题目 #24044

过点作OOC ⊥ AB于点连接,OA, C 则AC = BC =\(\frac{1}{2}\)AB = 4, 在RtOAC中,OC = =\(\sqrt{5^{2}−4^{2}} = 3, \frac{\sqrt{OA^{2}−AC^{2}}}{OP}\) 为半径时最长, 为垂线段最短,\(\frac{OP}{OP}\) 的取值范围为3  OP  5 设圆弧所在圆的圆心为，连结O， ，\(\frac{OA}{OA} \frac{OA}{OP}\)如图所示设半径为 ()则OA =  = = x(m)\(\frac{xm}{AM}\)由垂径定理可知 = BM，A N = B N  = 60，\(\frac{AB}{AM} \frac{m}{m}\) = 30，且OM = OP − PM = (x − m 18)_{2} 在Rt AOM中由勾股定理可得, = OM + AM\(2 \frac{AO^{2}}{34}\)即 = (x − +，解得x =\(fracx^2\)ON frac18)^2PN \(frac30^2\)34 = OP − = − 4 = 30(m) 在 A ON中，由勾股定理可 得  =\(\sqrt{(OA^{2}−ON^{2})} = \sqrt{(34^{2}−30^{2})} = 16(m) \frac{A N}{A B}\)  = 32m  30m 不需要采取紧急措施.