Thursday, January 9, 2025
Google search engine
HomeData Modelling & AILargest square that can be inscribed within a hexagon which is inscribed...

Largest square that can be inscribed within a hexagon which is inscribed within an equilateral triangle

Given here is an equilateral triangle of side length a, which inscribes a hexagon which in turn inscribes a square. The task is to find the side length of the square.
Examples: 
 

Input:  a = 6
Output: 2.538

Input: a = 8
Output: 3.384

 

 

Approach
We know the, side length of a hexagon inscribed within an equilateral triangle is h = a/3. Please refer Largest hexagon that can be inscribed within an equilateral triangle .
Also, the side length of the square that can be inscribed within a hexagon is x = 1.268h Please refer Largest Square that can be inscribed within a hexagon.
So, side length of the square inscribed within a hexagon which in turn is inscribed within an equilateral triangle, x = 0.423a.
Below is the implementation of the above approach:
 

C++




// C++ program to find the side of the largest square
// that can be inscribed within the hexagon which in return
// is incsribed within an equilateral triangle
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the side
// of the square
float squareSide(float a)
{
 
    // Side cannot be negative
    if (a < 0)
        return -1;
 
    // side of the square
    float x = 0.423 * a;
    return x;
}
 
// Driver code
int main()
{
    float a = 8;
    cout << squareSide(a) << endl;
 
    return 0;
}


Java




// Java program to find the side of the
// largest square that can be inscribed
// within the hexagon which in return is
// incsribed within an equilateral triangle
class cfg
{
     
// Function to find the side
// of the square
static float squareSide(float a)
{
 
    // Side cannot be negative
    if (a < 0)
        return -1;
 
    // side of the square
    float x = (0.423f * a);
    return x;
}
 
// Driver code
public static void main(String[] args)
{
    float a = 8;
    System.out.println(squareSide(a));
 
}
}
 
// This code is contributed by
// Mukul Singh.


Python3




# Python 3 program to find the side of the
# largest square that can be inscribed
# within the hexagon which in return
# is incsribed within an equilateral triangle
 
# Function to find the side of the square
def squareSide(a):
     
    # Side cannot be negative
    if (a < 0):
        return -1
 
    # side of the square
    x = 0.423 * a
    return x
 
# Driver code
if __name__ == '__main__':
    a = 8
    print(squareSide(a))
 
# This code is contributed by
# Sanjit_Prasad


C#




// C# program to find the side of the
// largest square that can be inscribed
// within the hexagon which in return is
// incsribed within an equilateral triangle
using System;
 
class GFG
{
     
// Function to find the side
// of the square
static float squareSide(float a)
{
 
    // Side cannot be negative
    if (a < 0)
        return -1;
 
    // side of the square
    float x = (0.423f * a);
    return x;
}
 
// Driver code
public static void Main()
{
    float a = 8;
    Console.WriteLine(squareSide(a));
}
}
 
// This code is contributed by
// shs


PHP




<?php
// PHP program to find the side of the
// largest square that can be inscribed
// within the hexagon which in return is
// incsribed within an equilateral triangle
 
// Function to find the side of the square
function squareSide($a)
{
 
    // Side cannot be negative
    if ($a < 0)
        return -1;
 
    // side of the square
    $x = 0.423 * $a;
    return $x;
}
 
// Driver code
$a = 8;
echo squareSide($a);
 
// This code is contributed by ajit.
?>


Javascript




<script>
// javascript program to find the side of the
// largest square that can be inscribed
// within the hexagon which in return is
// incsribed within an equilateral triangle
 
// Function to find the side
// of the square
function squareSide(a)
{
 
    // Side cannot be negative
    if (a < 0)
        return -1;
 
    // side of the square
    var x = (0.423 * a);
    return x;
}
 
// Driver code
var a = 8;
document.write(squareSide(a));
 
// This code is contributed by Princi Singh
</script>


Output: 

3.384

 

Time Complexity: O(1) since no loop is used the algorithm takes constant time to finish its execution

Auxiliary Space: O(1) since no extra array is used the space required by the algorithm to complete is constant.

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!

Last Updated :
07 Aug, 2022
Like Article
Save Article


Previous

<!–

8 Min Read | Java

–>


Next


<!–

8 Min Read | Java

–>

RELATED ARTICLES

Most Popular

Recent Comments