Analysis: As shown in the figure, extend OK intersection MF to point M ′, PQ intersection BC to point G, intersection FN to point N ′, and let the radius of the circular hole be R. In Rt△KBG, according to Pythagorean theorem, R = 16 (cm) According to the meaning of the question, if the center O is on the diagonal of the rectangle EFGH, then KN ′ = 65438+. om′= km′+r = 1/2cb = 65cm。 Then, according to the sum-difference relation of relevant line segments in the figure, CN = QG-QN ′ = 44-26 =18 (cm), AM = BC-PD-km ′ =130.

Hope to adopt, thank you.