5. Let $a, b$ be positive integers, and satisfy $56 \leqslant a+b \leqslant 59, 0.9<\frac{a}{b}<0.91$. Then $b^{2}-a^{2}$ equals ( ).
(A) 171
(B) 177
(C) 180
(D) 182
5.B.

From the given, we have
$$
0.9 b+b56 \text {, }
$$

then $29<b<32$. Therefore, $b=30,31$.
When $b=30$, from $0.9 b<a<0.91 b$, we get $27<a<28$, there is no such positive integer $a$;

When $b=31$, from $0.9 b<a<0.91 b$, we get $27<a<29$, so $a=28$.
Thus, $b^{2}-a^{2}=177$.