The diagram below shows an indoor adventure park in the shape of a trapezoidal prism with a square base ABCD. The indoor adventure park is positioned on horizontal ground and the walls ADHE and BCGF are slanted. The top of the prism, EFGH, is the ceiling of the adventure park which is also horizontal. EFGH is a square and the centre of EFGH lies vertically above the centre of ABCD. $AB = 120\ \text{m},\ AE = BF = CG = DH = 10\ \text{m}$ and $\angle ABF = 60^\circ$.

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.