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

Neu ba mat phang cSt nhau theo ba giao tuyen phan biet t h i ba giao tuyen ay hoac dong quy hoac doi mot song song... T a t ca cac canh khong thuoc hai day deu song song vdri nhau.. Goi

mm i I tiJHG THIG VA HJT PHIB TRDNE KHONG emii

B A I T O A N 1 QUAN H E SONG SONG

Hai diTofng t h a n g phan biet cung song song vdfi mot di/dng t h a n g thuf

ba t h i song song vdi nhau

Neu ba mat phang cSt nhau theo ba giao tuyen phan biet t h i ba giao tuyen ay hoac dong quy hoac doi mot song song

He qua

Neu hai mat phang phan biet Ian lirgt d i qua hai du'&ng t h a n g song song t h i giao tuyen ciia chiing (neu c6) song song vdi hai diTdng t h a n g

do (hoac trijng vdi mot trong hai diTdng thSng doj

2 Dvfcfng t h ^ n g song song vdfi m a t p h ^ n g

g a song

song v

di mo

t m at p hd ng (P) th

g (Q ) chiif

a a m

a c

at (P ) t

hi cS

t (P ) t he

o gia

o t uy en song song

i m at p hS ng p ha

n bie

t cu ng song song v

di mo

n cii

a ch iin

g (n eu c6) cu ng song song v

di diTdn

g t ha

ng

do

3 Ha

i m at p ha ng s on

g so

t p ha ng g

oi la song song k

hi ch un

g kh on

g c

6 d ie

m chung

b) Dieu

kien de hai mat

phdng song song :

Dinh li :

Ne

u mo

t m at p ha ng ( P) chijf

a ha

i d iid ng t ha ng

a va

b ca

di ma

t p ha ng ( Q) th

i ( P) son

g son

g v

di (Q )

m ng oa

i mo

t m at p ha ng c6 m ot v

a c

hi mo

t m at

p

ha

ng

song song v

di ma

t p ha ng

g th an

g a song song v

di ma

t p ha ng ( Q) th

i q ua

t m at p ha ng ( P) son

g son

g v

di ma

t p ha ng ( Q)

He qua

2

Ha

i ma

t p hd ng p ha

n bie

t cii ng song song v

di mo

t m at

ha ng

di nh au

Tinh chat

2

Ne

u ha

i m at

ha ng (P) va (Q) son

g son

g th

i m

oi ma

t p ha ng (R) d

a ca

t

(P) th

i p ha

i c

at (Q ) v

a ca

c gia

o t uy en ciia chun

g son

g song

d) Dinh

li Ta-let trong khong

gian :

Dinh li thuan

Ba m at

ha ng d

oi mo

t son

g son

g ch an r

a t re

n ha

i c

at tu ye

tin

g

ti le

Dinh li ddo

Gia sijf tr en h

ai dii dn

g th an

g a v

a a ' I an li id

t l ay

ai b

o d ie

A

B BC C

A'B ' B'

C CA '

Kh

i d

o ba d ud ng t ha ng A A' , B B' , C

C ctin

g son

g son

g vd

g

e) Hinh

Idng tru

va hinh hop :

Binh nghla hinh Idng

tru

Hi nh l an

g tr

u l

a h in

h c

6 h

ai da

y l

a h

ai da giac na

m t ro ng

u Th6 ' Ha

u NguySn VTn

-h CJ

n

phSng song song T a t ca cac canh khong thuoc hai day deu song song vdri nhau

H i n h lang t r u c6 :

- Cac canh ben b&ng nhau

- Cac mat ben la cac h i n h h i n h hanh

- Hai day la hai da giac bang nhau

Dinh nghia hinh hop

H i n h lang t r u c6 day l a h i n h b i n h hanh diTcfc goi la 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 hanh, cac m a t doi 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 diem ciia moi 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 trong muc nay, chii yeu la chijfng m i n h hai dudng t h a n g cheo nhau, h a i di/dng thang 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 giai cac b a i t a p n a y ta thifdng suf dung cac t i n h chat ciia quan he song song va t r o n g nhieu

triidng hgfp t a suf dung phiCang phdp chiing minh phdn chiJcng

b) Ngoai ra cac t i n h chat ciia h i n h lang t r u , h i n h hop cung giup t a giai cac bai t a p ve quan he song song (cac canh ciia h i n h lang t r u , h i n h hop, cac mat ben ciia h i n h hop v.v )

^ B a i t ^ i p

du'cfng thang sau la cac cap dudng thang cheo nhau :

A B va CD A D va BC AC va BD

Chi dan : Suf dung phi/cfng phap phan chufng

diem ciia cac canh SA, SB, SC

a) Chufng m i n h mp (MNP) // mp (ABCD)

b) Chufng m i n h giao diem ciia mp (MNP) vdi SD l a t r u n g diem Q cua

doan thang SD

c) Goi O la giao diem ciia A C va BD, O' l a giao diem cua MQ va N P

Chufng m i n h ba diem S, O, O' thSng hang

Chi dan : Doc gia tir giai

theo thuf tir la t r u n g diem ciia hai canh SB, SD

Hpc va on luyen theo CTDT mon Toan THPT 195

^^^^^^^^^

a) Chirn

g min

h M

N / / (ABCD)

b) Ne

u eac

h xa

c din

h gia

o die

m P cua m

p (AMN) vdf

i can

h SC

e) Xa

c din

Ch

i dS

n : Doc gi

a tii

giai

4 Ch

o hin

h ho

p ABCD.A'B'C'D'

; O la giao diem eiia A

a) Chufn

g min

h m

p (AB'D') /

/ mp(C'BD)

b) Chijfn

g min

h AO/

/CO'

Chi da

n : Doc gi

h ho

p ABCD.A'B'C'D': go

i O, O' the

a) Chufn

g min

h mp(AB'D') /

/ mp(C'BD)

b) Chijfn

g min

h A'

O / / CO'

Chi da

n : Doc gi

h tij

f mp

t di em d en mpt ma

t ph an

g

Cho ma

t phan

g (P) v

n (P)

Khoang each

tii die

m O den

m

at

phang (P) l

a d

p da

i doa

n thang

OH /p\

2

d(d, (P)) = d(0

e d, (P))

196 tJ.

; TS , V

Q Th

e Hu

u 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

cheo nhau la do d a i doan vuong goe

ehung eiia hai di/ofng t h ^ n g

diidng t h a n g cheo nhau a; b t h i bSng

khoang each tCr mot diem A e a den

mot m a t phang (P) qua b va song

song vdri difcfng t h a n g a

Chuy :

a) Ngoai ba trUcfng hcrp t r e n day, ta con c6 k h o a n g each giiJa hai diTcrng

thang song song, khoang each giiJa h a i m a t phang song song Cac

khong nhae l a i of day

b) De dung mat phang (P) qua b va song song vdfi a, ta lam nhii sau :

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 d a n g b a i t a p

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 tren (P), sau do t i n h do dai doan thang OH

Bai toan nay lien quan den v a n de difofng thang vuong goe vdfi m a t phang va k h i t i n h OH, thiTcfng ta phai sijf dung den cac kien thiife ve he thufc Itfofng trong tam giac, nhat la he thijfe lUgfng trong tam 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 mot mat phang (P) song

song vdi A bang each t i m khoang each tiT mot diem M G A den m a t

phang (P) Dieu quan t r o n g trong viee giai bai toan nay la of cho chon diem M e A mot each thich 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 khoang each giOfa hai diTcfng t h a n g cheo nhau a; b, t a thiidfng lam nhif sau :

Hoc va on luyen theo CTDT mon Toan THPT , 1 9 7

- Ti

m mo

t ma

t phan

g (P) chuf

a a va song

s on g

vdi b (hoac chijf

at phan

g (P) thiicfng

- Tim mot die

m M thich

hop thup

c a

va tim khoang eac

h ti

r M den

c on g

thijfc, h

e thijfc

liTpfng

trong ca

c ta

m giac

2 Ca

c b

ai ta

p

6.

Cho hin

h cho

p tam giac S.ABC, ha

i m

at phan

g (SAB) v

a (SAC) vuon

h A

t re n

mat phan

g (SBC)

b) Bie

t A

B = 13cm, B

C = 14em, C

A = 15cm va SA = 16em Tin

h A den m

at phan

g (SBC)

CH

I DA

N

(SAB) 1

(ABC)

a) Ta

CO

: (SAC

) 1

(ABC) ^ SA

± (ABC)

(SAB) n

(SAC) =

SA

Trong tam giac AB

C, ke

diiomg ca

o A

M

AM ± BC

SA

1

(ABC)

AM ±

BC

Tir SM

1

BC

(Dinh

li

3

diTotng v uo ng g oc )

Trong

t am g ia c

SAM ke

M

BC ± (SAM)'

AH

cz

(SAM)J

TCr (1) v

a (2) t

at phan

g (SBC),

b) Trirdt

c het, t

a tin

h die

n tic

h tam giac AB

C the

o cong

1

suy r

a p

- a = 8

; p

- b = 7

; p

- c = 6

SABC =

V21.8.7.6 =

84 2S M = o A a e y t Tir da

^ABC _

u Th

e Hi;

u Ng

-u ye

n V

T nh CJn

Tarn giac S A M vuong t a i A va A H la di/dng cao thuoc canh huyen

Theo he thiic li/gfng trong tarn giac vuong ta c6 :

1 Ta CO the t i n h A H theo each khac

Trirdrc het, ta t i n h S M tiT tam giac vuong S A M

2 Ta cung c6 the dirng diem H theo each l i luan sau :

Do SA 1 (ABC) =^ SA 1 BC Dirng qua SA mot mat phang (P) vuong goc vdri BC; mSt phang nay cat BC tai M Trong mat phang (P), ta ke

nen A H 1 (SBC) hay H la h i n h chieu ciia A tren mat phang (SBC)

7 Cho h i n h chop S.ABC; hai mat phang (SAC) va (SAB) vuong goc v6i

mat day (ABC) va SA = a V 2 T i n h khoang each tCr dinh A den m a t

phang (SBC) trong cac trudng hop :

a) Day ABC la tam giac deu canh a

b) Day ABC la tam giac can dinh A, goc A - 120° va A B = AC = a

c) Day ABC la tam giac vuong t a i B; AC = 5a, BC = 4a

AH'' (aV2)'

• +

AH =

iV66

11

b) Go

i M la

trung die

m cii

a BC

Trong tam giac SA

1 (SBC)

Trong tam giac SA

M t

hi

11

1

AH

^ AS' AM'

vdi A

S aV2 v

-a A

M = AB.coseO" =

-2

ta tin

C O

: S

A ± (ABC)

BC ± (SAB)

Trong tam giac SAB, k

e A

H 1

SB

^ AH ± (SBC) v

De tha

y AB' = 9a'

13

8 Ch

o hin

h cho

p S.ABCD, chie

u ca

o bSn

g a\/3, ha

A = 60"

SB

CH

I DA

N

Ta C O

: Go

i O la giao die

200 ,

.' TS V

u Th

e Hi/

u Nguy§n VTn

-h Ca

n

9 Cho tuf dien A B C D H a i m a t ben ABC va DBC n a m trong hai m a t

phang hop \6i nhau mot goc 60° M a t ben ABC la mot t a m giac deu

con m a t ben DBC la mot t a m giac vuong can, dinh D Biet D B = a Goi M la trung diem cua canh BC

=>

AM

D 60°

-CH

I DA

N

1.

M l

a tru

C

Tam giac DB

2

Ta

m gia

c ABC deu, can

h

li cosi

n va

o ta

m gia

c AMD :

AD^

= AM' + DM' 2AM.D

M.C

0S AM D

y d(A, (DBC)) =

A

H

Trong tam giac vuon

g AHM, A

M = ; A

MH = 60

°

AH = AM.sin60

° =

> A

H = 3a

V2

Ta cung c6 B

C 1 (AMD) ^

(ABC) 1

(AMD) Tron

g tam giac AM

K 1 (ABC)

DK = d(D, (ABC))

Trong tam giac vuon

g DKM, D

M =

1V2

va AMD = 60

°

DK = DM.sin60

° ^ D

K =

3 Go

i J la giao die

D t

ai I hay MI

1 A

D

Ta l

ai c

6 B

C ± (AMD) m

a MI

c :

(AMD) MI ± B

C

=> MI la doan vuong

AAIM

c/5

AAK

D IM A

D

AM K

D IM = AM.A

D

KD

=> IM = aV8-2V

3

Trong m

at phan

g (DBC), ti

r C ta

ke C

x // DM t

hi ma

t phan

g

(A, Cx) l

M, (A, Cx))

Ta lay mot die

m tu

y y tren D

M, die

m H chang ha

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

-n

Tong hop cac t h o n g t i n t r e n , ta c6 :

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

1 1 1 Trong t a m giac vuong A H K t h i

HK^

Suy ra H P =

10

Chii y : V i A H i (DBC) =^ A H 1 Cx, do do ta c6 the xac dinh mat phang

(AHK) bSng each diing qua A H mot m a t phSng vuong goc vdi Cx, m a t

phang nay cat Cx t a i K va ta c6 A K _L Cx, H K 1 Cx

a) T i m khoang each tii d i n h C den m a t phang (C'BD)

b) T i m khoang each giiifa hai difcirng t h a n g A D ' va CB'

c) * T i m khoang each giiifa hai dLforng t h a n g A B ' va B C

a) Ti^ CB = C C = CD va A ' C = A'B' = A D t a suy ra CA' 1 (C'BD)

Goi H la giao diem ciia CA' vdti mp

Tam giac deu C'BD c6 canh la aV2

nen chieu cao

Chii y

: C

o th

e tin

h C

H the

o eac

h khac

6 can

h aV

2 nen

S CB

U = =

4 2

Ma

t khac

V

c.c BD =

^.

CH S CB

D CH

= ^iXtoe =

e) T

a C O

AB' // C

D ne

n m

at phSn

g (C'BD) l

i I la giao die

m cu

a AB' v

a BA' t

hi I thuoe AB

a B

C th

i b^n

g khoan

g

each giCfa difcfn

g than

g AB' vdr

i ma

t phSn

g (CBD) v

a cun

g bSn

g

khoang each

i\i die

m I den mp (C'BD) :

d(AB', BC) = d(AB', (C'BD)

) = d(I, (C'BD))

Ta xac din

h hin

h chie

u J ciia die

m I tren mp (CBD)

Vi CA' i (CBD) m

a (A'BC) 3

CA' ne

n t

a e

o (AB

C) L

(C'BD)

Ta la

i C O (A'BC) n

c (A'BC) ne

n hin

h chie

u J ciia I tren mp

(C'BD)

phai na

m tre

n diicfn

g thang B

H

A'C

L

(CB

IJ // A'C

Trong tam giac A'B

H t

hi IJ

diicrng trun

g bin

h nen

J = —

AH 2

Vi A'

C = aV3;

CH = ^ nen

A'H

D c

6 h

ai ma

n ma

t

ben DBC la tam giac vuon

g ca

n ta

i din

h D va DB = a

a) Tin

at phSn

g (DBC) v

b) Tin

D v

a BC, khoan

C v

a DM; M la trung die

m cii

a can

h BC

204

'; TS , V

u Th

e Hif

u NguySn Vin

-h Ca

n

CHI D A N

a) Goi M la trung diem ciia canh BC t h i A M 1 BC; D M 1 BC

=:> A M D = 60"

Ta cung c6 BC ± (AMD) ^ (AMD) ± (DBC)

Trong tam giac A M D , ke A H 1 M D x> A H 1 (DBC)

=> A H la khoang each tii diem A den mp<DBC)

Trong tam giac DBC t h i DB = a => BC = aV2

Tam giac ABC la tam giac deu canh

Trong tam giac vuong A H M ,

T i n h A D tCr tam giac A M D vdri

AD^ = A M ' + M D ' - 2 A M M D C 0 S A M D =0 A D = -Vs - 2^3

2

Til da

y t

a tin

h dirg

e

MI 3a

V8-2V3

Chu y : C

o th

e tin

h MI theo each MI

AD = AH

MD = 2SAMD-

TCr C ta

ke C

x / / MD t

hi ma

t phan

g (A, Cx) l

a

vi H

e MD nen ta c6

d(

MD, AC) = d(M

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

, Cx))

Ke HF ± C

g tam giae AHF

ke HE ± A

F ^ H

E J (A, Cx)

a e

o d(D

M, AC) = d(H, (A

, Cx)) =

H

Trong tam giae vuong

AHF, t

a e

o :

H E*

- ^ HA' HF'

^ , „

3aV

2 -

.^

aV2 _ 2V5

BAITOAN 3.

CAC BAITOAN

VE G

OC

I- CA

C KIE

N THLT

C C d BAN

De xa

e din

t die

m O thuoc a

a v

a b' l

g a, b

b) Goc

giUa duang thdng va

a v

a ma

g a

va hin

h

chieu a' eii

a a tre

n m

at phSn

g (P)

r sa

u :

+ La

y mo

t die

m A thuoc a

mat

phang (P) : A

H ± (P)

+ Gi

a sij

f a cat (P) t

ai die

m O t

hi go

e AO

at phSn

g (P)

c) Goc

giQa hai mat phang

Gia

SLf

hai ma

t phan

g (P), (Q) gia

dinh goc giufa hai mat phang (P), (Q) nhu sau :

- Lay mot diem O thuoc giao tuyen d

- Trong mat phing (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

(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°]

2 Chii 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- CAC DANG BAI TAP VA PHLfdNG PHAP OIAI

l ] C a c d a n g b a i t|lp v a phifofng phap c h u n g de giai

- 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

y de

n ma

t ben

CH

I DA

N

1 a) Hinh cho

SO

1 (ABCD) =

b) S

O 1 (ABCD)

c go

c SAO, SB

B, SD vdfi ma

t phan

g day ^

c) Go

i M la trung die

C, OM ± B

C

=> goc

S 'M O

la go

c giiJf

a ma

t ben

(SBC)

voi ma

t phan

g day

(A BC

D)

d) Goc

O SM

la goc giCif

a chie

u cao

O

va ma

t ben

( SB C)

e) Tron

g ta

m gia

c SOM, k

e O

H ± S

M =

^ O

H 1 (SBC)

=> H la hinh chieu cua O tren m

p (SBC) =

S A BC

la

hinh chop ta

la

tam giac deu, ta

m O la trong ta

m cu

a

day Go

i M la trung die

m can

h BC t

hi

AM la trung tuyen

AM

=

^5 :^

;A 0=

^ AM

AM

1

B C;

SM ± B

C

a) SO J- (ABC) ^ S

b) SA

c) SM

i ma

t phan

g day

d)

O SM

la goc giOr

a diTofn

g cao

S O

va ma

t ben

( SB C)

e) Tron

g ta

m gia

c SOM ke

H ± S

M t

hi OH ± (SBC)

each tir diem

O

den mp

( SB C)

Chii y : K

n th

OH la

khoang

20

8

;T S.

Vu The Hij

u -

N gu

y gn Vfnh Ca

n

YtNfini.TH'E'TiCHCIlCKil

BAI TOAN 4 TINH THE TICH CAC KHOI DA DIEN

The tich h i n h hop chOf n h a t V = abc

Cong thufc (**) cung diing vdi triTorng hop k h o i lang t r u xien

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 l a t i n h the t i c h va dien t i c h xung quanh cac k h o i

da dien, doi k h i k e t hop vdi viec t i n h mot so yeu to ve khoang each

Cung can biet t h e m cong thijfc : Stp = S d + Sxq

Dien t i c h toan phan = 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 phai l a m hai viec :

- Xac d i n h va t i n h chieu cao ciia k h o i da dien

h i n h dac biet de tiet k i e m thcfi gian k h i l a m toan

HQC va on luyen theo CTOT mon Toan THPT 2 0 9

la :

S = ^pCp -

a)(p b)(p - c) (con

-g thuf

c Heron)

vdri a + b + c = 2

c ma

t be

n SAB, SA

2

SsBc = -SB.SC.sinS

2

^ -aV2.aV2.sinS

= =

> sin

S = — ^

S = 60°

^B

C =

AM' = a' -

1 V2 2a^

4

>AM-

Vs.AB C

Vu The Huu

- Nguyi

n VTn

h C0

n

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 trong h i n h chop A.SBC

VA.SBC = — A H S S B C = VS.ABC

Goi M la t r u n g diem ciaa canh BC

S M I B C

=> (SAM) 1 (SBC)

Trong m a t phang (SAM) ke A H 1 SM A H ± (SBC)

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

Trong t a m giac vuong D ' l A : A D ' =

•ong t a m giac vuong D'DA : D'D'

a' = ( 2 x ) 2 - x ^ ^ a ' = 3 x 2 ^ x =

D ' l

= 2x sin45°

Trong t a m giac vuong D'DA : D'D^ = D'A^ - DA^

6 can

h be

n AA' =

aV? Ti

SMN

CO ch

u v

i la 54a v

a ca

c can

h SM, MN, N

N = 12x v

CO

: 25

x + 12x + 17x = 54a x

CO

: S

M = 25a, MN = 12a, N

) 1

(B CC 'B ')

Ke

diXfSng

cao

S H

cua ta

m giac

S MN

thi

H =

(S MN )

1

B B'

^ (BCC'B'

) 3

B B'

(S MN ) ± (BCC'B')

Ta la

i C

O

(S MN

1

(B CC 'B ')

hay

S H

S.BCC'B'

Tuf giac

B B'

= aV

7 v

a chie

u cao

Suy r

a VS.BCCB'

= -S BC CB -S

a) Ta

m gia

c AB

C c6 canh A

B = 3a, B

C = 4a, C

A = 5a tron

- Nguyi

n VTn

h CJ

n

BC ± ABJ

Trong tarn giac vuong SAB : SA = ABtan30°

B

SA = aVs (dvtt)

Goi M la trung diem ciia BC t h i A M = — — va A M ± BC

Tam giac ABC can, vuong dinh A cho ta => AC = a

Tam giac vuong SAC cho ta :

Goi H la hinh chieu ciia A tren m a t phSng (SBC) t h i A H l a khoang

each tii A den mp (SBC) hay A H la chieu cao ciia h i n h chop A.SBC

2aS

Ma

t khac

VA.SBC = VS.AB C

= a'

-TCr da

y t

a c

6 : -

a' Vs

nhiT

sau :

j

AH^

AS^

AM^

c) D

e tha

y tam giac AB

C can, din

h A Go

i M l

1

BC va BAM = 60

°

a

AB =

BM

iV3

sin60° V S

3

Trong tam giac vuon

g SA

B :

SA' = SB' - AB'

=i>

SA = •

Die

n tic

h tam giac AB

C :

= -AB.AC.sin

Goi H l

a hin

h chie

u cii

a A tren m

p (SBC) th

A S

S BC

la tam giac de

u can

h a nen

SSBC =

VA.SBC = VS.AB

C =

^V2

36

(1) (2)

TCr (1) v

a (2) t

a c

6 : -AH.-

o th

e tin

M, k

e A

H ± SM ^ AH

J (SBC)

Trong ta

m gia

c vuon

g SAM t

hi AH la difofng ca

a C

O : AM=

AH^ A H ' =

6a^

d) Ke dircyng cao A D cua tarn giac ABC: A D i BC

Tarn giac vuong SBC vdti BC = a

SDA = 30°

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 la khoang each tiT A den mp (SBC)

8 3a Trong t a m giac vuong A H D , A D = — va D = 30° n e n

(ABC) doi k h i dtfoc cho dirdri dang :

- H a i m a t ben (SAB) va (SAC) vuong goc vdfi m a t phang day

- Hai t a m giac SAB va SAC la cac t a m giac vuong goc t a i d i n h A

phang vuong goc vdi nhau, goc nhon A ciia h i n h thoi bang 60° Goi I ,

J , K theo thur tiT la trung diem ciia cac canh A B , A D va SD

a) T i n h the tich khoi chop K C I J

b) T i n h goc tao bofi m a t ben (SCD) va day (ABCD)

c) T i n h khoang each giOfa hai dLfomg t h a n g SD va A B

HQC va on luyen theo CTDT mon Toan THPT 2 1 5

SI 1

(ABCD)

CH

I DA

u = SABCD

- (SAI

J + SCBI+

SCDJ)

vdi SABC

D = a^Vs

g (ABCD) t

hi H

e

DI;

H

3a

2

VK CI

a (ABCD

)

SI = I

c) Ma

t phan

g (SDC)

la m

1

(SCD)

d(AB, SD) = d(AB, (SCD)

) = d(I, (SCD)) =

K

aVe ) = d(I, (SCD)

18 Cho hin

g vuon

g t

ai A va D, c

6 A

D = A

B = 2a va CD = a M

CH

I DA

N

Tam giac SA

D can, din

h S ma

D ne

n t

a

suy ra S

I ± (ABCD) ha

y S

I l

a

chieu cao cu

a hin

h cho

p S.ABCD

Ttr I

ke IK ± CB, t

a c

6 :

D

216 :.': TS

V u The ' Hi/

u Nguyin Vin

-h Ca

n

SI 1 (ABCD)l

Ta tinh IK Vi I K J BC nen Smc = - I K B C

=> SiBc = SABCD - (SIAB + SIDC) = — — - => SIBC = — —

Goi E la trung diem ciia AB ^ CE = AD = 2a va E B = a

Trong tam giac vuong CEB ta c6 :

hay IH la khoang each ti^ diem I den mat phang (SBC)

Trong tam giac vuong SIK, IH la difdng cao thupc canh huyen nen

1 1

IH^ I S ' IK^

Ta tinh difoc IH =

1 , 3V15 3V5 vcfi IS = a v a IK = a

la ta

m gia

c de

u A'B = A'

C

=> AA BC

la ta

m gia

c can

S AB

C =

A'l ±

BC x^V3

AB =

AC

A'lA = 30

°

A'I = A'I =

x^ = 3a^

3aV3

>AA = 3a

AI = A'Icos30°

AI =

Vdri B

C =

X = aVs v

a A

I =

SA BC

= ^BC.A

I

8 3aV3

thi

'A BC =

16 9a=

(A 'B C)) = dd' , ( A'B C))

Hai ma

t phang (AA'I'I)

va

(A 'B C)

F = AA') v

a go

c I'lH

= 60

°

I'H

= ri sin60" =>r

H

3aV3

Chti y : C

u Th

e Hii

u Nguyin

-V Fn

h Ca

n

20 Cho l a n g t r u t a r n giac deu A B C A ' B ' C Duorng t h a n g A ' B tao v d i m a t

p h a n g (BCC'B') m o t goc 30° G o i I ' l a t r u n g d i e m cua c a n h B ' C ; t a m giac A ' l ' B CO d i e n t i c h l a ^

8 a) T i n h t h e t i c h k h o i l a n g t r u theo a

b) T i n h k h o a n g each giiifa h a i difdng t h a n g A ' B va B ' C

b) M a t p h a n g ( A ' B C ) l a m a t p h a n g chufa A ' B v a song song v6i B ' C

Goi I l a t r u n g d i e m ciia canh BC T r o n g t a m giac A T I ke I ' H J A ' l t h i

2 1 Cho h i n h l a n g t r u A B C D A ' B ' C ' D ' , day A B C D l a h i n h t h o i , canh a, goc

n h o n A = 60° D i n h A ' ciia day t r e n each deu ba d i n h A , B , D thuoc day

dii6i C a n h ben ciia l a n g t r u tao vdti m a t p h a n g day cac goc 45°

I DA

N

a) Go

i H la hin

h chie

u cii

a din

h A' tre

n

mat da

y (ABCD)

Ta

CO

A'A = A'

B = A'D ^ H la tam

ciia ta

m gia

c ABD Ta

m gia

c ABD

la

tam giac de

u

3 2 AH

=

AH ± (ABCD)=

> A'A

b)

Ta la

i c

6 SAB

D =

A'H ± (ABCD)

i^V3

^ VAAB

D =

A'H S

-ABD

12 (dvtt)

DB

1

A'H

ABCD la

Ti^ (1)

va (2) suy r

a hin

1

BB'

c) BB' =

AA' =

DB

1 (ACC'A'

) cosA'AH

(BDD'B) = 00' ne

n ne

u ti

T A' k

e A'l

= A'l.OO' =

AO.A'H=>

A'l = AO.A'H

00'

o hin

M,

N, P theo thu

a) Tin

b) Tin

c) Tin

g S

C v

a AB

u The ' H

i /u

- Nguyg

n V

T nh Can

ke M J 1 S N thi d(AB, SC) = d(AB, (SCD)) = d(M, (SCD)) = M J

b) Chil y : Trong cac cong thOfc tren, ta k i hieu h, r, I theo thuf tyt la

chieu cao, ban kinh day va diicfng sinh, vdfi dieu can nhd la trong mot hinh tru tron xoay thi diicfng cao va dirdng sinh thi b^ng nhau

C A C BAI TAP VA CACH GIAI

[l^ Cac dang bai tap

a) Phan lorn bai tap trong muc nay tap trung vao viec tinh the tich cac khoi va tinh dien tich xung quanh cac hinh

u ca

o h, diTcfn

g sin

h / B

2] Ca

c ba

6 die

n tic

h

la 4a^V3

Tinh the tic

a c

6 : ^ ^ = 4a^V

3 x = 4a

T Cf day su

x ^

r = 2a =

^ V = -

n r\

dLTorng

sinh /, go

c giufa

difcfng

sinh va

h no

n :

h = Zcos30° =

Ban kin

h da

y :

r = /sin30° =

7:; V = (dvtt)

; S

.q

= (dvtt)

7

AB : B

C : CA = 9 : 1

Suf dun

g con

g thuf

c Heron, t

a tin

h

d iTg

r c x = 2

a 18a, 20a, 34a

Ap dun

g con

g thuf

c S = pr, t

222

O J TS V

Mot hinh tru c6 thiet dien qua true la mot hinh chiif nhat c6 di§n tich

40a'^; hai canh ciia hinh chijf nhat t i le vdi 2 va 5

Tinh dien tich xung quanh hinh tru va the tich khoi tru

Goi canh cua hinh chuf nhat la 2x va 5x Ta c6 : 2x.5x = 40a^

=> x^ = 4a^ =>x = 2a => cac canh cua hinh chOf nhat la 4a va 10a

Ta xet hai trifdng hcfp :

a) 2r = 4a va h = 10a => r = 2a, h = 10a

b) 2r =10a va h = 4a => r = 5a, h = 4a

Sxq = 40Tia^ (dvtt); V = lOOna^ (dvtt)

28 Cho mot tam nhom hinh chOf nhat, dien tich 18a^, c6 chieu rgng va

chieu dai t i le vdi nhau theo t i so 2 : 3 Ttr tam nhom nay ngiTcri ta cuon

tron theo mot canh ciia hinh chiif nhat de tao thanh mot chiec hop tron,

hinh tru (khong ke nap va day) Hoi phai uon theo canh nao de difcrc

chiec hop tron c6 the tich lorn nhat ?

C H I D A N

Goi cac canh cua hinh chiif nhat la 2x, 3x ta tinh dxiac hai canh ciia

hinh cha nhat la :

chieu dai : 3a\/3 ; chieu rong : 2aV3

aVs

29

71 n

Neu uon theo chieu dai, ta di/gfc hinh tru chieu cao h2 = 2a\f3 va ban

kinh V2 = va duac thi tich V2 = ^ ^ ^ ' ^ (dvtt)

271 • n

Vi V2 > Vi nen uon theo chieu dai cua hinh chijf nhat ta se dUtfc mot

chiec hop hinh tru c6 the tich I6n nhat

Chufng minh rSng neu ban kinh cua ba hinh cau t i le v6i nhau theo

1 : 2 : 3 thi the tich cua khoi cau Idrn se gap 3 Ian tong the tich cua

hai khoi cau nho

C H I D A N

hai khoi cau thi bang lap phtfofng cua hai ban kinh Goi cac ban kinh

=> = SC

30.

Trong m

at phan

g (P), ch

o ta

m gia

c ABC, bie

t A

B = 10cm, B

C =

6cm,

CA = 8cm Mo

t ma

t ca

u c

6 ta

m I each

mat phan

g (P) mo

CH

I DA

o difcfn

g tro

a tin

p tam giac AB

C l

a R = 5cm

h ma

t ca

u :

r^ = d*^

+ =

^ r

^ = 12''^ + 5^ =

169 =

^ r = 1

(cm'); V =

^

^7 r

(c m=

^)

3

31.

Cho hin

h vuon

g ABCD, can

h 2a Diin

g du'orn

g tron ta

h vuong Ch

o hin

B, t

a difo

c mo

t hin

h tru

va mo

t hin

h cau

N

Vt , = 27ia^

; Ve^

u = ^ Tia

^ V = Vt

™ V cS = ^ rr

32.

Cho ta

m gia

c ca

n ABC, AB = A

C =

a v

a A = 2a K

h non

a) Tin

h non

b) Tin

h th

e tic

h kho

i ca

u d

o diiofn

g tro

i qua

y hin

h v

e xun

g quan

h AH

CH

I DA

a =

h CH = asin

a =

r

Vn on =

— J rr

'h

= — a^

sin'a

COS

a = — a

^ sin2a.sin

b) Go

i r' l

a ba

n kin

K = r'

Tacd: B = 15°

l^

= 90 A;

224

El TS V

D Th

e Hu

u Nguygn Vin

-h Ca

n

- 1 - A

B i = B = 4 5 °

-2 4 Trong tarn giac vuong B I H :

33 Cho h i n h chop S.ABC, c6 canh ben SA vuong goc v6i m a t phang day,

SA = a\l2, day A B C l a t a m giac vuong can t a i dinh B Canh

canh SB, SC

a) T i n h the tich khoi chop

b) Chufng m i n h n a m diem A, B, C, D, E nSm t r e n mot m a t cau

c) T i m tam va ban k i n h mat cau Tinh the tich va dien tich h i n h cau

=> Tam giac ADC vuong t a i D

Gpi M la trung diem ciia canh AC

AABC vuong t a i M nen B M = A M = C M

AADC vuong t a i D nen D M = A M = C M

AAEC vuong t a i C nen E M = A M = C M

TCr cac ket qua t r e n suy ra A M = B M = C M = D M = E M hay n a m

c) Trong t a m giac vuong SAC, c6 SA = a\[2, SC = a\f5 n e n

- V EC TO CUNG

1 Ca

c q uy t dc c an nhcf

a) Quy

tdc tam giac doi

voi phep cgng vecto

AC = A

B + B

C

Qu

y tS

c ta

quy tdc

ba diem

b) Quy

tdc hinh binh hdnh

2

AB CD la

h in

h bi nh h an

h

AB + AD

-

AC

Nh ie

u kh

h ha

nh

cu ng d if

Oc go i

la

quy tdc dudng cheo

hinh binh hdnh

c) Quy

tdc ba diem doi

vai phep trii vecto

Vdri b

a di em

A ,

B, C

b at

i, ta l uo

n c

6 : A

B = C

B C

-A

d) Nhdn

mot vecto voi mot

so

Ti ch

c ua

v ec to

u (

u ? t

0) la

mot vecto,

k

i

hi eu

k

u :

+ k > 0

t

hi

ku

C li ng h ii dn

g vd

i u

+ k < 0

t

hi

ku

n gif cf

c hi Td ng v

di u

+ ku

CO do d

ai !ku

| |k|.|u

-e) Vecto

cung phuang

Di eu k ie

n ca

n va

d

ii de

h

ai ve ct

o a , b (b 0) cu ng p hi/

g) Trung

diem cua doan thdng

M la

t ru ng d ie

m cu

a do an t ha ng

B =

0

h) Trong

tam cua tam giac

G la

t ro ng t am c ua t am g ia

B + G

C =

0

Ve ct

cf va t oa d

p

Cdc phep todn

Ch

o ha

i v ec to

u =

(Uj; U2) , v

= v^)

226 :.':

TS Vu The Hi;u

- Nguyg

n Wn

h Ca

n

+ Tga do vecta tong : ( u + v ) = ( u i + v f , U2 + V2)

+ Tga do vecta hieu : ( u - v ) = ( u i - V i ; U2 - V2)

+ Toa do vecta tick ciia mot vecta vai mot so : k u = ( k u i ; k u 2 )

b) Toa do cua vecta : A ( X A ; Y A ) , B ( X B ; Y B ) = > A B = ( X B X A ; ye Y A )

-c) Dp dai cua vecta u = ( U j ; U 2 ) = >

d) Tga do trung diem cua doqn thdng

M la trung diem ciia doan thang A , B :

M A + M B = 0 o

y.M = ^ ^ ( y A + y B )

e) Tga do trgng tdm ciia tam gidc

G la trong t a m cua tam giac ABC :

- ( x ^ + X 3 + X c )

C A C D A N G BAI TAP V A PHOONG PHAP GIAI

y C a c b a i tap

ta suf dung cac quy tac ba diem, quy tdc h i n h b i n h hanh, cong thufc trung diem

b) Tim toa do cac vectcf, cac diem : Ket hcfp giufa viec chufng m i n h cac dang thufc vecto v6i toa do ciia cac tong, hieu, tich mot so v6i mot vecto

c) Trong cac bai tap ve chufng m i n h cac difcfng thang song song hoac chufng minh cac diem thang hang, ta suf dung k i ^ thufc ve vecto ciing phiJcfng

2] B a i tap

34 a) Chufng m i n h rang neu M la trung diem cua doan thang A B t h i v6i

moi diem N trong mat phang, ta luon c6 he thufc 2 N M = N A + N B b) Chufng m i n h rang neu G la t r o n g t a m ciia t a m giac ABC t h i vdfi moi diem N trong mat phang, ta luon c6 he thifc 3NG = N A + N B + NC

M =

0 (2)

TiT (1) v

a (2) su

y r

a dpcm

b) Do

c gi

a ti

T giai

35.

Cho tij

f gia

c ABCD Go

i M, N, P, Q theo thii

D v

a DA Chufn

D + BC + B

A +

CH

I DA

N

Doc gi

a t

ii giai

A +

MB 3MC, tron

b) Dirn

g die

m I sao ch

o v = CI

c) M' l

CMng min

h :

v = 2M'A

+ SWC

WB-CH

I DA

N

a)

S iJf dung : MA = M

C + CA ; M

B MC + CB

va CB

c) Ta

CO :

2MA = 2MM' +

2WA

MB = MM' + M'B

-3MC = -3MM' - 3M'C

V

= 2M'

A + M'B - 3M'C

; 1), C(2

; 4)

g hang

CH

I DA

N

a) Ta

CO

:AB = (-1

; -4), A

C = (-4; -1)

De tha

y A

B ^ k.AC , V

k

b) Ta

CO

: AB

^ = 17; AC

C can, din

h A

38.

Cho na

m die

m A(l;

-2), B(3

; 0), C(4

; 3), D(2

; 1), E(-2

; -3)

CH

I DA

N

a) Ta

a die

m C, D, E thang hang

c) F(-3;-2)

-n

BAI TOAN 7 TICH V6 Hl/CfNG CUA H A I VECTCl

I- CAC KIEN THLfC C d BAN

1 T i c h v 6 hvtdng c i i a h a i vectcf

a) Binh nghla H a i vectof a, b deu khac vectcf 0 Tich vd hicdng cua a va

b la mot so, k i hieu a.b diiac xac d i n h bofi cong thufc :

- Goc giOfa hai vectcf : cos(a, b) =

- Dieu k i e n can va dii de hai vectof vuong goc vdi nhau :

a l b o a b = 0

2 B i e u thiic t o a dp c i i a t i c h v 6 hifdng

a) Bieu thiic toa do :

Cho hai vectcf a = (ai; a2), b = (hi; b2) t h i : a.b = aibi + a2b2

b) He qua

= V ^ [ 7 b f

- Do dai ciia vectof

- Khoang each giOfa hai diem A(XA; YA), B(XB; ys)

A B = I A B I = V ( x B - x J ^ + { y 3 - y J ^

ajbj + agb^

- Goc giCfa hai vectof : cos(a, b) = , ,

- Dieu kien can va dii de hai vectof vuong goc : a _L b <=> aibi + a2b2 = 0

li- CAC DANG BAi TAP VA PHLfdNG PHAP GiAl

t i n h chat cua tuf giac De giai cac bai tap nay, t a cung thi/cfng difa ve chufng m i n h cac vectcf bSng nhau, cac vectof cung phiTofng va cac vectcf vuong goc vdfi nhau

Hoc va on Itiyen theo CTDT m6n Toan THPT Jl 229

2 Ba

t A(6

; 5), B(5

; 1), C(2

; 4) Chufn

N

Cdch 1.

Ta tirth dUgfc, the

g A

B = AC

= 1

7

=> AA BC can, din

h A

Ta CO

: BA =

(1; 4), B

C (-3; +3)

-l.(-3) +

VF 74^.VG

^37^

V34

CA = (4

; 1), C

B = (3; -3)

cosC = cos(CA, CB

) = 4.3 +

(-l).(-3)

V34

Do d

o : cosB = cosC, v

i 0 < B < 90°

, 0 < C < 90° ne

C can, din

h A

40 Ch

o ha

i die

m A(-2

; 1), B(4

; 3) v

a die

m C(-l; y) Xa

N

a) Ta CO : AB = (6

; 2), A

C = (1; y

- 1)

^ 6

1 + 2(y - 1) =

0 =: > y = -

2 ha

y C(-l; 2)

C = (-5; y

- 3)

2(y 3) =

0 =

> y = 1

8 =

> C(-l; 18

)

c) Ta CO : C

A = (-1; 1

- y), C

B = (5; 3

- y)

i C

A i C

n die

u kie

n AAB

C vuong

Ci(-1; 2 + V6), C2(-l; 2

- V6)

CH

I DA

m gia

c A BC v uo ng can, din

h C Nhu

f vay, t

a ca

n c

6

fCA I

CB = C CA

B

CA IC

B

CA

| = |CB

CA.CB

0

CB CA (*)

230 L'

; TS V

D Th

e Hu

u Nguygn VTn

-h Ca

n

Ta C O : CA = (4 - x; 3 - y), CB = (2 - x; 5 - y)

CA Tif dieu kien (*), ta c6 :

42 Cho tam giac ABC v6i A(3; 1), B(0; 7), C(5; 2)

a) Chufng minh tam giac ABC vuong

b) Tinh dien tich S cua tam giac

c) Tinh do dai diidng cao ke tU dinh A

a) AABC vuong tai C

b) AABC la tam giac c6 ba goc nhon

c) AABC CO goc A la goc tti

Vdi cac goc B, C ta cung xet tiiofng tif

Trong cau c) vi AB.AC = -15 < 0 ^cosA < 0 =>A > 90°

44 Chufng minh tuf giac ABCD v6i A(2; 6), B(5; 0), C(-3; -4), D(-6; 2) la

hinh chuf nhat