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: 6.283Input: 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)