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.
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
C++
// CPP program to find minimum distance to be travlled// by lizard.#include <bits/stdc++.h>#define ll long long intusing 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;} |
Java
//Java program to find minimum//distance to be travelled by lizardimport 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);}} |
Python3
# Python3 program to find minimum # distance to be travelled by lizardimport math#side of cubeif __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 |
C#
// C# program to find minimum// distance to be travelled by lizardusing 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 |
PHP
<?php// PHP program to find minimum distance // to be travlled by lizard.// side of cube$a = 5;// understand from diagram$AC = $a;// understand from diagram$CE = 2 * $a;// minimum distance$shortestDistance = (double)(sqrt($AC * $AC + $CE * $CE));echo $shortestDistance . "\n";// This code is contributed// by Akanksha Rai?> |
Javascript
<script>// Javascript program to find minimum distance to be travlled// by lizard.// side of cubevar a = 5;// understand from diagramvar AC = a;// understand from diagramvar CE = 2 * a;// minimum distancevar shortestDistace = Math.sqrt(AC * AC + CE * CE);document.write( shortestDistace.toFixed(4));</script> |
11.1803
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