Area of the circle that has a square and a circle inscribed in it

Given the side of a square that 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 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;
}

Java

// Java implementation of the approach
public 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));
    }
}

Python3

# 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)))

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

Javascript

<script>

// Javascript implementation of the approach

// Function to return the required area
function getArea(a)
{

    // Calculate the area
    var area = (Math.PI * a * a) / 4;
    return area;
}

// Driver Code
var a = 3;

document.write(getArea(a));

// This code is contributed by Kirti

</script>

Output

7.06858

Time Complexity: O(1)
Auxiliary Space: O(1)