Let the side length of the original square be \( s \) meters. The area of the original square is \( s^2 \).
When one dimension is increased by 5 meters and the other by 3 meters, the new dimensions become \( s + 5 \) and \( s + 3 \).
The area of the resulting rectangle is given as 98 square meters. Therefore, we can set up the equation:
\[ (s + 5)(s + 3) = 98 \]
Expanding the left side:
\[ s^2 + 3s + 5s + 15 = 98 \]
This simplifies to:
\[ s^2 + 8s + 15 = 98 \]
Next, we subtract 98 from both sides:
\[ s^2 + 8s + 15 - 98 = 0 \]
This simplifies to:
\[ s^2 + 8s - 83 = 0 \]
Now, we can use the quadratic formula to solve for \( s \):
\[ s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this case, \( a = 1 \), \( b = 8 \), and \( c = -83 \):
\[ s = \frac{-8 \pm \sqrt{8^2 - 4 \times 1 \times -83}}{2 \times 1} \] \[ s = \frac{-8 \pm \sqrt{64 + 332}}{2} \] \[ s = \frac{-8 \pm \sqrt{396}}{2} \] \[ s = \frac{-8 \pm 2\sqrt{99}}{2} \] \[ s = -4 \pm \sqrt{99} \]
Since we are looking for a positive dimension, we take the positive root:
\[ s = -4 + \sqrt{99} \]
To find the area of the original square:
\[ \text{Area} = s^2 = (-4 + \sqrt{99})^2 \] \[ = (-4)^2 + 2 \cdot (-4) \cdot \sqrt{99} + (\sqrt{99})^2 \] \[ = 16 - 8\sqrt{99} + 99 \] \[ = 115 - 8\sqrt{99} \]
Thus, the area of the original square is:
\[ \boxed{115 - 8\sqrt{99}} \text{ square meters.} \]
To approximate, \( \sqrt{99} \approx 9.95 \):
\[ 115 - 8 \cdot 9.95 \approx 115 - 79.6 \approx 35.4 \text{ square meters.} \]
So the area of the original square is approximately \( 35.4 \text{ square meters.} \)