一道数学题...急已知正五边形ABCDE的边心距为r,P为正五边形内任一点,过P向各边作高PF,PG,PM,PN,PQ,求证:PF+PG+PM+DN+PQ=5r.急...)求证:PF+PG+PM+PN+PQ=5r,上面的不对

Question

一道数学题...急
已知正五边形ABCDE的边心距为r,P为正五边形内任一点,过P向各边作高PF,PG,PM,PN,PQ,求证:PF+PG+PM+DN+PQ=5r.急...)
求证:PF+PG+PM+PN+PQ=5r,上面的不对

Answer 1

连接PA,PB,PC,PD,PE,
设正五边形的边长为a,则
S正五边形ABCDE＝S△PAB+S△PBC+S△PCD+S△PDE+S△PEA,
又∵正多边形的面积等于1/2周长×边心距,
∴1/2×5a×r=1/2×a×PF+1/2×a×PG+1/2×a×PM+1/2×a×PN+1/2×a×PQ,
左右两边同时约去1/2a,得：
5r＝PF+PG+PM+PN+PQ,
即PF+PG+PM+PN+PQ＝5r．