Skip to content
Related Articles

Related Articles

Find the side of the squares which are lined in a row, and distance between the centers of first and last square is given

View Discussion
Improve Article
Save Article
  • Last Updated : 31 May, 2022
View Discussion
Improve Article
Save Article

Given here are n squares which touch each other externally, and are lined up in a row. The distance between the centers of the first and last square is given. The squares have equal side length. The task is to find the side of each square.
Examples: 
 

Input: d = 42, n = 4
Output: The side of each square is 14

Input: d = 36, n = 5
Output: The side of each square is 9

 

Approach: 
Suppose there are n squares each having side of length a
Let, the distance between the first and last squares = d 
From the figure, it is clear, 
a/2 + a/2 + (n-2)*a = d 
a + na – 2a = d 
na – a = d 
so, a = d/(n-1) 
side of the square = distance between the first and last squares/(no. of squares given - 1)
 

C++




// C++ program to find side of the squares
// which are lined in a row and distance between the
// centers of first and last squares is given
 
#include <bits/stdc++.h>
using namespace std;
 
void radius(int n, int d)
{
    cout << "The side of each square is "
         << d / (n - 1) << endl;
}
 
// Driver code
int main()
{
    int d = 42, n = 4;
    radius(n, d);
    return 0;
}


Java




// Java program to find side of the squares
// which are lined in a row and distance between the
// centers of first and last squares is given
import java.io.*;
 
class GFG
{
 
static void radius(int n, int d)
{
    System.out.print( "The side of each square is "
        + d / (n - 1));
}
 
// Driver code
public static void main (String[] args)
{
    int d = 42, n = 4;
    radius(n, d);
}
}
 
// This code is contributed by vt_m.


Python3




     
# Python program to find side of the squares
# which are lined in a row and distance between the
# centers of first and last squares is given
 
def radius(n, d):
 
    print("The side of each square is ",
          d / (n - 1));
 
 
d = 42; n = 4;
radius(n, d);
 
 
 
# This code contributed by PrinciRaj1992


C#




// C# program to find side of the squares
// which are lined in a row and distance between the
// centers of first and last squares is given
using System;
 
class GFG
{
 
static void radius(int n, int d)
{
    Console.Write( "The side of each square is "
        + d / (n - 1));
}
 
// Driver code
public static void Main ()
{
    int d = 42, n = 4;
    radius(n, d);
}
}
 
// This code is contributed by anuj_67..


Javascript




<script>
// javascript program to find side of the squares
// which are lined in a row and distance between the
// centers of first and last squares is given
function radius(n , d)
{
    document.write( "The side of each square is "
        + d / (n - 1));
}
 
// Driver code
var d = 42, n = 4;
radius(n, d);
 
// This code is contributed by 29AjayKumar
</script>


Output: 

The side of each square is 14

 

Time Complexity: O(1)

Auxiliary Space: O(1)


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!