# 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 = 4Output:The side of each square is 14Input:d = 36, n = 5Output: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 â€“ 2a = d ****na â€“ a = d**

so, **a = d/(n-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)