Let's form a right-angled triangle - with legs $a$ and $b$, and hypotenuse $c$ - by rotating around the lines passing through the vertices and parallel to the opposite sides. Calculate the surface area and volume of the 3 resulting solids of revolution. How do the surface areas and volumes of the 3 solids compare to each other when $a=b$?
Let's denote the surface areas of the resulting solids of revolution as $F_{a}, F_{b}, F_{c}$, and their volumes as $K_{a}, K_{b}, K_{c}$, respectively.

a) When the right-angled triangle is rotated around an axis passing through point $A$ and parallel to the leg $a$, the leg $a$ forms a cylindrical surface, the leg $b$ forms a circular base, and the hypotenuse $c$ forms a conical surface.

Thus, $F_{a}=2 b \pi \cdot a + b^{2} \pi + b c \pi = b \pi (2 a + b + c)$.

The volume of the resulting solid of revolution is the difference between a cylinder and a cone. The cone is $\frac{1}{3}$ of the cylinder, so the sought volume of the solid of revolution is $\frac{2}{3}$ of the cylinder's volume, i.e.,

$$
K_{a}=\frac{2}{3} b^{2} \pi \cdot a = \frac{2}{3} a b^{2} \pi
$$

b) Similarly, we obtain

$$
F_{b}=2 a \pi \cdot b + a^{2} \pi + a c \pi = a \pi (a + 2 b + c)
$$

and

$$
K_{b}=\frac{2}{3} a^{2} b \pi
$$

c) When the triangle is rotated around an axis passing through point $C$ and parallel to the hypotenuse $c$, the legs $a$ and $b$ form conical surfaces, and the hypotenuse $c$ forms a cylindrical surface, with the radius of the base of the cones and the cylinder being the height $m$ to the hypotenuse. Thus,

$$
F_{c}=a m \pi + b m \pi + 2 m \pi \cdot c = m \pi (a + b + 2 c)
$$

Since the area of our right-angled triangle is $\frac{1}{2} c m = \frac{1}{2} a b$, we have $m = \frac{a b}{c}$, and thus

$$
F_{c}=\frac{a b \pi}{c} (a + b + 2 c)
$$

The volume $K_{c}$ is again $\frac{2}{3}$ of the volume of the cylinder formed by the side $c$, i.e.,

$$
\begin{gathered}
K_{c}=\frac{2}{3} m^{2} \pi \cdot c = \frac{2}{3} \cdot \frac{a^{2} b^{2}}{c^{2}} \cdot \pi c = \frac{2}{3} \cdot \frac{a^{2} b^{2} \pi}{c} \\
\text{If } a=b, \text{ then } c=a \sqrt{2}, \text{ and thus} \\
F_{a}=F_{b}=a \pi (3 a + a \sqrt{2}) = a^{2} \pi (3 + \sqrt{2})
\end{gathered}
$$

and

$$
F_{c}=\frac{a^{2} \pi}{a \sqrt{2}} (2 a + 2 a \sqrt{2}) = 2 a^{2} \pi \frac{1 + \sqrt{2}}{\sqrt{2}} = a^{2} \pi (2 + \sqrt{2})
$$

Thus,

$$
F_{a}: F_{b}: F_{c}=1: 1: \frac{2 + \sqrt{2}}{3 + \sqrt{2}}
$$

But

$$
\frac{2 + \sqrt{2}}{3 + \sqrt{2}}=\frac{(2 + \sqrt{2})(3 - \sqrt{2})}{9 - 2}=\frac{4 + \sqrt{2}}{7}
$$

and thus

$$
F_{a}: F_{b}: F_{c}=1: 1: \frac{4 + \sqrt{2}}{7} (\sim 0.7735)
$$

Similarly, if $a=b$,

$$
K_{a}=K_{b}=\frac{2}{3} a^{3} \pi
$$

and

$$
K_{c}=\frac{2}{3} \frac{a^{4} \pi}{a \sqrt{2}} = \frac{\sqrt{2}}{3} a^{3} \cdot \pi
$$

Thus,

$$
K_{a}: K_{b}: K_{c}=1: 1: \frac{\sqrt{2}}{2} (\sim 0.7071)
$$