13. (3 points) There is a rectangle, the lengths of whose sides are natural numbers less than 10. If the width is used as the numerator and the length as the denominator, then the value of the resulting fraction is greater than $\frac{3}{10}$ and less than $\frac{1}{2}$. The sum of the areas of all such rectangles is . $\qquad$
【Solution】Solution: According to the problem, we know that $\frac{3}{10}<\frac{\text { width }}{\text { length }}<\frac{1}{2}$, after transformation we get: $2 \times$ width $<$ length $<\frac{10}{3} \times$ width,
Therefore: (1) If width $=1$, then $2<$ length $<10 / 3$, length $=3$;
(2) If width $=2$, then $4<$ length $<20 / 3$, length $=5$ or 6;
(3) If width $=3$, then $6<$ length $<10$, length $=7$ or 8 or 9;
(4) If width $=4$, then $8<$ length $<10<40 / 3$, length $=9$.

Therefore, the sum of the areas of all rectangles that meet the conditions is $1 \times 3+2 \times 5+2 \times 6+3 \times 7+3 \times 8+3 \times 9+4 \times 9=133$.