The problem states the following
The figure shows a small iron sphere hanging from a spoon-like structure by a flexible wire labeled as $\textrm{QH}$. The sphere is pushed in the indicated direction in such a way that the wire always stays taut so that the sphere reaches $\textrm{MN}$. Calculate the length traveled by the sphere. Consider these lengths $\textrm{MN=NP=PQ=9 inch}$.
The figure is attached below:
What I did to tackle this problem was to separate the distance traveled by the hanging sphere into three different arc segments to which I labeled as $\textrm{L}_{1}$, $\textrm{L}_{2}$, $\textrm{L}_{3}$. See figure below of the proposed solution
Taken into consideration I noticed that the angle of all circle segments were $60^\circ$, $30^\circ$ and again $60^\circ$.
However for this to be consistent I took the fact that the gravity acting on the iron sphere would make that part such as a leveling instrument and therefore the angle on $\angle HQ\textrm{wall on the right side}=90^\circ$. From then on I calculated them as this way:
$$\textrm{I define the point where the structure is fixed on the right wall=R}$$
$$\angle PQR = 150^\circ$$ $$\angle PQH= 150^\circ-90^\circ=60^\circ$$
$$\angle HPN= 180^\circ-150^\circ=30^\circ$$ $$\angle HNM= 180^\circ-120^\circ=60^\circ$$
With these angles in mind I proceeded to calculate the lengths by using the formula $L=\theta\times R$
The first radius is calculated directly as is the same as the length of the wire thus is $21\,\textrm{inches}$.
$$R_{1}=21$$
The second radius is calculated as the difference between the 9 inches of the right wall which is on the back of the spoon and the total 21, hence the acting radius will be:
$$R_{2}=21-9=12$$
The third radius same as the previous one, will be the difference between those acting 12 inches and the other 9 inches from the base of the spoon becoming:
$$R_{3}=12-9=3$$
The the last step will be just replacing those ones with the angles already found as (taken into consideration transforming the sexagesimal angles in radians):
$$L_{1}=\theta \times R=(60^\circ\times\frac{\pi}{180^{\circ}})\times 21=7\pi$$
$$L_{2}=\theta \times R=(30^\circ\times\frac{\pi}{180^{\circ}})\times 12=2\pi$$
$$L_{3}=\theta \times R=(60^\circ\times\frac{\pi}{180^{\circ}})\times 3=1\pi$$
Hence the total distance would be the sum of each distance:
$$\textrm{Distance traveled}=7\pi + 2\pi + \pi= 10 \pi$$
Would this approach be okay?. I'm not very sure if my assumptions are correct as I'm just assuming that the structure is leveled with the ground and forming parallel lines.