# Smallest square formed with given rectangles

Given a rectangle of length l and breadth b, we need to find the area of the smallest square which can be formed with the rectangles of these given dimensions.
Examples:

```Input : 1 2
Output : 4
We can form a 2 x 2 square
using two rectangles of size
1 x 2.

Input : 7 10
Output :4900
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Let’s say we want to make a square of side length a from rectangles of length l & b. This means that a is a multiple of both l & b. Since we want the smallest square, it has to be the lowest common multiple (LCM) of l & b.

Program 1:

## C++

 `// C++ Program to find the area ` `// of the smallest square which ` `// can be formed with rectangles ` `// of given dimensions ` `#include ` `using` `namespace` `std; ` `// Recursive function to return gcd of a and b ` `int` `gcd(``int` `a, ``int` `b) ` `{ ` `    ``// Everything divides 0 ` `    ``if` `(a == 0 || b == 0) ` `        ``return` `0; ` ` `  `    ``// Base case ` `    ``if` `(a == b) ` `        ``return` `a; ` ` `  `    ``// a is greater ` `    ``if` `(a > b) ` `        ``return` `gcd(a - b, b); ` `    ``return` `gcd(a, b - a); ` `} ` ` `  `// Function to find the area ` `// of the smallest square ` `int` `squarearea(``int` `l, ``int` `b) ` `{ ` ` `  `    ``// the length or breadth or side ` `    ``// cannot be negative ` `    ``if` `(l < 0 || b < 0) ` `        ``return` `-1; ` ` `  ` `  `        ``// LCM of length and breadth  ` `        ``int` `n = (l * b) / gcd(l, b);  ` ` `  `        ``// squaring to get the area ` `        ``return` `n * n;  ` `     `  `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `l = 6, b = 4; ` `    ``cout << squarearea(l, b) << endl; ` `    ``return` `0; ` `} `

## Java

 `// JavaProgram to find the area ` `// of the smallest square which ` `// can be formed with rectangles ` `// of given dimensions ` `class` `GFG  ` `{ ` ` `  `// Recursive function to  ` `// return gcd of a and b ` `static` `int` `gcd(``int` `a, ``int` `b) ` `{ ` `// Everything divides 0 ` `if` `(a == ``0` `|| b == ``0``) ` `    ``return` `0``; ` ` `  `// Base case ` `if` `(a == b) ` `    ``return` `a; ` ` `  `// a is greater ` `if` `(a > b) ` `    ``return` `gcd(a - b, b); ` `return` `gcd(a, b - a); ` `} ` ` `  `// Function to find the area ` `// of the smallest square ` `static` `int` `squarearea(``int` `l, ``int` `b) ` `{ ` ` `  `// the length or breadth or side ` `// cannot be negative ` `if` `(l < ``0` `|| b < ``0``) ` `    ``return` `-``1``; ` ` `  ` `  `    ``// LCM of length and breadth  ` `    ``int` `n = (l * b) / gcd(l, b);  ` ` `  `    ``// squaring to get the area ` `    ``return` `n * n;  ` ` `  `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `l = ``6``, b = ``4``; ` `    ``System.out.println(squarearea(l, b)); ` `} ` `} ` ` `  `// This code is contributed  ` `// by ChitraNayal `

## Python 3

 `# Python3 Program to find the area ` `# of the smallest square which ` `# can be formed with rectangles ` `# of given dimensions ` ` `  `# Recursive function to return gcd of a and b ` `def` `gcd( a, b): ` ` `  `    ``# Everything divides 0 ` `    ``if` `(a ``=``=` `0` `or` `b ``=``=` `0``): ` `        ``return` `0` ` `  `    ``# Base case ` `    ``if` `(a ``=``=` `b): ` `        ``return` `a ` ` `  `    ``# a is greater ` `    ``if` `(a > b): ` `        ``return` `gcd(a ``-` `b, b) ` `    ``return` `gcd(a, b ``-` `a) ` ` `  ` `  `# Function to find the area ` `# of the smallest square ` `def` `squarearea( l, b): ` ` `  ` `  `    ``# the length or breadth or side ` `    ``# cannot be negative ` `    ``if` `(l < ``0` `or` `b < ``0``): ` `        ``return` `-``1` ` `  ` `  `        ``# LCM of length and breadth  ` `    ``n ``=` `(l ``*` `b) ``/` `gcd(l, b)  ` ` `  `        ``# squaring to get the area ` `    ``return` `n ``*` `n  ` `     `  ` `  ` `  `# Driver code ` `if` `__name__``=``=``'__main__'``: ` `    ``l ``=` `6` `    ``b ``=` `4` `    ``print``(``int``(squarearea(l, b))) ` ` `  `#This code is contributed by ash264 `

## C#

 `// C# Program to find the area ` `// of the smallest square which ` `// can be formed with rectangles ` `// of given dimensions ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Recursive function to  ` `// return gcd of a and b ` `static` `int` `gcd(``int` `a, ``int` `b) ` `{ ` `// Everything divides 0 ` `if` `(a == 0 || b == 0) ` `    ``return` `0; ` ` `  `// Base case ` `if` `(a == b) ` `    ``return` `a; ` ` `  `// a is greater ` `if` `(a > b) ` `    ``return` `gcd(a - b, b); ` `return` `gcd(a, b - a); ` `} ` ` `  `// Function to find the area ` `// of the smallest square ` `static` `int` `squarearea(``int` `l, ``int` `b) ` `{ ` ` `  `// the length or breadth or side ` `// cannot be negative ` `if` `(l < 0 || b < 0) ` `    ``return` `-1; ` ` `  ` `  `    ``// LCM of length and breadth  ` `    ``int` `n = (l * b) / gcd(l, b);  ` ` `  `    ``// squaring to get the area ` `    ``return` `n * n;  ` ` `  `} ` ` `  `// Driver code ` `public` `static` `void` `Main()  ` `{ ` `    ``int` `l = 6, b = 4; ` `    ``Console.Write(squarearea(l, b)); ` `} ` `} ` ` `  `// This code is contributed  ` `// by ChitraNayal `

## PHP

 ` ``\$b``) ` `        ``return` `gcd(``\$a` `- ``\$b``, ``\$b``); ` `    ``return` `gcd(``\$a``, ``\$b` `- ``\$a``); ` `} ` ` `  `// Function to find the area ` `// of the smallest square ` `function` `squarearea(``\$l``, ``\$b``) ` `{ ` ` `  `    ``// the length or breadth or side ` `    ``// cannot be negative ` `    ``if` `(``\$l` `< 0 || ``\$b` `< 0) ` `        ``return` `-1; ` ` `  ` `  `        ``// LCM of length and breadth  ` `        ``\$n` `= (``\$l` `* ``\$b``) / gcd(``\$l``, ``\$b``);  ` ` `  `        ``// squaring to get the area ` `        ``return` `\$n` `* ``\$n``;  ` `     `  `} ` ` `  `// Driver code ` `\$l` `= 6; ` `\$b` `= 4; ` `echo` `squarearea(``\$l``, ``\$b``).``"\n"``; ` ` `  `// This code is contributed  ` `// by ChitraNayal ` `?> `

Output:

```144
```

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