# Program to find the Volume of an irregular tetrahedron

Last Updated : 04 Aug, 2022

Given the length of edges of an irregular tetrahedron. The task is to determine the volume of that tetrahedron. Let the Edge length of pyramids be u, U, v, V, w, W.

Examples:

Input: u = 1000, v = 1000, w = 1000, U = 3, V = 4, W = 5
Output: 1999.9947

Input: u = 2000, v = 2000, w = 2000, U = 3, V = 4, W = 5
Output: 3999.9858

Formula to calculate the Volume of an irregular Tetrahedron in terms of its edge lengths is:

A =

Volume = sqrt(A/288) =
sqrt(4*u*u*v*v*w*w – u*u*(v*v + w*w – U*U)^2 – v*v(w*w + u*u – V*V)^2 – w*w(u*u + v*v – W*W)^2 + (u*u + v*v – W*W) * (w*w + u*u – V*V) * (v*v + w*w – U*U)) / 12

Below is the implementation of the above approach:

## C++

 // C++ implementation of above approach#include using namespace std;#define db double// Function to find the volumevoid findVolume(db u, db v, db w, db U, db V, db W, db b){     // Steps to calculate volume of a    // Tetrahedron using formula    db uPow = pow(u, 2);    db vPow = pow(v, 2);    db wPow = pow(w, 2);    db UPow = pow(U, 2);    db VPow = pow(V, 2);    db WPow = pow(W, 2);     db a = 4 * (uPow * vPow * wPow)        - uPow * pow((vPow + wPow - UPow), 2)        - vPow * pow((wPow + uPow - VPow), 2)        - wPow * pow((uPow + vPow - WPow), 2)        + (vPow + wPow - UPow) * (wPow + uPow - VPow)        * (uPow + vPow - WPow);    db vol = sqrt(a);    vol /= b;     cout << fixed << setprecision(4) << vol;} // Driver codeint main(){     // edge lengths    db u = 1000, v = 1000, w = 1000;    db U = 3, V = 4, W = 5;    db b = 12;     findVolume(u, v, w, U, V, W, b);     return 0;}

## Java

 // Java implementation of above approach import java.util.*;import java.lang.*;import java.io.*;  class GFG{ // Function to find the volumestatic void findVolume(double u, double v, double w, double U,                       double V, double W, double b){     // Steps to calculate volume of a    // Tetrahedron using formula    double uPow = Math.pow(u, 2);    double vPow = Math.pow(v, 2);    double wPow = Math.pow(w, 2);    double UPow = Math.pow(U, 2);    double VPow = Math.pow(V, 2);    double WPow = Math.pow(W, 2);     double a = 4 * (uPow * vPow * wPow)        - uPow * Math.pow((vPow + wPow - UPow), 2)        - vPow * Math.pow((wPow + uPow - VPow), 2)        - wPow * Math.pow((uPow + vPow - WPow), 2)        + (vPow + wPow - UPow) * (wPow + uPow - VPow)         * (uPow + vPow - WPow);    double vol = Math.sqrt(a);    vol /= b;     System.out.printf("%.4f",vol);} // Driver codepublic static void main(String args[]){     // edge lengths    double u = 1000, v = 1000, w = 1000;    double U = 3, V = 4, W = 5;    double b = 12;     findVolume(u, v, w, U, V, W, b);}     }

## Python3

 # Python 3 implementation of above approach  # from math lib import everythingfrom math import * # Function to find the volume def findVolume(u, v, w, U, V, W, b) :        # Steps to calculate volume of a     # Tetrahedron using formula     uPow = pow(u, 2)     vPow = pow(v, 2)     wPow = pow(w, 2)     UPow = pow(U, 2)     VPow = pow(V, 2)     WPow = pow(W, 2)        a = (4 * (uPow * vPow * wPow)         - uPow * pow((vPow + wPow - UPow), 2)         - vPow * pow((wPow + uPow - VPow), 2)         - wPow * pow((uPow + vPow - WPow), 2)         + (vPow + wPow - UPow) * (wPow + uPow - VPow)         * (uPow + vPow - WPow))             vol = sqrt(a)    vol /= b       print(round(vol,4))  # Driver code     if __name__ == "__main__" :     # edge lengths     u, v, w = 1000, 1000, 1000    U, V, W = 3, 4, 5    b = 12       findVolume(u, v, w, U, V, W, b)   # This code is contributed by ANKITRAI1

## C#

 // C# implementation of above approachusing System;  class GFG{ // Function to find the volumestatic void findVolume(double u, double v,                        double w, double U,                        double V, double W,                        double b){     // Steps to calculate volume of a    // Tetrahedron using formula    double uPow = Math.Pow(u, 2);    double vPow = Math.Pow(v, 2);    double wPow = Math.Pow(w, 2);    double UPow = Math.Pow(U, 2);    double VPow = Math.Pow(V, 2);    double WPow = Math.Pow(W, 2);     double a = 4 * (uPow * vPow * wPow) -                     uPow * Math.Pow((vPow + wPow - UPow), 2) -                     vPow * Math.Pow((wPow + uPow - VPow), 2) -                     wPow * Math.Pow((uPow + vPow - WPow), 2) +                     (vPow + wPow - UPow) *                     (wPow + uPow - VPow) * (uPow + vPow - WPow);    double vol = Math.Sqrt(a);    vol /= b;     Console.Write(System.Math.Round(vol, 4));} // Driver codepublic static void Main(){     // edge lengths    double u = 1000, v = 1000, w = 1000;    double U = 3, V = 4, W = 5;    double b = 12;     findVolume(u, v, w, U, V, W, b);}} // This code is contributed// by ChitraNayal

## PHP

 

## Javascript

 

Output:
1999.9947

Time Complexity: O(logn) as using inbuilt pow function

Auxiliary Space: O(1)

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