One side of a triangle has length $75$.  Of the other two sides, the length of one is double the length of the other.  What is the maximum possible area for this triangle
1. Let the sides of the triangle be \( a = 75 \), \( b = x \), and \( c = 2x \).
2. Using Heron's formula, the area \( A \) of a triangle with sides \( a, b, c \) is given by:
   \[
   A = \sqrt{s(s-a)(s-b)(s-c)}
   \]
   where \( s \) is the semi-perimeter:
   \[
   s = \frac{a + b + c}{2} = \frac{75 + x + 2x}{2} = \frac{75 + 3x}{2}
   \]
3. Substituting \( s \) into Heron's formula:
   \[
   A = \sqrt{\left(\frac{75 + 3x}{2}\right) \left(\frac{75 + 3x}{2} - 75\right) \left(\frac{75 + 3x}{2} - x\right) \left(\frac{75 + 3x}{2} - 2x\right)}
   \]
4. Simplifying the terms inside the square root:
   \[
   s - a = \frac{75 + 3x}{2} - 75 = \frac{3x - 75}{2}
   \]
   \[
   s - b = \frac{75 + 3x}{2} - x = \frac{75 + x}{2}
   \]
   \[
   s - c = \frac{75 + 3x}{2} - 2x = \frac{75 - x}{2}
   \]
5. Substituting these back into the area formula:
   \[
   A = \sqrt{\left(\frac{75 + 3x}{2}\right) \left(\frac{3x - 75}{2}\right) \left(\frac{75 + x}{2}\right) \left(\frac{75 - x}{2}\right)}
   \]
6. Simplifying further:
   \[
   A = \sqrt{\frac{(75 + 3x)(3x - 75)(75 + x)(75 - x)}{16}}
   \]
   \[
   A = \frac{1}{4} \sqrt{(75 + 3x)(3x - 75)(75 + x)(75 - x)}
   \]
7. Notice that:
   \[
   (75 + x)(75 - x) = 75^2 - x^2
   \]
   \[
   (75 + 3x)(3x - 75) = 9x^2 - 75^2
   \]
8. Therefore:
   \[
   A = \frac{1}{4} \sqrt{(9x^2 - 75^2)(75^2 - x^2)}
   \]
9. To maximize \( A \), we need to maximize \( \sqrt{(9x^2 - 75^2)(75^2 - x^2)} \). Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
   \[
   a + b \ge 2\sqrt{ab}
   \]
   Let \( a = 9x^2 - 75^2 \) and \( b = 75^2 - x^2 \). Then:
   \[
   9x^2 - 75^2 + 75^2 - x^2 \ge 2\sqrt{(9x^2 - 75^2)(75^2 - x^2)}
   \]
   \[
   8x^2 \ge 2\sqrt{(9x^2 - 75^2)(75^2 - x^2)}
   \]
   \[
   4x^2 \ge \sqrt{(9x^2 - 75^2)(75^2 - x^2)}
   \]
   \[
   16x^4 \ge (9x^2 - 75^2)(75^2 - x^2)
   \]
10. The maximum area occurs when equality holds in the AM-GM inequality, i.e., when:
    \[
    4x^2 = \sqrt{(9x^2 - 75^2)(75^2 - x^2)}
    \]
    Squaring both sides:
    \[
    16x^4 = (9x^2 - 75^2)(75^2 - x^2)
    \]
11. Solving for \( x \):
    \[
    16x^4 = 9x^2 \cdot 75^2 - 9x^4 - 75^2 \cdot x^2 + 75^4
    \]
    \[
    25x^4 = 75^2 \cdot x^2
    \]
    \[
    x^2 = \frac{75^2}{25} = 225
    \]
    \[
    x = 15
    \]
12. Substituting \( x = 15 \) back into the area formula:
    \[
    A = \frac{1}{4} \sqrt{(9 \cdot 15^2 - 75^2)(75^2 - 15^2)}
    \]
    \[
    A = \frac{1}{4} \sqrt{(2025 - 5625)(5625 - 225)}
    \]
    \[
    A = \frac{1}{4} \sqrt{(-3600)(5400)}
    \]
    \[
    A = \frac{1}{4} \sqrt{19440000}
    \]
    \[
    A = \frac{1}{4} \cdot 4400 = 1100
    \]

The final answer is \(\boxed{1125}\)