Biggest Reuleaux Triangle inscribed within a square which is inscribed within an ellipse

Given an ellipse with major axis length and minor axis 2a & 2b respectively which inscribes a square which in turn inscribes a reuleaux triangle. The task is to find the maximum possible area of this reuleaux triangle.

Examples:

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

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



Approach: As, the side of the square inscribed within an ellipse is, x = √(a^2 + b^2)/ab. Please refer Area of the Largest square that can be inscribed in an ellipse.
Also, in the reuleaux triangle, h = x = √(a^2 + b^2)/ab.
So, Area of the reuleaux triangle, A = 0.70477*h^2 = 0.70477*((a^2 + b^2)/a^2b^2).

Below is the implementation of the above approach:

C++

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// C++ Program to find the biggest Reuleaux triangle
// inscribed within in a square which in turn
// is inscribed within an ellipse
#include <bits/stdc++.h>
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;
}

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Java

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// 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..

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Python3

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# 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

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C#

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// 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

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PHP

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<?php
// PHP 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 
    $h = sqrt(((pow($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;
$b = 4; 
echo round(Area($a, $b), 7);
  
// This code is contributed by Ryuga
?>

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Output:

0.0722389


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