# Sum of lengths of all 12 edges of any rectangular parallelepiped

• Difficulty Level : Easy
• Last Updated : 25 Sep, 2022

Given the area of three faces of the rectangular parallelepiped which has a common vertex. Our task is to find the sum of lengths of all 12 edges of this parallelepiped.
In geometry, a parallelepiped is a three-dimensional figure formed by six parallelograms. By analogy, it relates to a parallelogram just as a cube relates to a square or as a cuboid to a rectangle. A picture of a rectangular parallelepiped is shown below. Examples:

Input: 1 1 1
Output: 12

Input: 20 10 50
Output: 68

Approach: The area given are s1, s2 and s3 . Let a, b and c be the lengths of the sides that have one common vertex. Where   . It’s easy to find the length in terms of faces areas:   . The answer will be the summation of all the 4 sides, there are four sides that have lengths equal to a, b and c.
In the first example the given area s1 = 1, s2 = 1 and s3 = 1. So with the above approach, the value of a, b, c will come out to be 1. So the sum of the length of all 12 edges will be 4 * 3 = 12.
Below is the implementation of the above approach:

## C++

 // C++ program to illustrate// the above problem#include using namespace std; // function to find the sum of// all the edges of parallelepipeddouble findEdges(double s1, double s2, double s3){    // to calculate the length of one edge    double a = sqrt(s1 * s2 / s3);    double b = sqrt(s3 * s1 / s2);    double c = sqrt(s3 * s2 / s1);     // sum of all the edges of one side    double sum = a + b + c;     // net sum will be equal to the    // summation of edges of all the sides    return 4 * sum;} // Driver codeint main(){    // initialize the area of three    // faces which has a common vertex    double s1, s2, s3;    s1 = 65, s2 = 156, s3 = 60;     cout << findEdges(s1, s2, s3);     return 0;}

## Java

 // Java program to illustrate// the above problem import java.io.*; class GFG {   // function to find the sum of// all the edges of parallelepipedstatic double findEdges(double s1, double s2, double s3){    // to calculate the length of one edge    double a = Math.sqrt(s1 * s2 / s3);    double b = Math.sqrt(s3 * s1 / s2);    double c = Math.sqrt(s3 * s2 / s1);     // sum of all the edges of one side    double sum = a + b + c;     // net sum will be equal to the    // summation of edges of all the sides    return 4 * sum;}        // Driver code     public static void main (String[] args) {            // initialize the area of three    // faces which has a common vertex    double s1, s2, s3;    s1 = 65; s2 = 156; s3 = 60;     System.out.print(findEdges(s1, s2, s3));    }}  // this code is contributed by anuj_67..

## Python3

 import math # Python3 program to illustrate# the above problem # function to find the sum of# all the edges of parallelepipeddef findEdges(s1, s2, s3):     # to calculate the length of one edge    a = math.sqrt(s1 * s2 / s3)    b = math.sqrt(s3 * s1 / s2)    c = math.sqrt(s3 * s2 / s1)     # sum of all the edges of one side    sum = a + b + c     # net sum will be equal to the    # summation of edges of all the sides    return 4 * sum  # Driver codeif __name__=='__main__':     # initialize the area of three# faces which has a common vertex    s1 = 65    s2 = 156    s3 = 60     print(int(findEdges(s1, s2, s3)))         # This code is contributed by# Shivi_Aggarwal

## C#

 // C# program to illustrate// the above problemusing System; public class GFG{     // function to find the sum of// all the edges of parallelepipedstatic double findEdges(double s1, double s2, double s3){    // to calculate the length of one edge    double a = Math.Sqrt(s1 * s2 / s3);    double b = Math.Sqrt(s3 * s1 / s2);    double c = Math.Sqrt(s3 * s2 / s1);     // sum of all the edges of one side    double sum = a + b + c;     // net sum will be equal to the    // summation of edges of all the sides    return 4 * sum;} // Driver code     static public void Main (){    // initialize the area of three    // faces which has a common vertex    double s1, s2, s3;    s1 = 65; s2 = 156; s3 = 60;     Console.WriteLine(findEdges(s1, s2, s3));    }}  // This code is contributed by anuj_67..

## PHP

 

## Javascript

 

Output:

120

Time Complexity: O(logn) because the inbuilt sqrt function is being used
Auxiliary Space: O(1)

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