Given a regular Hexagon with side length a, the task is to find the area of the circle inscribed in it, given that, the circle is tangent to each of the six sides.
Examples:
Input: a = 4 Output: 37.68 Input: a = 10 Output: 235.5
Approach:
From the figure, it is clear that, we can divide the regular hexagon into 6 identical equilateral triangles.
We take one triangle OAB, with O as the centre of the hexagon or circle, & AB as one side of the hexagon.
Let M be mid-point of AB, OM would be the perpendicular bisector of AB, angle AOM = 30 deg
Then in right angled triangle OAM,
tanx = tan30 = 1/?3
So, a/2r = 1/?3
Therefore, r = a?3/2
Area of circle, A =?r²=?3a^2/4
Below is the implementation of the approach:
C++
// C++ Program to find the area of the circle// which can be inscribed within the hexagon#include <bits/stdc++.h>using namespace std;// Function to find the area// of the inscribed circlefloat circlearea(float a){ // the side cannot be negative if (a < 0) return -1; // area of the circle float A = (3.14 * 3 * pow(a, 2)) / 4; return A;}// Driver codeint main(){ float a = 4; cout << circlearea(a) << endl; return 0;} |
Java
//Java program to find the //area of the circle //which can be inscribed within the hexagonimport java.util.*;class solution{static double circlearea(double a){// the side cannot be negative if (a < 0) return -1;// area of the circle double A = (3.14 * 3 * Math.pow(a,2) ) / 4; return A;}public static void main(String arr[]){ double a = 4; System.out.println(circlearea(a));}} |
Python 3
# Python 3 program to find the # area of the circle # which can be inscribed within the hexagon# Function to find the area # of the inscribed circle def circlearea(a) : # the side cannot be negative if a < 0 : return -1 # area of the circle A = (3.14 * 3 * pow(a,2)) / 4 return A# Driver code if __name__ == "__main__" : a = 4 print(circlearea(a))# This code is contributed by ANKITRAI1 |
C#
// C# program to find // the area of the circle // which can be inscribed // within the hexagonusing System;class GFG{static double circlearea(double a){ // the side cannot be negative if (a < 0) return -1; // area of the circle double A = (3.14 * 3 * Math.Pow(a, 2)) / 4; return A;}// Driver Codepublic static void Main(){ double a = 4; Console.WriteLine(circlearea(a));}}// This code is contributed// by inder_verma |
PHP
<?php// PHP Program to find the area of // the circle which can be inscribed // within the hexagon// Function to find the area// of the inscribed circlefunction circlearea($a){ // the side cannot be negative if ($a < 0) return -1; // area of the circle $A = (3.14 * 3 * pow($a, 2)) / 4; return $A;}// Driver code$a = 4;echo circlearea($a) . "\n";// This code is contributed // by Akanksha Rai(Abby_akku) |
Javascript
<script>// javascript program to find the //area of the circle //which can be inscribed within the hexagonfunction circlearea(a) { // the side cannot be negative if (a < 0) return -1; // area of the circle var A = (3.14 * 3 * Math.pow(a, 2)) / 4; return A;}var a = 4;document.write(circlearea(a));// This code is contributed by 29AjayKumar </script> |
Output:
37.68
Time complexity: O(1)
Auxiliary Space: O(1)
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