Geometry

Using Law of Cosine in $\triangle PQR$ gives

$8^2 = 5^2 + 11^2 - 2(5)(11) \cos \angle Q\\ \cos \angle Q = \dfrac{5^2 + 11^2 - 8^2}{2(5)(11)}\\ \cos \angle Q = \dfrac{41}{55}$

Using Law of Cosine in $\triangle PQM$ gives

$PM^2 = 5^2 + \left(\dfrac{11}2\right)^2 - 2(5)\left(\dfrac{11}2\right) \cos \angle Q\\ PM^2 = 5^2 + \left(\dfrac{11}2\right)^2 - 2(5)\left(\dfrac{11}2\right)\left(\dfrac{41}{55}\right)\\ PM^2 = \dfrac{57}4\\ PM = \dfrac{\sqrt{57}}2$