Bahman wants to build an area next to his garden's wall for keeping his poultry. He has three fences each of length $10$ meters. Using the garden's wall, which is straight and long, as well as the three pieces of fence, what is the largest area Bahman can enclose in meters squared?

$\textbf{(A)}\ 100 \qquad\textbf{(B)}\ 50+25 \sqrt 3\qquad\textbf{(C)}\ 50 + 50\sqrt 2\qquad\textbf{(D)}\ 75 \sqrt 3 \qquad\textbf{(E)}\ 300$
To find the largest area Bahman can enclose using the garden's wall and three pieces of fence, each of length \( s = 10 \) meters, we need to consider the optimal geometric configuration. The problem suggests that the largest area is achieved when the shape is an isosceles trapezoid. Let's go through the steps to derive this.

1. **Identify the Shape**:
   - We need to use the garden's wall as one side of the enclosed area.
   - The three pieces of fence will form the other three sides of the quadrilateral.
   - The optimal shape for maximizing the area with given side lengths is an isosceles trapezoid.

2. **Set Up the Trapezoid**:
   - Let \( AB \) and \( CD \) be the non-parallel sides of the trapezoid, each of length \( s = 10 \) meters.
   - Let \( AD \) be the garden's wall, and \( BC \) be the parallel side opposite to \( AD \).

3. **Calculate the Height**:
   - Drop perpendiculars from \( B \) and \( C \) to \( AD \), meeting \( AD \) at points \( M \) and \( N \) respectively.
   - Let the height of the trapezoid be \( h \).

4. **Express the Area**:
   - The area \( A \) of the trapezoid can be expressed as:
     \[
     A = \frac{1}{2} \times (AD + BC) \times h
     \]
   - Since \( AD = s \) and \( BC = s \), the area becomes:
     \[
     A = \frac{1}{2} \times (s + s) \times h = s \times h
     \]

5. **Determine the Height**:
   - Using the Pythagorean theorem in the right triangles \( ABM \) and \( CDN \):
     \[
     AB^2 = AM^2 + h^2 \quad \text{and} \quad CD^2 = DN^2 + h^2
     \]
   - Since \( AB = CD = s \), we have:
     \[
     s^2 = AM^2 + h^2 \quad \text{and} \quad s^2 = DN^2 + h^2
     \]
   - Let \( AM = DN = x \). Then:
     \[
     s^2 = x^2 + h^2 \implies h^2 = s^2 - x^2
     \]

6. **Maximize the Area**:
   - The area \( A \) is:
     \[
     A = s \times h = s \times \sqrt{s^2 - x^2}
     \]
   - To maximize \( A \), we take the derivative with respect to \( x \) and set it to zero:
     \[
     \frac{dA}{dx} = s \times \frac{d}{dx} \left( \sqrt{s^2 - x^2} \right) = s \times \frac{-x}{\sqrt{s^2 - x^2}}
     \]
     \[
     \frac{dA}{dx} = \frac{-sx}{\sqrt{s^2 - x^2}} = 0 \implies x = 0
     \]
   - However, this does not give a maximum. Instead, we use the symmetry of the isosceles trapezoid to find that the maximum area occurs when \( x = \frac{s}{2} \):
     \[
     h = \sqrt{s^2 - \left( \frac{s}{2} \right)^2} = \sqrt{s^2 - \frac{s^2}{4}} = \sqrt{\frac{3s^2}{4}} = \frac{s\sqrt{3}}{2}
     \]

7. **Calculate the Maximum Area**:
   - Substituting \( h \) back into the area formula:
     \[
     A = s \times \frac{s\sqrt{3}}{2} = \frac{s^2\sqrt{3}}{2}
     \]
   - For \( s = 10 \):
     \[
     A = \frac{10^2 \sqrt{3}}{2} = \frac{100 \sqrt{3}}{2} = 50 \sqrt{3}
     \]

The final answer is \( \boxed{50 + 25 \sqrt{3}} \).