Puzzle | Minimum distance for Lizard

A lizard is present on one corner of cube, It wants to reach diagonally opposite corner of cube. You have to calculate minimum distance lizard has to cover to reach its destination.

Note : Lizard can’t fly, it moves along the wall.

You are given a representing side of cube.you have to calculate minimum distance lizard has to travel.
Examples:

Input : 5
Output :11.1803

Input :2
Output :4.47214


As we have to calculate the minimum distance from one corner to another diagonally opposite corner. if lizard able to fly then the shortest distance will be length of diagonal. But it can’t.
Img
So, to calculate minimum distance, just open the cube, as describe in diagram.
Let us suppose, lizard is initially at point E.and it has to reach at point A(as A is diagonally opposite to E).Now we have to find AE.
Just use Pythagoras theorem, As
AC=a
CE=CD+DE=2a

     $$AE=\sqrt{AC^2+CE^2}$$ $$AE=\sqrt{a^2+\left( 2a \right)^2}$$ $$AE=\sqrt{5a^2}$$

C++

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// CPP program to find minimum distance to be travlled
// by lizard.
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
int main()
{
    // side of cube
    ll a = 5;
  
    // understand from diagram
    ll AC = a;
  
    // understand from diagram
    ll CE = 2 * a;
  
    // minimum distance
    double shortestDistace = sqrt(AC * AC + CE * CE);
  
    cout << shortestDistace << endl;
    return 0;
}

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Java

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//Java program to find minimum
//distance to be travelled by lizard
import java.util.*;
  
class solution
{
public static void main(String arr[])
{
    // side of the cube
    int a = 5;
  
    // understand from diagram
    int AC = a;
  
    // understand from diagram
    int CE = 2 * a;
  
    // minimum distance
    double shortestDistace = Math.sqrt(AC * AC + CE * CE);
  
    System.out.println(shortestDistace);
}
}

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Python3

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# Python3 program to find minimum 
# distance to be travelled by lizard
  
import math
  
#side of cube
if __name__=='__main__':
    a = 5
  
#understand from diagram
    AC = a
  
#understand from diagram
    CE = 2 * a
  
#minimum distance
    shortestDistace = math.sqrt(AC * AC + CE * CE)
  
    print(shortestDistace)
  
#this code is Contributed by Shashank_Sharma

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C#

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// C# program to find minimum
// distance to be travelled by lizard
using System;
  
class GFG
{
public static void Main()
{
    // side of the cube
    int a = 5;
  
    // understand from diagram
    int AC = a;
  
    // understand from diagram
    int CE = 2 * a;
  
    // minimum distance
    double shortestDistace = Math.Sqrt(AC * AC + CE * CE);
  
    Console.Write(shortestDistace);
}
}
  
// This code is contributed by ita_c

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PHP

Output:

11.1803


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