Let m.n be two points on the radius op of the ball 0 and NP = Mn = OM. Make three circles that are perpendicular to the surface of OP through m.n.o. respectively, then the area ratio of these circles is
Let OP = R, then PN = nm = Mo = R / 3,
Through O, m and N, make the vertical truncated surface of OP respectively, and get the circle O, m and N, whose radii are R, R √ (8 / 9) and R √ (5 / 9), respectively,
Its area ratio is 1: (8 / 9): (5 / 9) = 9:8:5
(Nantong, 2011) as shown in the figure, ⊙ O's chord AB = 8, M is the midpoint of AB, and OM = 3, then the radius of ⊙ o is equal to ()
A. 8B. 4C. 10D. 5
Connecting OA, ∵ m is the midpoint of AB, ∵ om ⊥ AB, and am = 4. In the right angle △ OAM, OA = AM2 + om2 = 5, so D is selected