17・159 As shown in the figure, extend each side of $\triangle ABC$ by 2 times to $A^{\prime}, B^{\prime}, C^{\prime}$. Then the area of $\triangle A^{\prime} B^{\prime} C^{\prime}$ is 
(A) 13 times.
(B) 4 times.
(C) 19 times.
(D) 7 times.
[Solution] Connect $B C^{\prime}$, and let $S_{\triangle A B C}=S$. By the formula for the area of a triangle and the given conditions, we have
$$
S_{\triangle B C C^{\prime}}=2 S_{\triangle A B C}=2 S.
$$

Similarly, $S_{\triangle C^{\prime} B^{\prime} B}=2 S_{\triangle B C C^{\prime}}=4 S$.
Then $S_{\triangle O C^{\prime} B^{\prime}}=S_{\triangle B C C^{\prime}}+S_{\triangle C^{\prime} B^{\prime} B}=6 S$.
Similarly, by the same method, we have $S_{\triangle A A^{\prime} C^{\prime}}=S_{\triangle B B^{\prime} A^{\prime}}=$
$6 S$.
Thus,
$$
S_{\triangle A^{\prime} B^{\prime} C^{\prime}}=3 \cdot 6 S+S=19 S.
$$

Therefore, the answer is $(C)$.