CD是RT△ABC斜边上AB的高,若AB=1,AC:BC=4:1,则CD的长为?
问题描述：
CD是RT△ABC斜边上AB的高,若AB=1,AC:BC=4:1,则CD的长为?
答
由AC:BC=4:1得AC=4BC
再由AC^2+BC^2=AB^2得(4BC)^2+BC^2=1,BC^2=1/17,AC^2=AB^2-BC^2=16/17
S△ABC=AC*BC/2=AB*CD/2
∴AC*BC=AB*CD,AC^2*BC^2=AB^2*CD^2
(16/17)*(1/17)=CD^2,CD^2=16/17^2
CD=4/17