20. Two isosceles triangles $T$ and $T'$ are not congruent, but they have equal perimeters and equal areas. The side lengths of triangle $T$ are $5, 5, 8$, and the side lengths of triangle $T'$ are $a, a, b$. Then the number ( ) that is closest to $b$ is:
(A) 3
(B) 4
(C) 5
(D) 6
(E) 8
20. A.

From the problem, we know that the perimeter of triangle $T$ is 18, and the area is 12, i.e., $\square$
$$
\begin{array}{l}
\left\{\begin{array}{l}
2 a+b=18, \\
\frac{1}{2} b \sqrt{a^{2}-\left(\frac{b}{2}\right)^{2}}=12
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
2 a+b=18, \\
b \sqrt{4 a^{2}-b^{2}}=48
\end{array}\right. \\
\Rightarrow b \sqrt{(18-b)^{2}-b^{2}}=48 \\
\Rightarrow b \sqrt{324-36 b}=48 \\
\Rightarrow b^{3}-9 b^{2}+64=0 \\
\Rightarrow(b-8)\left(b^{2}-b-8\right)=0 .
\end{array}
$$

Since $b>0$, we have $b=8$ or $\frac{1+\sqrt{33}}{2}$. When $b=8$, triangles $T$ and $T^{\prime}$ are congruent (discard this case). Therefore, $b=\frac{1+\sqrt{33}}{2}<\frac{1+\sqrt{36}}{2}=3.5$.