1. Set up coordinates

Let the square floor lie in the plane $z=0$ with side length $10$ m. Choose coordinates:

$A(0,0,0),\quad B(10,0,0),\quad C(10,10,0),\quad D(0,10,0)$

Since the walls are vertical and the height of the hall is $h$, the corresponding top vertices are:

$E(0,0,h),\ F(10,0,h),\ G(10,10,h),\ H(0,10,h)$

We need the angle between diagonals $AG$ and $BH$, i.e. between vectors:

$\vec{AG}=G-A=(10,10,h)$ $\vec{BH}=H-B=(-10,10,h)$

2. Use the dot product formula

Given:

$\cos \angle GPH = \frac{1}{5}$

From the geometry, this is the angle between the diagonals $AG$ and $BH$. So,

$\cos \theta = \frac{\vec{AG}\cdot\vec{BH}}{|\vec{AG}|\,|\vec{BH}|}$

Compute the dot product:

$\vec{AG}\cdot\vec{BH}=(10)(-10)+(10)(10)+h\cdot h$ $=-100+100+h^2=h^2$

Now compute magnitudes:

$|\vec{AG}|=\sqrt{10^2+10^2+h^2}=\sqrt{200+h^2}$ $|\vec{BH}|=\sqrt{(-10)^2+10^2+h^2}=\sqrt{200+h^2}$

Hence,

$\cos\theta=\frac{h^2}{200+h^2}$

Given $\cos\theta=\frac{1}{5}$, so

$\frac{h^2}{200+h^2}=\frac{1}{5}$

3. Solve for $h$

Cross-multiplying:

$5h^2=200+h^2$ $4h^2=200$ $h^2=50$ $h=\sqrt{50}=5\sqrt{2}$

Since height is positive,

$\boxed{h=5\sqrt{2}\text{ m}}$

4. Check options

• A: $5$ ✗
• B: $2\sqrt{10}$ ✗
• C: $5\sqrt{3}$ ✗
• D: $5\sqrt{2}$ ✓

Therefore, the correct option is D.