F5 maths about circle

Let the mid point of PQ be m1 and the mid point of RS is m2 ,Om1 = Om2 , since PQ = RS .PO^2 = (Om1)^2 + (Pm1)^213^2 = (Om1)^2 + (24/2)^225 = (Om1)^2Om1 = 5 , also Om2 = 5 Consider O m1 A m2 is a rectangle , therefore m1 A = O m2 = 5 ,We have OA^2 = (Om1)^2 + (m1 A)^2OA^2 = 5^2 + 5^2OA^2 = 50OA = √50 = 5√2 cmAnswer is (B)