使用中值的等边三角形的圆周面积

给定等边三角形M的中位数，任务是使用中位数 M 找到该等边三角形的外接圆面积。
例子：

Input: M = 3 
Output: 12.5664
Input: M = 6 
Output: 50.2655 
 

方法：问题中的关键观察是等边三角形的质心、外心、垂心和内心都在同一点上。

因此，内接等边三角形的给定中值的圆的半径可以推导出为：
$\text{Radius of Circumcircle =} \frac{2}{3}*M$
然后可以使用本文中使用的方法计算圆的面积
下面是上述方法的实现：

C++

// C++ implementation to find the
// equation of circle which
// inscribes equilateral triangle
// of median M
 
#include 
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 r
void circleArea(double r)
{
    cout << (pi * r * r);
}
 
// Function to find the
// equation of circle which
// inscribes equilateral triangle
// of median M
void findCircleAreaByMedian(double m)
{
    double r = 2 * m / 3;
 
    // Util Function to find the
    // circle equation
    circleArea(r);
}
 
// Driver code
int 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 M
 
pi = 3.14159265358979323846
 
# Function to find the equation
# of circle whose center is (x1, y1)
# and the radius of circle is r
def circleArea(r):
     
    print(round(pi * r * r, 4))
 
# Function to find the
# equation of circle which
# inscribes equilateral triangle
# of median M
def findCircleAreaByMedian(m):
     
    r = 2 * m /3
 
    # Function to find the
    # circle equation
    circleArea(r)
 
# Driver code
if __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

// 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

输出：

12.5664