The width of the space is\(w = 8\;{\rm{m}}\).

The relation of force on the pointed arch and the round arch is \({F'_{\rm{H}}} = \frac{1}{3}{F_{\rm{H}}}\).

In this problem, the load force is kept on the same horizontal location on the round and pointed arch, and for each half arch, consider torque about the lower right-hand corner.

The following is the free body diagram of the round arch.

The following is the free body diagram of the pointed arch.

The relation to calculate the net torque on the round arch can be written as:

\(\begin{array}{c}\sum \tau = 0\\\left( {{F_{\rm{L}}} \times R - x} \right) - \left( {{F_{\rm{H}}} \times R} \right) = 0\\{F_{\rm{H}}} = {F_{\rm{L}}}\left( {\frac{{R - x}}{R}} \right)\end{array}\)

Here, \({F_{\rm{L}}}\) is the force due to the load, \({F_{\rm{H}}}\) is the force in the round arch, R is the radius, and \(x\) is the distance from the end point to the load.

The relation to calculate the net torque on the pointed arch can be written as:

\(\begin{array}{c}\sum \tau = 0\\\left( {{F_{\rm{L}}} \times \left( {R - x} \right)} \right) - \left( {{{F'}_{\rm{H}}} \times y} \right) = 0\\{{F'}_{\rm{H}}} = {F_{\rm{L}}}\left( {\frac{{R - x}}{y}} \right)\end{array}\)

Here, \({F'_{\rm{H}}}\) is the force in the pointed arch and \(y\) is the height of the arch.

The relation to calculate the height can be written as:

\({F'_{\rm{H}}} = \frac{1}{3}{F_{\rm{H}}}\)

On plugging the values in the above relation, you get:

\(\begin{array}{c}{F_{\rm{L}}}\left( {\frac{{R - x}}{y}} \right) = \frac{1}{3}{F_{\rm{L}}}\left( {\frac{{R - x}}{R}} \right)\\3\left( {\frac{{R - x}}{y}} \right) = \left( {\frac{{R - x}}{R}} \right)\\y = 3R\\y = 3\left( {\frac{{8\;{\rm{m}}}}{2}} \right)\\y = 12\,{\rm{m}}\end{array}\)

Thus, \(y = 12\;{\rm{m}}\) is the height of the pointed arch.