Given a square of side length ‘a’, the task is to find the side length of the biggest octagon that can be inscribed within it.
Examples:
Input: a = 4 Output: 1.65685 Input: a = 5 Output: 2.07107
Approach:
=> From the figure, it can be seen that, side length of the Octagon = b => Also since the polygons are regular, therefore 2*x + b = a => From the right angled triangle, x^2 + x^2 = b^2 => Hence, x = b/?2, => So, ?2b + b = a => Therefore, b = a/(?2 +1)
Below is the implementation of the above approach:
C++
// C++ Program to find the side of the octagon// which can be inscribed within the square#include <bits/stdc++.h>using namespace std;// Function to find the side// of the octagonfloat octaside(float a){ // side cannot be negative if (a < 0) return -1; // side of the octagon float s = a / (sqrt(2) + 1); return s;}// Driver codeint main(){ // Get he square side float a = 4; // Find the side length of the square cout << octaside(a) << endl; return 0;} |
Java
// Java Program to find the side of the octagon // which can be inscribed within the square import java.io.*;class GFG { // Function to find the side // of the octagon static double octaside(double a) { // side cannot be negative if (a < 0) return -1; // side of the octagon double s = a / (Math.sqrt(2) + 1); return s; } // Driver code public static void main (String[] args) { // Get he square side double a = 4; // Find the side length of the square System.out.println( octaside(a)); }}//This Code is contributed by ajit |
Python3
# Python 3 Program to find the side # of the octagon which can be # inscribed within the squarefrom math import sqrt# Function to find the side# of the octagondef octaside(a): # side cannot be negative if a < 0: return -1 # side of the octagon s = a / (sqrt(2) + 1) return s# Driver codeif __name__ == '__main__': # Get he square side a = 4 # Find the side length of the square print("{0:.6}".format(octaside(a))) # This code is contributed# by Surendra_Gangwar |
C#
// C# Program to find the side // of the octagon which can be // inscribed within the square using System;class GFG{ // Function to find the side // of the octagon static double octaside(double a) { // side cannot be negative if (a < 0) return -1; // side of the octagon double s = a / (Math.Sqrt(2) + 1); return s; } // Driver code static public void Main (){ // Get he square side double a = 4; // Find the side length // of the square Console.WriteLine( octaside(a)); } } // This code is contributed // by akt_mit |
PHP
<?php// PHP Program to find the side of the octagon// which can be inscribed within the square// Function to find the side// of the octagonfunction octaside($a){ // side cannot be negative if ($a < 0) return -1; // side of the octagon $s = $a / (sqrt(2) + 1); return $s;}// Driver code // Get he square side $a = 4; // Find the side length of the square echo octaside($a);// This code is contributed by ajit?> |
Javascript
<script>// javascript Program to find the side of the octagon // which can be inscribed within the square // Function to find the side // of the octagon function octaside(a) { // side cannot be negative if (a < 0) return -1; // side of the octagon var s = a / (Math.sqrt(2) + 1); return s; } // Driver code // Get he square side var a = 4; // Find the side length of the square document.write( octaside(a).toFixed(5)); // This code is contributed by shikhasingrajput </script> |
Output:
1.65685
Time Complexity: O(1) since no loop is used the algorithm takes up constant time to perform the operations
Space Complexity: O(1) since no extra array is used so the space taken by the algorithm 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!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
