22 Let the maximum side length of an isosceles trapezoid be 13, and the perimeter be 28.
(1) If the area of the trapezoid is 27, find its side lengths;
(2) Can the area of such a trapezoid be 27.001?
22. As shown in the figure, let $A D$ be the larger base, and $B H$ be the height of the given trapezoid $A B C D$. If $A B=C D=13$, then $A D+B C=2$, and $S_{\text {trapezoid } A B C D}=B H \cdot \frac{A D+B C}{2} \leqslant 13 \cdot \frac{2}{2}=$ $13<27$, which is impossible. Therefore, $A D=13$. Let $A B=x$, then $B C=28-13-2 x=15-2 x$, $A H=x-1, B H=\sqrt{2 x-1}$. By the AM-GM inequality, we get $S_{\text {trapezoid } A B C D}=\sqrt{2 x-1} \cdot \frac{28-2 x}{2}$ $=\sqrt{(2 x-1)(14-x)^{2}} \leqslant \sqrt{\left[\frac{(2 x-1)+(14-x)+(14-x)}{3}\right]^{3}}=27$. Equality holds if and only if $2 x-1=14-x$, i.e., $x=5$, which also means $A B=B C=C D=5$ when $S_{\text {trapezoid } A B C D}=27$, and the equation $S_{\text {trapezoid } A B C D}=27.001$ is impossible to hold.