如图1,在△ABC中,∠C=90°,AC=3,BC=4,CD是斜边AB上的高,点E在斜边AB上,过点E作直线与△ABC的直角

2025-05-04 19:45:19
推荐回答(1个)
回答1:

(1)∵△ABC中,∠C=90°,AC=3,BC=4,
∴AB=

32+42
=5,
∵CD⊥AB,
∴∠CDA=∠ACB,
又∠CAD=∠CAD,
∴Rt△ADC∽Rt△ACB,
AD
AC
=
AC
AB
,即
AD
3
=
3
5
,AD=
9
5


(2)①由于E的位置不能确定,故应分两种情况讨论:
如图A:当0<x≤AD,即0<x≤
9
5
时,
∵EF⊥AB,
∴Rt△AEF∽Rt△ACB,即
AE
AC
=
EF
BC

∵AC=3,BC=4,AE=x,
x
3
=
EF
4
,EF=
4
3
x,
S△AEF=y=
1
2
AE?EF=
1
2
x?
4
3
x=
2
3
x2
如图B:当AD<x≤AB,即
9
5
<x≤5时,
∵EF⊥AB,
∴Rt△BEF∽Rt△BCA,
EB
BC
=
EF
AC

∵AE=x,△AEF的面积为y,
5?x
4
=
EF
3

∴EF=
15?3x
4

y=
1
2
×AE×EF=
1
2
x?