Let the original side length of the square be 's'.
The original area of the square is \( A_{original} = s^2 \).
When the side length is increased by 3 times, the new side length becomes \( s_{new} = 3s \).
The new area of the square is \( A_{new} = (s_{new})^2 = (3s)^2 \).
\[ A_{new} = (3s) \times (3s) = 9s^2 \]
Now, we compare the new area to the original area:
\[ A_{new} = 9s^2 = 9 \times A_{original} \]
This means the new area is 9 times the original area.
Ответ: Yuzasi 9 marta ortadi.