下图展示了一个形状为梯形柱体（底面为正方形 ABCD）的室内冒险乐园。该室内冒险乐园建在水平地面上，墙面 ADHE 和 BCGF 是倾斜的。 柱体的顶部 EFGH 是冒险乐园的天花板，它也是水平的。EFGH 是一个正方形，且 EFGH 的中心垂直位于 ABCD 的中心正上方。 $AB = 120\ \text{m},\ AE = BF = CG = DH = 10\ \text{m}$ 且 $\angle ABF = 60^\circ$。

1. Find the perpendicular height of trapezium ABFE.

In $\triangle ABF$, $BF=10$ m and $\angle ABF=60^\circ$.

The perpendicular height of ABFE is
$10\sin 60^\circ=8.660\ldots\text{ m}.$

2. Find the horizontal offset from each bottom vertex to the corresponding top vertex.

The horizontal component of $BF$ along $BA$ is
$10\cos 60^\circ=5\text{ m}.$

Since $AE=BF$ and the ceiling square is centred above the base square, the offset at $A$ is also $5$ m.

3. Find the length of the other parallel side EF.

Since there is a $5$ m horizontal offset from each end of $AB$,
$EF=120-5-5=110\text{ m}.$

4. Calculate the area of trapezium ABFE.

$\text{Area of ABFE}=\frac12(AB+EF)(\text{height})$
$=\frac12(120+110)(10\sin60^\circ)$
$=995.929\ldots$
$\approx 996\text{ m}^2.$

1. Find the horizontal offset of each top vertex from the corresponding base vertex.

From part (a), $EF=110\text{ m}$ while $AB=120\text{ m}$. Since the centres of the two squares are vertically aligned, the top square is inset equally on both sides.

$\text{offset on each side}=\frac{120-110}{2}=5\text{ m}$

2. Find the vertical height of H above the ground.

For the slanted edge $BF=10\text{ m}$, the horizontal displacement from $B$ to the point vertically below $F$ is $5\text{ m}$ in each horizontal direction.

Let the vertical height of the ceiling be $h$.

$h^2+5^2+5^2=10^2$

$h^2=100-25-25=50$

$h=\sqrt{50}\text{ m}$

3. Calculate the angle of depression from H to B and compare with the safety limit.

The point vertically below $H$ is $5\text{ m}$ from two adjacent sides of the base square, so its horizontal distance from $B$ is

$\sqrt{115^2+115^2}=115\sqrt{2}\text{ m}$

Let the angle of depression from $H$ to $B$ be $\theta$.

$\tan \theta=\frac{\sqrt{50}}{115\sqrt{2}}$

$\theta=\tan^{-1}\left(\frac{\sqrt{50}}{115\sqrt{2}}\right)$

$\theta\approx 2.49^\circ$

Since $2.49^\circ<5^\circ$, the safety requirement is met.