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


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++

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// 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;
}

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Java

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// 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.

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Python3

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# 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

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C#

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// 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..

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Output:

The side of each square is 14


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Improved By : vt_m, princiraj1992



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