Minimum volume of cone that can be circumscribed about a sphere of radius R
Given a sphere of radius R, The task is to find out the minimum volume of the cone that can be circumscribed about it.
Examples:
Input: R = 10 Output: Volume of cone = 8373.33 Explanation: Radius of cone = 14.14 and Height of cone = 40, Volume of cone =So, volume = 8373.33 Input: R = 4 Output: Volume of cone = 535.89
Approach:
we have given a sphere of radius R inscribed in Cone. We need to find out the radius and height of the cone to find out the volume of the cone.
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In triangle AOE and ALC compute sin(X) i.e. For triangle AOE
and for triangle ALC
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Now, From equating both we get
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Insert the value of H in Volume i.e.
and for volume to be minimum
.
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From the above equation we get
and putting this value in H we get
-
Hence, applying the formula of volume of cone and putting
and
we get the desired result.
C++
// C++ program to find the minimum // volume of the cone that can be // circumscribed about a sphere // of radius R #include<bits/stdc++.h> using namespace std; // Function to find the volume // of the cone float Volume_of_cone( float R) { // r = radius of cone // h = height of cone // Volume of cone = (1 / 3) * (3.14) * (r*r) * (h) // we get radius of cone from the derivation // is root(2) times multiple of R // we get height of cone from the derivation // is 4 times multiple of R float V = (1 / 3.0) * (3.14) * (2 * ( R * R ) ) * (4 * R); return V; } // Driver code int main() { float R = 10.0; cout << Volume_of_cone(R); } // This code is contributed by Samarth |
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Java
// Java program to find the minimum // volume of the cone that can be // circumscribed about a sphere // of radius R import java.util.*; class GFG{ // Function to find the volume // of the cone static double Volume_of_cone( double R) { // r = radius of cone // h = height of cone // Volume of cone = (1 / 3) * (3.14) * (r*r) * (h) // we get radius of cone from the derivation // is root(2) times multiple of R // we get height of cone from the derivation // is 4 times multiple of R double V = ( double )(( 1 / 3.0 ) * ( 3.14 ) * ( 2 * (R * R)) * ( 4 * R)); return V; } // Driver code public static void main(String[] args) { double R = 10.0 ; System.out.print(Volume_of_cone(R)); } } // This code is contributed by sapnasingh4991 |
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Python3
# Python3 program to find the minimum # Volume of the cone that can be circumscribed # about a sphere of radius R import math # Function to find the volume # of the cone def Volume_of_cone(R): # r = radius of cone # h = height of cone # Volume of cone = (1 / 3) * (3.14) * (r**2) * (h) # we get radius of cone from the derivation # is root(2) times multiple of R # we get height of cone from the derivation # is 4 times multiple of R V = ( 1 / 3 ) * ( 3.14 ) * ( 2 * ( R * * 2 ) ) * ( 4 * R) return V # Driver code if __name__ = = "__main__" : R = 10 print (Volume_of_cone(R)) |
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C#
// C# program to find the minimum // volume of the cone that can be // circumscribed about a sphere // of radius R using System; class GFG{ // Function to find the volume // of the cone static double Volume_of_cone( double R) { // r = radius of cone // h = height of cone // Volume of cone = (1 / 3) * (3.14) * (r*r) * (h) // we get radius of cone from the derivation // is root(2) times multiple of R // we get height of cone from the derivation // is 4 times multiple of R double V = ( double )((1 / 3.0) * (3.14) * (2 * (R * R)) * (4 * R)); return V; } // Driver code public static void Main() { double R = 10.0; Console.Write(Volume_of_cone(R)); } } // This code is contributed by Nidhi_biet |
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Output:
8373.333333333332
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