Bai Tap Hinh Hoc Khong Gian Lop 12 Co Loi Giai

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TH TCH LNG TRDng 1: Khi lng tr ng c chiu cao hay cnh y

V d 1: y ca lng tr ng tam gic ABC.ABC l tam gic ABC vung cn ti A c

cnh BC = a v bit A'B = 3a. Tnh th tch khi lng tr.

Li gii:

Ta c

vung cn ti A nn AB = AC = a

ABC A'B'C' l lng tr ng

Vy V = B.h = SABC .AA' =

V d 2:Cho lng tr t gic u ABCD.ABCD' c cnh bn bng 4a v ng cho 5a. Tnh th tch khi lng tr ny

Li gii:

ABCD A'B'C'D' l lng tr ng nn

BD2 = BD'2 - DD'2 = 9a2

ABCD l hnh vung

Suy ra B = SABCD =

Vy V = B.h = SABCD.AA' = 9a3

V d 3: y ca lng tr ng tam gic ABC.ABC l tam gic u cnh a = 4 v bit din tch tam gic ABC bng 8. Tnh th tch khi lng tr.

+ Phn tch V= B.h tm B v h

trong hnh l cc i tng no ?

+ Tm din tch B = SABC bng cng thc no ?

+ T din tch suy ra cnh no ? ti sao ?

+ Tm h = AA' dng tam gic no v nh l g ?

Li gii:

Gi I l trung im BC .Ta cABC u nn

Vy : VABC.ABC = SABC .AA'=

V d 4: Cho hnh hp ng c y l hnh thoi cnh a v c gc nhn bng 600 ng cho ln ca y bng ng cho nh ca lng tr. Tnh th tch hnh hp . + Phn tch V= B.h tm B v h trong hnh l cc i tng no ?

+ Tm din tch B ca hnh thoi ABCD bng cch no ?

+ Tm h = DD' trong tam gic vung no ? v nh l g ?

Li gii:

Ta c tam gic ABD u nn : BD = a

v SABCD = 2SABD =

Theo bi BD' = AC =

Vy V = SABCD.DD' =

V d 5: Mt tm ba hnh vung c cnh 44 cm, ngi ta ct b i mi gc tm ba mt hnh vung cnh 12 cm ri gp li thnh mt ci hp ch nht khng c np. Tnh th tch ci hp ny.

+ Phn tch V= B.h tm B v h trong hnh l cc i tng no ?

+ Tm h = AA' ? Ti sao ?

+ Tm AB ? Suy ra B = SABCD = AB2 ?

Gii

Theo bi, ta c

AA' = BB' = CC' = DD' = 12 cm nn ABCD l hnh vung c AB = 44 cm - 24 cm = 20 cm

v chiu cao hp h = 12 cm

Vy th tch hp l V = SABCD.h = 4800cm3

BI TP T RN LUYNBi 1: Cho lng tr ng c y l tam gic u bit rng tt c cc cnh ca lng tr bng a. Tnh th tch v tng din tch cc mt bn ca lng tr. S: ; S = 3a2Bi 2: Cho lng tr ng ABCD.A'B'C'D' c y l t gic u cnh a bit rng . Tnh th tch ca lng tr. s: V = 2a3Bi 3.Lng tr ng t gic c y l hnh thoi m cc ng cho l 6cm v 8cm bit rng chu vi y bng 2 ln chiu cao lng tr.Tnh th tch v tng din tch cc mt ca lng tr. s:V = 240cm3 v S = 248cm2Bi 4: Cho lng tr ng tam gic c di cc cnh y l 37cm ; 13cm ;30cm v bit tng din tch cc mt bn l 480 cm2 . Tnh th tch lng tr .

s: V = 1080 cm3Bi 5: Cho lng tr ng tam gic ABC A'B'C' c y ABC l tam gic vung cn ti A ,bit rng chiu cao lng tr l 3a v mt bn AA'B'B c ng cho l 5a . Tnh th tch lng tr.

s: V = 24a3Bi 6:Cho lng tr ng t gic u c tt c cc cnh bng nhau v bit tng din tch cc mt ca lng tr bng 96 cm2 .Tnh th tch lng tr.

s: V = 64 cm3Bi 7.Cho lng tr ng tam gic c cc cnh y l 19,20,37 v chiu cao ca khi lng tr bng trung bnh cng cc cnh y. Tnh th tch ca lng tr.

s: V = 2888Bi 8. Cho khi lp phng c tng din tch cc mt bng 24 m2 .Tnh th tch khi lp phng

s: V = 8 m3Bi 9:Cho hnh hp ch nht c 3 kch thc t l thun vi 3,4,5 bit rng di mt ng cho ca hnh hp l 1 m.Tnh th tch khi hp ch nht.

s: V = 0,4 m3Bi 10. Cho hnh hp ch nht bit rng cc ng cho ca cc mt ln lt l . Tnh th tch khi hp ny.

s: V = 6

Dng 2: Lng tr ng c gc gia ng thng v mt phngV d 1: Cho lng tr ng tam gic ABC A'B'C' c y ABC l tam gic vung cn ti B vi BA = BC = a ,bit A'B hp vi y ABC mt gc 600 . Tnh th tch lng tr.

*) Tm hnh chiu ca A'B trn y ABC. Suy ra gc [A'B,(ABC)] = ?

*) Phn tch V= B.h tm B v h trong hnh l cc i tng no ?

*) Tm din tch B ca tam gic ABC bng cng thc no ?

*) Tm h = AA' trong tam gic vung no ? v dng h thc lng gic no ?

Li gii:

Ta c l hnh chiu ca A'B trn y ABC .

Vy

SABC =

Vy V = SABC.AA' =

V d 2: Cho lng tr ng tam gic ABC A'B'C' c y ABC l tam gic vung ti A vi

AC = a , = 60 o bit BC' hp vi (AA'C'C) mt gc 300. Tnh AC' v th tch lng tr.

Phn tch *) Tm hnh chiu ca BC' trn (AA'C'C). Suy ra gc [BC',(AA'C'C)] = ?

*) Tm AC' trong tam gic no?Dng h thc lng gic g ?




Li gii:

.

Ta c:

nn AC' l hnh chiu ca BC' trn (AA'C'C).

Vy gc[BC';(AA"C"C)] = = 30o

V = B.h = SABC.AA'

l na tam gic u nn . Vy V =

V d 3: Cho lng tr ng ABCD A'B'C'D' c y ABCD l hnh vung cnh a v ng cho BD' ca lng tr hp vi y ABCD mt gc 300. Tnh th tch v tng din tch ca cc mt bn ca lng tr .

Phn tch

*) Dng hnh vung ABCD hay A'B'C'D' v cc cnh bn ca hnh lng tr .

*) Dng BD' v BD ?

phn tch yu cu ca bi ra cc yu cu nh: *) Tm hnh chiu ca BD' trn y ABCD. Suy ra gc [BD',(ABCD)] = ?


*) Tm din tch B ca hnh vung ABCD bng cng thc no ?

*) Tm h = DD' trong tam gic vung no ? v dng h thc lng gic no ?

Gii: Ta c ABCD A'B'C'D' l lng tr ng nn ta c: v BD l hnh chiu ca BD' trn ABCD . Vy gc [BD';(ABCD)] =

Vy V = SABCD.DD' = S = 4SADD'A' =

V d 4: Cho hnh hp ng ABCD A'B'C'D' c y ABCD l hnh thoi cnh a v = 60o bit AB' hp vi y (ABCD) mt gc 30o .Tnh th tch ca hnh hp.

Phn tch yu cu ca bi ra cc yu cu nh:*) Tm hnh chiu ca AB' trn (ABCD). Suy ra gc [AB',(ABCD)] = ?


*) Dng BD. Suy ra ABD c hnh tnh g ? Suy ra din tch B ca ABCD bng cch no?

+Tnh h = BB' trong tam gic no ? Dng h thc lng gic no ?

Gii u cnh a

vung tiB

Vy

BI TP T RN LUYNBi 1. Cho lng tr ng ABC A'B'C' c y ABC vung cn ti B bit A'C = a v A'C hp vi mt bn (AA'B'B) mt gc 30o . Tnh th tch lng tr

S:

Bi 2. Cho lng tr ng ABC A'B'C' c y ABC vung ti B bit BB' = AB = a v B'C hp vi y (ABC) mt gc 30o . Tnh th tch lng tr.

S:

Bi 3. Cho lng tr ng ABC A'B'C' c y ABC l tam gic u cnh a bit AB' hp vi mt bn (BCC'B') mt gc 30o . Tnh di AB' v th tch lng tr .

S: ; Bi 4. Cho lng tr ng ABC A'B'C' c y ABC vung ti A bit AC = a v bit BC' hp vi mt bn (AA'C'C) mt gc 30o .Tnh th tch lng tr v din tch tam gic ABC'. S

Bi 5. Cho lng tr tam gic u ABC A'B'C' c khong cch t A n mt phng (A'BC) bng a v AA' hp vi mt phng (A'BC) mt gc 300 . Tnh th tch lng tr

S:

Bi 6. Cho hnh hp ch nht ABCD A'B'C'D' c ng cho A'C = a v bit rng A'C hp vi (ABCD) mt gc 30o v hp vi (ABB'A') mt gc 45o .Tnh th tch ca khi hp ch nht. s:

Bi 7. Cho hnh hp ng ABCD A'B'C'D' c y ABCD l hnh vung . Gi O l tm ca ABCD v OA' = a Tnh th tch ca khi hp khi:1) ABCD A'B'C'D' l khi lp phng .S

2) OA' hp vi y ABCD mt gc 60o .S

3) A'B hp vi (AA'CC') mt gc 30o.S

Bi 8. Cho lng tr ng ABCD A'B'C'D' c y ABCD l hnh vung v BD' = a . Tnh th tch lng tr trong cc trng hp sau y:

1) BD' hp vi y ABCD mt gc 60o .S

2) BD' hp vi mt (AA'D'D) mt gc 30o . S

Bi 9. Chiu cao ca lng tr t gic u bng a v gc ca 2 ng cho pht xut t mt nh ca 2 mt bn k nhau l 60o.Tnh th tch lng tr v tng din tch cc mt ca lng tr . s: V = a3 v S = 6a2Bi 10.Cho hnh hp ch nht ABCD A'B'C'D' c AB = a ;AD = b;AA' = c v BD' = AC' = CA' =

1) Chng minh ABCD A'B'C'D' l hp ch nht.

2) Gi x,y,z l gc hp bi mt ng cho v 3 mt cng i qua mt ng thuc ng cho. Chng minh rng .

Dng 3: Lng tr ng c gc gia 2 mt phng

V d 1. Cho lng tr ng tam gic ABC A'B'C' c y ABC l tam gic vung cn ti B vi BA = BC = a ,bit (A'BC) hp vi y (ABC) mt gc 600 .Tnh th tch lng tr.

Phn tch yu cu ca bi ra cc yu cu nh: *) Nhn xt AB v A'B c vung gc vi BC khng ? ti sao?

*) Suy ra gc[(A'BC);(ABC)] = ?




Li gii:

Ta c

Vy

.

SABC = Vy V = SABC.AA' =

V d 2: y ca lng tr ng tam gic ABC.ABC l tam gic u . Mt (ABC) to vi y mt gc 300 v din tch tam gic ABC bng 8. Tnh th tch khi lng tr.

Phn tch yu cu ca bi ra cc yu cu nh: *) Nhn xt c hnh tnh g ? Suy ra I l trung im ca BC cho ta v tr AI v A'I th no vi

BC? Suy ra gc[(A'BC);(ABC)] = ?


*) t BC = 2x . Suy ra A'I bi tam gic no ?

*) T din tch tam gi A"BC suy ra x bi cng thc no?


Gii. u m AA' nn A'I(l 3). Vy gc[(A'BC);)ABC)] = = 30o

Gi s BI = x .Ta c

AA = AI.tan 300 =

Vy VABC.ABC = CI.AI.AA = x3

M SABC = BI.AI = x.2x = 8.Do VABC.ABC = 8

V d 3. Cho lng tr t gic u ABCD A'B'C'D' c cnh y a v mt phng (BDC') hp vi y (ABCD) mt gc 60o. Tnh th tch khi hp ch nht.

Phn tch yu cu ca bi ra cc yu cu nh:*) Xc nh gc[BDC');(ABCD)] = ?


*) Tm din tch B ca ABCD bng cng thc no ?

*) Tm h = CC' trong tam gic vung no ? v dng h thc lng gic no ?

Gii.

Gi O l tm ca ABCD . Ta c

ABCD l hnh vung nn

CC'(ABCD) nn OC'BD (l 3).

Vy gc[(BDC');(ABCD)] = = 60o

Ta c V = B.h = SABCD.CC'

ABCD l hnh vung nn SABCD = a2

vung nn CC' = OC.tan60o =

Vy V =

V d 4. Cho hnh hp ch nht ABCD A'B'C'D' c AA' = 2a ; mt phng (A'BC) hp vi y (ABCD) mt gc 60o v A'C hp vi y (ABCD) mt gc 30o .Tnh th tch khi hp ch nht.

Phn tch yu cu ca bi ra cc yu cu nh:*) Nhn xt AB v A'B c vung gc vi BC khng ? ti sao?

*) Suy ra gc[(A'BC);(ABCD)] = ?

*) Tm hnh chiu ca A'C trn (ABCD) ? Suy ra gc[A'C,(ABCD)] = ?



*) Tm AB v AC bi tam gic vung no? Dng h thc lng gic no ?


Ta c AA' AC l hnh chiu ca A'C trn (ABCD) .

Vy gc[A'C,(ABCD)] =

BC AB BC A'B (l 3) . [(A'BC),(ABCD)] =

AC = AA'.cot30o =

AB = AA'.cot60o =

Vy V = AB.BC.AA' =

BI TP T RN LUYN

Bi 1. Cho hp ch nht ABCD A'B'C'D' c AA' = a bit ng cho A'C hp vi y ABCD mt gc 30o v mt (A'BC) hp vi y ABCD mt gc 600 .Tnh th tch hp ch nht.

s:

Bi 2. Cho lng tr ng ABCD A'B'C'D' c y ABCD l hnh vung v cnh bn bng a bit rng mt (ABC'D') hp vi y mt gc 30o.Tnh th tch khi lng tr.

s: V = 3a3Bi 3. Cho lng tr ng ABCA'B'C' c y ABC l tam gic vung cn ti B v AC = 2a bit rng (A'BC) hp vi y ABC mt gc 45o. Tnh th tch lng tr.

s:

Bi 4. Cho lng tr ng ABCA'B'C' c y ABC l tam gic cn ti A vi AB = AC = a v bit rng (A'BC) hp vi y ABC mt gc 45o. Tnh th tch lng tr.

s:

Bi 5. Cho lng tr ng ABCA'B'C' c y ABC l tam gic vung ti B v BB' = AB = h bit rng (B'AC) hp vi y ABC mt gc 60o. Tnh th tch lng tr.

s:

Bi 6. Cho lng tr ng ABC A'B'C' c y ABC u bit cnh bn AA' = a.Tnh th tch lng tr trong cc trng hp sau y:

1) Mt phng (A'BC) hp vi y ABC mt gc 60o .

s:

2) A'B hp vi y ABC mt gc 45o.

S:

3) Chiu cao k t A' ca tam gic A'BC bng di cnh y ca lng tr.S:

Bi 7. Cho lng tr t gic u ABCD A'B'C'D' c cnh bn AA' = 2a .Tnh th tch lng tr trong cc trng hp sau y:

1) Mt (ACD') hp vi y ABCD mt gc 45o .

S : V = 16a32) BD' hp vi y ABCD mt gc 600 .

S : V = 12a33) Khong cch t D n mt (ACD') bng a .

S :

Bi 8. Cho lng tr ng ABCD A'B'C'D' c y ABCD l hnh vung cnh a. Tnh th tch lng tr trong cc trng hp sau y:1)Mt phng (BDC') hp vi y ABCD mt gc 60o

S : .

2)Tam gic BDC' l tam gic u.

S : V =

3)AC' hp vi y ABCD mt gc 450

S : V =

Bi 9. Cho lng tr ng ABCD A'B'C'D' c y ABCD l hnh thoi cnh a v gc nhn A = 60o .Tnh th tch lng tr trong cc trng hp sau y:1) (BDC') hp vi y ABCD mt gc 60o .S:

2)Khong cch t C n (BDC') bng

S :

3)AC' hp vi y ABCD mt gc 450S :

Bi 10. Cho hnh hp ch nht ABCD A'B'C'D' c BD' = 5a ,BD = 3a.Tnh th tch khi hp trong cc trng hp sau y:

1) AB = a

S :

2) BD' hp vi AA'D'D mt gc 30o

S :

3) (ABD') hp vi y ABCD mt gc 300

S :

Dng 4. Khi lng tr xin

V d 1. Cho lng tr xin tam gic ABC A'B'C' c y ABC l tam gic u cnh a , bit cnh bn l v hp vi y ABC mt gc 60o . Tnh th tch lng tr.

Phn tch yu cu ca bi ra cc yu cu nh:*) Xc nh gc gia cnh bn vi y : Hnh chiu ca CC' trn (ABC) l g?

*) Suy ra gc[CC';(ABC)] = ?



*) Tm h = CC' trong tam gic vung no ? v dng h thc lng gic no ?

Li gii:

Ta c l hnh chiu ca CC' trn (ABC)

Vy

SABC = .Vy V = SABC.C'H =

V d 2. Cho lng tr xin tam gic ABC A'B'C' c y ABC l tam gic u cnh a . Hnh chiu ca A' xung (ABC) l tm O ng trn ngoi tip tam gic ABC bit AA' hp vi y ABC mt gc 60 .

1) Chng minh rng BB'C'C l hnh ch nht.

2) Tnh th tch lng tr .

Phn tch yu cu ca bi ra cc yu cu nh:*) Xc nh gc gia cnh bn AA' vi y ABC : Hnh chiu ca AA' trn (ABC) l g? Suy ra gc[AA'';(ABC)] = ?

*) Chng minh BC AA' bng cch Chng minh BC mt phng no ? T c th BCCC' khng ?

ti sao? Vy BB'C'C l hnh g?



*) Tm h = AA'' trong tam gic vung no ? v dng h thc lng gic no ?

Li gii:

1) Ta c l hnh chiu ca AA' trn (ABC)

Vy

Ta c BB'CC' l hnh bnh hnh ( v mt bn ca lng tr)

ti trung im H ca BC nn (l 3 )

m AA'//BB' nn .Vy BB'CC' l hnh ch nht.

2) u nn

Vy V = SABC.A'O =

V d 3. Cho hnh hp ABCD.ABCD c y l hnh ch nht vi AB = AD =.Hai mt bn (ABBA) v (ADDA) ln lt to vi y nhng gc 450 v 600. . Tnh th tch khi hp nu bit cnh bn bng 1.

Phn tch yu cu ca bi ra cc yu cu nh: *) Xc nh gc gia mt bn vi y.Dng ng cao A'H v HNAD

HMAB Suy ra gc[(ABB'A');(ABCD)] =? gc[(ADD'A');(ABCD)] = ?



*) Tm h = A'H khng dng trc tip tam gic vung no c ? t x = A'H

*) Dng hai tam gic no bi nh l g to ra phng trnh theo x ?

Li gii:

K AH ,HM (l 3)

t AH = x . Khi AN = x : sin 600 =

AN =

M HM = x.cot 450 = x

Ngha l x =

Vy VABCD.ABCD = AB.AD.x =

BI TP T RN LUYNBi 1. Cho lng tr ABC A'B'C'c cc cnh y l 13;14;15v bit cnh bn bng 2a hp vi y ABCD mt gc 45o . Tnh th tch lng tr.

s: V =

Bi 2. Cho lng tr ABCD A'B'C'D'c y ABCD l hnh vung cnh a v bit cnh bn bng 8 hp vi y ABC mt gc 30o.Tnh th tch lng tr.

s: V =336

Bi 3. Cho hnh hp ABCD A'B'C'D'c AB =a;AD =b;AA' = c v v bit cnh bn AA' hp vi y ABC mt gc 60o.Tnh th tch lng tr.

s: V =

Bi 4. Cho lng tr tam gic ABC A'B'C' c y ABC l tam gic u cnh a v im A' cch u A,B,C bit AA' = .Tnh th tch lng tr.

s: Bi 5. Cho lng tr ABC A'B'C' c y ABC l tam gic u cnh a , nh A' c hnh chiu trn (ABC) nm trn ng cao AH ca tam gic ABC bit mt bn BB'C'C hp vio y ABC mt gc 60o .

1) Chng minh rng BB'C'C l hnh ch nht.

2) Tnh th tch lng tr ABC A'B'C'.

s:

Bi 6. Cho lng tr ABC A'B'C' c y ABC l tam gic u vi tm O. Cnh b CC' = a hp vi y ABC 1 gc 60o v C' c hnh chiu trn ABC trng vi O .

1) Chng minh rng AA'B'B l hnh ch nht. Tnh din tch AA'B'B.

S :

2) Tnh th tch lng tr ABCA'B'C'.

S: Bi 7. Cho lng tr ABC A'B'C' c y ABC l tam gic u cnh a bit chn ng vung gc h t A' trn ABC trng vi trung im ca BC v AA' = a.

1) Tm gc hp bi cnh bn vi y lng tr.

S : 30o. 2) Tnh th tch lng tr

S:

Bi 8. Cho lng tr xin ABC A'B'C' c y ABC l tam gic u vi tm O. Hnh chiu ca C' trn (ABC) l O.Tnh th tch ca lng tr bit rng khong cch t O n CC' l a v 2 mt bn AA'C'Cv BB'C'C hp vi nhau mt gc 90o

s:

Bi 9. Cho hnh hp ABCD A'B'C'D' c 6 mt l hnh thoi cnh a,hnh chiu vung gc ca A' trn(ABCD) nm trong hnh thoi,cc cnh xut pht t A ca hp i mt to vi nhau mt gc 60o .

1) Chng minh rng H nm trn ng cho AC ca ABCD.

2) Tnh din tch cc mt cho ACC'A' v BDD'B'.

S:

3) Tnh th tch ca hp.

s:

Bi 10. Cho hnh hp ABCD A'B'C'D' c y ABCD l hnh thoi cnh a v gc A = 60o chn ng vung gc h t B' xung ABCD trng vi giao im 2 ng cho y bit BB' = a.

1) Tm gc hp bi cnh bn v y.

S : 60o 2) Tnh th tch v tng din tch cc mt bn ca hnh hp.

S:

TH TCH KHI CHP

Dng 1: Khi chp c cnh bn vung gc vi y

V d 1. Cho hnh chp SABC c SB = SC = BC = CA = a . Hai mt (ABC) v (ASC) cng vung gc vi (SBC). Tnh th tch hnh chp .

Phn tch yu cu ca bi ra cc yu cu nh: *) Phn tch V= B.h tm B v h trong hnh l cc i tng no ?

*) Tm din tch B ca SBC bng cng thc no ?

Li gii:

Ta c

Do

V d 2: Cho hnh chp SABC c y ABC l tam gic vung cn ti B vi AC = a bit SA vung gc vi y ABC v SB hp vi y mt gc 60o.

1) Chng minh cc mt bn l tam gic vung .

2) Tnh th tch hnh chp .

Phn tch yu cu ca bi ra cc yu cu nh: *) Xc nh gc[SB,(ABC)] = ? Ti sao?


*) Tm din tch B ca ABC bng cng thc no ? Tnh BA ?

*) Tm h = SA qua tam gic no bi cng thc g ?

Li gii:

1)

m ( l 3 ).

Vy cc mt bn chp l tam gic vung.

2) Ta c l hnh chiu ca SB trn (ABC).

Vy gc[SB,(ABC)] = .

vung cn nn BA = BC =

SABC = ;

Vy

V d 3. Cho hnh chp SABC c y ABC l tam gic u cnh a bit SA vung gc vi y ABC v (SBC) hp vi y (ABC) mt gc 60o. Tnh th tch hnh chp .

Phn tch yu cu ca bi ra cc yu cu nh: *) Xc nh gc[(SBC),(ABC)] = ? Ti sao?


*) Tm din tch B ca ABC bng cng thc no ?

*) Tm h = SA qua tam gic no v cng thc g ?

Li gii:

M l trung im ca BC,v tam gic ABC u nn

AM BCSABC (l3) .[(SBC);(ABC)] = .

Ta c V =

Vy V =

V d 4. Cho hnh chp SABCD c y ABCD l hnh vung c cnh a v SA vung gc y ABCD v mt bn (SCD) hp vi y mt gc 60o.

1) Tnh th tch hnh chp SABCD.

2) Tnh khong cch t A n mt phng (SCD).

Phn tch bi dng hnh :

*) Dng t gic ABCD v cnh bn SA(ABCD) ? .

Hng dn hc sinh phn tch yu cu ca bi ra cc yu cu nh: *) Xc nh gc[(SCD),(ABCD)] = ? Ti sao?



*) Tm h = SA qua tam gic no bi cng thc g ?

Li gii:

1)Ta c v ( l 3 ).(1)

Vy gc[(SCD),(ABCD)] = = 60o .

vung nn SA = AD.tan60o =

Vy

2) Ta dng AH ,v CD(SAD) (do (1) ) nn CD AH

EMBED Equation.DSMT4

Vy AH l khong cch t A n (SCD).

. Vy AH =

BI TP T RN LUYNBi 1: Cho hnh chp SABC c y ABC l tam gic vung cn ti B vi BA=BC=a bit SA vung gc vi y ABC v SB hp vi (SAB) mt gc 30o. Tnh th tch hnh chp .

s: V =

Bi 2. Cho hnh chp SABC c SA vung gc vi y (ABC) v SA = h ,bit rng tam gic ABC u v mt (SBC) hp vi y ABC mt gc 30o .Tnh th tch khi chp SABC

s:

Bi 3. Cho hnh chp SABC c y ABC vung ti A v SB vung gc vi y ABC bit SB = a,SC hp vi (SAB) mt gc 30o v (SAC) hp vi (ABC) mt gc 60o .Chng minh rng SC2 = SB2 + AB2 + AC2 Tnh th tch hnh chp.

s: Bi 4: Cho t din ABCD c AD(ABC) bit AC = AD = 4 cm,AB = 3 cm, BC = 5 cm.

1) Tnh th tch ABCD.

s: V = 8 cm32) Tnh khong cch t A n mt phng (BCD).

s: d =

Bi 5: Cho khi chp SABC c y ABC l tam gic cn ti A vi BC = 2a , , bit v mt (SBC) hp vi y mt gc 45o . Tnh th tch khi chp SABC.

s:

Bi 6: Cho khi chp SABCD c y ABCD l hnh vung bit SA (ABCD),SC = a v SC hp vi y mt gc 60o Tnh th tch khi chp.

s:

Bi 7: Cho khi chp SABCD c y ABCD l hnh ch nht bit rng SA (ABCD) , SC hp vi y mt gc 45o v AB = 3a , BC = 4a. Tnh th tch khi chp. s: V = 20a3Bi 8: Cho khi chp SABCD c y ABCD l hnh thoi cnh a v gc nhn A bng 60o v SA (ABCD) Bit rng khong cch t A n cnh SC = a.Tnh th tch khi chp SABCD.

s:

Bi 9: Cho khi chp SABCD c y ABCD l hnh thang vung ti A v B bit AB = BC = a , AD = 2a , SA (ABCD) v (SCD) hp vi y mt gc 60o Tnh th thch khi chp SABCD. s:

Bi 10 :Cho khi chp SABCD c y ABCD l na lc gic u ni tip trong na ng trn ng knh AB = 2R bit (SBC) hp vi y ABCD mt gc 45o.Tnh th tch khi chp SABCD.s:

Dng 2 : Khi chp c mt mt bn vung gc vi y

V d 1: Cho hnh chp S.ABCD c y ABCD l hnh vung c cnh a. Mt bn SAB l tam gic u nm trong mt phng vung gc vi yABCD,

1) Chng minh rng chn ng cao khi chp trng vi trung im cnh AB.

2) Tnh th tch khi chp SABCD.

Phn tch yu cu ca bi ra cc yu cu nh: *) H l trung im ca AB. Chng minh SH (ABCD) ?



*) Tm h = SH qua tam gic no bi cng thc g ?

Li gii:

1) Gi H l trung im ca AB.

u

m

Vy H l chn ng cao ca khi chp.

2) Ta c tam gic SAB u nn SA =

suy ra

V d 2: Cho t din ABCD c ABC l tam gic u ,BCD l tam gic vung cn ti D , (ABC)(BCD) v AD hp vi (BCD) mt gc 60o .Tnh th tch t din ABCD.

Phn tch yu cu ca bi ra cc yu cu nh: *) Xc nh gc[AD,(BCD)] = ? Tm hnh chiuca AD trn (BCD) ?


*) Tm din tch B ca BCD bng cng thc no ?

*) Tm h = AH qua tam gic no bi cng thc g ?

Li gii:

Gi H l trung im ca BC.Ta c tam gic ABC u nn AH(BCD) , m (ABC) (BCD) AH .

Ta c AHHDAH = AD.tan60o =

& HD = AD.cot60o =

BC = 2HD = suy ra

V =

V d 3: Cho hnh chp S.ABC c y ABC l tam gic vung cn ti B, cBC = a. Mt bn SAC vung gc vi y, cc mt bn cn li u to vi mt y mt gc 450.

a. Chng minh rng chn ng cao khi chp trng vi trung im cnh AC.

b. Tnh th tch khi chp SABC.

Phn tch bi dng hnh :

*) Dng tam gic ABC v SAC da vo (SAC)(ABC) ? .

Phn tch yu cu ca bi ra cc yu cu nh: *) Xc nh gc[(SAB),(ABC)] = ? v gc[(SBC),(ABC)] = ?

*) So snh tam gic SHI v SHJ cho g ? Suy ra AH l g ca tam gic ABC ?



*) Tm h = SH qua cc tam gic no bi tch cht g ?

Li gii:

a) K SH BC v mp(SAC)mp(ABC) nn SHmp(ABC).

Gi I, J l hnh chiu ca H trn AB v BC SIAB, SJBC, theo gi thit

Ta c: nn BH l ng phn gic ca suy ra H l trung im ca AC.

b) HI = HJ = SH =

EMBED Equation.DSMT4 VSABC=

BI TP T RN LUYNBi 1: Cho hnh chp SABC c y ABC u cnh a, tam gic SBC cn ti S v nm trong mt phng vung gc vi (ABC).

1) Chng minh chn ng cao ca chp l trung im ca BC.

2) Tnh th tch khi chp SABC.

s:

Bi 2: Cho hnh chp SABC c y ABC vung cn ti A vi AB = AC = a bit tam gic SAB cn ti S v nm trong mt phng vung gc vi (ABC) ,mt phng (SAC) hp vi (ABC) mt gc 45o. Tnh th tch ca SABC.

s:

+Bi 3: Cho hnh chp SABC c ; SBC l tam gic u cnh a v (SAB) (ABC). Tnh th tch khi chp SABC.

s:

Bi 4: Cho hnh chp SABC c y ABC l tam gic u;tam gic SBC c ng cao SH = h v (SBC) (ABC). Cho bit SB hp vi mt (ABC) mt gc 30o .Tnh th tch hnh chp SABC.s:

Bi 5: T din ABCD c ABC v BCD l hai tam gic u ln lt nm trong hai mt phng vung gc vi nhau bit AD = a.Tnh th tch t din.

s:

Bi 6 :Cho hnh chp S.ABCD c y ABCD l hnh vung .Mt bn SAB l tam gic u c ng cao SH = h ,nm trong mt phng vung gc vi ABCD,

1) Chng minh rng chn ng cao khi chp trng vi trung im cnh AB.

2) Tnh th tch khi chp SABCD .

s:

Bi 7: Cho hnh chp SABCD c ABCD l hnh ch nht , SAB u cnh a nm trong mt phng vung gc vi (ABCD) bit (SAC) hp vi (ABCD) mt gc 30o .Tnh th tch hnh chp SABCDs:

Bi 8: Cho hnh chp SABCD c ABCD l hnh ch nht c AB = 2a , BC = 4a, SAB (ABCD) , hai mt bn (SBC) v (SAD) cng hp vi y ABCD mt gc 30o .Tnh th tch hnh chp SABCD. s:

Bi 9:Cho hnh chp SABCD c y ABCD l hnh thoi vi AC = 2BD = 2a v SAD vung cn ti S , nm trong mt phng vung gc vi ABCD. Tnh th tch hnh chp SABCD.

s:

Bi 10: Cho hnh chp SABCD c y ABCD l hnh thang vung ti A v D; AD = CD = a ; AB = 2a,SAB u nm trong mt phng vung gc vi (ABCD). Tnh th tch khi chp SABCD . s: Dng 3 : Khi chp u

V d 1: Cho chp tam gic u SABC cnh y bng a v cnh bn bng 2a. Chng minh rng chn ng cao k t S ca hnh chp l tm ca tam gic u ABC.Tnh th tch chp u SABC .

? Dng tam gic u ABC , t tm O dng SO (ABC) . Ti sao ?

Phn tch yu cu ca bi ra cc yu cu nh: *) So snh SA,SB,SC suyra OA,OB,OC bi tch cht no ?



*) Tm h = SO qua tam gic no bi nh l g ?

Li gii:

Dng SO(ABC) Ta c SA = SB = SC suy ra OA = OB = OC

Vy O l tm ca tam gic u ABC.

Ta c tam gic ABC u nn

AO =

.Vy

V d 2:Cho khi chp t gic SABCD c tt c cc cnh c di bng a .

1) Chng minh rng SABCD l chp t gic u.

2) Tnh th tch khi chp SABCD.

? Dng hnh thoi ABCD v t cu hi 1, dng SO (ABCD) . Ti sao ?

Phn tch yu cu ca bi ra cc yu cu nh: *) Hnh thoi ABCD c ni tip trong ng trn khng? Suy ra g t gi thit?



*) Tm h = SO qua tam gic no bi nh l g ?

Li gii:

Dng SO (ABCD)

Ta c SA = SB = SC = SD nn

OA = OB = OC = ODABCD l hnh thoi c ng trn gnoi tip nn ABCD l hnh vung .

Ta c SA2 + SB2 = AB2 +BC2 = AC2 nn vung ti S


Vy

V d 3: Cho khi t din u ABCD cnh bng a, M l trung im DC.

a) Tnh th tch khi t din u ABCD.

b)Tnh khong cch t M n mp(ABC).Suy ra th tch hnh chp MABC.

? Dng tam gic u ABC ,t tm O dng DO (ABC) . Ti sao ?

Phn tch yu cu ca bi ra cc yu cu nh:*) Phn tch V= B.h tm B v h trong hnh l cc i tng no ?


*) Tm h = DO qua tam gic no bi nh l g ?

*) Mt phng (DCO)(ABC) ? Dng MHOC suy ra iu g ?Tnh MH ?

Li gii:

a) Gi O l tm ca


,


b) K MH// DO, khong cch t M n mp(ABC) l MH

.Vy

BI TP T RN LUYN

Bi 1: Cho hnh chp u SABC c cnh bn bng a hp vi y ABC mt gc 60o . Tnh th tch hnh chp.

s:

Bi 2: Cho hnh chp tam gic u SABC c cnh bn a, gc y ca mt bn l 45o.

1) Tnh di chiu cao SH ca chp SABC .

s: SH =

2) Tnh th tch hnh chp SABC.

s:

Bi 3: Cho hnh chp tam gic u SABC c cnh y a v mt bn hp vi y mt gc 60o. Tnh th tch hnh chp SABC.

s:

Bi 4 : Cho chp tam gic u c ng cao h hp vi mt mt bn mt gc 30o .

Tnh th tch hnh chp.

s:

Bi 5 : Cho hnh chp tam gic u c ng cao h v mt bn c gc nh bng 60o.

Tnh th tch hnh chp.

s:

Bi 6 : Cho hnh chp t gic u SABCD c cnh y a v .

1) Tnh tng din tch cc mt bn ca hnh chp u.

s:

2) Tnh th tch hnh chp.

s:

Bi 7 : Cho hnh chp t gic u SABCD c chiu cao h ,gc nh ca mt bn bng 60o. Tnh th tch hnh chp.

s:

Bi 8: Cho hnh chp t gic u c mt bn hp vi y mt gc 45o v khong cch t chn ng cao ca chp n mt bn bng a.Tnh th tch hnh chp . s:

Bi 9: Cho hnh chp t gic u c cnh bn bng a hp vi y mt gc 60o.

Tnh th tch hnh chp.

s:

Bi 10: Cho hnh chp SABCD c tt c cc cnh bng nhau. Chng minh rng SABCD l chp t gic u.Tnh cnh ca hnh chp ny khi th tch ca n bng .

s: AB = 3a Dng 4 : Khi chp & phng php t s th tch

V d 1.Cho hnh chp S.ABC c tam gic ABC vung cn B, ,SA vung gc vi y ABC ,

1) Tnh th tch ca khi chp S.ABC.

2) Gi G l trng tm tam gic ABC, mt phng () qua AG v song song

vi BC ct SC, SB ln lt ti M, N. Tnh th tch ca khi chp S.AMN

Phn tch:

*) Dng tam gic ABC vung cn ti B v SA (ABC).

*) Dng mt phng qua G v // BC , cho MN //BC . Ti sao ?

Phn tch yu cu ca bi ra cc yu cu nh: *) Phn tch V= B.h tm B v h trong hnh l cc i tng no ?


*) Tm h = SA qua tam gic no bi nh l g ?

*) Tnh trc tip th tch SAMN qu phc tp ta phi lm sao ? Lp t s th tch ca SAMN v SABC ? Suy ra iu g ?

Li gii:a)Ta c: v

+

b) Gi I l trung im BC. G l trng tm,ta c:

// BC MN// BC

. Vy:

V d 2. Cho tam gic ABC vung cn A v . Trn ng thng qua C v vung gc vi (ABC) ly im D sao cho . Mt phng qua C vung gc vi BD, ct BD ti F v ct AD ti E.

a) Tnh b) Chng minh c) Tnh th tch khi t din CDEF.

Phn tch: *) Dng tam gic ABC vung cn ti A v SC (ABC) *) Dng mt phng qua C v BD cho thit din CEF.

Phn tch yu cu ca bi ra cc yu cu nh: *) Phn tch V= tm B v h ca ABCD l cc i tng no ?


*) Chng minh CE vung gc vi 2 ng thng no trong mt phng (ABD)?

*) Tnh trc tip th tch CDEF phc tp ta phi lm sao ? Lp t s th tch ca DCEF v DABC bng t s cc i lng hnh hc trong tam gic vung no ?

Li gii:

a)Tnh :

b)Tac:


EMBED Equation.DSMT4 Ta c:

c) Tnh :Ta c:

M , chia cho

Tng t:

T(*) .Vy

V d 3. Cho khi chp t gic u SABCD. Mt mt phng qua A, B v trung im M ca SC . Tnh t s th tch ca hai phn khi chp b phn chia bi mt phng .

Phn tch.

*) Dng t gic u ABCD v SO (ABCD)

*) Dng (ABM) // CD c im N ?

*) Dng BD v BN . Ti sao ?

*) Phn tch yu cu ca bi ra cc yu cu nh: *) Phn tch hai chp t gic thnh cc chp tam gic no lp t s ?

*) Hy so snh th tch ca SABD v SBCD vi SABCD ?

*) Lp t s th tch ca SABN vi SABD ; SAMN vi SABC ?

Li gii:

K MN // CD (N th hnh thang ABMN l thit din ca khi chp khi ct bi mt phng (ABM).

*)

*)

M VSABMN = VSANB + VSBMN = . VABMN.ABCD =

Do :

V d 4. Cho hnh chp t gic u S.ABCD, y l hnh vung cnh a, cnh bn to vi y gc . Gi M l trung im SC. Mt phng i qua AM v song song vi BD, ct SB ti E v ct SD ti F. a) Hy xc nh mp(AEMF)

b) Tnh th tch khi chp S.ABCD

c) Tnh th tch khi chp S.AEMF

Phn tch:

*)Xc nh gc gia SA v ABCD l gc no ?

*)Phn tch tm B v h ca SABCD l cc i tng no ?

*)Tm h = SO qua tam gic v h thc lng gic no?

*)Phn tch hai chp t gic thnh cc chp tam gic no lp t s ?

*)Tnh th tch ca SAEMF qu phc tp th sao ?Lp t s th tch ca SAEMF v SABCD bng cch no ?

*) Hy so snh th tch ca SABD v SBCD vi SABCD ?

*) Lp t s th tch ca SAMF vi SACD ?

Li gii:

a) Gi . Ta c (AEMF) //BD EF // BD

b) vi

+ c: . Vy:

c) Phn chia chp t gic ta c = VSAMF + VSAME =2VSAMF = 2VSACD = 2 VSABCXt khi chp S.AMF v S.ACD Ta c:

c trng tm I, EF // BD


V d 5. Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, SA vung gc y, . Gi B, D l hnh chiu ca A ln lt ln SB, SD. Mt phng (ABD) ct SC ti C.

a) Tnh th tch khi chp S.ABCD. b) Chng minh

c) Tnh th tch khi chp S.ABCD

Phn tch yu cu ca bi ra cc bi ton nh: *) Phn tch tm B v h ca SABCD l cc i tng no ?

*) Chng minh SC vung gc 2 ng thng no trong (AB'D') ?

*) Phn tch hai chp t gic thnh cc chp tam gic no lp t s ?

*) Hy so snh th tch ca SABC v SACD vi SABCD ?

*) Hy so snh th tch ca SAB'C' v SAC'D' vi SAB'C'D' ?

*) Lp t s th tch ca SAB'C' vi SABC . Suy ra iu g ?

Li gii:

a) Ta c:

b) Ta c & Suy ra:

nn AB'SC .Tng t AD'SC.Vy SC (AB'D')

c) Tnh : Ta c:

vung cn nn

Ta c:

T

EMBED Equation.DSMT4+

BI TP T RN LUYNBi 1. Cho t din ABCD. Gi B' v C' ln lt l trung im ca AB v AC. Tnh t s th tch ca khi t din AB'C'D v khi t din ABCD.

s:

Bi 2. Cho t din ABCD c th tch 9m3 ,trn AB,AC,AD ln lt ly cc im B',C',D' sao cho AB = 2AB'

2AC = 3AD' ;AD = 3AD'. Tnh t tch t din AB'C'D'.

s: V = 2 m3Bi 3. Cho t din u ABCD c cnh a. Ly cc im B';C' trn AB v AC sao cho . Tnh th tch t din AB'C'D .

s:

Bi 4. Cho t dinABCD c th tch 12 m3 .Gi M,P l trung im ca AB v CD v ly N trn AD sao cho DA = 3NA. Tnh th tch t din BMNP.

s: V = 1 m3Bi 5. Cho hnh chp SABC c y ABC l tam gic u cnh ,ng cao SA = a.Mt phng qua A v vung gc vi SB ti H v ct SC ti K. Tnh th tch hnh chp SAHK.

s:

Bi 6. Cho hnh chp SABCD c th tch bng 27m3 .Ly A'trn SA sao cho

SA = 3SA'. Mt phng qua A' v song song vi y hnh chp ct SB,SC,SD ln lt ti B',C',D' .Tnh th tch hnh chp SA'B'C'D'.

s: V = 1 m3

Bi 7. Cho hnh chp SABCD c th tch bng 9m3, ABCD l hnh bnh hnh , ly M trn SA sao cho 2SA = 3SM. Mt phng (MBC) ct SD ti N.Tnh th tch khi a din ABCDMN . s: V = 4m3

Bi 8. Cho hnh chp SABCD c y l hnh vung cnh a, chiu cao SA = h. Gi N l trung im SC. Mt phng cha AN v // BD ln lt ct SB,SDF ti M v P. Tnh s:

Bi 9 : Cho hnh chp SABCD c y ABCD l hnh bnh hnh v I l trung im ca SC.Mt phng qua AI v song song vi BD chia hnh chp thnh 2 phn.Tnh t s th tch 2 phn ny.s: Bi 10: Cho hnh chp SABCD c y ABCD l hnh bnh hnh v ly M trn SA sao cho Tm x mt phng (MBC) chia hnh chp thnh 2 phn c th tch bng nhau.

s:

5) Dng 5 : n tp khi chp v lng tr

V d 1: Cho hnh chp S.ABCD c ABCD l hnh vung cnh 2a, SA vung

gc y. Gc gia SC v y bng v M l trung im ca SB.

1) Tnh th tch ca khi chp S.ABCD.

2) Tnh th tch ca khi chp MBCD.

+ Dng t gic ABCD v SA(ABCD)

+ Dng H trung im AB. Nhn xt MH vi AB ? Ti sao ?

+ Xc nh gc[SC,(ABCD)] = ? Ti sao ?

+ Phn tch V= B.h tm B v h ca SABCD l cc i tng no ?

+ Tnh h = SA trong tam gic no v h thc lng gic no ?

+ Phn tch V= B.h tm B v h ca SABCD l cc i tng no ?

+ MABCD c ng cao l g ? ti sao ? Tnh MH bi tnh cht g ?

. Li gii:

a)Ta c

+

+

b) K

Ta c: ,

V d 2: Cho hnh chp tam gic S.ABC c AB = 5a, BC = 6a, CA = 7a. Cc mt

bn SAB, SBC, SCA to vi y mt gc 60o .Tnh th tch khi chp.

+ Dng tam gic ABC v SH(ABC) vi H (ABC) v H cch u 3 cnh tam gic ABC.

phn tch yu cu ca bi ra cc yu cu nh: + Xc nh gc hp bi 3 mt bn vi y chp ?

+ Phn tch V= B.h tm B v h ca SABC l cc i tng no ?

+ Tnh B = SABC bng cng thc no ?

+ Tnh h = SH trong tam gic no v h thc lng gic no ?

Li gii:

H SH, k HEAB, HFBC, HJAC

suy ra SEAB, SFBC, SJAC . Ta c


EMBED Equation.DSMT4 nn HE =HF = HJ = r

( r l bn knh ng trn ngai tip )

Ta c SABC =

vi p = Nn SABC =

Mt khc SABC = p.r

Tam gic vung SHE:

SH = r.tan 600 =

Vy VSABC = .

V d 3: Cho hnh hp ch nht ABCD.ABCD c , AD = a,

AA = a, O l giao im ca AC v BD.

a) Tnh th tch khi hp ch nht, khi chp OABCD

b) Tnh th tch khi OBBC.

c) Tnh di ng cao nh C ca t din OBBC.

+ Dng hp ch nht , hnh chp OA'B'C'D' v OBB'C' . phn tch yu cu ca bi ra cc yu cu nh: + Phn tch V= B.h tm B v h ca OA'B'C'D' l cc i tng no ?

+ Phn tch V= B.h tm B v h ca OBB'C' l cc i tng no ?

+ Tnh B = SBB'C' bng cng thc no ?

+ Tnh h = OM ? Dng tam gic no v tnh cht g ?

+ i vi chp OBB'C' chn nh C' v y l ta c chiu cao yu cu v

dng cng thc no tm n ?

Li gii:

a) Gi th tch khi hp ch nht l V.

Ta c:


* Khi OABCD c y v ng cao ging khi hp nn:

b) M l trung im BC

c) Gi CH l ng cao nh C ca t

din OBBC. Ta c:


V d 4: Cho hnh lp phng ABCD.ABCDc cnh bng a. Tnh th tch khi t din ACBD.+ Phn tch V= B.h tm B v h ca ACB'D' l cc i tng no ?

+ Tnh trc tip th tch ACB'D' phc tp ? Ta phn tch lp phng thnh 4 khi t din c th tch bng nhau no ?

+ Khi nhn xt v ? Suy ra iu g ?

Li gii:

Hnh lp phng c chia thnh: khi ACBD v bn khi CBDC, BBAC, DACD, ABAD.

+Cc khi CBDC, BBAC, DACD, ABAD c din tch y v chiu cao bng nhau nn c cng th tch.

Khi CBDC c

+Khi lp phng c th tch:

V d 5: Cho hnh lng tr ng tam gic c cc cnh bng a.a) Tnh th tch khi t din AB BC.

b) E l trung im cnh AC, mp(ABE) ct BC ti F. Tnh th tch khi CABFE.

+ Phn tch V= B.h tm B v h ca A'B'BC l cc i tng no ?

+ Tnh trc tip th tch CA'B'FE phc tp ? Ta phn tch khi chp thnh 2

khi t din no m tnh th tch n gin hn ?

Li gii:

a) Khi AB BC:Gi I l trung im AB,

EMBED Equation.DSMT4b)Khi CABFE: phn ra hai khi CEFA v CFAB.

+Khi ACEFc y l CEF, ng cao AA nn

EMBED Equation.DSMT4+Gi J l trung im BC. Ta c khi ABCF c y l CFB, ng cao JA nn

+ Vy:

Bi tp tng t:

Bi 1: Cho lng tr ng ABCA1B1C1 c ABC vung. AB = AC = a; AA1 = a. M l trung im AA1. Tnh th tch lng tr MA1BC1 s:V =

Bi 2: Hnh chp SABCD c ABC vung ti B, SA(ABC). = 60o,

BC = a, SA = a,M l trung im SB.Tnh th tch MABC . s: VMABC =

Bi 3: SABCD c y ABCD l hnh thang vi y ln AB = 2, = 90o. SAC v SBD l cc tam gic u c cnh bng . Tnh th tch khi chp SABCD. s: VSABCD =

Bi 4: Tnh th tch hnh chp tam gic u SABC trong cc trng hp sau:

a) Cnh y bng 1, gc ABC = 60o . s: V =

b) AB = 1, SA = 2 . s: V =

Bi 5. Cho lng tr ABCABC c di cnh bn = 2a, ABC vung ti A,

AB = a, AC = a. Hnh chiu vung gc ca A trn (ABC) l trung im BC.

Tnh VAABC theo a? s: V =

Bi 6: Cho hnh chp SABC c y ABCD l hnh bnh hnh v SABCD = v gc gia 2 ng cho bng 60o, cc cnh bn nghing u vi y 1 gc 45o.

Tnh VSABCD . s:

Bi 7: Cho hnh chp SABC c SA = SB = SC = a. ASB = 60o, BSC = 90o,

CSA = 120o.Chng minh rng ABC vung .Tnh VSABC . s: Bi 8: Cho hnh chp S.ABCD c y ABCD l hnh vung cnh 2a, SA = a ,SB=v mt phng (SAB) vung gc mt phng y. Gi M,N ln lt l trung im ca cc cnh AB.BC.Tnh theo a th tch khi chp S.BMDN

s:

Bi 9: Cho lng tr ng tam gic u ABCABC c cnh y v cnh bn u bng a. M, N, E ln lt l trung im ca BC, CC, CA. Tnh t s th tch hai phn lng tr do (MNE) to ra. s: k = 1

Bi 10: Cho hnh chp S.ABCD c y l hnh vung cnh a,mt bn SAD l tam gic u v nm trong mt phng vung gc vi y .Gi M,N ln lt l trung im ca cc cnh SB,BC,CD.Chng minh AM vung gc vi BP v tnh th tch ca khi t din CMNP.

s :

_1316400400.unknown

_1317128275.unknown

_1317278597.unknown

_1317365530.unknown

_1317368915.unknown

_1317376300.unknown

_1317376324.unknown

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Documents

Bai Tap Hinh Hoc Khong Gian Lop 12 Co Loi Giai