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>