Geometry - pls help i don't understand how to do it

Note that PX bisects $\angle QPR$. Then, we can suppose $\angle XPQ = \angle XPR = \theta$.

Recall the formula:

If two sides of a triangle are of length a and b respectively, and the angle included by these two sides is $\phi$, then the area of the triangle is $\dfrac12 ab \sin \phi$.

Now, draw the diagram and look at triangle PMY. Its height is MY (which is 8) and its base is PM (which is 18). So the area of triangle PMY is $\dfrac12 \cdot 8 \cdot 18 = 72$. But on the other hand, using the formula stated above, the same area is also equal to $\dfrac12 (PM)(PY) \sin \theta = 9\cdot PY \sin \theta$.

Therefore, we know that $PY \sin \theta = 8$ by comparing.

Now, we look at triangle PYR. By the formula, the area is $\dfrac12 (PY)(PR) \sin \theta$, but we know PR = 22. Using $PY \sin \theta = 8$, the required area is $\dfrac12 (22) (PY \sin \theta) = 11(8) = \boxed{88}$.

I managed to solve it myself and got the same thing!!