Problem 149

Given that point $P(1, \sqrt{2})$ lies on the graph of the "Pythagorean reciprocal function" $y=\frac{a+b}{cx}$, we can substitute $x=1$ and $y=\sqrt{2}$ into the equation, obtaining:

$$\sqrt{2} = \frac{a+b}{c} \implies a+b = \sqrt{2}c \quad \text{(1)}$$

Given that the area of right triangle $\triangle ABC$ is $4$, and knowing the formula for the area of a right triangle is $\frac{1}{2}ab$, we have:

$$\frac{1}{2}ab = 4 \implies ab = 8 \quad \text{(2)}$$

Since $\triangle ABC$ is a right triangle with $\angle C = 90^{\circ}$, by the Pythagorean theorem, we have:

$$a^2 + b^2 = c^2 \quad \text{(3)}$$

Substituting equation (1) into equation (3) and using equation (2), we get:

$$c^2 = (\sqrt{2}c)^2 - 2ab = 2c^2 - 16 \implies c^2 = 16 \implies c = \pm 4$$

Since $c$ represents a length, we discard $c = -4$ and keep $c = 4$. Substituting $c = 4$ into equation (1), we find:

$$a+b = \sqrt{2} \cdot 4 = 4\sqrt{2}$$

Therefore, the perimeter of $\triangle ABC$, which is $a+b+c$, is:

$$a+b+c = 4\sqrt{2} + 4$$

Hence, the perimeter of $\triangle ABC$ is $\boxed{4\sqrt{2} + 4}$.