由題得：OB=r=1 ,  PO=2倍根號2
    如圖：在Rt△POB中,由勾股定理：PB²=PO²+OB²=(2倍根號2)²+1²=9
                    ∴ PB=3  ,
                     ∴在扇形圖中,R=P'A'=3
                       ∵弧ABA'的長=圓錐的底面周長=2πr=2π*1=2π
                        根據(jù)扇形的弧長公式：L=(2πRn)/360°
                                             ∴2π=(2π*3n)/360°
                                             ∴n=120°  即,∠AP'A'=120°
                      根據(jù)兩點之間的直線為最短,連接AA', 則最短繩長=AA'
                           如圖：  過點P‘作P'B⊥AA’交AA'于D'
                              則,∠D'P'A'=60°
                          在Rt△P'D'A'中,∵∠D'A'P'=30°
                             ∴ P'D'=P'A'/2=3/2 ,  D'A'=3/2(根號3)
                            ∴AA'=3倍根號3  ,  PD=P'D'=3/2
              所以,最短繩長必經(jīng)過圓錐的母線PB的中點D,這時,最短的繩長為：3倍根號3