图 1 显示了一个等腰梯形棱柱 ABCDEFGH 形状的无盖容器，其正方形底面 ABCD 的边长为 $x$ cm，其中 $x$ 是常数。侧面 ABFE 和 DCGH 垂直于底面 ABCD。然后将该容器沿 BC 倾斜，使得侧面 BCGF 和底面 ABCD 分别与水平面成 30° 角，如图 2 所示。接着向这个空容器中以 $x^3$ cm^3/min 的恒定速率注水。在本题中，假设容器的厚度忽略不计。

1. Find the vertical height to the overflow level

Since the relevant edge of length $x$ makes an angle of $30^\circ$ with the horizontal,
$\text{height}=x\sin 30^\circ=\frac{x}{2}.$

1. Find the area of the triangular cross-section of water

From part (a), the vertical height of the water is $\dfrac{x}{2}$ cm.

In the tilted cross-section, each sloping side makes $30^\circ$ with the horizontal. Since
$x\sin 30^\circ=\frac{x}{2},$
the water surface reaches the two upper edges just before overflow.

The horizontal width of the water surface is
$2x\cos 30^\circ=2x\left(\frac{\sqrt3}{2}\right)=\sqrt3x.$

So the cross-sectional area of water is
$\frac12(\sqrt3x)\left(\frac{x}{2}\right)=\frac{\sqrt3}{4}x^2.$

2. Find the volume of water filled up to this height

The prism has length $BC=x$ cm.

$\text{Volume}=\text{cross-sectional area}\times \text{length}$
$=\frac{\sqrt3}{4}x^2\times x=\frac{\sqrt3}{4}x^3\text{ cm}^3.$

3. Divide volume by filling rate and convert to seconds

Water is filled at a rate of $x^3\text{ cm}^3/\text{min}$.

$\text{Time}=\frac{\frac{\sqrt3}{4}x^3}{x^3}=\frac{\sqrt3}{4}\text{ min}.$

Converting to seconds,
$\frac{\sqrt3}{4}\times 60=25.980\ldots\text{ s}.$

To the nearest second, the time is
$26\text{ s}.$