What is the largest possible area of an isosceles trapezoid in which the largest side is $13$ and the perimeter is $28$?

$ 
\textbf{(A)}\ 13
\qquad\textbf{(B)}\ 24
\qquad\textbf{(C)}\ 27
\qquad\textbf{(D)}\ 28
\qquad\textbf{(E)}\ 30
$
1. **Define the variables and constraints:**
   - Let the lengths of the bases of the isosceles trapezoid be \(a\) and \(b\) with \(a > b\).
   - Let the lengths of the legs be \(c\).
   - Given that the largest side is 13, we have \(a = 13\).
   - The perimeter of the trapezoid is 28, so we have:
     \[
     a + b + 2c = 28
     \]
     Substituting \(a = 13\), we get:
     \[
     13 + b + 2c = 28 \implies b + 2c = 15
     \]

2. **Express \(b\) in terms of \(c\):**
   \[
   b = 15 - 2c
   \]

3. **Calculate the height \(h\) of the trapezoid:**
   - The height \(h\) can be found using the Pythagorean theorem in the right triangle formed by the height, half the difference of the bases, and the leg \(c\).
   - The difference of the bases is \(13 - b = 13 - (15 - 2c) = 2c - 2\).
   - Half of this difference is \(\frac{2c - 2}{2} = c - 1\).
   - Using the Pythagorean theorem:
     \[
     h = \sqrt{c^2 - (c - 1)^2} = \sqrt{c^2 - (c^2 - 2c + 1)} = \sqrt{2c - 1}
     \]

4. **Express the area \(A\) of the trapezoid:**
   - The area \(A\) of a trapezoid is given by:
     \[
     A = \frac{1}{2} \times (a + b) \times h
     \]
   - Substituting \(a = 13\), \(b = 15 - 2c\), and \(h = \sqrt{2c - 1}\):
     \[
     A = \frac{1}{2} \times (13 + 15 - 2c) \times \sqrt{2c - 1} = \frac{1}{2} \times (28 - 2c) \times \sqrt{2c - 1}
     \]
     \[
     A = (14 - c) \times \sqrt{2c - 1}
     \]

5. **Maximize the area \(A\):**
   - To maximize \(A = (14 - c) \sqrt{2c - 1}\), we use the AM-GM inequality:
     \[
     A \leq \sqrt{\left(\frac{14 - c + 14 - c + 2c - 1}{3}\right)^3} = \sqrt{\left(\frac{27}{3}\right)^3} = \sqrt{27^3} = 27
     \]
   - Equality in AM-GM holds when all terms are equal:
     \[
     14 - c = 2c - 1 \implies 14 + 1 = 3c \implies 15 = 3c \implies c = 5
     \]

6. **Verify the values:**
   - If \(c = 5\), then:
     \[
     b = 15 - 2c = 15 - 2 \times 5 = 5
     \]
   - The height \(h\) is:
     \[
     h = \sqrt{2c - 1} = \sqrt{2 \times 5 - 1} = \sqrt{9} = 3
     \]
   - The area \(A\) is:
     \[
     A = (14 - 5) \times 3 = 9 \times 3 = 27
     \]

Thus, the maximum possible area of the isosceles trapezoid is \(\boxed{27}\).