Then the area of the ring is **s1 – s2**.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to return the required area ` `float` `getArea(` `int` `a) ` `{ ` ` ` ` ` `// Calculate the area ` ` ` `float` `area = (M_PI * a * a) / 4.0; ` ` ` `return` `area; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `a = 3; ` ` ` ` ` `cout << getArea(a); ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the approach ` `class` `GFG { ` ` ` ` ` `// Function to return the required area ` ` ` `static` `float` `getArea(` `int` `a) ` ` ` `{ ` ` ` ` ` `// Calculate the area ` ` ` `float` `area = (` `float` `)(Math.PI * a * a) / ` `4` `; ` ` ` `return` `area; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main(String args[]) ` ` ` `{ ` ` ` `int` `a = ` `3` `; ` ` ` `System.out.println(getArea(a)); ` ` ` `} ` `} ` |

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## Python

`# Python3 implementation of the approach ` `import` `math ` ` ` `# Function to return the required area ` `def` `getArea(a): ` ` ` ` ` `# Calculate the area ` ` ` `area ` `=` `(math.pi ` `*` `a ` `*` `a) ` `/` `4` ` ` `return` `area ` ` ` `# Driver code ` `a ` `=` `3` `print` `(` `'{0:.6f}'` `.` `format` `(getArea(a))) ` |

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## C#

`// C# implementation of the approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` ` ` `// Function to return the required area ` ` ` `static` `float` `getArea(` `int` `a) ` ` ` `{ ` ` ` ` ` `// Calculate the area ` ` ` `float` `area = (` `float` `)(Math.PI * a * a) / 4; ` ` ` `return` `area; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `a = 3; ` ` ` `Console.Write(getArea(a)); ` ` ` `} ` `} ` ` ` `// This code is contributed by mohit kumar 29 ` |

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**Output:**

7.06858

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