In a quadrilateral pyramid $S A B C D$, the base is a trapezoid $A B C D (B C \| A D), B C=\frac{4}{5} A D, \angle A S D=\angle$ $C D S=\frac{\pi}{2}$. All vertices of the pyramid lie on the circles of the bases of a cylinder, the height of which is 2, and the radius of the base is $\frac{\frac{5}{3}}{3}$. Find the volume of the pyramid.
If vertices $A$ and $D$ are located on different bases of the cylinder, then by the theorem of the intersection of two parallel planes by a third plane, $A B \| C D$, i.e., $A B C D$ is a parallelogram, not a trapezoid. Let vertices $A$ and $D$ be located on the circumference of one base of the cylinder. Then the median $S M$ of the right triangle $A S D$ is equal to half the hypotenuse $A D$, and thus is less than the radius of the base circle of the cylinder, i.e., $S M < \frac{5}{3}$. If $S M > \frac{5}{3}$, then point $S$ cannot lie on the circumference of the other base. Therefore, in this case, the right triangle $A S D$ is inscribed in the base circle of the cylinder, and then $A D$ is the diameter of the circle. In this case, points $B$ and $C$ lie on the circumference of the other base of the cylinder. Let $B_1$ and $C_1$ be the orthogonal projections of points $B$ and $C$ onto the plane of the base of the cylinder containing points $A, D$, and $S$.

Since $C D \perp S D$, by the theorem of three perpendiculars, $C_1 D \perp S D$, and since $A S \perp S D$, then $C_1 D \| A S$, so $A S D C_1$ is a rectangle. Since $B_1 C_1 \| B C$, the quadrilateral $A B_1 C_1 D$ is an isosceles trapezoid, so $A B_1 = D C_1 = S A$, which means the diameter $A D$ is perpendicular to the chord $S B_1$ and passes through its midpoint $K$. The segment $S C_1$ is the diagonal of the rectangle $A S D C_1$, so $S C_1$ is the diameter of the base circle of the cylinder, $S C_1 = A D = \frac{10}{3}$, and since

$$
B_1 C_1 = B C = \frac{4}{5} A D = \frac{4}{5} \cdot \frac{10}{3} = \frac{8}{3}
$$

then

$$
S K = \frac{1}{2} S B_1 = \frac{1}{2} \sqrt{S C_1^2 - B_1 C_1^2} = \frac{1}{2} \sqrt{\left(\frac{10}{3}\right)^2 - \left(\frac{8}{3}\right)^2} = 1
$$

By the theorem of three perpendiculars, $B K \perp A D$, i.e., $B K$ is the height of the trapezoid $A B C D$. From the right triangle $B K B_1$, we find that

$$
B K = \sqrt{B B_1^2 + K B_1^2} = \sqrt{4 + 1} = \sqrt{5}
$$

Therefore,

$$
S_{A B C D} = \frac{1}{2} (A D + B C) \cdot B K = \frac{1}{2} \left(\frac{10}{3} + \frac{8}{3}\right) \cdot \sqrt{5} = 3 \sqrt{5}
$$

Let $H$ be the foot of the perpendicular dropped from point $S$ to the line $B K$. Then $S H$ is the height of the pyramid $S A B C D$, since $S H \perp B K$ and $S H \perp A D$. Writing the area of triangle $B K S$ in two ways, we get

$$
S H = \frac{S K \cdot B B_1}{B K} = \frac{1 \cdot 2}{\sqrt{5}} = \frac{2}{\sqrt{5}}
$$

Therefore,

$$
V_{S A B C D} = \frac{1}{3} S_{A B C D} \cdot S H = \frac{1}{3} \cdot 3 \sqrt{5} \cdot \frac{2}{\sqrt{5}} = 2
$$

## Answer

2.00