18.25 For a convex quadrilateral $A B C D$, extend the sides $A B, B C, C D$, and $D A$ to points $B^{\prime}, C^{\prime}, D^{\prime}$, and $A^{\prime}$, respectively. Given that $A B=B B^{\prime}=6, B C=C C^{\prime}=7$, $C D=D D^{\prime}=8$, and $D A=A A^{\prime}=9$. If the area of $A B C D$ is 10, then the area of $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is
(A) 20.
(B) 40.
(C) 45.
(D) 50.
(E) 60.
(29th American High School Mathematics Examination, 1978)
[Solution] Since the base $A A^{\prime}$ of $\triangle A A^{\prime} B^{\prime}$ is the same length as the base $A D$ of $\triangle A B D$, and the corresponding height of $\triangle A A^{\prime} B^{\prime}$ is twice that of $\triangle A B D$, then
$$
S_{\triangle A A^{\prime} B^{\prime}}=2 S_{\triangle A B D} .
$$

Similarly,
$$
S_{\triangle B B^{\prime} C^{\prime}}=2 S_{\triangle B A C} \text {, }
$$

and
$$
S_{\triangle C C^{\prime} D^{\prime}}=2 S_{\triangle C B D},
$$

and
$$
S_{\triangle D D^{\prime} A^{\prime}}=2 S_{\triangle D C A} \text {. }
$$
$$
\begin{array}{l}
\therefore S_{A^{\prime} B^{\prime} C^{\prime} D^{\prime}} \\
=\left(S_{\triangle A A^{\prime} B^{\prime}}+S_{\triangle B B^{\prime} C^{\prime}}\right)+\left(S_{\triangle O C^{\prime} D^{\prime}}+S_{\triangle D D^{\prime} A^{\prime}}\right)+S_{A B C D} \\
=2\left(S_{\triangle A B D}+S_{\triangle B A C}\right)+2\left(S_{\triangle C B D}+S_{\triangle D C A}\right)+S_{A B C D} \\
=5 S_{A B C D} \\
=50 .
\end{array}
$$

Therefore, the answer is $(D)$.