Figure 1 shows an open container in the form of an isosceles trapezoidal prism $ABCDEFGH$ with square base $ABCD$ of sides $x\text{ cm}$, 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 $30^\circ$ each with the horizontal surface, as shown in Figure 2. Water is then filled into this empty container at a constant rate of $x^3\text{ cm}^3/\text{min}$. Assume that the thickness of the container is negligible in this question.

1. Identify the lowest point of the container's opening from which water will overflow.

The container is open at the top face $EFGH$. When tilted, the water will overflow from the lowest point on the rim $EFGH$. Since the container is tilted along $BC$, the edge $FG$ is horizontal and lower than $EH$. Thus, water overflows over the edge $FG$.

2. Calculate the vertical height of the edge $FG$ above the horizontal surface.

The face $BCGF$ makes an angle of $30^\circ$ with the horizontal. The length of $BF$ is $x$ (since $ABFE$ is an isosceles trapezoid and $AE=x$, by symmetry $BF=x$).

The height of $F$ (and $G$) above the horizontal surface at $B$ is given by:
$\text{Height} = BF \sin 30^\circ$
$\text{Height} = x \left(\frac{1}{2}\right) = 0.5x$

1. Find the cross-sectional area of the water (triangle ABF)

Just before overflowing, the horizontal water surface passes through $A$ and $F$, both at height $\tfrac{x}{2}$ (from part (a)). The water cross-section is the isosceles triangle $ABF$ with $AB = BF = x$ and included angle $\angle ABF = 30^\circ+30^\circ = 120^\circ$.

Using the surface $AF$ as base with height $\tfrac{x}{2}$, where $AF = 2\bigl(x\cos 30^\circ\bigr) = x\sqrt{3}$:
$\text{Area} = \tfrac{1}{2}\times AF \times \text{height} = \tfrac{1}{2}\times x\sqrt{3}\times \tfrac{x}{2} = \frac{\sqrt{3}}{4}x^2$
(Equivalently $\tfrac{1}{2}x\cdot x\sin120^\circ = \tfrac{\sqrt3}{4}x^2$.)

2. Find the volume of water in the container

The container is a prism of depth $BC = x$, so the volume of water is the cross-sectional area multiplied by the depth:
$V = \frac{\sqrt{3}}{4}x^2 \times x = \frac{\sqrt{3}}{4}x^3 \text{ cm}^3$

3. Find the filling time and convert to seconds

Water is added at a constant rate of $x^3\text{ cm}^3/\text{min}$, so
$\text{Time} = \frac{V}{\text{rate}} = \frac{\tfrac{\sqrt{3}}{4}x^3}{x^3} = \frac{\sqrt{3}}{4}\text{ min}$
Converting to seconds:
$\frac{\sqrt{3}}{4}\times 60 = 15\sqrt{3} \approx 25.98\text{ s} \approx 26\text{ s}$