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++ 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;
} |
// 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 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# 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 |
<?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 ?> |
<script> // 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 </script> |
0.0722389
Time Complexity: O(logn) as it is using inbuilt sqrt function
Auxiliary Space: O(1)