3. Given real numbers $a, b$ are constants, the system of inequalities about $x$
$$
\left\{\begin{array}{l}
x+a^{2} \leqslant 2 a, \\
x+6 \geqslant \sqrt{b}
\end{array}\right.
$$

has only 8 integer solutions. Then the value of $a+b$ is ( ).
(A) 1
(B) 0
(C) -1
(D) 2
3.A.

The solution set of the system of inequalities is $\sqrt{b}-6 \leqslant x \leqslant 2 a-a^{2}$. From the problem, we have $\left(2 a-a^{2}\right)-(\sqrt{b}-6)+1 \geqslant 8$, which simplifies to $(a-1)^{2}+\sqrt{b} \leqslant 0$.
Solving this, we get $a=1, b=0$.
It is easy to see that quadrilateral $B C D F$ is a square.
Thus, $B C=B F$
$=D F=6$,
$\angle F B C=90^{\circ}$.
Furthermore, $\angle B F G=\angle C$
$=90^{\circ}, F G=C E$,