Figure 1 shows an open container in the form of an isosceles trapezoidal prism ABCDEFGH with square base ABCD of sides $x$ cm, where $x$ is a constant. The faces ABFE and DCGH are perpendicular to the base ABCD. The container is then tilted along BC such that the face BCGF and the base ABCD make an angle of 30° each with the horizontal surface, as shown in Figure 2. Water is then filled into this empty container at a constant rate of $x^3$ cm^3/min. Assume that the thickness of the container is negligible in this question.

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
$26\text{ s}.$