Find area of the Circle when the area of inscribed Square is given

Given the area of a square inscribed in a circle as N, the task is to calculate the area of a circle in which the square is inscribed.

Examples: 

Input: N = 4 
Output: 6.283

Input: N = 10
Output:  15.707

Approach:

Consider the below image:

• Let the area of the square is ‘A’
• The side of the square is given by = A**(1/2)
• A right-angled triangle is formed by the two sides of the square and the diameter of the circle
• The hypotenuse of the triangle will be the diameter of the circle
• The diameter of circle ‘D’ is calculated as ((A * A) + (A * A))**(1/2)
• The radius of circle ‘r’ is given by D/2
• The resultant area of the circle is pi*r*r

Below is the implementation of the above approach: 

C++

#include <iostream>
#include<math.h>
#include <iomanip>
using namespace std;

// Function to calculate the area of circle
double areaOfCircle(double a)
{
    // declaring pi
    double pi=2*acos(0.0);

    // Side of the square
    double side = pow(a,(1.0 / 2));

    // Diameter of circle
    double D =pow( ((side * side) + (side * side)) ,(1.0 / 2));

    // Radius of circle
    double R = D / 2;

    // Area of circle
    double Area = pi * (R * R);

    return Area;

}
 //Driver Code
int main() {

    double areaOfSquare = 4;
    cout<<setprecision(15)<<areaOfCircle(areaOfSquare);
    return 0;
}
// This code is contributed by ANKITKUMAR34

Java

// Java code for the above approach
import java.util.*;

class GFG
{

    // Function to calculate the area of circle
    static double areaOfCircle(double a)
    {

        // Side of the square
        double side = Math.pow(a, (1.0 / 2));

        // Diameter of circle
        double D = Math.pow(((side * side) + (side * side)),
                            (1.0 / 2));

        // Radius of circle
        double R = D / 2;

        // Area of circle
        double Area = Math.PI * (R * R);

        return Area;
    }

    // Driver Code
    public static void main(String[] args)
    {
        double areaOfSquare = 4;
        System.out.println(areaOfCircle(areaOfSquare));
    }
}

// This code is contribute by Potta Lokesh

Python3

# Python program for the above approach
import math

# Function to calculate the area of circle
def areaOfCircle(a):

    # Side of the square
    side = a**(1 / 2)

    # Diameter of circle
    D = ((side * side) + (side * side))**(1 / 2)

    # Radius of circle
    R = D / 2

    # Area of circle
    Area = math.pi * (R * R)

    return Area

# Driver Code
areaOfSquare = 4
print(areaOfCircle(areaOfSquare))

C#

// C# code for the above approach
using System;

class GFG
{

    // Function to calculate the area of circle
    static double areaOfCircle(double a)
    {

        // Side of the square
        double side = Math.Pow(a, (1.0 / 2));

        // Diameter of circle
        double D = Math.Pow(((side * side) + (side * side)),
                            (1.0 / 2));

        // Radius of circle
        double R = D / 2;

        // Area of circle
        double Area = Math.PI * (R * R);

        return Area;
    }

    // Driver Code
    public static void Main()
    {
        double areaOfSquare = 4;
        Console.Write(areaOfCircle(areaOfSquare));
    }
}

// This code is contribute by Samim Hossain Mondal.

Javascript

<script>

// Function to calculate the area of circle
function areaOfCircle(a) {
  // declaring pi
  let pi = 2 * Math.acos(0.0);

  // Side of the square
  let side = Math.pow(a, (1.0 / 2));

  // Diameter of circle
  let D = Math.pow(((side * side) + (side * side)), (1.0 / 2));

  // Radius of circle
  let R = D / 2;

  // Area of circle
  let Area = Math.PI * (R * R);

  return Area;

}
//Driver Code

let areaOfSquare = 4;
document.write(areaOfCircle(areaOfSquare));

// This code is contributed by gfgking
</script>

Output

6.283185307179588

Time Complexity: O(1) as it would take constant time to perform operations
Auxiliary Space: O(1)