\(x\) and \(y\) are 2 different digits. If the sum of the two digit numbers formed by using both the digits is a perfect square, then find \(x+y\).

Let’s analyze the given information to solve the problem:

Given that \(x\) and \(y\) are two different digits forming two-digit numbers \(xy\) and \(yx\), the sum of these two numbers can be represented as:

\[
10x + y + 10y + x = 11x + 11y
\]

This sum is stated to be a perfect square. Given that \(x\) and \(y\) are digits, the sum \(11(x + y)\) suggests that \(x + y\) itself must be a number that, when multiplied by 11, results in a perfect square.

Looking at the options and considering that \(x + y\) must be small enough to fit the constraints of single digits (1 through 9), let’s analyze the options directly:

– (a) 10: \(11 \times 10 = 110\), not a perfect square.
– (b) 11: \(11 \times 11 = 121\), which is a perfect square (\(11^2\)).
– (c) 12: \(11 \times 12 = 132\), not a perfect square.
– (d) 13: \(11 \times 13 = 143\), not a perfect square.

The only option where \(11(x + y)\) forms a perfect square is when \(x + y = 11\), making \(11 \times 11 = 121\), which is indeed a perfect square (\(11^2\)).

Therefore, the correct answer is (b) **11**.