# Find volume of Icosahedron

An icosahedron is a three-dimensional geometric solid with 20 equilateral triangle faces. Finding the volume of an icosahedron involves calculating the amount of space enclosed within its boundaries. This calculation is important in geometry and architecture, where icosahedrons are used to model certain structures and crystals.

## Problem Statement

Given the length of the side of the icosahedron, the task is to calculate its volume. The formula for finding the volume of an icosahedron in terms of its side length 's' is:

Volume = (5/12) * (3 + √5) * s³

## Example Scenario

Imagine you are a crystallographer studying the molecular arrangement of certain compounds. To understand the internal space occupied by the molecule, you need to calculate the volume of the icosahedral crystal structure. This calculation helps you analyze the molecule's characteristics and interactions.

## Idea to Solve the Problem

To solve this problem, we can follow these steps:

- Accept the side length of the icosahedron as input.
- Use the formula to calculate the volume of the icosahedron.
- Display the calculated volume.

## Pseudocode

```
function icosahedron_volume(side):
volume = (5/12) * (3 + √5) * (side * side * side)
return volume
main:
side1 = 4
side2 = 8
side3 = 13.6
volume1 = icosahedron_volume(side1)
volume2 = icosahedron_volume(side2)
volume3 = icosahedron_volume(side3)
print("Icosahedron [ Side :", side1, "]")
print("Volume :", volume1)
print("Icosahedron [ Side :", side2, "]")
print("Volume :", volume2)
print("Icosahedron [ Side :", side3, "]")
print("Volume :", volume3)
```

## Algorithm Explanation

- Define a function
`icosahedron_volume`

that takes the side length as input. - Inside the function, use the provided formula to calculate the volume of the icosahedron.
- In the
`main`

function, set three test cases with different values for the side length. - Calculate the volumes for each test case by calling the
`icosahedron_volume`

function. - Display the calculated volumes along with their respective side length values.

## Code Solution

```
//C Program
//Find volume of Icosahedron
#include <stdio.h>
#include <math.h>
//Calculate volume of Icosahedron by given side
void icosahedron_volume(double side)
{
//Calculate Formula
//5/12 (3 + √5)(side³)
double volume = ((double) 5 / 12) * (3 + sqrt(5)) * (side * side * side);
//Display the result of icosahedron volume
printf(" Icosahedron [ Side : %lf ] ", side);
printf("\n Volume : %lf\n\n", volume);
}
int main()
{
//Test Cases
icosahedron_volume(4);
icosahedron_volume(8);
icosahedron_volume(13.6);
return 0;
}
```

#### Output

```
Icosahedron [ Side : 4.000000 ]
Volume : 139.628479
Icosahedron [ Side : 8.000000 ]
Volume : 1117.027835
Icosahedron [ Side : 13.600000 ]
Volume : 5487.957754
```

```
// Java Program
// Find volume of Icosahedron
class IcosahedronVolume
{
//Calculate volume of Icosahedron by given side
public void icosahedron_volume(double side)
{
//Calculate Formula
//5/12 (3 + √5)(side³)
double volume = ((double) 5 / 12) * (3 + Math.sqrt(5)) * (side * side * side);
//Display the result of icosahedron volume
System.out.print(" Icosahedron [ Side : " + side + " ] ");
System.out.print("\n Volume : " + volume + "\n\n");
}
public static void main(String[] args)
{
IcosahedronVolume obj = new IcosahedronVolume();
//Test Cases
obj.icosahedron_volume(4);
obj.icosahedron_volume(8);
obj.icosahedron_volume(13.6);
}
}
```

#### Output

```
Icosahedron [ Side : 4.0 ]
Volume : 139.6284793999944
Icosahedron [ Side : 8.0 ]
Volume : 1117.0278351999552
Icosahedron [ Side : 13.6 ]
Volume : 5487.95775433738
```

```
// C++ Program
// Find volume of Icosahedron
#include<iostream>
#include <math.h>
using namespace std;
class IcosahedronVolume
{
public:
//Calculate volume of Icosahedron by given side
void icosahedron_volume(double side)
{
//Calculate Formula
//5/12 (3 + √5)(side³)
double volume = ((double) 5 / 12) *(3 + sqrt(5)) *(side *side *side);
cout << " Icosahedron [ Side : " << side << " ] ";
cout << "\n Volume : " << volume << "\n\n";
}
};
int main()
{
IcosahedronVolume obj = IcosahedronVolume();
//Test Cases
obj.icosahedron_volume(4);
obj.icosahedron_volume(8);
obj.icosahedron_volume(13.6);
return 0;
}
```

#### Output

```
Icosahedron [ Side : 4 ]
Volume : 139.628
Icosahedron [ Side : 8 ]
Volume : 1117.03
Icosahedron [ Side : 13.6 ]
Volume : 5487.96
```

```
// C# Program
// Find volume of Icosahedron
using System;
class IcosahedronVolume
{
//Calculate volume of Icosahedron by given side
public void icosahedron_volume(double side)
{
//Calculate Formula
//5/12 (3 + √5)(side³)
double volume = ((double) 5 / 12) * (3 + Math.Sqrt(5)) * (side * side * side);
Console.Write(" Icosahedron [ Side : " + side + " ] ");
Console.Write("\n Volume : " + volume + "\n\n");
}
public static void Main(String[] args)
{
IcosahedronVolume obj = new IcosahedronVolume();
//Test Cases
obj.icosahedron_volume(4);
obj.icosahedron_volume(8);
obj.icosahedron_volume(13.6);
}
}
```

#### Output

```
Icosahedron [ Side : 4 ]
Volume : 139.628479399994
Icosahedron [ Side : 8 ]
Volume : 1117.02783519996
Icosahedron [ Side : 13.6 ]
Volume : 5487.95775433738
```

```
<?php
// Php Program
// Find volume of Icosahedron
class IcosahedronVolume
{
//Calculate volume of Icosahedron by given side
public function icosahedron_volume($side)
{
//Calculate Formula
//5/12 (3 + √5)(side³)
$volume = (5 / 12) *(3 + sqrt(5)) *($side *$side *$side);
//Display the result of icosahedron volume
echo(" Icosahedron [ Side : ". $side ." ] ");
echo("\n Volume : ". $volume ."\n\n");
}
}
function main()
{
$obj = new IcosahedronVolume();
//Test Cases
$obj->icosahedron_volume(4);
$obj->icosahedron_volume(8);
$obj->icosahedron_volume(13.6);
}
main();
```

#### Output

```
Icosahedron [ Side : 4 ]
Volume : 139.62847939999
Icosahedron [ Side : 8 ]
Volume : 1117.0278352
Icosahedron [ Side : 13.6 ]
Volume : 5487.9577543374
```

```
// Node Js Program
// Find volume of Icosahedron
class IcosahedronVolume
{
//Calculate volume of Icosahedron by given side
icosahedron_volume(side)
{
//Calculate Formula
//5/12 (3 + √5)(side³)
var volume = (5 / 12) *(3 + Math.sqrt(5)) *(side *side *side);
//Display the result of icosahedron volume
process.stdout.write(" Icosahedron [ Side : " + side + " ] ");
process.stdout.write("\n Volume : " + volume + "\n\n");
}
}
function main(args)
{
var obj = new IcosahedronVolume();
//Test Cases
obj.icosahedron_volume(4);
obj.icosahedron_volume(8);
obj.icosahedron_volume(13.6);
}
main();
```

#### Output

```
Icosahedron [ Side : 4 ]
Volume : 139.6284793999944
Icosahedron [ Side : 8 ]
Volume : 1117.0278351999552
Icosahedron [ Side : 13.6 ]
Volume : 5487.95775433738
```

```
# Python 3 Program
# Find volume of Icosahedron
import math
class IcosahedronVolume :
# Calculate volume of Icosahedron by given side
def icosahedron_volume(self, side) :
# Calculate Formula
# 5/12 (3 + √5)(side³)
volume = (5 / 12) * (3 + math.sqrt(5)) * (side * side * side)
# Display the result of icosahedron volume
print(" Icosahedron [ Side : ", side , end = " ]")
print("\n Volume : ", volume ,"\n")
def main() :
obj = IcosahedronVolume()
# Test Cases
obj.icosahedron_volume(4)
obj.icosahedron_volume(8)
obj.icosahedron_volume(13.6)
if __name__ == "__main__": main()
```

#### Output

```
Icosahedron [ Side : 4 ]
Volume : 139.6284793999944
Icosahedron [ Side : 8 ]
Volume : 1117.0278351999552
Icosahedron [ Side : 13.6 ]
Volume : 5487.95775433738
```

```
# Ruby Program
# Find volume of Icosahedron
class IcosahedronVolume
# Calculate volume of Icosahedron by given side
def icosahedron_volume(side)
# Calculate Formula
# 5/12 (3 + √5)(side³)
volume = (5.0 / 12) * (3 + Math.sqrt(5)) * (side * side * side)
# Display the result of icosahedron volume
print(" Icosahedron [ Side : ", side ," ] ")
print("\n Volume : ", volume ,"\n\n")
end
end
def main()
obj = IcosahedronVolume.new()
# Test Cases
obj.icosahedron_volume(4)
obj.icosahedron_volume(8)
obj.icosahedron_volume(13.6)
end
main()
```

#### Output

```
Icosahedron [ Side : 4 ]
Volume : 139.6284793999944
Icosahedron [ Side : 8 ]
Volume : 1117.0278351999552
Icosahedron [ Side : 13.6 ]
Volume : 5487.95775433738
```

```
// Scala Program
// Find volume of Icosahedron
class IcosahedronVolume
{
//Calculate volume of Icosahedron by given side
def icosahedron_volume(side: Double): Unit = {
//Calculate Formula
//5/12 (3 + √5)(side³)
var volume: Double = (5.0 / 12) * (3 + Math.sqrt(5)) * (side * side * side);
//Display the result of icosahedron volume
print(" Icosahedron [ Side : " + side + " ] ");
print("\n Volume : " + volume + "\n\n");
}
}
object Main
{
def main(args: Array[String]): Unit = {
var obj: IcosahedronVolume = new IcosahedronVolume();
//Test Cases
obj.icosahedron_volume(4);
obj.icosahedron_volume(8);
obj.icosahedron_volume(13.6);
}
}
```

#### Output

```
Icosahedron [ Side : 4.0 ]
Volume : 139.6284793999944
Icosahedron [ Side : 8.0 ]
Volume : 1117.0278351999552
Icosahedron [ Side : 13.6 ]
Volume : 5487.95775433738
```

```
// Swift Program
// Find volume of Icosahedron
import Foundation
class IcosahedronVolume
{
//Calculate volume of Icosahedron by given side
func icosahedron_volume(_ side: Double)
{
//Calculate Formula
//5/12 (3 + √5)(side³)
let volume: Double = (5 / 12) * (3 + sqrt(5)) * (side * side * side);
//Display the result of icosahedron volume
print(" Icosahedron [ Side : ", side , terminator: " ]");
print("\n Volume : ", volume ,"\n");
}
}
func main()
{
let obj: IcosahedronVolume = IcosahedronVolume();
//Test Cases
obj.icosahedron_volume(4);
obj.icosahedron_volume(8);
obj.icosahedron_volume(13.6);
}
main();
```

#### Output

```
Icosahedron [ Side : 4.0 ]
Volume : 139.628479399994
Icosahedron [ Side : 8.0 ]
Volume : 1117.02783519996
Icosahedron [ Side : 13.6 ]
Volume : 5487.95775433738
```

## Output Explanation

The code calculates the volume for each test case and displays the results. For an icosahedron with a side length of 4, the volume is approximately 139.628479. Similarly, for a side length of 8, the volume is approximately 1117.027835. The third test case follows the same pattern with a different set of values.

## Time Complexity

The time complexity of this code is constant O(1) because the calculations involve basic arithmetic operations and the value of the square root of 5, which are calculated in constant time regardless of the input size. The program performs a fixed number of operations for each test case, making it efficient.

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