Largest sphere that can be inscribed in a right circular cylinder inscribed in a frustum
Given here is a frustum of height h, top-radius r & base-radius R, which inscribes a right circular cylinder which in turn inscribes a sphere . The task is to find the largest possible volume of this sphere.
Examples:
Input: r = 5, R = 8, h = 11 Output: 523.333 Input: r = 9, R = 14, h = 20 Output:3052.08
Approach: Let the height of the cylinder = H, radius of the sphere = x
We know, the height and radius of the cylinder inscribed within the frustum is equal to the height and top-radius of the frustum respectively(Please refer here).So the height of the cylinder = h, radius of the cylinder = r.
Also, radius of the sphere inscribed within a cylinder is equal to radius of the cylinder(Please refer here), so x = r.
So, volume of the sphere, V = 4*π*r^3/3.
Below is the implementation of the above approach:
C++
// C++ Program to find the biggest sphere // that can be inscribed within a right // circular cylinder which in turn is inscribed // within a frustum #include <bits/stdc++.h> using namespace std; // Function to find the biggest sphere float sph(float r, float R, float h) { // the radii and height cannot be negative if (r < 0 && R < 0 && h < 0) return -1; // radius of the sphere float x = r; // volume of the sphere float V = (4 * 3.14 * pow(r, 3)) / 3; return V; } // Driver code int main() { float r = 5, R = 8, h = 11; cout << sph(r, R, h) << endl; return 0; } |
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Java
// Java Program to find the biggest sphere // that can be inscribed within a right // circular cylinder which in turn is inscribed // within a frustum import java.lang.Math; class gfg { // Function to find the biggest sphere static float sph(float r, float R, float h) { // the radii and height cannot be negative if (r < 0 && R < 0 && h < 0) return -1; // radius of the sphere float x = r; // volume of the sphere float V = (float)(4 * 3.14f * Math.pow(r, 3)) / 3; return V; } // Driver code public static void main(String[] args) { float r = 5, R = 8, h = 11; System.out.println(sph(r, R, h)); } } // This Code is contributed by Code_Mech. |
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Python3
# Python3 Program to find the biggest sphere # that can be inscribed within a right # circular cylinder which in turn is inscribed # within a frustum import math as mt # Function to find the biggest sphere def sph(r, R, h): # the radii and height cannot # be negative if (r < 0 and R < 0 and h < 0): return -1 # radius of the sphere x = r # volume of the sphere V = (4 * 3.14 * pow(r, 3)) / 3 return V # Driver code r, R, h = 5, 8, 11print(sph(r, R, h)) # This code is contributed by # Mohit kumar 29 |
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C#
// C# Program to find the biggest sphere // that can be inscribed within a right // circular cylinder which in turn is // inscribed within a frustum using System; class gfg { // Function to find the biggest sphere static float sph(float r, float R, float h) { // the radii and height // cannot be negative if (r < 0 && R < 0 && h < 0) return -1; // radius of the sphere float x = r; // volume of the sphere float V = (float)(4 * 3.14f * Math.Pow(r, 3)) / 3; return V; } // Driver code public static void Main() { float r = 5, R = 8, h = 11; Console.WriteLine(sph(r, R, h)); } } // This code is contributed by Ryuga |
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PHP
<?php // PHP Program to find the biggest sphere // that can be inscribed within a right // circular cylinder which in turn is // inscribed within a frustum Function // to find the biggest sphere function sph($r, $R, $h) { // the radii and height // cannot be negative if ($r < 0 && $R < 0 && $h < 0) return -1; // radius of the sphere $x = $r; // volume of the sphere $V = (4 * 3.14 * pow($r, 3)) / 3; return $V; } // Driver code $r = 5; $R = 8; $h = 11; echo sph($r, $R, $h); #This Code is contributed by ajit.. ?> |
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Output:
523.333
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