$19 \cdot 41$ As shown in the figure, $\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+$ $\angle 7$ is
(A) $450^{\circ}$.
(B) $540^{\circ}$.
(C) $630^{\circ}$.
(D) $720^{\circ}$.
(Tianjin Middle School Mathematics Competition, 1983; Anhui Middle School Mathematics Competition, 1997)
[Solution 1] As shown in the figure (let $\angle 8, \angle 9, \angle 10$ )
$$
\left\{\begin{array}{l}
\angle 5+\angle 6+\angle 7+\angle 8=360^{\circ}, \\
\angle 2+\angle 3+\angle 4+\left(180^{\circ}-\angle 9\right)=360^{\circ}, \\
\angle 9=\angle 10, \\
\angle 8=\angle 10+\angle 1 .
\end{array}\right.
$$

From (3) and (4) substituted into (1), we have
$$
\angle 5+\angle 6+\angle 7+\angle 9+\angle 1=360^{\circ},
$$
(2) + (5) gives
$$
\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+180^{\circ}=720^{\circ},
$$

which means $\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7=540^{\circ}$.
Therefore, the answer is $(B)$.
[Solution 2] Consider the extreme case (for easier calculation or when the angles formed are special angles, note that the lengths of the segments are not limited) as shown in the right figure, then the sum of the required angles is
$$
3 \cdot 60^{\circ}+4 \cdot 90^{\circ}=540^{\circ}.
$$

Therefore, the answer is $(B)$.