Học và ôn luyện theo cấu trúc đề thi môn toán phần 2

?H4N II

HIJVHHOC

1 9 2 k'': TS, Vu The Huu - Nguygn Vinh Can


mm i I. tiJHG THIG VA HJT PHIB TRDNE KHONG emii
B A I T O A N 1 . QUAN H E SONG SONG
I. KIEN THLfC QUAN TRQNG
1.

Dvfcfng t h ^ n g s o n g s o n g vofi diiofng t h a n g
a) Dinh nghia :
- H a i dudng t h a n g goi l a cheo nhau neu chung k h o n g cung n a m t r o n g
mot m a t phang.
- H a i di/ofng t h a n g goi l a song song neu chung dong phSng va k h o n g c6
diem chung.
b) Cdc tinh chat :
Tinh chat 1
Trong k h o n g gian, qua mot diem ngoai mot diicfng t h a n g c6 m o t va
chi mot diiorng t h a n g song song vdi dudng thSng do.
Tinh chat 2
Hai diTofng t h a n g p h a n b i e t cung song song vdfi m o t di/dng t h a n g thuf
ba t h i song song v d i nhau.
Dinh li (ve giao tuyen ciia ba m a t phang)
Neu ba m a t phang cSt nhau theo ba giao tuyen p h a n b i e t t h i ba giao
tuyen ay hoac dong quy hoac doi mot song song.
He qua
Neu hai m a t phang p h a n b i e t I a n lirgt d i qua h a i du'&ng t h a n g song
song t h i giao tuyen ciia chiing (neu c6) song song v d i h a i diTdng t h a n g

do (hoac t r i j n g vdi mot t r o n g hai diTdng thSng doj.

2.

Dvfcfng t h ^ n g s o n g s o n g vdfi m a t p h ^ n g
a) Dinh nghia :
M o t diidng thSng va mot m a t phSng goi l a song song v d i nhau neu
chiing k h o n g c6 d i e m chung.
b) Dieu kien de mot dudng thang song song vdi mot mat phang :
Dinh li 1
Neu diidng thang a song song vdi mot dudng thang b nao do n a m t r e n
mot mat phang (P) khong chufa a t h i a song song vdi m a t phang (P;.
Dinh li 2
Neu mot dddng t h a n g song song vdi mot m a t phang t h i no song song
vdi mot di/dng t h a n g nao do n a m t r o n g m a t phang.

HQC va on luyen theo CTDT mon Toan THPT

193


3.

He qua 1
N e u diTcfng t h a n g a s o n g s o n g v d i m o t m a t p h d n g ( P ) t h i m o i m a t
p h a n g ( Q ) chiifa a m a c a t ( P ) t h i cSt ( P ) t h e o g i a o t u y e n s o n g s o n g v d i a.
He qua 2
N e u h a i m a t p h S n g p h a n b i e t c u n g s o n g s o n g v d i m o t difofng t h a n g
t h i g i a o t u y e n c i i a c h i i n g ( n e u c6) c u n g s o n g s o n g v d i diTdng t h a n g do.
H a i m a t p h a n g song song

a) Dinh

nghia :

H a i m a t p h a n g g o i l a s o n g s o n g k h i c h u n g k h o n g c6 d i e m c h u n g .
b ) Dieu
Dinh

kien de hai mat phdng

song song :

li :

N e u m o t m a t p h a n g ( P ) chijfa h a i d i i d n g t h a n g a v a b c a t n h a u va
cung song song v d i m a t p h a n g (Q) t h i (P) song song v d i (Q).
c) Tinh
Tinh

chat :
chat 1

Q u a m o t d i e m n g o a i m o t m a t p h a n g c6 m o t v a c h i m o t m a t p h a n g
song song v d i m a t p h a n g do.
He qua 1
N e u d i i d n g t h a n g a s o n g s o n g v d i m a t p h a n g ( Q ) t h i q u a a cd m o t v a
chi m o t m a t p h a n g (P) song song v d i m a t p h a n g (Q).
He qua 2
H a i m a t p h d n g p h a n b i e t c i i n g s o n g s o n g v d i m o t m a t p h a n g t h d ba
t h i song song v d i nhau.

Tinh

chat 2

N e u h a i m a t p h a n g ( P ) v a ( Q ) s o n g s o n g t h i m o i m a t p h a n g ( R ) d a cat
(P) t h i p h a i c a t ( Q ) v a cac giao t u y e n ciia c h u n g s o n g song.
d)

Dinh
Dinh

li Ta-let

trong khong

gian :

li thuan

Ba m a t p h a n g d o i m o t song song chan r a t r e n h a i cat tuyen b a t k i
cac d o a n t h a n g ti/cfng t i n g t i l e .
Dinh

li ddo

G i a sijf t r e n h a i d i i d n g t h a n g a v a a' I a n l i i d t l a y h a i bo d i e m ( A , B , C)
.
^,
,
v a ( A , B , C ) sao c h o

AB

=

A'B'

BC

=

B'C

CA
CA'

K h i do b a d u d n g t h a n g A A ' , B B ' , C C c t i n g s o n g s o n g v d i m o t m a t
phang.
e) Hinh
Binh

Idng tru va hinh hop :
nghla

hinh Idng tru

H i n h l a n g t r u l a h i n h c6 h a i d a y l a h a i d a g i a c n a m t r o n g h a i m a t
1 9 4 j - ' l TS. Vu Th6' Hau - NguySn VTnh CJn

phSng song song. T a t ca cac canh k h o n g thuoc h a i day deu song song
vdri nhau.
H i n h l a n g t r u c6 :
- Cac canh ben b&ng nhau.
-

Cac m a t ben l a cac h i n h h i n h h a n h .

-

H a i day l a h a i da giac bang nhau.
Dinh nghia hinh hop
H i n h l a n g t r u c6 day l a h i n h b i n h h a n h diTcfc goi l a h i n h hop.
H i n h hop c6 sau m a t deu l a h i n h b i n h h a n h , cac m a t d o i dien t h i
song song vdri nhau.
Cac diJcfng cheo ciia h i n h hop cdt nhau t a i t r u n g d i e m ciia m o i di/dfng.

II. CAC DANG BAI TAP VA PHl/dNG PHAP GIAt
1.

C a c d a n g b a i t^p
a) Cac bai t a p t r o n g muc nay, chii yeu l a chijfng m i n h h a i dudng t h a n g
cheo nhau, h a i di/dng t h a n g song song, diicJng t h a n g va m a t phSng
song song, h a i m a t phang song song. De g i a i cac b a i t a p n a y t a
thifdng suf dung cac t i n h chat ciia quan he song song va t r o n g n h i e u
t r i i d n g hgfp t a suf dung phiCang phdp chiing minh phdn chiJcng.
b) Ngoai r a cac t i n h chat ciia h i n h l a n g t r u , h i n h hop cung giup t a g i a i
cac b a i t a p ve quan he song song (cac canh ciia h i n h l a n g t r u , h i n h
hop, cac m a t ben ciia h i n h hop v.v...).

^

Bait^ip

1.

Cho bon d i e m k h o n g dong phang A , B, C, D. Chufng m i n h r a n g cac
du'cfng t h a n g sau l a cac cap dudng t h a n g cheo nhau :
A B va C D
A D va BC
AC va B D .
Chi dan : Suf dung phi/cfng phap p h a n chufng.

2.

Cho h i n h chop tuf giac S.ABCD. Goi M , N , P theo thuf t\i l a t r u n g
diem ciia cac canh SA, SB, SC.
a) Chufng m i n h mp ( M N P ) // m p (ABCD).
b) Chufng m i n h giao d i e m ciia m p ( M N P ) v d i SD l a t r u n g d i e m Q cua
doan t h a n g SD.
c) Goi O l a giao diem ciia A C va B D , O' l a giao d i e m cua MQ va N P .
Chufng m i n h ba d i e m S, O, O' thSng hang.
Chi dan : Doc gia tir g i a i .

3.

Cho h i n h chop tuf giac S.ABCD; day A B C D l a h i n h vuong. G o i M , N
theo thuf tir la t r u n g d i e m ciia h a i canh SB, SD.

Hpc va on luyen theo CTDT mon Toan THPT

195


4.

^^^^^^^^^

a) Chirng m i n h M N // (ABCD).
b) Neu each xac d i n h giao d i e m P cua m p ( A M N ) vdfi canh SC.
e) Xac d i n h giao t u y e n eiia m p (ABCD) \6i mp ( A M N ) .
C h i dSn : Doc gia tii g i a i .
Cho h i n h hop ABCD.A'B'C'D';

O la giao diem eiia AC va B D ; O' la

giao d i e m eiia A ' C va B D ' .
a) Chufng m i n h m p (AB'D') // mp(C'BD).
b) Chijfng m i n h A O / / C O ' .
C h i dan : Doc gia tiT g i a i .
5 . Cho h i n h hop ABCD.A'B'C'D': goi O, O' theo thiif tiT la giao diem cua AC
v d i B D va ciia A ' C v d i B'D'
a) Chufng m i n h mp(AB'D') // mp(C'BD)
b) Chijfng m i n h A'O // CO'
C h i dan : Doc gia t\i g i a i .

BAITOAN 2. QUAN HE VUONG GOC
NH0NG BAI TOAN VE KHOANG CACH

1.

K h o a n g e a c h tijf m p t d i e m d e n mpt m a t p h a n g
Cho m a t p h a n g (P) va d i e m O; H la
h i n h chieu vuong goc ciia O t r e n (P).
K h o a n g each tii d i e m O den m a t
p h a n g (P) l a dp d a i doan t h a n g O H .

/p\

2.

d(d, (P)) = d ( 0 e d, (P)).
196

tJ.; TS, VQ The Huu - Nguygn VTnh Can


3.

K h o a n g e a c h givia h a i dxXdng t h a n g c h e o n h a u
Khoang each

giCifa

h a i difcfng thSng

cheo nhau l a do d a i doan vuong goe
ehung eiia h a i di/ofng t h ^ n g .
Tinh

chat

: K h o a n g each

giijfa

hai

diidng t h a n g cheo nhau a; b t h i bSng
khoang each tCr m o t d i e m A e a den
mot m a t p h a n g (P) qua b va song
song vdri difcfng t h a n g a.
Chuy

:

a) Ngoai ba

t r U c f n g hcrp

t r e n day, t a con c6 k h o a n g each giiJa h a i

diTcrng

t h a n g song song, k h o a n g each giiJa h a i m a t phang song song. Cac
kien

thufc

n a y dofn g i a n va eung i t gap t r o n g cac b a i t o a n n e n t a

k h o n g nhae l a i of day.
b) De dung mat phang (P) qua b va song song vdfi a, ta l a m n h i i sau :
Lay mot diem M e b. Q u a M diing difofng t h a n g a' // a. M a t phang xae
dinh bofi hai dUcfng t h i n g b va a ' chinh la mat phang (P).

II. CAC BAI TAP VA PHl/dNG PHAP GlAi
Lj

C a c dang bai tap

a) Bai tap ve tim khoang

each tit mot diem den mot mat

phang

De t i m khoang each ttr diem O den mat phang (P), triTdrc het ta t i m h i n h
chieu H eiia O t r e n (P), sau do t i n h do dai doan thang O H .
Bai toan nay lien quan den v a n de difofng t h a n g vuong goe vdfi m a t
phang va k h i t i n h O H , thiTcfng t a phai sijf dung den cac k i e n thiife ve he
thufc Itfofng trong t a m giac, n h a t la he thijfe lUgfng trong t a m giae vuong.
b) Bai tap ve khoang each til mot diCang thang den mot mat phang song song
Ta t i n h khoang each gi-iifa ducfng t h a n g A den m o t m a t phang (P) song
song vdi A bang each t i m k h o a n g each tiT m o t diem M G A d e n m a t
phang (P). Dieu quan t r o n g t r o n g viee g i a i b a i toan nay l a of cho chon
diem M e A m o t each t h i c h hgp, giup t a suf dung het cac gia t h i e t ciia
bai toan
t i n h duoe d ( M , (P)).
e) Bai toan ve khoang each giita hai duang thdng cheo nhau

De t i m k h o a n g each giOfa h a i diTcfng t h a n g cheo nhau a; b , t a thiidfng
l a m nhif sau :
Hoc va on luyen theo CTDT mon Toan THPT , 1 9 7


T i m mot diem M thich hop thupc a va t i m khoang each tir M den (P). Diem
M can chon la diem c6 Hen he vcfi cac gia thiet ciia bai toan de giup t a van
dung dirge cac c o n g thijfc, he thijfc liTpfng trong cac tam giac.

-

T i m m o t m a t p h a n g (P) chufa a va song
song song v d i a). M a t p h a n g (P) thiicfng
dirofng t h a n g cAt a va song song v 6 i b.

-

song
diTgrc

v d i b (hoac chijfa b va
xac d i n h b o i a va mot

Cho h i n h chop t a m giac S.ABC, h a i m a t phang (SAB) va (SAC) vuong

6.

C a c b a i tap

2.

goc vdfi m a t phang day (ABC).
a) N e u each xac d i n h h i n h chieu H cua dinh A t r e n mat phang (SBC).
b) B i e t A B = 13cm, B C = 14em, CA = 15cm va SA = 16em. T i n h khoang
each tiT d i n h A den m a t phang (SBC).
CHI DAN
(SAB) 1 (ABC)
a) Ta

CO

: (SAC) 1 (ABC)

^

SA ± (ABC)

(SAB) n (SAC) = SA
T r o n g t a m giac A B C , k e diiomg cao A M
A M ± BC.
Tir

SA 1 (ABC)
AM ±BC

S M 1 BC

( D i n h li 3 diTotng v u o n g g o c ) .
T r o n g t a m g i a c S A M ke A H 1 S M
BC 1 S M

• BC 1 (SAM)
Tir

(1)

BC ± A M

BC ± ( S A M ) '
A H cz (SAM)J
TCr (1) va (2) t a diroc

• A H ± BC

(2)

AHXBC

A H 1 (SBC)

AH I S M

=> H l a h i n h chieu cua A t r e n m a t phang (SBC),
b) Trirdtc h e t , t a t i n h d i e n t i c h t a m giac A B C theo cong

thuTc

H e r o n , t a c6

: 2p = 42 => p = 2 1
suy r a p - a = 8; p - b = 7; p - c = 6

SABC =

V21.8.7.6 = 84

Tir day t a eo A M =

2S.
^ABC
BC

_ 19

198 ei TS. Vu The Hi;u - Nguyen VTnh CJn


Tarn giac S A M vuong t a i A v a A H l a di/dng cao thuoc canh huyen.
Theo he thiic li/gfng t r o n g tarn giac vuong t a c6 :
1

1

AH^

1
SA^

=> A H ^ =

M A ^
^

400

1

1

A H ^

16^

A H = 9,6

+

1

1

12^

256

TT =

1

+

144

(cm).

Chiiy :
1. Ta CO the t i n h A H theo each khac.
Trirdrc het, t a t i n h S M tiT t a m giac vuong S A M .
S M ' = S A ' + A M ' = > S M = 20

T i n h A H t\X hai t a m giac vuong dong dang :
.c^TTA

A T

A S H A CO A S A M = >

AH

A H
SA

=

A M
S M

.^^
=^AH =

AM.SA
S M

=1^=9,6
20

T i n h A H tiT dien t i c h t a m giac S A M .
2SsAM = A H . S M
2SsAM = S A . A M => A H . S M = S A . A M =^ A H =

A M

OA

SM

= 9,6.

2. Ta cung c6 t h e dirng d i e m H theo each l i luan sau :
Do SA 1 (ABC) =^ SA 1 BC. Dirng qua SA m o t m a t p h a n g (P) vuong
goc vdri BC; mSt p h a n g nay cat BC t a i M . T r o n g m a t p h a n g (P), t a ke
AH

1

SM t h i v i (SAM) ± (SBC) va (SAM) n (SBC) = SM ma A H 1 A M

nen A H 1 (SBC) hay H l a h i n h chieu ciia A t r e n m a t p h a n g (SBC).
7.

Cho h i n h chop S.ABC; h a i m a t p h a n g (SAC) va (SAB) vuong goc v6i
m a t day (ABC) va SA = a V 2 . T i n h k h o a n g each tCr d i n h A den m a t

phang (SBC) t r o n g cac t r u d n g hop :
a) Day ABC l a t a m giac deu canh a.
b) Day ABC l a t a m giac can d i n h A, goc A - 1 2 0 ° va A B = A C = a.
c) Day ABC l a t a m giac vuong t a i B; A C = 5a, BC = 4 a .

CHI D A N
a) T a CO SA 1 (ABC). Goi M l a t r u n g d i e m ciia canh BC. T r o n g t a m giac
S A M ke A H ± S M t h i A H 1 (SBC)
T r o n g t a m giac S A M t h i

AH'

AS' A M '
Hgc va on luyen theo CTDT mon Toan THPT

199


8.

AH''

AH =

(aV2)' •

+

iV66

11

b) Goi M la t r u n g d i e m ciia BC.
T r o n g t a m giac S A M ke A H ± S M
=^ A H 1 (SBC)
T r o n g t a m giac S A M t h i
1
AH^

1

1

AS'

AM'

vdi AS - aV2 va A M = AB.coseO" =

CO

2

l^f2

ta t i n h duoc A H =
c) Ta

: SA ± (ABC)
BC ± (SAB)

T r o n g t a m giac SAB, ke A H 1 SB
^ A H ± (SBC) va t a cung c6
1
AH'

1
1
AS' + AB'

De t h a y A B ' = 9 a '
AH =

6aVl3
13

Cho h i n h chop S.ABCD, chieu cao bSng a\/3, hai m a t phang (SAC) va
(SBD) vuong goc v d i m a t p h a n g day (ABCD); day A B C D la h i n h thoi,
canh a, goc n h o n A = 60". T i n h khoang each giOfa h a i dUcfng t h a n g
A D va SB.

CHI DAN
Ta CO : Goi O la giao diem ciia hai diTdng cheo AC va BC ciia h i n h
thoi A B C D :
200

,.'. TS. Vu The Hi/u - Nguy§n VTnh Can


(SAC) 1 (ABCD)

(SBD) ± (ABCD)
(SAC) n (SBD) = SO
o

SO 1 (ABCD) =:> SO = aVs

BC // A D => m a t p h a n g (SBC)
la m a t p h a n g chijfa (SB) va
song song vdfi A D .

^

^

Qua SO ta difng mot mat phang vuong goc vofi canh BC, mat phang nay cat
AD d P va cat BC d Q. Ttr P ke P K 1 SQ t h i

P K 1 (SBC)

do d6d(AD, SB) = d(AD, (SBC)) = d(P, (SBC)) = P K
TCr O ke O H 1 SQ =o O H 1 (SBC)
De t h a y PQ = AB.sin60° =

IN/3

Trong t a m giac vuong SOQ t h i

OH'

^aV3^'

=> O H // P K

(aVS)^

OQ-

OH^

OQ*-^

OH^ =

3a^
17

+ •

OS'
OH =

17

Trong t a m giac P Q K t h i O H l a diTcfng t r u n g b i n h nen
PK = 2 0 H =

aVsi

17
Chii y : Co t h e t i n h O H t i ^ h a i t a m giac vuong dong d a n g SOQ va

SHO.
OH
SO
SO.OQ
ASOQ
ASHO
OH =
OQ
SQ
SQ
vdi
9.

SQ-" = SO' + O Q l

Cho tuf dien A B C D . H a i m a t b e n A B C va DBC n a m t r o n g h a i m a t
phang hop \6i nhau m o t goc 60°. M a t ben A B C l a m o t t a m giac deu
con m a t ben DBC l a m o t t a m giac vuong can, d i n h D. B i e t D B = a.
Goi M l a t r u n g d i e m cua canh BC.
1. T i n h canh A D .
2. T i n h khoang each tCr d i n h A den m a t phang (DBC) va k h o a n g each tCf
d i n h D den m a t phang (ABC).
3. T i n h khoang each giufa difdng t h a n g A D va diTcfng t h a n g BC, k h o a n g
each giuTa h a i diidng t h a n g AC va D M .
H Q C va on luyen theo CTDT mon Toan THPT.'.'; 201


CHI DAN
1. M l a t r u n g d i e m cua BC t h i
A M 1 BC

D M ± BC

=> A M D - 60°

T a m giac D B C can vuong t a i D

#D

va D B = a, cho t a BC = a^f2
^ D M =

2

^

T a m giac A B C deu, canh aV2 => A M =
A p dung d i n h l i cosin vao t a m giac A M D :
AD^ = A M ' + D M '

2AM.DM.C0SAMD

AD =

-^8-2^.

2

2. BC 1 ( A M D )

(DBC) ± (AMD). Trong t a m giac A M D , ke A H 1 M D

==> A H 1 (DBC) hay d(A, (DBC)) = A H
T r o n g t a m giac vuong A H M , A M =
A H = AM.sin60° => A H =
T a cung c6 BC 1 ( A M D ) ^
DK 1 A M ^

D K 1 (ABC)

3a

; A M H = 60°

V2

(ABC) 1 ( A M D ) . T r o n g t a m giac A M D ke
D K = d(D, (ABC))

T r o n g t a m giac vuong D K M , D M =
D K = DM.sin60° ^

1V2

va A M D = 60°

DK =

3. Goi J la giao diem cua A H va D K t h i M J ± A D t a i I hay M I 1 A D .
Ta l a i c6 BC ± ( A M D ) ma M I

c:

(AMD)

M I ± BC

=> M I la doan vuong goc chung ciia A D va BC.
A A I M c/5 A A K D

KD

AM

AD

IM

IM =

AM.AD
KD

=> I M = a V 8 - 2 V 3 .
T r o n g m a t p h a n g (DBC), tir C t a ke

Cx // D M t h i m a t phang

(A, Cx) l a m a t p h a n g chufa A C va song song vdi D M , do do t a c6
d(AC, D M ) = d ( D M , (A, Cx))
Ta lay m o t d i e m t u y y t r e n D M , diem H chang h a n t h i v i H G D M nen

d(AC, D M ) = d ( D M , (A, Cx)) = d ( H , (A, Cx))

202

IS Vu The Hiju - Nguyin VTnh Can


Tii H t a ke H K 1 Cx, do A H 1 (DBC)
=> theo d i n h h' 3 difofng vuong goc t a
suy r a A K 1 Cx va Cx 1 ( A H K ) , cho
ta (AHK) 1 (A, Cx). Til day, suy ra r k n g
neu ta ke H P 1 A K t h i H P ± (A, Cx)
hay H P la k h o a n g each ti^ H de'n m a t
phang (A, Cx).
Tong hop cac t h o n g t i n t r e n , ta c6 :
d ( D M , AC) = d ( D M , (A, Cx)) = d ( H , (A, Cx)) = H P
1
1
1
T r o n g t a m giac vuong A H K t h i
HP2 H A ' HK^
D i thay H K = M C =
Suy r a H P =

10.

a-j2

v-a H A =

3a V2

3a V5

10
Chii y : V i A H i. (DBC) =^ A H 1 Cx, do do t a c6 the xac d i n h m a t p h a n g
( A H K ) bSng each diing qua A H m o t m a t phSng vuong goc v d i Cx, m a t
phang n a y cat Cx t a i K va t a c6 A K _L Cx, H K 1. Cx.
Cho h i n h l a p phuofng ABCD.A'B'C'D', canh a.

a) T i m khoang each tii d i n h C d e n m a t phang (C'BD).
b) T i m khoang each giiifa h a i difcirng t h a n g A D ' va CB'.
c) * T i m khoang each giiifa h a i dLforng t h a n g A B ' va B C .
CHI DAN
a) Ti^ CB = C C = CD va A ' C = A'B' = A D t a suy r a CA' 1 (C'BD).
Goi H la giao d i e m ciia CA' vdti m p
(C'BD) t h i H l a h i n h chieu eiia C t r e n
mp (C'BD) va C H l a k h o a n g each t i f C
den mp (C'BD).
H i n h chop C.C'BD l a h i n h chop t a m
giac deu n e n H l a t a m ciia t a m giac
deu C'BD.
T a m giac deu C'BD c6 canh la aV2
nen chieu cao
CO =

a^.Vs

aVe

Ta l a i c6 C'H = - C O n e n C H =
2
2
3
T a m giac C H C vuong t a i H cho t a
CH^ = CC'2 - C'H^ ^ CH^ = a' -

V3
>CH = a

Hoc va on luy?n theo CTDT mon Toan THPT L'i

203


Chii y : Co t h e t i n h C H theo each khac.
1 a^
a'^
: Vc.c BD = - • y . a = —

Ta

CO

(aV2)^V3
T a m giac C'BD deu, c6 canh aV2 n e n

M a t khac

Vc.cBD = ^ . C H . S C B D

C H

SCBU =

a^Vs
=

4

2

= ^iXtoe. = _ 6

3

bcBD

a

^ £ ^

v3

3

2

b) D a p so': a.
e) T a CO A B ' // C D n e n m a t phSng (C'BD) la m a t phang chufa B C va

song song v d i A B ' . Goi I la giao d i e m cua A B ' va BA' t h i I thuoe A B '
va k h o a n g each giOfa h a i diicfng thSng A B ' va B C t h i b^ng khoang
each giCfa difcfng t h a n g A B ' vdri m a t phSng ( C B D ) va cung bSng
khoang each i\i d i e m I den m p (C'BD) :
d(AB', B C ) = d(AB', (C'BD)) = d ( I , (C'BD))
Ta xac d i n h h i n h chieu J ciia d i e m I t r e n m p ( C B D ) .
V i CA' i . ( C B D ) m a (A'BC) 3 CA' n e n t a eo ( A B C ) L (C'BD)
Ta l a i CO (A'BC) n (C'BD) = B H
V i I e A ' B m a A ' B c (A'BC) n e n h i n h chieu J ciia I t r e n m p (C'BD)
p h a i n a m t r e n diicfng t h a n g B H .
Ta cung c6

A'C L ( C B D )
IJKC'BD) J

I J // A'C.

T r o n g t a m giac A ' B H t h i I J l a diicrng t r u n g b i n h n e n I J = — A H
2
V i A'C = aV3; C H = ^

=> I J =
11.

3

2

2aV3

1

3

=> I J =

nen A ' H =
aV3
3

3

; d(AB , BC ) =

aVs
3

.

Cho tuf d i e n A B C D c6 h a i m a t ben A B C va D B C n a m t r o n g h a i m a t
phang tao vdri nhau m o t goc 60°. M a t ben ABC la t a m giac deu, con mat
ben DBC l a t a m giac vuong can t a i dinh D va D B = a.

a) T i n h k h o a n g each tiT d i n h A den m a t phSng (DBC) va khoang each tir
d i n h D den m a t phSng (ABC).
b) T i n h k h o a n g each giuTa h a i diTcfng t h a n g A D va BC, khoang each giCfa
h a i diTcfng t h a n g A C va D M ; M l a t r u n g d i e m ciia canh BC.
204

. . ' ; TS, Vu The Hifu - NguySn Vinh Can

CHI D A N
a) Goi M la t r u n g d i e m ciia canh BC t h i A M 1 BC; D M 1 BC
=:> A M D = 60"
Ta cung c6 BC ± ( A M D ) ^ ( A M D ) ± (DBC)
Trong t a m giac A M D , ke A H 1 M D x> A H 1 (DBC)
=> A H l a khoang each tii diem A den m p < D B C ) .
Trong t a m giac DBC t h i D B = a => BC = aV2
Tam giac ABC l a t a m giac deu canh

aV^ n e n A M -

aVe

Trong t a m giac vuong A H M ,
AM =

A M D = 60° n e n

A H = AM.sin60° ^ A H =

3aV2

Trong t a m giac A M D , ke D K _L A M .
Ta da c6 BC ± ( A M D )
i:^ B C l DK, suy ra D K 1 (ABC)

D K la khoang each tU D den mp (ABC).

Trong t a m giac vuong D M K t h i D M =
D K = DM.sin60°

aV2

va K M D = 60°

DK =

b) Ta CO, t r o n g t a m giac A M D ke M I 1 A D
BC 1 ( A M D )
BC ± M I
=> M I l a doan vuong goc chung ciia h a i diiorng t h d n g A D , B C h a y
chinh la khoang each giCfa hai di/cfng thSng A D , BC.
H a i t a m giac vuong M I D va A H D dong dang :
MI MD
MD.AH
AMID ^ AAHD
MI =
AH AD
AD
3a V2
Ta da c6 M D AH =
2
2
T i n h A D tCr t a m giac A M D vdri
iV2
A M D = 60", A M MD
bang each ap dung d i n h l i cosin :
AD^ = A M ' + M D ' - 2 . A M . M D . C 0 S A M D =0 A D = -Vs

- 2^3.

2
Hoc va on luyen theo CTDT mon Toan THPT ' 2 0 5


1.

Til day ta t i n h dirge

MI

3a
V8-2V3

Chu y : Co the t i n h M I theo each M I . A D = A H . M D = 2SAMDTCr C ta ke Cx // M D t h i m a t phang (A, Cx) la m a t ph^ng chufa AC va
song song vdfi M D , va v i H e M D n e n ta c6
d ( M D , AC) = d ( M D , (A, Cx)) = d ( H , (A, Cx))
Ke H F ± Cx => A F ± Cx
T r o n g t a m giae A H F ke H E ± A F ^ H E J. (A, Cx)
K e t hop v6i cac k e t qua t r e n , ta eo d ( D M , AC) = d ( H , (A, Cx)) = H E
1

T r o n g t a m giae vuong A H F , ta eo :

1

HE*-^

1

HA'

HF'

^
,
„ .
3aV2
.
-.^
aV2
_
2V5a
Ta eo : H A =
va H F = M C =
, suy r a H E =
.

BAITOAN 3. CAC BAITOAN VE G O C
I- C A C KIEN THLTC C d BAN
X a c dinh c a c loai goc
a) Goc giUa hai duang

thdng

De xae d i n h goe giOfa h a i diTcfng thSng a
va b, ta lay m o t d i e m O thuoc a va qua
O, ve diidng t h a n g b' // b. Goc tao bofi

hai di/cfng t h a n g a va b' la goc giiifa h a i
difdng t h a n g a, b.
b) Goc giUa duang thdng va mat
-

phang

Goc giOfa di/crng t h d n g a va m a t phang
(P) la goc giOfa difcfng t h a n g a va h i n h
chieu a' eiia a t r e n m a t phSng (P).

-

Ta xae d i n h goc giOfa diTorng thang a va mat
phang (P) nhir sau :
+ L a y m o t d i e m A thuoc a va ha diTdng vuong goc A H xuong mat
p h a n g (P) : A H ± (P)
+ Gia sijf a cat (P) t a i d i e m O t h i goe A O H la goc giiifa di/cfng t h i n g a
va m a t phSng (P).

c) Goc giQa hai mat

phang

Gia SLf h a i m a t p h a n g (P), ( Q ) giao nhau theo giao tuyen d. Ta xac
206

;

TS. Vu Thg Huu - Nguyin Vinh CJn

dinh goc giufa hai mat phang (P), (Q) nhu sau :
- Lay mot diem O thuoc giao tuyen d.
- Trong mat p h i n g (P), ke tia Ox vuong goc
vdfi giao tuyen d va trong mat phSng (Q), ke
tia Oy vuong goc vdi giao tuyen d. Goc
3cOy

6

[0°; 90°] la goc giaa hai mat phang

(P), (Q). Doi khi ta cung noi goc xOy la goc
mat phang (P) tao vdi mat phang (Q). T a
cung CO the phat bieu each xac dinh goc
xOy nhif sau :
- Qua mot diem O thuoc giao tuyen d cua hai mat phang (P), (Q) ta
difng mat phang (R) vuong goc vdi d. Mat phang (R) cat (P) theo giao
tuyen Ox va cSt mat phang (Q) theo giao tuyen Oy. Goc xOy e [0°; 90°]
CO

2.

di/Oc la goc

giUa

hai mat phSng (P), (Q).

C h i i y. Cac bai toan ve goc, trong phan Idn trtfcfng htfp, c6 lien he
den quan he vuong goc giOfa diXofng thang va mat phang.

II- C A C DANG BAI TAP VA PHLfdNG PHAP OIAI
l]

C a c d a n g b a i t|lp v a phifofng p h a p c h u n g de g i a i
-

Viec xac dinh cac goc, nhat la xac dinh goc giOfa difdng thang va mat
phang, goc giOra hai mat phang, thifcfng la mot trong nhufng cong viec
dau tien khi bat dau giai bai toan hinh, khong gian va cong viec nay
nhieu khi c6 tinh chat quyet dinh vi neu khong xac dinh dMc cac goc
hoac xac dinh sai cac goc thi ta khong the tinh difcfc hoac tinh sai cac
ket qua khac.

-

Viec tinh toan cac yeu to c6 lien quan den cac goc, cac do dai thtfcfng
lien quan den :
+ Cac he thufc Itfqng trong tam giac vuong.
+ Cac ti so ItfOng giac cua goc nhon.
+ Dinh li sin, cosin.
Do vay, co gang tim ra tren hinh ve cac tam giac vuong c6 chufa cac
goc da cho va nhieu khi ta phai ve them cac diTcfng phu de tao ra cac
tam giac vuong ay.

2.

C a c b a i tap

12.

Cho hinh chop tuf giac deu S . A B C D ; O la giao diem cua hai difdng cheo
AC, BD.
1. Hay chi r5 :
HQC

6n luyOn theo C T D T mOn Jo&n

THPT

£3

207


a) Chieu cao cua hinh chop.
b) Goc giiJfa canh ben va mat phang day.
c) Goc giOfa mat ben va mat phSng day.
d) Goc giufa chieu cao va mat ben.
e) Khoang each tCr tam O cua day den mat ben.
2. Cung v6i cac cau hoi tren, giai bai toan trong trifcfng hop hinh chop
dacho la hinh chop tam giac deu (O la tam cua day).
CHI D A N

1. a) Hinh chop tijf giac deu c6 day la hinh vuong va O la tam ciia day.
SO 1 (ABCD) => SO la di/orng cao ciia hinh chop.
S
b) SO 1 (ABCD) => OC la hinh chieu ciia

SC tren day :
Goc SCO la goc giOfa canh ben SC vdri
day. Tiiong tu, ta c6 cac goc S A O , S B O ,
S D O theo thijf tiT la cac goc giiJa cac canh
ben SA, S B , S D vdfi mat phang day.
^
c) Goi M la trung diem ciia canh B C t h i S M 1 B C , O M ± B C
=> goc S ' M O la goc giiJfa mat ben ( S B C ) voi mat phang day ( A B C D ) .
d) Goc O S M la goc giCifa chieu cao S O va mat ben ( S B C ) .
e) Trong tam giac S O M , ke O H ± S M =^ O H 1 ( S B C )
=> H la hinh chieu cua O tren mp ( S B C ) => O H la khoang each t i i tam
O den mat ben ( S B C ) .

2. Trong trirdng hop hinh chop S . A B C la
hinh chop tam giac deu t h i day A B C la
tam giac deu, tam O la trong tam cua
day. Goi M la trung diem canh B C t h i
A M la trung tuyen.

A M = ^5:^;A0=

^AM;

2
3
A M 1 BC;S M ± BC.
a) S O J- ( A B C ) ^ S O la chieu cao hinh chop.
b) S A M la goc giOfa canh ben S A vdi mat phang day.
c) S M O la goc giOfa mat ben ( S B C ) v(Ji mat phang day.
d) O S M la goc giOra diTofng cao S O va mat ben ( S B C ) .

e) Trong tam giac S O M ke O H ± S M t h i O H ± ( S B C )
OH la khoang
each tir diem O den mp ( S B C ) .
Chii y : K h i can t i n h toan cac yeu to' tren t h i ta nen diTa vao cac
he thiirc liigrng trong tam giac, dac biet la he thufc liitfng trong tam
giac vuong.
2 0 8 ; T S . Vu The Hiju - Nguygn Vfnh Can


YtNfini.TH'E'TiCHCIlCKil
B A I TOAN 4. T I N H THE T I C H CAC K H O I DA DIEN

I- CAC KIEN THlfC C6 BAN
1.

C o n g thiJc t i n h V v a Sxq c a c k h o i d a d i ^ n
Hinh

The t i c h V

Chop

V=

-Bh
3

D i e n t i c h xung quanh Sxq
Sxq=^pd(*)

L a n g t r y dufng

V = Bh

(**)

Sxq =

2p.h

H i n h hop diifng

V = Bh

(**)

Sxq =

2p.h

B l a dien t i c h day, h l a chieu cao, 2p l a chu v i day.
Chiiy
Cong thufc (*), Sxq = — pd chi diing cho h i n h chop deu, t r o n g do d l a
2
t r u n g doan ciia h i n h chop (chieu cao ciia m a t ben).
The t i c h h i n h hop chOf n h a t V = abc.
Cong thufc (**) cung diing v d i triTorng hop k h o i l a n g t r u x i e n .

II- CAC BAI TAP VA CACH GIAI
1.

C a c l o a i b a i t^p
a) Cac bai tap chu yeu
da dien, doi k h i k e t
Cung can biet t h e m
D i e n t i c h toan p h a n

l a t i n h the t i c h va dien t i c h xung quanh cac k h o i
hop v d i viec t i n h mot so yeu to ve k h o a n g each.
cong thijfc : Stp = S d + Sxq
= dien t i c h day + dien t i c h xung quanh.

b) De t i n h the t i c h cac k h o i da dien, t a thiTcfng p h a i l a m h a i viec :
- Xac d i n h v a t i n h chieu cao ciia k h o i da dien.
- T i n h dien t i c h day.
Viec xac d i n h va t i n h di/orng cao ciia k h o i da dien l i e n quan den cac
quan he vuong goc t r o n g k h o n g gian.
Viec t i n h dien t i c h day l i e n quan den viec t i n h dien t i c h cac t a m giac,
tuf giac, cac da giac. Can ghi nhdf mot so cong thufc t i n h cho mot so
h i n h dac biet de t i e t k i e m thcfi gian k h i l a m toan.

HQC va on luyen theo CTOT mon Toan THPT

209


+ Dien tich tarn giac deu canh a : S =
(chieu cao tarn giac deu canh a la h =
+ Dien tich tarn giac c6 hai canh a, b va goc xen giOfa hai canh ay la a :
S = —ab.sina

2
+ Dien tich tarn giac biet ba canh a, b, c la :
S = .^pCp - a)(p - b)(p - c) (cong thufc Heron)
vdri a + b + c = 2p (2p la chu vi tarn giac).
+ Dien tich tarn giac biet chu v i 2p va ban kinh diicfng tron noi tiep r :
S = pr.
+ Lien he giuTa dien tich tarn giac vdfi ban kinh diicfng tron ngoai tiep :
abc
S=
4R'
c) De tinh the tich khoi chop tarn giac, ta con c6 the suf dung cong thiic
ti so the tich.

^
13.

Cac bai t^p
Cho hinh chop S.ABC, cac mat ben SAB, SAC la cac tarn giac vuong
can tai A, con mat ben SBC c6 dien tich la —a'Vs. Chieu cao hinh
2

chop la a.
a) Tinh the tich khoi chop.
b) Tinh khoang each t\i dinh A den mat phang (SBC).
CHI DAN
a)

SA 1 AB
SA 1 AC

SA L (ABC)

=> SA la chieu cao hinh chop, SA = a.
De thay SB = SC = aV2
SsBc = -SB.SC.sinS
2
^

-aV2.aV2.sinS =

2

2

=> sinS = — ^ S = 6 0 ° ^ B C = a>^

2

Goi M la trung diem cua BC : AM^ = AB^ - MB^
A M ' = a' -

1V2

2a^
4

>AM-

Vs.ABC = ^SA.S,3c = \.^\ ^ . a V ^

^ V, S.ABC =

a" (dvtt)

210 ta TS. Vu The Huu - Nguyin VTnh C0n


b) Goi H l a h i n h chieu cua A t r e n m a t phSng (SBC) => A H l a chieu cao
ke tif d i n h A t r o n g h i n h chop A.SBC.
VA.SBC = — A H . S S B C = VS.ABC

o
3
2
Chii y : Ta CO the t i n h A H theo each khac.
Goi M la t r u n g d i e m ciaa canh BC

6

SMIBC

>AH =

BC 1 (SAM)

=> (SAM) 1 (SBC)
Trong m a t p h a n g (SAM) ke A H 1 S M

A H ± (SBC)

Trong t a m giac vuong S A M t h i
1

AH'

14.

1

3 =>AH^ = —

• + •

SA' A M '

AH =

raV2

Cho l a n g t r u duTng A B C D . A ' B ' C ' D ' , chieu cao a; day A B C D l a h i n h
vuong. DLfdng t h a n g A D ' tao vdri m a t phang ( A A ' C ) m o t goc 45°. T i n h
the tich cua k h o i l a n g t r u va t i n h goc giufa h a i di/cfng t h a n g A I va
BD', t r o n g do I la t a m cua h i n h vuong A'B'C'D'.

CHI DAN
Goi I la giao d i e m cua A ' C va B D ' .
De t h a y :

D'l L A'C'l
D'

D ' l l A'A

=> D ' l 1 ( A A ' C ) ^ D ' A I = 45°

B

Goi canh day l a n g t r u la x,
A'D' = x ^ D ' I = i ^

D
D'l

T r o n g t a m giac vuong D ' l A : A D ' =

= 2x
sin45°
D'D' = D'A^ - DA^
T r•ong
o n g t a m giac vuong D ' D A : D'D^
a' = ( 2 x ) 2 - x ^ ^ a ' = 3 x 2 ^ x =
ABCD.ABCD' =

aVs

(dvtt)

V i B B ' // AA' nen goc giufa h a i difcfng t h S n g A I va B B ' cung bSng goc
giufa h a i di/cfng t h a n g A I va AA'.
T r o n g t a m giac vuong A A ' I t a c6 :

cosA'AI =

A I

lA'

cosA'AI = —
2

>A'AI = 60°.

Hoc va 6n luyen theo CTDT mSn To4n THPT £0 211


15.

Mot lang try xien tam giac ABC.A'B'C, c6 canh ben AA' = aV?. TiT
mot diem S thuoc AA', ta dUng mot mat phang vuong goc vdi AA';
mat phang nay cat BB', CC theo thuf tiX tai cac diem M , N . Tam giac
SMN CO chu v i la 54a va cac canh SM, M N , NS c6 do dai t i le vdi cac
so 25, 12 va 17.
SM : M N : NS = 25 : 12 : 17
Tinh the tich khoi chop S.BCC'B'.

CHI DAN

Goi cac canh ciia tam giac SMN la : SM = 25x, M N = 12x va
NS = 17x ta CO : 25x + 12x + 17x = 54a
x= a
va ta CO : SM = 25a, M N = 12a, NS = 17a

SiJf dung cong thufc Heron, ta tinh diTcfc SSMN
Ke

diXfSng

(SMN)

cao
1

SH

BB'

^ (BCC'B') 3

cua tam giac

BB'

SMN

thi S H

= 1080a^.
2S,-"SMN
= 15a
MN

=

( S M N ) 1 (BCC'B')

( S M N ) ± (BCC'B')
Ta l a i CO ( S M N ) n ( B C C ' B ' ) = M N

1

SH

M N

=^ S H 1 ( B C C ' B ' ) hay S H la chieu cao
cua hinh chop S . B C C ' B ' .
Tuf giac B C C ' B ' la hinh binh hanh c6 B B ' = aV7 va chieu cao
nen SBCCB = 12a'VT"
Suy ra
16.

VS.BCCB'=

-SBCCB-SH

M N

= 12a

Vs.BCCB' = 60^/^a^ (dvtt).

Cho hinh chop tam giac S.ABC, c6 canh ben SA vuong goc vdfi mat

phang (ABC). Hay tinh the tich khoi chop S.ABC va khoang each tif
d i n h A den mat p h S n g (SBC)

trong cac trUofng hgfp :

a) Tam giac ABC c6 canh AB = 3a, BC = 4a, CA = 5a trong do a la mot do dai
cho tnfdc va mat ben (SBC) tao vcfi mat day mot goc 30°.
b) Tam giac ABC vuong goc t a i dinh A va mat ben SBC la tam giac deu
^V3
CO dien tich la

c) Tam giac ABC c6 A = 120°; mat ben SBC la tam giac deu canh a.
d) Tam giac SBC vuong goc t a i dinh S; SBC = 60°, canh BC = a va mat
phang (SBC) tao vdi mat phang day mot goc 30°.
212

IS. Vu The Huu - Nguyin VTnh CJn


S

CHI D A N

a) De t h a y tarn giac A B C vuong goc t a i d i n h B .
A B 1 BC
Ta CO :

SA 1 (ABC)l
BC ± A B

SB 1 BC

(dinh h' 3 diidng vuong goc)
B

BC ± ABJ

SA = aVs ( d v t t )

Trong tarn giac vuong SAB : SA = ABtan30°
Vs.ABC =

^•i3a.4a.a>/3=>

VS.ABC = 2 a '

Trong tarn giac SAB ke A H 1 SB. Ta c6 :

BCl(SAB)
(ABC)

3

BC

(ABC) 1 (SAB)

(ABC) 1 (SAB)
(ABC)

n

(SAB) - SB

A H 1 (SBC)

AH I S B
T r o n g tarn giac vuong SHB, v d i A B = 3a, B = 30°
^ A H = AB.sin30° ^ A H = - a .
2
Chii y : Co the t i n h A H theo cong thiJfc :
b) D a t BC = X, t a c6 : Ssac =

^V3

1

1

AH'
AS' A B '
x'V3 a'V3

= aV2

Goi M l a t r u n g d i e m ciia BC t h i A M = — — va A M ± BC
T a m giac A B C can, vuong d i n h A cho t a => A C = a
T a m giac vuong SAC cho t a :
SA' = S C ' - A C '
^

SA' = (aA^)'.a' = a '
Vs.ABC=

SA = a

i-^^B.AC.SA

=> Vs.ABC = - a ' ( d v t t )
O

Goi H l a h i n h chieu ciia A t r e n m a t phSng (SBC) t h i A H l a khoang
each tii A d e n m p (SBC) hay A H l a chieu cao ciia h i n h chop A.SBC.
Ta CO :

VA.SBC = ^ A H . S , ^ , , = -a'.Vs . A H

Hoc va on luyen theo CTDT mon Toan THPT l-L 213


M a t khac

VA.SBC = VS.ABC

= -a'

2aS

TCr day t a c6 : - a ' V s . A H = - a ' => A H =
6

3
3
Chu y : Co the xac d i n h d i e m H va t i n h A H n h i T sau :
Trong tarn giac SAM ke A H 1 SM. Ta de dang chilng minh

diTOc

A H 1 (SBC)

T a t i n h A H theo cong thufc :
=
+ —J-j.
AH^
AS^
AM^
c) De t h a y t a m giac A B C can, d i n h A. Goi M l a t r u n g d i e m ciia canh BC
t h i A M 1 BC va B A M = 60°
AB =

BM
sin60°

a
VS

iV3
3

T r o n g t a m giac vuong SAB :
S A ' = S B ' - A B ' =i> SA = •

D i e n t i c h t a m giac A B C :
3 ABC = - A B . A C . s i n A =>S
ABC
12
i%/2
(dvtt)
V s . A B c = — S A . S ^ B c = > VS.ABC =
36
Goi H l a h i n h chieu ciia A t r e n m p ( S B C ) t h i A H l a chieu cao ciia
h i n h chop A . S B C
VA.SBC = — A H . S s B c
o

T a m giac

SBC

la t a m giac deu canh a n e n

VASBC = ^ A H . -

3

M a t khac

V3

SSBC =

^V3

(1)

^

4

VA.SBC = VS.ABC =

^V2
36

(2)

AH =
TCr (1) va (2) t a c6 : - A H . - ^ =
^
36
9
3
4
Chil y : Co the t i n h A H theo each sau :
T r o n g t a m giac S A M , ke A H ± S M ^ A H J . (SBC)
Trong t a m giac vuong S A M t h i A H la difofng cao thuoc canh huyen nen:
-J— = -l+ -l-.
Ta CO : A M = l c o t 6 0 ° = —
AH'
AS' A M '
2
6

4 L'; TS. Vu The Hi^u - Nguyin VTnh Can


AH' =

AH^

6a^

>AH =

81

SDA = 30°

d) Ke dircyng cao A D cua tarn giac ABC: A D i . BC
Tarn giac vuong SBC vdti BC = a
va B = 60"

SB = BCcos60° = 2
Tarn giac SDB vuong t a i D
va CO B = 60° => SD = SB.sin60° =
Trong t a m giac SAD, ke A H 1 SD.
Ta chOfng m i n h de dang A H 1 (SBC)
^ A H l a khoang each tiT A d e n m p (SBC)
Trong t a m giac vuong SAD, SD =
A D = SD.cos30° =>AB =

va D = 30° n e n

3a
8

3a
Trong t a m giac vuong A H D , A D = — va D = 30° n e n
8
3a
A H = AD.sin30° ^ A H =
16
3a
Khoang each iii A d e n m p ( S B C ) : d(A, ( S B C ) ) =
16

T a m giac S B C c6 B C = a, S B = - va B = 60°
2
V3
S s B C = - B C . S B . s i n 6 0 " => SSBC =

2

-

8

Vs.ABC - VA.SBC = — AH.SSBC => V;S.ABC =

^V3

128
Chu y : Trong b a i t a p nay, gia t h i e t SA vuong goc vdfi m a t phang

(ABC) doi k h i dtfoc cho dirdri dang :
- H a i m a t b e n (SAB) va (SAC) vuong goc vdfi m a t p h a n g day.
- H a i t a m giac SAB va SAC l a cac t a m giac vuong goc t a i d i n h A .
1 7 . Cho t a m giac deu SAB, canh a va h i n h thoi A B C D n ^ m t r o n g h a i m a t
phang vuong goc v d i nhau, goc n h o n A ciia h i n h thoi b a n g 60°. Goi I ,
J , K theo thur tiT l a t r u n g d i e m ciia cac canh A B , A D va SD.
a) T i n h t h e t i c h k h o i chop K . C I J .
b) T i n h goc tao bofi m a t b e n (SCD) va day (ABCD).
c) T i n h khoang each giOfa h a i dLfomg t h a n g SD va A B .
HQC va on luyen theo CTDT mon Toan THPT

215


CHI DAN
(SAB) 1 (ABCD)
a) (SAB) n (ABCD) = A B

S I 1 (ABCD)

S I l AB
Ta

K

c6: Scu = SABCD - (SAIJ + S C B I + SCDJ)

vdi

SABCD =

SAIJ =

a^Vs

1

—a^Vs
o

A

J

3a'V3
Sf'Ri = Scnj - - —S
= 4a'V3 =^Scu =
4
Goi H la hinh chieu ciia K tren mat phSng (ABCD) t h i H e D I ; K H // SI va
K H

= -SI
2

K H

1V3

VK.CIJ = — S c i j . K H

3a 2

VKCIJ =

—

(dvtt).

b) De thay CD 1 ( S D I ) ^ S D I la goc giiTa hai mat phang (SCD) va (ABCD)
SI = I D va S I 1 I D ^ S D I = 45".
c) M a t p h a n g (SDC) l a m a t p h a n g chufa SD va song song v d i A B . Tarn
giac S I D vuong can, K la t r u n g d i e m ciia SD suy ra
I K 1 SD
I K 1 (SCD)
d(AB, SD) = d ( A B , (SCD)) = d ( I , (SCD)) = I K
d ( I , (SCD)) =

aVe

18. Cho h i n h chop S.ABCD c6 m a t ben SAD la t a m giac can, d i n h S, nam
t r o n g m o t m a t p h a n g vuong goc vdfi m a t phSng (ABCD); day A B C D la
h i n h t h a n g vuong t a i A va D , c6 A D = A B = 2a va CD = a. M a t ben
(SBC) tao vdri m a t day goc 60°.
T i n h the t i c h kho'i chop S.ABCD theo a va t i n h k h o a n g each ti^ t r u n g
d i e m I ciia canh A D den m a t phSng (SBC).
CHI DAN
T a m giac SAD can, d i n h S m a I l a
t r u n g d i e m cua A D n e n S I _L A D
Theo gia thiet (SAD) ± (ABCD) va
do (SAD) n (ABCD) = A D n e n t a
suy r a S I ± (ABCD) h a y S I l a
chieu cao cua h i n h chop S.ABCD.

Ttr I ke I K ± CB, t a c6 :

D

216 :.': TS. Vu The' Hi/u - Nguyin Vinh Can