17-153 The number of non-congruent right triangles whose perimeter (in $\mathrm{cm}$) and area (in $\mathrm{cm}^{2}$) are numerically equal is
(A) none.
(B) 1 .
(C) 2 .
(D) 4 .
(E) infinitely many.
(34th American High School Mathematics Examination, 1983)
[Solution 1] Let $a, b, c$ be the lengths (in $\mathrm{cm}$) of the two legs and the hypotenuse of a right triangle, respectively. Then the ratio of the perimeter (in $\mathrm{cm}$) to the area $\left(\mathrm{cm}^{2}\right)$ is
$$
r=\frac{2(a+b+c)}{a b} .
$$

Now consider a right triangle with side lengths $k a, k b, k c$. The ratio then becomes
$$
\frac{2(k a+k b+k c)}{k a k b}=\frac{1}{k} \cdot \frac{2(a+b+c)}{a b}=\frac{r}{k} \text {. }
$$

By choosing $k=r$, the ratio becomes 1, meaning they are equal.
This shows that the ratio of the perimeter to the area of a right triangle can be arbitrarily changed by substituting a similar triangle to the original right triangle, and there is always one among them whose perimeter (regardless of the unit of length) and area (in the corresponding unit of area) are numerically equal.

Since there are infinitely many non-similar right triangles, there are also infinitely many non-similar right triangles for which their perimeter equals their area.
Therefore, the answer is $(E)$.
[Solution 2] Let $a, b$ be the lengths of the two legs of a right triangle, then the length of the hypotenuse is
$$
c=\sqrt{a^{2}+b^{2}} \text {. }
$$

The perimeter and area of this right triangle are numerically equal if and only if
$$
\frac{1}{2} a b=a+b+\sqrt{a^{2}+b^{2}},
$$

which is an equation with two non-negative real variables. Generally, such an equation has infinitely many solutions.
Therefore, the answer is $(E)$.