Let's verify with specific numbers.
Example 1: $\(a=b=10\)$
- $\(\sqrt{x}=10, \sqrt{y}=10\)$
- $\(\sqrt{x y}=10 \times 10=100\)$
- Quantity $A=100$, Quantity $B=100$
- So, Quantity A = Quantity B

Example 2: $a=15, b=5$
- $\(\sqrt{x}=15, \sqrt{y}=5\)$
- $\(\sqrt{x y}=15 \times 5=75\)$
- Quantity A $=75$, Quantity B $=100$
- So, Quantity A < Quantity B

Example 3: $a=19, b=1$
- $\(\sqrt{x}=19, \sqrt{y}=1\)$
- $\(\sqrt{x y}=19 \times 1=19\)$
- Quantity A = 19, Quantity B = 100
- So, Quantity A < Quantity B


Step 6: Considering Edge Cases
Are there any other possibilities where the relationship might differ?
- If $a$ or $b$ is zero: But $\(\sqrt{x}\)$ and $\(\sqrt{y}\)$ are real numbers, and $\(\sqrt{x} \geq 0, \sqrt{y} \geq 0\)$. If $a=0$, then $b=20$, but $a b=0$, which is less than 100 . Similarly, if $b=0, a=20, a b=0$.
- Negative values: $\(\sqrt{x}\)$ and $\(\sqrt{y}\)$ are principal (non-negative) square roots, so negative values aren't considered here.

Thus, the only possibilities are:
1. $\(\sqrt{x y}=100\)$ (when $\(\sqrt{x}=\sqrt{y}=10\)$ )
2. $\(\sqrt{x y}<100\)$ (all other cases)

Step 7: Final Determination
Given that:
- There exists at least one case where Quantity A equals Quantity B (when $\(\sqrt{x}=\sqrt{y}=10\)$ ).
- There exist cases where Quantity A is less than Quantity B (when $\(\sqrt{x} \neq \sqrt{y}\)$ ).

Therefore, the relationship between Quantity A and Quantity B cannot be universally determined from the given information alone; it depends on the specific values of $x$ and $y$.