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 circledouble 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 Codeint main() { double areaOfSquare = 4; cout<<setprecision(15)<<areaOfCircle(areaOfSquare); return 0;}// This code is contributed by ANKITKUMAR34 |
Java
// Java code for the above approachimport 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 approachimport math# Function to calculate the area of circledef 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 CodeareaOfSquare = 4print(areaOfCircle(areaOfSquare)) |
C#
// C# code for the above approachusing 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 circlefunction 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 Codelet 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)
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