圆锥底面半径为1,高为二倍根号二,轴截面积为PAB,从A点垃一绳子绕远追侧面一周回到A点,求最短绳长

由题得：OB=r=1 ,  PO=2倍根号2
    如图：在Rt△POB中,由勾股定理：PB²=PO²+OB²=(2倍根号2)²+1²=9
                    ∴ PB=3  ,
                     ∴在扇形图中,R=P'A'=3
                       ∵弧ABA'的长=圆锥的底面周长=2πr=2π*1=2π
                        根据扇形的弧长公式：L=(2πRn)/360°
                                             ∴2π=(2π*3n)/360°
                                             ∴n=120°  即,∠AP'A'=120°
                      根据两点之间的直线为最短,连接AA', 则最短绳长=AA'
                           如图：  过点P‘作P'B⊥AA’交AA'于D'
                              则,∠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(根号3)
                            ∴AA'=3倍根号3  ,  PD=P'D'=3/2
              所以,最短绳长必经过圆锥的母线PB的中点D,这时,最短的绳长为：3倍根号3