Radius of the biggest possible circle inscribed in rhombus which in turn is inscribed in a rectangle
Give a rectangle with length l & breadth b, which inscribes a rhombus, which in turn inscribes a circle. The task is to find the radius of this circle.
Examples:
Input: l = 5, b = 3 Output: 1.28624 Input: l = 6, b = 4 Output: 1.6641
Approach: From the figure, it is clear that diagonals, x & y, are equal to the lenght and breadth of the rectangle.
Also radius of the circle, r, inside a rhombus is = xy/2√(x^2+y^2).
So, radius of the circle in terms of l & b is = lb/2√(l^2+b^2).
Below is the implementation of the above approach:
C++
// C++ implementation of above approach #include <bits/stdc++.h> using namespace std; // Function to find the radius // of the inscribed circle float circleradius(float l, float b) { // the sides cannot be negative if (l < 0 || b < 0) return -1; // radius of the circle float r = (l * b) / (2 * sqrt((pow(l, 2) + pow(b, 2)))); return r; } // Driver code int main() { float l = 5, b = 3; cout << circleradius(l, b) << endl; return 0; } |
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Java
// Java implementation of above approach import java.io.*; class GFG { // Function to find the radius // of the inscribed circle static float circleradius(float l, float b) { // the sides cannot be negative if (l < 0 || b < 0) return -1; // radius of the circle float r = (float)((l * b) / (2 * Math.sqrt((Math.pow(l, 2) + Math.pow(b, 2))))); return r; } // Driver code public static void main (String[] args) { float l = 5, b = 3; System.out.print (circleradius(l, b)) ; } } // This code is contributed by inder_verma.. |
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Python3
# Python 3 implementation of # above approach from math import sqrt # Function to find the radius # of the inscribed circle def circleradius(l, b): # the sides cannot be negative if (l < 0 or b < 0): return -1 # radius of the circle r = (l * b) / (2 * sqrt((pow(l, 2) + pow(b, 2)))); return r # Driver code if __name__ == '__main__': l = 5 b = 3 print("{0:.5}" . format(circleradius(l, b))) # This code is contribute # by Surendra_Gagwar |
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C#
// C# implementation of above approach using System; class GFG { // Function to find the radius // of the inscribed circle static float circleradius(float l, float b) { // the sides cannot be negative if (l < 0 || b < 0) return -1; // radius of the circle float r = (float)((l * b) / (2 * Math.Sqrt((Math.Pow(l, 2) + Math.Pow(b, 2))))); return r; } // Driver code public static void Main () { float l = 5, b = 3; Console.WriteLine(circleradius(l, b)); } } // This code is contributed // by inder_verma |
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PHP
<?php // PHP implementation of above approach // Function to find the radius // of the inscribed circle function circleradius($l, $b) { // the sides cannot be negative if ($l < 0 || $b < 0) return -1; // radius of the circle $r = ($l * $b) / (2 * sqrt((pow($l, 2) + pow($b, 2)))); return $r; } // Driver code $l = 5; $b = 3; echo circleradius($l, $b), "\n"; // This code is contributed by ajit ?> |
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Output:
1.28624
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