Given the median of the Equilateral triangle M, the task is to find the area of the circumcircle of this equilateral triangle using the median M.
Examples:
Input: M = 3
Output: 12.5664Input: M = 6
Output: 50.2655
Approach: The key observation in the problem is that the centroid, circumcenter, orthocenter and incenter of an equilateral triangle all lie at the same point.
Therefore, the radius of the circle with the given median of the equilateral triangle inscribed in the circle can be derived as: 
Then the area of the circle can be calculated using the approach used in this article
Below is the implementation of the above approach:
C++
// C++ implementation to find the// equation of circle which// inscribes equilateral triangle// of median M#include <iostream>const double pi = 3.14159265358979323846;using namespace std;// Function to find the equation// of circle whose center is (x1, y1)// and the radius of circle is rvoid circleArea(double r){ cout << (pi * r * r);}// Function to find the// equation of circle which// inscribes equilateral triangle// of median Mvoid findCircleAreaByMedian(double m){ double r = 2 * m / 3; // Util Function to find the // circle equation circleArea(r);}// Driver codeint main(){ double m = 3; // Function Call findCircleAreaByMedian(m); return 0;} |
Java
// Java implementation to find the // equation of circle which // inscribes equilateral triangle // of median M import java.util.*;class GFG{ // Function to find the equation // of circle whose center is (x1, y1) // and the radius of circle is r static double circleArea(double r){ double pi = 3.14159265358979323846; return (pi * r * r);} // Function to find the // equation of circle which // inscribes equilateral triangle // of median M static double findCircleAreaByMedian(int m){ double r = 2 * m / 3; // Function call to find // the circle equation return circleArea(r);}// Driver code public static void main(String args[]){ int m = 3; System.out.printf("%.4f", findCircleAreaByMedian(m));}}// This code is contributed by virusbuddah_ |
Python3
# Python3 implementation to find the# equation of circle which inscribes # equilateral triangle of median Mpi = 3.14159265358979323846# Function to find the equation# of circle whose center is (x1, y1)# and the radius of circle is rdef circleArea(r): print(round(pi * r * r, 4))# Function to find the# equation of circle which# inscribes equilateral triangle# of median Mdef findCircleAreaByMedian(m): r = 2 * m /3 # Function to find the # circle equation circleArea(r)# Driver codeif __name__ == '__main__': m = 3 # Function call findCircleAreaByMedian(m)# This code is contributed by mohit kumar 29 |
C#
// C# implementation to find the // equation of circle which // inscribes equilateral triangle // of median M using System;class GFG{ // Function to find the equation // of circle whose center is (x1, y1) // and the radius of circle is r static double circleArea(double r) { double pi = 3.14159265358979323846; return (pi * r * r); } // Function to find the // equation of circle which // inscribes equilateral triangle // of median M static double findCircleAreaByMedian(int m) { double r = 2 * m / 3; // Function call to find // the circle equation return circleArea(r); } // Driver code public static void Main(string []args) { int m = 3; Console.WriteLine("{0:f4}", findCircleAreaByMedian(m)); } } // This code is contributed by AnkitRai01 |
Javascript
<script>// javascript implementation to find the // equation of circle which // inscribes equilateral triangle // of median M // Function to find the equation // of circle whose center is (x1, y1) // and the radius of circle is r function circleArea(r) { var pi = 3.14159265358979323846; return (pi * r * r); } // Function to find the // equation of circle which // inscribes equilateral triangle // of median M function findCircleAreaByMedian(m) { var r = 2 * m / 3; // Function call to find // the circle equation return circleArea(r); } // Driver code var m = 3; document.write(findCircleAreaByMedian(m).toFixed(4));// This code is contributed by Rajput-Ji </script> |
Output
12.5664
Time Complexity: O(1), as we are not using any looping statements.
Auxiliary Space: O(1), as we are not using any extra space.
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