# Ratio of area of a rectangle with the rectangle inscribed in it

Given two rectangles, X with a ratio of length to width **a:b** and Y with a ratio of length to width **c:d** respectively. Both the rectangles can be resized as long as the ratio of sides remains the same. The task is to place the second rectangle inside the first rectangle such that at least 1 side is equal and that side overlaps of both the rectangles and find the ratio of (space occupied by a 2nd rectangle) : (space occupied by the first rectangle).

**Examples:**

Input:a = 1, b = 1, c = 3, d = 2Output:2:3 The dimensions can be 3X3 and 3X2.Input:a = 4, b = 3, c = 2, d = 2Output:3:4 The dimensions can be 4X3 and 3X3

**Approach:** If we make one of the sides of rectangles equal then the required ratio would be the ratio of the other side.

Consider 2 cases:

- a*d < b*c : We should make a and c equal.
- b*c < a*d : We should make b and d equal.

Since multiplying both sides of a ratio does not change its value. First try to make a and c equal, it can be made equal to their lcm by multiplying (a:b) with lcm/a and (c:d) with lcm/c. After multiplication, the ratio of (b:d) will be the required answer. This ratio can be reduced by dividing b and d with gcd(b, d).

Below is the implementation of the above approach:

## C++

`// C++ implementation of above approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to find the ratio ` `void` `printRatio(` `int` `a, ` `int` `b, ` `int` `c, ` `int` `d) ` `{ ` ` ` `if` `(b * c > a * d) { ` ` ` `swap(c, d); ` ` ` `swap(a, b); ` ` ` `} ` ` ` ` ` `// LCM of numerators ` ` ` `int` `lcm = (a * c) / __gcd(a, c); ` ` ` ` ` `int` `x = lcm / a; ` ` ` `b *= x; ` ` ` ` ` `int` `y = lcm / c; ` ` ` `d *= y; ` ` ` ` ` `// Answer in reduced form ` ` ` `int` `k = __gcd(b, d); ` ` ` `b /= k; ` ` ` `d /= k; ` ` ` ` ` `cout << b << ` `":"` `<< d; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `a = 4, b = 3, c = 2, d = 2; ` ` ` ` ` `printRatio(a, b, c, d); ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of above approach ` ` ` `import` `java.io.*; ` ` ` `class` `GFG { ` `// Recursive function to return gcd of a and b ` ` ` `static` `int` `__gcd(` `int` `a, ` `int` `b) ` ` ` `{ ` ` ` `// Everything divides 0 ` ` ` `if` `(a == ` `0` `) ` ` ` `return` `b; ` ` ` `if` `(b == ` `0` `) ` ` ` `return` `a; ` ` ` ` ` `// 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 ratio ` ` ` `static` `void` `printRatio(` `int` `a, ` `int` `b, ` `int` `c, ` `int` `d) ` `{ ` ` ` `if` `(b * c > a * d) { ` ` ` `int` `temp = c; ` ` ` `c =d; ` ` ` `d =c; ` ` ` `temp =a; ` ` ` `a =b; ` ` ` `b=temp; ` ` ` ` ` `} ` ` ` ` ` `// LCM of numerators ` ` ` `int` `lcm = (a * c) / __gcd(a, c); ` ` ` ` ` `int` `x = lcm / a; ` ` ` `b *= x; ` ` ` ` ` `int` `y = lcm / c; ` ` ` `d *= y; ` ` ` ` ` `// Answer in reduced form ` ` ` `int` `k = __gcd(b, d); ` ` ` `b /= k; ` ` ` `d /= k; ` ` ` ` ` `System.out.print( b + ` `":"` `+ d); ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main (String[] args) { ` ` ` `int` `a = ` `4` `, b = ` `3` `, c = ` `2` `, d = ` `2` `; ` ` ` ` ` `printRatio(a, b, c, d); ` ` ` `} ` `} ` ` ` `// This code is contributed by inder_verma.. ` |

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## Python3

`import` `math ` `# Python3 implementation of above approach ` ` ` `# Function to find the ratio ` `def` `printRatio(a, b, c, d): ` ` ` `if` `(b ` `*` `c > a ` `*` `d): ` ` ` `swap(c, d) ` ` ` `swap(a, b) ` ` ` ` ` `# LCM of numerators ` ` ` `lcm ` `=` `(a ` `*` `c) ` `/` `math.gcd(a, c) ` ` ` ` ` `x ` `=` `lcm ` `/` `a ` ` ` `b ` `=` `int` `(b ` `*` `x) ` ` ` ` ` `y ` `=` `lcm ` `/` `c ` ` ` `d ` `=` `int` `(d ` `*` `y) ` ` ` ` ` `# Answer in reduced form ` ` ` `k ` `=` `math.gcd(b,d) ` ` ` `b ` `=` `int` `(b ` `/` `k) ` ` ` `d ` `=` `int` `(d ` `/` `k) ` ` ` ` ` `print` `(b,` `":"` `,d) ` ` ` `# Driver code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` `a ` `=` `4` ` ` `b ` `=` `3` ` ` `c ` `=` `2` ` ` `d ` `=` `2` ` ` ` ` `printRatio(a, b, c, d) ` ` ` `# This code is contributed by ` `# Surendra_Gangwar ` |

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## C#

`// C# implementation of above approach ` ` ` `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) ` ` ` `return` `b; ` ` ` `if` `(b == 0) ` ` ` `return` `a; ` ` ` ` ` `// 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 ratio ` `static` `void` `printRatio(` `int` `a, ` `int` `b, ` `int` `c, ` `int` `d) ` `{ ` ` ` `if` `(b * c > a * d) { ` ` ` `int` `temp = c; ` ` ` `c =d; ` ` ` `d =c; ` ` ` `temp =a; ` ` ` `a =b; ` ` ` `b=temp; ` ` ` ` ` `} ` ` ` ` ` `// LCM of numerators ` ` ` `int` `lcm = (a * c) / __gcd(a, c); ` ` ` ` ` `int` `x = lcm / a; ` ` ` `b *= x; ` ` ` ` ` `int` `y = lcm / c; ` ` ` `d *= y; ` ` ` ` ` `// Answer in reduced form ` ` ` `int` `k = __gcd(b, d); ` ` ` `b /= k; ` ` ` `d /= k; ` ` ` ` ` `Console.WriteLine( b + ` `":"` `+ d); ` `} ` ` ` `// Driver code ` ` ` ` ` `public` `static` `void` `Main () { ` ` ` `int` `a = 4, b = 3, c = 2, d = 2; ` ` ` ` ` `printRatio(a, b, c, d); ` ` ` `} ` `} ` ` ` `// This code is contributed by inder_verma.. ` |

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## PHP

$b)

return __gcd($a – $b, $b);

return __gcd($a, $b – $a);

}

// Function to find the ratio

function printRatio($a, $b, $c, $d)

{

if ($b * $c > $a * $d)

{

$temp = $c;

$c = $d;

$d = $c;

$temp = $a;

$a = $b;

$b = $temp;

}

// LCM of numerators

$lcm = ($a * $c) / __gcd($a, $c);

$x = $lcm / $a;

$b *= $x;

$y = $lcm / $c;

$d *= $y;

// Answer in reduced form

$k = __gcd($b, $d);

$b /= $k;

$d /= $k;

echo $b . “:” . $d;

}

// Driver code

$a = 4; $b = 3; $c = 2; $d = 2;

printRatio($a, $b, $c, $d);

// This code is contributed

// by Akanksha Rai

?>

**Output:**

3:4

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