tiet 39 luyen tap

KiÓm tra bµi cò

TiÕt 39: LuyÖn TËp

Bµi 1 : (Bµi 60 - T133 SGK)
Cho tam gi¸c nhän ABC kÎ AH vu«ng gãc víi BC
(H ∈BC). BiÕt AB = 13 cm, AH = 12 cm,
HC = 16 cm. TÝnh c¸c ®é dµi AC, BC.
A
B
C
H
13
12
16
GT
∆ABC nhän
AH ⊥ BC (H ∈BC)
AB = 13cm, AH = 12 cm
HC = 16 cm.
AC = ?
BC = ?
KL

§¸p ¸n:
* TÝnh AC:
V× AH ⊥ BC t¹i H nªn ∆AHC vu«ng t¹i H.
=> AC
2
= AH

2
+ HC
2
(§/l Pitago)
AC
2
= 12
2
+ 16
2

AC
2
= 144 + 256 = 400 = 20
2
AC = 20 (cm)
A
B
C
H
13
12
16
GT
∆ABC nhän
AH ⊥ BC (H ∈BC)
AB= 13cm, AH = 12 cm
HC = 16 cm.
AC = ? , BC = ?
KL

* TÝnh BC:
V× AH ⊥ BC t¹i H nªn ∆AHB vu«ng t¹i H.
=> AB
2
= AH
2
+ HB
2
(§/l Pitago)
13
2
= 12
2
+ HB
2
HB
2
= 13
2
-12
2
= 169 - 144 = 25 = 5
2
HB = 5 (cm).
V× ∆ABC nhän nªn H n»m gi÷a B vµ C.
=> BC = BH + HC = 5 + 16 = 21 (cm)
A
B
C

H
13
12
16
GT
∆ABC nhän
AH ⊥ BC (H ∈BC)
AB= 13cm, AH = 12 cm
HC = 16 cm.
AC = ? , BC = ?
KL