Chuyn :Luyn tp Hnh Hc Khng GianGV:L Minh Tin BhabcaaaBhCHUYN
:PHNG PHP LUYN TP TH TCH KHI A DIN
N TP 3KIN THC C BN HNH HC LP 12 A. TH TCH KHI A DIN I/ Cc cng
thc th tch ca khi a din: 1.THTCHKHILNG TR: V= B.hvi B: d i e n t c
h a yh: c h i e uc a o
a) Th tch khi hp ch nht: V = a.b.cvi a,b,c l ba kch thc b) Th
tch khi lp phng: V = a3
vi a l di cnh
2. TH TCH KHI CHP: V=13Bhvi B: dien tch ayh :chie u cao
3.TSTHTCHT DIN: ChokhitdinSABCvA, B,Clccimtylnlt thuc SA, SB, SC ta
c: SABCSA ' B' C 'VSA SBSCV SA ' SB' SC'= C'B'A'CBAS 4. TH TCH KHI
CHP CT: ( )hV B B' BB'3= + +
vi B,B' : dien tch hai ayh :chieu cao
B ACA'B'C' Chuyn :Luyn tp Hnh Hc Khng GianGV:L Minh Tin
a3aC'B'A'CBAII/ Bi tp: LOI 1: TH TCH LNG TR 1)Dng 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
ABCvung cn ti A c cnh BC = a 2vbitA'B = 3a. Tnh th tch khi lng tr.
a 2 Li gii: Ta c ABC Vvung cn ti A nn AB = AC = a ABC A'B'C' l lng
tr ngAA' AB 2 2 2 2AA' B AA' A' B AB 8a = = VAA' 2a 2 =Vy V = B.h =
SABC .AA' = 3a 2 V d 2:Cho lng tr t gic u ABCD.ABCD' c cnh bn bng4a
v ng cho 5a. Tnh th tch khi lng tr ny. 5a4aD'C'B'A'DCBA Li gii:
ABCD A'B'C'D' l lng tr ng nnBD2 = BD'2 - DD'2 = 9a2BD 3a = ABCD l
hnh vung 3aAB2 =Suy raB = SABCD = 29a4 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
vbitdin tch tam gic ABC bng 8. Tnh th tch khi lng tr.
A'C'B'ABCI
Li gii: Gi I l trung im BC .Ta c VABC u nn AB 33 &2AI 2 AI
BCA' I BC(dl3 )= = A'BCA'BC2S 1S BC.A' I A' I 42 BC= = =AA' (ABC)
AA' AI . 2 2A' AI AA' A' I AI 2 = = V Vy : VABC.ABC = SABC .AA'=8 3
Chuyn :Luyn tp Hnh Hc Khng GianGV:L Minh Tin A'DB'C'A'CD'C'B'
BD'A60D'C'B'A'DCBAo60C'B'A'CBAV d 4: 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 cnp. Tnh th tch ci hp ny. D'A'C'B'DACB Gii Theo bi,
ta cAA' = BB' = CC' = DD' = 12 cm nn ABCD l hnh vung cAB = 44 cm -
24 cm = 20 cm v chiu cao hp h = 12 cm Vy th tchhp lV = SABCD.h =
4800cm3 V d 5: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 . Li gii:
Ta c tam gic ABD u nn : BD = a vSABCD = 2SABD = 2a 32 Theo bi BD' =
AC = a 32 a 32=2 2DD' B DD' BD' BD a 2 = = VVy V = SABCD.DD' = 3a
62 2)Dng 2: Lng tr ng c gc gia ng thng v mt phng. V d 1: Cho lng tr
ng tam gic ABC A'B'C' c y ABC l tam gicvung cn ti B vi BA = BC = a
,bit A'B hp vi y ABC mt gc 600 . Tnh th tch lng tr. Li gii: Ta cA'
A (ABC) A' A AB&AB l hnh chiu ca A'B trn y ABC . Vy ogc[A' B,
(ABC)] ABA' 60 = =0ABA' AA' AB.tan60 a 3 = = VSABC = 21 aBA.BC2
2=Vy V = SABC.AA' = 3a 32 Chuyn :Luyn tp Hnh Hc Khng GianGV:L Minh
Tin V d 2: Cho lng tr ng tam gic ABC A'B'C' c y ABC l tam gicvungti
A vi AC = a , ACB= 60 o bit BC' hp vi (AA'C'C) mt gc 300. Tnh AC' v
th tch lng tr. ao60o30C'B'A'CBA Li gii: oa 3 ABC AB AC.tan60 = = V
. Ta c: AB AC;AB AA' AB (AA' C' C) nn AC' l hnh chiu ca BC' trn
(AA'C'C). Vy gc[BC';(AA"C"C)] = BC' A= 30o
oABAC' B AC' 3at an30 = = V V =B.h = SABC.AA' 2 2AA' C' AA' AC'
A' C' 2a 2 = = VABC Vl na tam gic u nn 2ABCa 3S2=Vy V = 3a 6 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 . o30aD'C'A'B'DC BA Gii: Ta c ABCD A'B'C'D' l lng tr
ng nn ta c:DD' (ABCD) DD' BD v BD l hnh chiu ca BD' trn ABCD . Vy
gc [BD';(ABCD)] = 0DBD' 30 =0a 6BDD' DD' BD.tan303 = = VVy V =
SABCD.DD' = 3a 63S = 4SADD'A' = 24a 63 V d 4: Cho hnh hp ng ABCD
A'B'C'D' c y ABCD l hnh thoi cnha v BAD = 60o bit AB' hp vi y
(ABCD) mt gc 30o . Tnh th tch ca hnh hp. ao30o60D'C'B'A'DCBA Gii
ABD V u cnh a 2ABDa 3S4 =2ABCD ABDa 3S 2S2 = =ABB' V vung tiBoBB'
ABt an30 a 3 = =Vy 3ABCD3aV B.h S .BB'2= = = Chuyn :Luyn tp Hnh Hc
Khng GianGV:L Minh Tin 3) 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 gc600 .Tnh th tch lng tr.
C'B'A'CBAo60 Li gii: Ta cA' A (ABC) &BC AB BC A' B Vy ogc[(A'
BC),(ABC)] ABA' 60 = =0ABA' AA' AB.tan60 a 3 = = VSABC = 21 aBA.BC2
2=Vy V = SABC.AA' = 3a 32 V d 2:y ca lng tr ng tam gic ABC.ABC l
tam gic u . Mt (ABC) to viy mt gc 300 v din tch tam gic ABC bng 8.
Tnh thtch khi lng tr. xo30IC'B'A'CBA Gii: ABC VuAI BC m AA' (ABC)
nnA'I BC (l 3 ). Vy gc[(A'BC);)ABC)] =A' IA = 30o Gi s BI = x 323
2xxAI = = .Ta cxx AIAI I A AI A 233 23230 cos : ' : '0= = = = A AA
= AI.tan 300 = x x =33. 3 Vy VABC.ABC = CI.AI.AA = x33M SABC =
BI.AI = x.2x = 8 2 = x Do VABC.ABC = 8 3 V d 3:Cho lng tr t gic u
ABCD A'B'C'D' c cnh y a vmt phng(BDC') hp vi y (ABCD) mt gc 60o.Tnh
th tch khi hp ch nht. Chuyn :Luyn tp Hnh Hc Khng GianGV:L Minh Tin
a060OA'D'B'C'CADB Gi O l tm ca ABCD . Ta c ABCD l hnh vung nnOC BD
CC' (ABCD)nnOC' BD(l3).Vy gc[(BDC');(ABCD)] = COC'= 60o
Ta c V = B.h = SABCD.CC' ABCD l hnh vung nn SABCD = a2
OCC' Vvung nn CC' = OC.tan60o =a 62 Vy V = 3a 62 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) mtgc 30o .Tnh th tch khi hp ch nht.
2ao30o60D'C'B'A'DCBA Ta c AA'(ABCD) AC l hnh chiu ca A'C trn (ABCD)
.Vy gc[A'C,(ABCD)] = oA' CA 30 =BCABBCA'B (l 3) .Vy
gc[(A'BC),(ABCD)] = oA' BA 60 =A' AC V AC = AA'.cot30o =2a 3A' AB V
AB = AA'.cot60o = 2a 33 2 24a 6ABC BC AC AB3 = = V Vy V = AB.BC.AA'
= 316a 23
4) Dng 4:Khi lng tr xin V d 1: Cho lng tr xin tam gic ABC A'B'C'
c y ABC l tam gicu cnh a , bit cnh bn la 3v hp vi y ABC mt gc 60o .
Tnh th tch lng tr. Ho60aB'A'C'CBA Li gii: Ta cC' H (ABC) CH l hnh
chiu ca CC' trn (ABC) Vy ogc[CC', (ABC)] C' CH 60 = =03aCHC' C' H
CC'.sin602 = = VSABC = 23 a4= .Vy V = SABC.C'H = 33a 38 Chuyn :Luyn
tp Hnh Hc Khng GianGV:L Minh Tin 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 .
HOo60C'AaB'A'CB Li gii: 1)Ta cA' O (ABC) OA l hnh chiu ca AA' trn
(ABC) Vy ogc[AA', (ABC)] OAA' 60 = =Ta c BB'CC' l hnh bnh hnh ( v
mt bn ca lng tr) AO BC ti trung im H ca BC nn BC A' H (l 3) BC (AA'
H) BC AA' m AA'//BB' nnBC BB' .Vy BB'CC' l hnh ch nht. 2)ABC Vu nn
2 2 a 3 a 3AO AH3 3 2 3= = =oAOA' A' O AOt an60 a = = VVy V =
SABC.A'O = 3a 34 V d 3: Cho hnh hp ABCD.ABCD c y l hnh ch nht vi AB
=3 AD = 7 .Hai mt bn (ABBA) v (ADDA) ln lt to vi ynhng gc 450 v
600.. Tnh th tch khi hp nubit cnh bn bng 1. HNMD'C'B'A'DCBA Li gii:
K AH) ( ABCD ,HM AD HN AB ,AD N A AB M A ' , ' (l 3) o oA' MH 45
,A' NH 60 = =t AH = x . Khi AN = x : sin 600 = 32x AN = HMxN A AA
== 34 3' '22 2 M HM = x.cot 450 = x Ngha l x = 7334 32= xx Vy
VABCD.ABCD = AB.AD.x= 33. 7. 37 = Chuyn :Luyn tp Hnh Hc Khng
GianGV:L Minh Tin LOI 2: TH TCH KHI CHP 1)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 .
_\//aBSCA Li gii: Ta c
(ABC) (SBC)(ASC) (SBC)AC (SBC) Do 2 3SBC1 1 a 3 a 3V S .AC a3 3
4 12= = = V d 2: Cho hnh chp SABC c y ABC l tam gic vung cn ti B
viAC = abit 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 . ao60SCBA Li
gii: 1) SA (ABC) SA AB&SA AC mBC AB BC SB ( l 3 ). Vy cc mt bn
chp l tam gic vung. 2) Ta cSA (ABC) AB l hnh chiu ca SB trn (ABC).
Vy gc[SB,(ABC)] = oSAB 60 = . ABC V vung cn nn BA = BC = a2 SABC =
21 aBA.BC2 4=oa 6SAB SA AB.t an602 = = VVy 2 3ABC1 1 a a 6 a 6V S
.SA3 3 4 2 24= = = V d 3:Cho hnh chp SABC c y ABC l tam gic u cnh a
bitSAvung gc vi y ABC v (SBC) hp vi y (ABC) mt gc 60o. Tnh th tch
hnh chp . Chuyn :Luyn tp Hnh Hc Khng GianGV:L Minh Tin ao60MCBAS Li
gii: Ml trung im ca BC,v tam gic ABC u nn AMBCSABC (l3) .
Vygc[(SBC);(ABC)] = oSMA 60 = . Ta c V = ABC1 1B.h S .SA3 3=o3aSAM
SA AMtan602 = = VVyV = 3ABC1 1 a 3B.h S .SA3 3 8= = V d 4: Cho hnh
chp SABCD c y ABCD l hnhvung c cnh av SAvung 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). HaDCBASo60 Li gii: 1)Ta cSA (ABC) v CD AD
CD SD ( l 3).(1) Vy gc[(SCD),(ABCD)] = SDA = 60o . SAD V vungnn SA
= AD.tan60o =a 3Vy 23ABCDa1 1 a 3V S .SA a 33 3 3= = = 2) Ta dng
AHSD ,v CD(SAD) (do (1) ) nn CDAH AH (SCD) Vy AH l khong cch t A n
(SCD). 2 2 2 2 2 21 1 1 1 1 4SADAH SA AD 3a a 3a = + = + = V Vy AH
= a 32 2)Dng 2 :Khi chp c mt mtbn vung gc vi y V d 1:Cho hnh chp
S.ABCD c y ABCD l hnh vung c cnh a Mt bnSAB l tam gic u nm trong mt
phngvung 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. aHDCBAS Li gii: 1)Gi
H l trung im ca AB. SAB VuSH AB m(SAB) (ABCD) SH (ABCD) Vy H l chn
ng cao ca khi chp. 2)Ta c tam gic SAB u nn SA = a 32 suy ra 3ABCD1
a 3V S .SH3 6= = Chuyn :Luyn tp Hnh Hc Khng GianGV:L Minh Tin 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.
o60aHDCBA Li gii: Gi H l trung im ca BC. Ta c tam gic ABC u nn AH
(BCD) , m (ABC)(BCD) AH(BCD) . Ta c AHHDAH = AD.tan60o =a 3& HD
= AD.cot60o =a 33 BCD V BC = 2HD = 2a 33suy ra V = 3BCD1 11 a 3S
.AH . BC.HD.AH3 32 9= = V d 3:Cho hnh chp S.ABC c y ABC l tam gic
vung cn ti B, c BC = a. Mt bnSAC 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. 45IJHACBS Li gii: a) K SHBC v
mp(SAC) mp(ABC) nn SHmp(ABC).Gi I, J l hnh chiu ca H trn AB v BC SI
AB, SJ BC, theo gi thit oSIH SJH 45 = = Ta c:HJ HI SHJ SHI = A = A
nn BH l ng phn gic caABC V suy ra H l trung im ca AC. b) HI = HJ =
SH =2aVSABC=12.313aSH SABC= Chuyn :Luyn tp Hnh Hc Khng GianGV:L
Minh Tin 3)Dng 3 :Khi chp u V d 1:Cho chp tam gic uSABC 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 gicu ABC.Tnh th tch chp u SABC . a2aHOCBAS Li gii: DngSO
(ABC)TacSA=SB=SC suy ra OA = OB = OC Vy O l tm ca tam gic u ABC. Ta
c tam gic ABC u nn AO = 2 2 a 3 a 3AH3 3 2 3= =22 2 211aSAO SO SA
OA3 = = Va11SO3 = .Vy 3ABC1 a 11V S .SO3 12= = V d 2:Cho khi chp t
gic SABCDc 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. aODCBAS Li gii: Dng SO(ABCD) Ta
c SA = SB = SC = SD nn OA=OB=OC=ODABCDl hnhthoicngtrngnoitip nn
ABCD l hnh vung . Ta c SA2 + SB2 = AB2 +BC2 = AC2 nnASC V vung ti S
22aOS = 321 1 2 2.3 3 2 6ABCDa aV S SO a = = =Vy 3a 2V6= 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.
Chuyn :Luyn tp Hnh Hc Khng GianGV:L Minh Tin aIHOMCBAD Li gii: a)
Gi O l tm caABC A ( ) DO ABC
1.3ABCV S DO = 234ABCaS =, 2 33 3aOC CI = = 2 2 : DOC vu ng c DO
DC OC A = 63a=
2 31 3 6 2.3 4 3 12a a aV = = b)KMH//DO,khongcchtMn mp(ABC) l
MH
1 62 6aMH DO = = 2 31 1 3 6 2. .3 3 4 6 24MABC ABCa a aV S MH =
= = Vy 3a 2V24=Bi tp tng t:Bi 1:Cho hnh chp uSABC c cnh bn bnga hp
vi y ABC mt gc 60o . Tnh th tch hnh chp.s: 33aV16= Bi 2:Cho hnh chp
tam gic u SABC c cnh bn a, gc y ca mt bnl 45o. 1) Tnh di chiu cao
SH ca chp SABC .s:SH = a3 2) Tnh th tch hnh chp SABC.s: 3aV6=Bi
3:Cho hnh chp tam gic u SABC c cnh y a v mt bn hp vi ymt gc 60o.
Tnh th tch hnh chp SABC.s: 3a 3V24=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: 3h 3V3=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: 3h 3V8=Bi 6 :Cho hnh chp t gic u SABCD c cnh y a v oASB
60 = .1) Tnh tng din tch cc mt bn ca hnh chp u.s: 2a 3S3=2) Tnh th
tch hnh chp. s: 3a 2V6= Chuyn :Luyn tp Hnh Hc Khng GianGV:L Minh
Tin 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: 32hV3=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: 38a 3V3=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: 3a 3V12=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 39a 2V2= . s: AB =
3a 4)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, 2 AC a =, SA vunggc vi y ABC ,SA a
=1) Tnh th tch ca khi chp S.ABC. 2) Gi G l trng tm tam gic ABC, mt
phng (o ) qua AG v song song vi BC ct SC, SB ln lt ti M, N. Tnh th
tch ca khi chp S.AMN GMNICBAS Li gii: a)Ta c:.1.3S ABC ABCV S SA
=vSA a =+ : 2 ABCc n c AC a AB a A = =212ABCS a =Vy: 321 1. .3 2
6SABCaV a a = = b) Gi I l trung im BC. G l trng tm,ta c : 23SGSI =
o // BC MN// BC23SM SN SGSB SC SI = = =
4.9SAMNSABCV SMSNV SB SC = = Vy: 34 29 27SAMN SABCaV V = = Chuyn
:Luyn tp Hnh Hc Khng GianGV:L Minh Tin V d 2: Cho tam gic ABC vung
cn A vAB a = . Trn ng thng qua C v vung gc vi mt phng (ABC) ly im D
sao choCD a = . Mt phng qua C vung gc vi BD, ct BD ti F v ct AD ti
E. a) Tnh th tch khi t din ABCD. b) Chng minh( ) CE ABD c) Tnh th
tch khi t din CDEF.
aaFEBACD Li gii: a)Tnh ABCDV: 3ABCD ABC1 aV S .CD3 6= =b)Tac: ,
AB AC AB CD ( ) AB ACD AB EC Ta c: DB EC ( ) EC ABD c) Tnh EF
DCV:Ta c: . (*)DCEFDABCV DE DFV DA DB= M 2. DE DA DC = , chia cho
2DA 2 22 212 2DE DC aDA DA a = = = Tng t: 2 22 2 213DF DC aDB DB DC
CB= = =+ T(*) 16DCEFDABCVV =.Vy316 36DCEF ABCDaV V = =
Vd3:ChokhichptgicuSABCD. Mt mt phng) (o quaA,Bv trung im Mca SC .
Tnh t s th tch ca hai phn khi chp bphn chia bi mt phng .
NSOMBDCA
Li gii: K MN // CD (N) SD e th hnh thang ABMN l thit din ca khi
chp khi ct bi mt phng (ABM). + SABCD SADB SANBSADBSANDV V
VSDSNVV412121= = = =SABCD SBCD SBMNSBCDSBMNV V
VSDSNSCSMVV81414121.21. = = = = = M VSABMN = VSANB + VSBMN =
SABCDV83.Suy ra VABMN.ABCD =SABCDV85 Do : 53.=ABCD ABMNSABMNVV
Chuyn :Luyn tp Hnh Hc Khng GianGV:L Minh Tin IOABCDSEFM V d 4: Cho
hnh chp t gic u S.ABCD, y l hnh vung cnh a, cnh bn toviy gc60o.GiMl
trungim SC.Mt phngiqua AMvsong 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 d) Li gii: a)GiI SO AM = .Tac(AEMF)//BD EF // BD b).
D D1.3S ABC ABCV S SO =vi 2D ABCS a = +SOA Vc : 6. tan 602aSO AOo=
=
Vy : 3. D66S ABCaV = c) Phn chia chp t gic ta c . EMF S AV=
VSAMF + VSAME=2VSAMF . S ABCDV= 2VSACD = 2 VSABC Xt khi chp S.AMF v
S.ACDTa c : 12SMSC = SAC A c trng tm I, EF // BD nn: 23SI SFSO SD =
=D1.3SAMFSACV SMSFV SCSD = =
3D D1 1 63 6 36SAMF SAC SACaV V V = = =3 3. EMF6 6236 18S Aa aV
= = V d 5: Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, SA vung
gcy,2 SA a = .GiB,DlhnhchiucaAlnltlnSB,SD.Mt phng (ABD) ct SC ti C.
a) Tnh th tch khi chp S.ABCD. b)Chng minh( ' ') SC AB D c) Tnh th
tch khi chp S.ABCD Chuyn :Luyn tp Hnh Hc Khng GianGV:L Minh Tin
ASIODBCC'D'B' Li gii: a) Ta c: 3.1 2.3 3S ABCD ABCDaV S SA = = b)
Ta c( ) ' BC SAB BC AB &' SB AB Suy ra: ' ( ) AB SBC nn AB' SC
.Tng t AD' SC. Vy SC(AB'D') c) Tnh . ' ' ' S AB C DV +Tnh . ' ' S
AB CV: Ta c: ' '' '. (*)SAB CSABCV SB SCV SB SC= SAC A vung cn nn '
12S CS C= Ta c: 2 2 22 2 2 2' 2 2 23 3SB SA a aSB SB SA AB a= = =
=+ T' '1(*)3SAB CSABCVV = 3 3' '1 2 2.3 3 9SAB Ca aV = = + 3. ' ' '
. ' '2 229S AB C D S AB CaV V = = 5)Dng 5 :n tpkhi 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 bng60o v M l trung im ca SB. 1) Tnh th tch ca khi chp
S.ABCD. 2)Tnh th tch ca khi chp MBCD. . 2ao60HDCBAS Li gii: a)Ta c
1.3ABCDV S SA = + 2 2(2 ) 4ABCDS a a = =+ : tan 2 6 SAC c SA AC C a
A = =321 8 64 .2 63 3aV a a = = b) K/ / ( ) MH SA MH DBC Ta c: 12MH
SA = , 12BCD ABCDS S = 3D1 2 64 3MBCaV V = = Chuyn :Luyn tp Hnh Hc
Khng GianGV:L Minh Tin 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. 60ACBHSF EJ Li gii: H SH ) ( ABC , k HEAB, HFBC, HJ AC
suy ra SEAB, SFBC, SJ AC . Ta c OSEH SFH SJH 60 = = = SJH SFH SAH A
= A = Ann HE =HF = HJ = r ( r l bn knh ng trn ngai tipABC A ) Ta c
SABC =) )( )( ( c p b p a p p vi p = ac b a92=+ +Nn SABC = 22 . 3 .
4 . 9 aMt khc SABC = p.r 36 2 apSr = = Tam gic vung SHE: SH = r.tan
600 = aa2 2 3 .36 2= Vy VSABC = 3 23 8 2 2 . 6 631a a a =. V d
3:Cho hnh hp ch nht ABCD.ABCD c3 AB a = , AD = a, AA = a, O l
giaoim 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.
MOD'C'B'A'DCB A Li gii: a) Gi th tch khi hp ch nht l V. Ta c : .
D.AA' V AB A =2 33. 3 a a a = = 2 2 : 2 ABD c DB AB AD a A = + = *
KhiOABCD c y v ng cao ging khi hp nn:3' ' ' '1 33 3OA B C DaV V = =
b) M l trung im BC ( ' ') OM BBC
2 3' ' ' '1 1 3 3. . .3 3 2 2 12OBB C BB Ca a aV S OM = = = c)
Gi CH l ng cao nh C ca tdin OBBC. Ta c : ' ''3'OBB COBBVC HS=
2 2 : 2 ABD c DB AB AD a A = + = 2'12OBBS a =' 2a 3 C H = Chuyn
:Luyn tp Hnh Hc Khng GianGV:L Minh Tin V d 4:Cho hnh lp phng
ABCD.ABCDc cnh bng a.Tnh th tch khi t din ACBD. aD'C'B'A'DCBA Li
gii: Hnhlpphngcchiathnh:khi ACBD v bn khi CBDC, BBAC, DACD, ABAD.
+Cc khi CBDC, BBAC, DACD, ABADcdintchyvchiucao bng nhau nn c cng th
tch. Khi CBDC c 2 311 1 1. .3 2 6V a a a = = +Khi lp phng c th tch:
32V a = 3 3 3' '1 14.6 3ACB DV a a a = = 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)
EltrungimcnhAC,mp(ABE)ctBCtiF.Tnhthtchkhi CABFE. JI FEC'B'A'CBA Li
gii: a) Khi AB BC:Gi I l trung im AB, ' ' ' '1.3A B BC A B BV S CI
=2 31 3 3.3 2 2 12a a a= = b)Khi CABFE: phn ra hai khi CEFA v CFAB.
+Khi ACEFcy l CEF, ng cao AA nn' EF EF1. '3A C CV S A A = 2EF1 34
16C ABCaS S = =3' EF348A CaV = +Gi J l trung im BC. Ta c khi ABCF
cy l CFB, ng cao JA nn ' ' F FB'1. '3A B C CV S A J =2FB' '12 4C
CBBaS S = =
2 3' ' F1 3 33 4 2 24A B Ca a aV = = + Vy :3A'B'FE316CaV =
