Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter.  The sides of $T$ have lengths $5$, $5$, and $8$, while those of $T'$ have lengths $a$, $a$, and $b$.  Which of the following numbers is closest to $b$?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }8$
The area of $T$ is $\dfrac{1}{2} \cdot 8 \cdot 3 = 12$ and the perimeter is 18.
The area of $T'$ is $\dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}$ and the perimeter is $2a + b$.
Thus $2a + b = 18$, so $2a = 18 - b$.
Thus $12 = \dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}$, so $48 = b \sqrt{4a^2 - b^2} = b \sqrt{(18 - b)^2 - b^2} = b \sqrt{324 - 36b}$.
We square and divide 36 from both sides to obtain $64 = b^2 (9 - b)$, so $b^3 - 9b^2 + 64 = 0$. Since we know $b = 8$ is a solution, we divide by $b - 8$ to get the other solution. Thus, $b^2 - b - 8 = 0$, so $b = \dfrac{1 + \sqrt{33}}{2} < \dfrac{1 + 6}{2} = 3.5.$ The answer is $\boxed{\textbf{(A) }3}$.

Solution 1.1
The area is $12$, the semiperimeter is $9$, and $a = 9 - \frac12b$. Using Heron's formula, $\sqrt{9\left(\frac{b}{2}\right)\left(\frac{b}{2}\right)(9-b)} = 12$. Squaring both sides and simplifying, we have $-b^3+9b^2-64=0$. Since we know $b = 8$ is a solution, we divide by $b - 8$ to get the other solution. Thus, $b^2 - b - 8 = 0$, so $b = \dfrac{1 + \sqrt{33}}{2} < \dfrac{1 + 6}{2} = 3.5.$ The answer is $\boxed{\textbf{(A) }3}$.