To find the area of the triangular seal, first, we need to calculate the area of the triangle in George's scale drawing.
The formula for the area \( A \) of a triangle is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
In George's drawing, the base of the triangle is 1 foot and the height is 2 feet. Plugging in these values:
\[ A = \frac{1}{2} \times 1 , \text{foot} \times 2 , \text{feet} = \frac{1}{2} \times 2 = 1 , \text{square foot} \]
Now, we need to find the actual area of the seal, which is a 1/25 scale of the drawing. To do this, we must account for the scale factor. The scale factor is:
\[ \text{Scale factor} = \frac{1}{25} \]
Since area scales with the square of the linear dimensions, we need to square the scale factor:
\[ \text{Scale factor for area} = \left( \frac{1}{25} \right)^2 = \frac{1}{625} \]
Now, we can find the area of the actual seal by scaling up the area of the drawing:
\[ \text{Area of the seal} = \text{Area of the drawing} \times \text{Scale factor for area} \]
Thus:
\[ \text{Area of the seal} = 1 , \text{square foot} \times 625 = 625 , \text{square feet} \]
The area of the seal is therefore:
\[ \boxed{625} , \text{square feet} \]