Let $ABC$ be a triangle such that midpoints of three altitudes are collinear. If the largest side of triangle is $10$, what is the largest possible area of the triangle?

$ 
\textbf{(A)}\ 20
\qquad\textbf{(B)}\ 25
\qquad\textbf{(C)}\ 30 
\qquad\textbf{(D)}\ 40
\qquad\textbf{(E)}\ 50
$
1. **Identify the largest side and midpoints:**
   Let \( BC = 10 \) be the largest side of the triangle \( \triangle ABC \). Let \( M_a \), \( M_b \), and \( M_c \) be the midpoints of the sides \( a = BC \), \( b = AC \), and \( c = AB \), respectively. Let \( H_a \), \( H_b \), and \( H_c \) be the midpoints of the altitudes from vertices \( A \), \( B \), and \( C \), respectively.

2. **Consider the collinearity of midpoints of altitudes:**
   The problem states that the midpoints of the altitudes are collinear. This implies that the triangle \( \triangle ABC \) must be a right triangle. This is because, in a right triangle, the midpoints of the altitudes are collinear (they lie on the nine-point circle's diameter).

3. **Determine the configuration of the right triangle:**
   Assume \( \triangle ABC \) is a right triangle with \( \angle C = 90^\circ \). Thus, \( BC \) is the hypotenuse, and \( AB \) and \( AC \) are the legs of the triangle.

4. **Calculate the area of the right triangle:**
   The area of a right triangle is given by:
   \[
   \text{Area} = \frac{1}{2} \times \text{leg}_1 \times \text{leg}_2
   \]
   Let \( AB = x \) and \( AC = y \). Since \( BC = 10 \) is the hypotenuse, by the Pythagorean theorem:
   \[
   x^2 + y^2 = 10^2 = 100
   \]

5. **Maximize the area:**
   To maximize the area, we need to maximize the product \( x \times y \). Using the AM-GM inequality:
   \[
   \frac{x^2 + y^2}{2} \geq \sqrt{x^2 y^2}
   \]
   \[
   \frac{100}{2} \geq \sqrt{x^2 y^2}
   \]
   \[
   50 \geq \sqrt{x^2 y^2}
   \]
   \[
   2500 \geq x^2 y^2
   \]
   \[
   50 \geq xy
   \]
   The maximum value of \( xy \) is 50 when \( x = y \). However, since \( x^2 + y^2 = 100 \), the maximum product \( xy \) occurs when \( x = y = \sqrt{50} \).

6. **Calculate the maximum area:**
   The maximum area of the triangle is:
   \[
   \text{Area} = \frac{1}{2} \times x \times y = \frac{1}{2} \times \sqrt{50} \times \sqrt{50} = \frac{1}{2} \times 50 = 25
   \]

Conclusion:
The largest possible area of the triangle is \( \boxed{25} \).