已知线段AB=m,在直线AB上取一点P,恰好使PB分之AP=n,点Q为PB的中点,求线段AQ的长.A.------------------.B
网友回答
P Q A--------------------B PB=X,AP=nX,PQ=X/2,AQ=AP+PQ=nX+X/2
AP+PB=AB=m=(n+1)X,X=m/(n+1)
AQ=mn/(n+1)+m/2(n+1)
======以下答案可供参考======
供参考答案1:
)∵AB=3, =2,
∴AP=3× =4,
PB=AB-AP=6-4=2,
∵点Q为PB的中点,
∴PQ= PB=1,
∴AQ=AP+PQ=4+1=5.