18.81 A square with a perimeter of 20 is inscribed in a square with a perimeter of 28. The maximum distance between a vertex of the inner square and a vertex of the outer square is 
(A) $\sqrt{58}$.
(B) $\frac{7 \sqrt{5}}{2}$.
(C) 8 .
(D) $\sqrt{65}$.
(E) $5 \sqrt{3}$.
(44th American High School Mathematics Examination, 1993)
［Solution］ The inscribed square intersects each side of the larger outer square at one point. Let these points divide the sides into segments of lengths $x$ and $y$.
Thus, $x+y=7, \quad x^{2}+y^{2}=25, \quad x \geqslant y$.
Solving these equations gives $x=4, y=3$. Therefore, the maximum distance from a vertex of the inner square to a vertex of the outer square is
$$
\begin{aligned}
A B & =\sqrt{x^{2}+(x+y)^{2}}=\sqrt{4^{2}+7^{2}} \\
& =\sqrt{65} .
\end{aligned}
$$

Hence, the answer is $(D)$.