2. A water channel with a length of $5 \mathrm{~m}$ can hold $1,440 \mathrm{~m}^{3}$ of water. The cross-section of the channel is an isosceles trapezoid with legs of $52 \mathrm{~cm}$ and a height of $48 \mathrm{~cm}$.

How much water can the channel hold if it is filled to half its height?
Solution. In the diagram, the trapezoid $ABCD$ represents the cross-section of the channel, $BB_{2} \| AD, B_{1}$ is the midpoint of $BB_{2}, B_{1}C_{1} \| AB$ and $C_{2}$ is the foot of the perpendicular dropped from $B$. Since $\triangle BCB_{2}$ is isosceles, point $C_{2}$ is the midpoint of $CB_{2}$. Therefore, $B_{1}C_{1} \| BC$ and

$$
\overline{CC_{2}}=\sqrt{\overline{CB}^{2}-{\overline{C_{2}B}}^{2}}=\sqrt{0.52^{2}-0.48^{2}}=0.2 \text{ m}
$$

Let $V$ be the volume of water when the channel is filled to half its height, and $V_{1}=1.440 \text{ m}^{3}$ when it is full. Since trapezoids $AB C_{1} D_{1}$ and $D_{1} B_{1} C_{2} D$ are congruent, $V_{1}=2V+P \cdot 5$ where $P$ is the area of parallelogram $B_{1} C_{1} C C_{2}$:

$$
P=\overline{CC_{2}} \cdot \frac{\overline{C_{2}B}}{2}=0.2 \cdot 0.24=0.048 \text{ m}^{2}
$$

Therefore,

$$
V=\frac{1.44-0.048 \cdot 5}{2}=0.6 \text{ m}^{3}
$$