内刻在椭圆内的正方形内的最大Reuleaux三角形

给定一个分别具有长轴长度和短轴2a和2b的椭圆，该椭圆内接一个正方形，而该正方形又内接一个reuleaux三角形。任务是找到该reuleaux三角形的最大可能面积。
例子：

Input: a = 5, b = 4
Output: 0.0722389

Input: a = 7, b = 11
Output: 0.0202076

方法：因此，椭圆内接的正方形的边为x =√(a ^ 2 + b ^ 2)/ ab。请参考可在椭圆上刻出的最大正方形的面积。
同样，在鲁洛三角形中， h = x =√(a ^ 2 + b ^ 2)/ ab 。
因此， reuleaux三角形的面积A = 0.70477 * h ^ 2 = 0.70477 *((a ^ 2 + b ^ 2)/ a ^ 2b ^ 2) 。
下面是上述方法的实现：

C++

// C++ Program to find the biggest Reuleaux triangle
// inscribed within in a square which in turn
// is inscribed within an ellipse
#include 
using namespace std;
 
// Function to find the biggest reuleaux triangle
float Area(float a, float b)
{
 
    // length of the axes cannot be negative
    if (a < 0 && b < 0)
        return -1;
 
    // height of the reuleaux triangle
    float h = sqrt(((pow(a, 2) + pow(b, 2))
                    / (pow(a, 2) * pow(b, 2))));
 
    // area of the reuleaux triangle
    float A = 0.70477 * pow(h, 2);
 
    return A;
}
 
// Driver code
int main()
{
    float a = 5, b = 4;
    cout << Area(a, b) << endl;
 
    return 0;
}

Java

// Java Program to find the biggest Reuleaux triangle
// inscribed within in a square which in turn
// is inscribed within an ellipse
import java.io.*;
 
class GFG
{
     
// Function to find the biggest reuleaux triangle
static float Area(float a, float b)
{
 
    // length of the axes cannot be negative
    if (a < 0 && b < 0)
        return -1;
 
    // height of the reuleaux triangle
    float h = (float)Math.sqrt(((Math.pow(a, 2) + Math.pow(b, 2))
                / (Math.pow(a, 2) * Math.pow(b, 2))));
 
    // area of the reuleaux triangle
    float A = (float)(0.70477 * Math.pow(h, 2));
 
    return A;
}
 
// Driver code
public static void main (String[] args)
{
    float a = 5, b = 4;
    System.out.println(Area(a, b));
}
}
 
// This code is contributed by anuj_67..

Python3

# Python3 Program to find the biggest Reuleaux
# triangle inscribed within in a square
# which in turn is inscribed within an ellipse
import math;
 
# Function to find the biggest
# reuleaux triangle
def Area(a, b):
 
    # length of the axes cannot
    # be negative
    if (a < 0 and b < 0):
        return -1;
 
    # height of the reuleaux triangle
    h = math.sqrt(((pow(a, 2) + pow(b, 2)) /
                   (pow(a, 2) * pow(b, 2))));
 
    # area of the reuleaux triangle
    A = 0.70477 * pow(h, 2);
 
    return A;
 
# Driver code
a = 5;
b = 4;
print(round(Area(a, b), 7));
 
# This code is contributed by chandan_jnu

C#

// C# Program to find the biggest Reuleaux triangle
// inscribed within in a square which in turn
// is inscribed within an ellipse
using System;
 
class GFG
{
     
// Function to find the biggest reuleaux triangle
static double Area(double a, double b)
{
 
    // length of the axes cannot be negative
    if (a < 0 && b < 0)
        return -1;
 
    // height of the reuleaux triangle
    double h = (double)Math.Sqrt(((Math.Pow(a, 2) +
                                    Math.Pow(b, 2)) /
                                   (Math.Pow(a, 2) *
                                   Math.Pow(b, 2))));
 
    // area of the reuleaux triangle
    double A = (double)(0.70477 * Math.Pow(h, 2));
 
    return A;
}
 
// Driver code
static void Main()
{
    double a = 5, b = 4;
    Console.WriteLine(Math.Round(Area(a, b),7));
}
}
 
// This code is contributed by chandan_jnu

Javascript

 
// Javascript Program to find the biggest Reuleaux triangle
// inscribed within in a square which in turn
// is inscribed within an ellipse
 
// Function to find the biggest reuleaux triangle
function Area(a, b)
{
 
    // length of the axes cannot be negative
    if (a < 0 && b < 0)
        return -1;
 
    // height of the reuleaux triangle
    let h = Math.sqrt(((Math.pow(a, 2) + Math.pow(b, 2))
                    / (Math.pow(a, 2) * Math.pow(b, 2))));
 
    // area of the reuleaux triangle
    let A = 0.70477 * Math.pow(h, 2);
 
    return A;
}
 
// Driver code
    let a = 5, b = 4;
    document.write(Area(a, b) + "<br>");
     
// This code is contributed by Mayank Tyagi
 

输出：

0.0722389