5. A four-digit number $\overline{a a b b}$ is a perfect square. Then the value of $a+b$ is ( ).
(A) 11
(B) 10
(C) 9
(D) 8
5.(A).

Since $\overline{a a b b}=1000 a+100 a+10 b+b$
$$
=11(100 a+b),
$$

By the problem, we can set $100 a+b=11 c^{2}$ (where $c$ is a positive integer),
Therefore, $101<100 a+b=11 c^{2}<999$,
That is, $9<c^{2}<90$.
Thus, $4 \leqslant c \leqslant 9$.
Upon inspection, when $c=8$, the condition is satisfied, at this time $a=7, b=4$. Hence $a+b=11$.