www.vnmath.com CHUYN TH TCH Bi 1 Cho hnh chp tam gic S.ABC c y l
tam gic vung B,cnh SA (ABC) . T A kAD SB vAE SC . Bit AB = a, BC =
b, SA = c.Tnh th tch ca khi chp S.ADE? - Phn tch - tm li gii AD,AE
l cc ng cao trong tam gic SAB,SAC
ABCDES Tnh ng cao: ABC Avung ti B nnAB BC Gi thit cho : SA (ABC)
SA BC BC (ABC) AD BC AD l ng cao trong tam gic SAB AD SB AD (SBC)
AD SC Mt khc :AE SC SC (ADE) Hay SE l ng cao ca hnh chp S.ADE di
SE:
2 2 2 2AS.AB AS.AB a.cADSBAS AB a c = = =+ + 2 22 2 2 2 2AS.AC
SA.AC c. a bAESBSA AC a b c+= = =+ + + p dng Pytago trong tam gic
SAE c:
2 2 22 2 22 2 2c (a b )SE AS AE ca b c+= = + += 22 2 2ca b c + +
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only.www.vnmath.com Din tch tam gic ADE: DE = 2 2AE AD + = 2 22 2 2
2 2c .b(a b c ).(a c ) + + + S = 1.AD.AE2 = 2 22 2 2 2 2 2 21 c .b
ac. .2(a b c ).(a c ) a c + + + + = 3 32 2 2 2 21 a.c .b.2(a b c
).(a c ) + + + Th tch: V =1 1.SE. .AD.DE3 2 = 3 32 2 2 2 2 2 2 21 c
1 a.c .b. .3 2a b c (a b c ).(a c ) + + + + + 2 42 2 2 2 21 a.b
.c.6 (a c )(a b c )=+ + + -Xt mt cch gii khc nh sau: DE (SAB)
BC(SAB) => DE // BC Pytago trong cc tam gic vung: SD2 = AS2 -
AD2; SE2 = AS2 - AE2 SB2 = SA2+AB2 SC2 = SA2+AC2 = SA2 + AB2 + AC2
Lp cc t s:
2 22 2 2cSA AEa b c+ =+ + 2 2 2 22 2 2 2 2SA AD SA AE.SA AB SA
SB SC =+ + +
2 2 2 2 22 22 2 2 2 22 2 2 2 2c .a c (a b )c ca c c a b.a c c a
b+ + + +=+ + +4 2 22 2 2 2 2 2 2c b .c.(a c ) (c a b )=+ + + =>
32 2 2 2 2SA SD SE b.c. .SA SB SC (c a b )(a c )=+ + + SADESABCV SA
SD SE. .V SA SB SC== 32 2 2 2 2b.c(c a b )(a c ) + + + =>SADEV=
32 2 2 2 2b.c(c a b )(a c ) + + +.SABCVGenerated by Foxit PDF
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only.www.vnmath.com = 32 2 2 2 2b.c(c a b )(a c ) + + +.1 1.SA.
.AB.BC3 2 = 2 42 2 2 2 21 a.b .c.6 (c a b )(a c ) + + + (vtt) Bi 2:
Cho hnh chp tam gic S.ABC c mt bn SBC l tam gic u cnh a.CnhSA (ABC)
, gc
0BAC 120 = .Tm th tch ca khi chp S.ABC?ABCDS
-Trnh by li gii: Xt hai tam gic vung SAB v SAC c: SA chung SB =
SC =>ASAB =ASAC (c.c) => AB = AC =>AABC l tam gic cn Gi D
l trung im ca BC ta c : tan CAD = CDAD => AD =
CD a2. 3 tanCAD =Din tch y: 2ABC1 a . 3S AD.BC2 4A= =Generated
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evaluation only.www.vnmath.com SD l ng cao trong tam gic u SBC cnh
a nn : SD =a 32 p dng nh l Pytago trong tam gic SAD ta c: SA2 =SD2
- AD2=2 2 23.a a 2a4 12 3 = =>SA = a 23 Th tch cn tnh: V =
ABC1SA.S3A = 3a . 236 (vtt) -Tng qut ha ta c bi ton sau: Cho hnh
chp S.ABC c mt bn SBC l tam gic u cnh a, gc
0BAC (0 90 ) = o s o s . Bit SA vung gc vi mt phng y. Tnh th tch
ca khi chp S.ABC theo a vo ?Mt cch hon ton tng t ta c li giinh sau:
AD =
AD a2.tantanCAD =o Din tch tam gic: 2ABC1 1 a aS AD.BC a2 2
2.tan 4.tanA= = =o o SD l ng cao trong tam gic u SBC nn SD =a 32 p
dng nh l Pytago trong tam gic SAD ta c: SA2 =SD2 - AD2=2 2 23.a a
2a4 12 3 = =>SA = a 23 Th tch cn tm:
ABC1V SA.S3A= =1 1 1.B.h .SA. .AD.BC3 3 2== 3a 212 3.tano (vtt)
Bi 3TrongkhnggianhtaOxyzchohnhchpS.ABCDcyl
hnhthoi.ACctBDtigctaO.imA(2;0;0),B(0;1;0), S(0;0;2 2 ). Gi M l
trung im ca SC v mt phng (ABCD) ct cnh SD ti N. Tnh th tch ca khi
chp S.ABMN? Generated by Foxit PDF Creator Foxit
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ABCSMNDO-Li gii Tanhnthymtphng(SBN)chiakhichpS.ABCDthnhhaikhi
chp S.ABN v S.MBN Theo nh ngha v th tch ta c: S.ABMNV= S.ABNV+
S.MBNV
S.ABNS.ABDV SN 1V SO 2= ==>S.ABNV= 12S.ABDV = 14S.ABCDVTng t
ta c: S.MBNS.BCDV SM SN 1.V SC SD 4= ==>S.MBNV= 14S.BCDV=
18S.ABCDVDo vy: S.ABMNV = 14S.ABCDV+ 18S.ABCDV = 83S.ABCDVTh tch
khi chp S.ABCD V = ABCD1 1.SO.S .SO.AC.BD3 6== 832 Th tch cn tnh:
S.ABMNV=2(vtt) - Nghin cu li gii Gi V1 l th tch khi a din nm di
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S.ABCDV= S.ABMNV + ABCMNVhayV= V1+S.ABMNVTa c :V1 = N.ABDV+
B.CDMNV
N.ABDV= S.ABD1V2= V4
ABCSMNDOHai hnh chp B.SCD v B.DCMN c chung nh v mt phng cha y
nn: DCMN SDC3S .S4A= Th tch ca chp S.ABCD l: V = ABCD1 1.SO.S
.SO.AC.BD3 6== 832 Th tch cn tnh:Bi 4 Generated by Foxit PDF
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only.www.vnmath.com Cho hnh vung ABCD c cnh a, cc na ng thng Ax v
Cy vung gc vimt phng (ABCD)v cng mt pha so vimt phng y.Ly imM A
=trn Ax, lyN C =trn Cy. t AM = m ; BN = n. Tnh th tch ca khi chp
B.AMNC theo a, m, n? -Trnh by li gii Theogithittac:CN (ABCD) CN CB
,Oltmynn OB AC OB (ACMN) hay OB l ng cao di OB =AC2= a 22. Mt khc
MA ACNC ACMA //NC nn tgic ACMN l hnh thang ACMNSMA NC 2AC a(m n)2
2+= = +Th tch khi chp:
2ACMN1 aV OB.S (m n)3 6= = +( vtt ) -Nghin cu li gii Nhnthydo AM
AB ,AM AD ,AB AD nnta avo h trc
taOxyzsaochoA(0;0;0),B(a;0;0),M(0;0;m),D(0;a;0)ttaxc
nhctanhC(a;a;0)saupdngcngthctnhthtchca khi hp: 1V AB, AC AM6(= =
2a(m n)6+(vtt) Bi 5: ChohnhchptamgicS.ABCcyltamgicucnha.Cnh SA
(ABC) ,SA=2a.GiM,NlhnhchiuvunggccaAlncc cnh SB, SC. Tnh th tch ca
khi chp ABCMN? Generated by Foxit PDF Creator Foxit
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only.www.vnmath.com ABCSMN -Trnh by li gii XtASAB vASAC c AB = AC,
SA chung, A = 090 ASAB =ASAC SB =SC mt bn SBC l tam gic cn. p dng
nh l ng cao trong cc tam gic SAB v SAC ta c:
2 2AB.ASAMAB AS=+ =2a5 2 2AC.ASANAC AS=+ = 2a5 p dung nh l
Pytago:SM = 2 24aSA AM5 =
2 24aSN SA AN5= =Ta c cc t s: SMSB = SNSC = 45 S.AMNS.ABCVV =
1625 S.AMNV= 1625S.ABCV= 38a 375 Th tch :
ABCNMV = S.ABCV - S.AMNV = 3a 36- 38a 375 =33a 350 (vtt) -Nghin
cu li gii Tacth giibitontrn bngphng phpta bngvicavo
htrctaOxyztrongA(0;0;0),B(a;0;0),S(0;0;2a).Taxcnh c ta ca C, M, N,
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0S.AMN1V AM, AN AS 60 BAC6(= = S.AMN1V AM, AN AS6(= ABCNMV =
S.ABCV - S.AMNV=33a 350(vtt) Bi 6: Cho hnh lng tr tam gic ABC.DEF c
BE = a, gc gia ng thng BEvimtphng(ABC)bng 060
.TamgicABCvungtiC,gc
060 BAC = , hnh chiu vung gc ca E ln (ABc) trng vi trng tm ca
tam gic ABC. Tnh th tch ca t din D.ABC? ABCDEFMG -Trnh by li gii Ta
c:EG (ABC) nn EG l ng cao ca chp p dng h thc lng trong tam gic vung
EGB ta c: EG = EBsinB = asin060= a 32 p dng pytago:
2 2BG BE EG = = a2 m BG = 23BM BM = 32BG = 3a4 p dung Pytago
trong tam gic BMC: Generated by Foxit PDF Creator Foxit
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only.www.vnmath.com MC = MBsin015=3a4sin015 , AC = 2MC = 3a2sin015
,BC = ACtan060=3a2sin015 3Th tch ca khi chp: V = 3ABC1 3EG.S a .3
4A=2 0sin 15(vtt) Bi 7: Cho t din ABCDgi dl khong cch gia hai
ngthngAB v CD,olgcgiahaingthng.Tnhthtchcatdin ABCD?ABCDE -Trnh by
li gii Dng hnh bnh hnh ABDE, do AE // (BCD) nnABCD E.BCD B.ECDV V V
= = = ECD1S .d(B, CDE)3A =
1 1 1CE.CD.sinECD AB.CD.d.sin3 2 6 = o (vtt) -Nghin cu li gii Ta
xt mt cch gii khc nh sau: Generated by Foxit PDF Creator Foxit
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only.www.vnmath.com BCDEFMNA DnghnhhpchnhtAEBF.MDNCngoitiptdinABCD,
AB (ABEF) e ,CD (CDMN) eV (ABEF) // (CDMN) nn chiu cao ca hp bng d
Th tch cn tnh: ABCD hp MDNC1 1V V S d3 3= == 1 1MN.CD.d.sin3 2 o =
1AB.CD.d.sin6oBi 8: Trong khng gian cho hnh hp ch nht ABCD.EFGH c
nh A trng vigcta,imB(a;0;0),D(0;a;0),E(0;0;b),Mltrungimca CG.tnh th
tch ca khi t din BDEM theo a v b? Generated by Foxit PDF Creator
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only.www.vnmath.com BCDEFGHM(0,a,0)(a,0,0)(0,0,b)A(0,0, 0) -Trnh by
li gii M l trung im ca CG nn: C GMC GMC GMx xx a2y yy a2z z bz2 2+
= =+ = =+ = = Ta cc vect:BM(0;a;b2),BD(-a;a;0),BE(-a;0;b) Xt tch hu
hng:
2b b0 a a 0 ab abMB, BD , , , , a2 2a a 2 2a 0 0 a| | | | | ( =
= | |\ . |\ . Tch v hng: MB, BD BE ( = 23a b2 Th tch: V = 1MB, BD
BE6 ( = 1623a b2 = 2a b4 (vtt) -Nghin cu li gii KCO BD , ko di EM
ct SO ti N, mt phng (BDM) chia khi chp thnh hai khi chp E.BDM v
N.BDM nn E.BDM E.BDN1V V2=Generated by Foxit PDF Creator Foxit
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only.www.vnmath.com V M l trung im ca SG nn: CN = CA Din tch tam
gic BDN: S = 1 1 3 2aBD.NO a 2.2 2 2== 23a2 ADFGHMNCBEO Th tch:
E.BDM E.BDN1V V2== 12 1623a22b = 2a b4 (vtt) Bi 9:
TnhthtchcakhitdinABCDccccnhibngnhauAB= CD = a, AC = BD = b, AD = BC
= c? -Trnh by li gii Dng t din APQR, y l t din vung ti nh A, tht
vy: AD = BC = PQ2 BC l ng trung bnh ca tam gic PQR BC = QD = DP AD
= QD = PDAQ AP Hon ton tng t ta c:AQ AR ,AR AP Ta c:APQR ADBQ ABCD
ACDP ACBRV V V V V = + + +
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only.www.vnmath.com ABCD APQR AQR1 1 1 1V V AP.S4 4 3 24A= = =
AP.AQ.AR p dng nh l Pytago trong tam gic APQ, AQR, APR
2 2 2 2 2AP AQ QP 4BC 4a + = = = 2 2 2 2 2AQ AR QR 4CD 4c + = =
= AP =2 2 22(a b c ) + , AQ = 2 2 22(b c a ) + , AR = 2 2 22(a c b
) + Th tch: ( )32 2 2 2 2 2 2 2 21V 2 (b c a )(a c b )(a b c )24= +
+ + = 2 2 2 2 2 2 2 2 22(b c a )(a c b )(a b c )2+ + + (vtt) Bi tp
nghBi 1 ( khi A - 2007) Cho hnh chp S.ABCD c y l hnh vung cnh a, mt
bn SAD l tam gic u,(SAD) (ABCD) , gi M, N, P l trung im ca
SB,BC,CD. Tnh th tch ca khi chp CMNP theo a? Bi 2 ( Khi A - 2009)
Chohnh chpS.ABCDcyl hnh thangvungtiAvD,AD = AB = a, gc gia hai mt
phng (SCD) v (ABCD) bng 060 , gi I l trung im AD, bit hai mt phng
(SBI) v (SDI) cng vung gc vi mt phng (ABCD). Tnh th tch ca khi chp
S.ABCD theo a? Bi 3 (Khi B - 2006)
ChohnhchpS.ABCDcylhnhchnhtAB=a,AD=a 2 , SA = a, SA vung gc vi mt
phng (ABCD). Gi M, N l trung im ca AD, SC, I l giao im ca AC v BM.
Tnh th tch ca t din ANIB? Bi 4 (Khi A - 2008)
ChohnhlngtrngABC.DEFcdicnhbnbng2a.y ABC l tam gic vung ti A,AB =
2a, AC = a 3. Hnh chiu vuong gc ca D ln mt phng (ABC) l trung im G
ca cnh BC. Tnh th tch ca khi chp G.ABC? 2. Th tch ca khi lng tr
Trongmcnytasdngnhlsau:Thtchcahnhlngtrbng mt phn ba din tch y nhn vi
chiu cao V B.h = trong : B l din tch y h l chiu cao Generated by
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evaluation only.www.vnmath.com Bi 1
ChohnhtgicuABCD.EFGHckhongcchgiahaing thng AD v ED bng 2. di ng cho
mt bn bng 5. Tnh th tch khi lng tr ? ABCEFGHKD -Trnh by li gii
GiKlhnhchiuvunggccaAlnEDAK ED ,AB// EF (EFD) edo AB // (EFD)nn
d(A,EFD) = d(AB,ED) MEF (EFDA)nnEF AKAB AK AK=d(A,EFD)= d(AB,ED) =
2 t EK = x ( 0s xs 5 ). Trong tam gic vung AED ta c: AK2 = KE.KD 4
= x(5-x) x2 - 5x + 4 = 0 x 1x 4=
= Vi x = 4 ta c AE = 2 2AK KE 5 + = V = AE.ABCDS=5( vtt) Vi x =
4 ta c AE = 2 5 V = 10 5( vtt) Bi 2 y cakhilngtr ngABC.DEF l tam
gic u. Mt phng y to vi mt phng (DBC) mt gc 030 . Tam gic DBC c din
tch bng 8. Tnh th tch khi lng tr ? Generated by Foxit PDF Creator
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only.www.vnmath.com ABCDEFK -Trnh by li gii t CK = x, DK vuong gc
vi BC nn
DKA = 030Xt tam gic ADK c: cos030= AKDK AK = x 3, DK = 2x Din
tch tam gic BCD: S = CK.Dk = x.2x = 8, do x = 2 AD = AK.tan030= x 3
33 = 2 Th tch khi lng tr: V = 13AD.CK.AK = 83(vtt) Bi 3 Cho khi lng
tr ng ABCD.EFGH c y l hnh bnh hnh v
BAD = 045 , cc ng cho EC v DF to vi y cc gc 045v 060 . Chiu cao
ca lng tr bng 2. Tnh th tch ca lng tr ? -Trnh by li gii T gi
thit:
GAC = 045 ,
BDF = 060 , AC = AG = 2, BD = 2.cot060= 23 p dng nh l cosin
trong tam gic: 2 2 2BD AB AD 2.AB.AD. = + cos045
2 2 2 0AC AD CD 2.AD.CD.cos135 = + 2 2 0 0BD AC 2.AD.AB.cos45
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only.www.vnmath.com = -2 2 AB.AD 43 - 4 = -2 2 AB.AD AB.AD = 43 2 -
Th tch cn tm: V = AB.AD.EA.sin045=43 2 22 2 (vtt) Bi 4
ChohnhlngtrtamgicABC.EGHcyABCltamgicvung cn, cnh huyn bng2 . Bit mt
phng (AED) vung gc vi mt phng (ABC),AE =3 .Gc
AEBlgcnhn,gcgia mtphng (AEC)vi (ABC) bng 060 . Tnh th tch ca lng
tr? ABCKEMHG -Trnh by li gii Generated by Foxit PDF Creator Foxit
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only.www.vnmath.com HEK AB(K AB) e EK (ABC) . V
AEB l gc nhn nn K thuc on ABK KM AC EM AC ( theo nh l ba ng vung
gc )
EMK = 060 . Gi s EK = x ,2 2EK EA EK = += 23 x MK = AK.sin
KAM = 22 23 x M MK = EKcot060= x3, do : 22 23 x = x3x =35 Vy V =
EK.ABCSA= 12AC.CB.EK = 3 510 ( vtt) Bi tp ngh Bi 1 Cho lng tr
tgicABCD.EFGH c yl hnhthoi c di cnh bng a. Gc
BAD = 060 ,AF BH . Tnh th tch ca khi lng tr ? Bi 2 Cho
lngtrABC.DEFcyl tamgicvung cn tiA,BC= 2a. M l trung im ca AD,
gc
BMC =o. Tnh th tch ca lng tr ? 3. Th tch ca khi hp ch nht Trong
mc ny ta s s dng nh l sau:th tch ca khi hp bng tch di ba kch thcV =
a.b.c = B.h trong : a, b, c l ba kch thc B l din tch y h l chiu cao
Bi 1 ChokhihpABCD.EFGHcttccccnhubnga,
BAD=
EAD =o ( 0sos 090). Tnh th tch khi hp ? Generated by Foxit PDF
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only.www.vnmath.com ABCEFGHKDM -Trnh by li gii HEM AC(M AC) e (1)
tam gic EBD cn ti E ( do EB = ED ) BD EO MBD AC BD (BAO) BD EM (2)
T (1) v (2) ta c:EM (ABCD) hay EM l ng cao t
EAO =, hEK AB MK AK (nh l ba ng vung gc) cos2ocos = AM AKAE AM=
AKAE = cos o cos = coscos2ooEM = a.sin = 22cosa 1cos2oo= 2 2acos
cos2cos2o oo Th tch cn tnh: V = AB.AD.EM.sino = 2a .sino2 2acos
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only.www.vnmath.com = 3a .sin2o2 2cos cos2o o (vtt) Bi 2: Cho khi
hp ch nht ABCD.EFGH c y l hnh ch nht c AB =3 , AD =7 , hai mt bn
(ABDE) v (ADEH) ln lt to vi y cc gc 045v 060 , di tt c cc cnh bn u
bng 1. Tnh th tch ca khi hp ? ABCEFGHMKDN -Trnh by li gii KEK
(ABCD), (K ABCD) e ,KM AD(M AD) e ,KN AB(N AB) eTheo nh l ba ng
vung gc ta c:AD EM, AB NK Ta c:
EMK = 060 ,
ENK= 045 ,t EK = x khi : EM = 0xsin60 = 2x3 AM = 2 2EA EM = 23
4.x3 = KN m KN = x.cot045 Nn x = 23 4.x3 do x = 37 Th tch khi hp ch
nht:Generated by Foxit PDF Creator Foxit
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only.www.vnmath.com V = AB.AD.x =7 . 3.37 = 3 (vtt) Bi 3 Cho hnh hp
ch nht c di ng cho bng d, ng cho to vi y gco, to vi mt bn ln
gc|,tnh th tch ca khi hp ? BCDEFGHA
-Trnh by li gii ng cho AG c hnh chiu ln (ABCD) l AC,ln mt phng
(BCGF0 l BG nn:
GAC = o,
AGB= | p dng nh l Pytago trong cc tam gic: ACG, GBA, ABC c CG =
d.sino , AC = d.cos o ,AB = d.sin|, BC = 2 2AC AB = d.2 2cos sin o
| ta c V = AB.BC.CG = d3.sino.sin|.2 2cos sin o | M: ( )1 cos2 1
cos2 1cos2 cos22 2 2+ o | = o + |= cos( o +|).cos( o -|) Vy V=
d3.sino .sin|. cos( ).cos( ) o + | o |(vtt) Bi tp ngh Bi 1 Cho hnh
hp ch nht ABCD.EFGH c AB = a, AD = b, gc
BAD = o , ng cho AD t vi y gc|. Tnh th tch khi hp ch nht ?
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only.www.vnmath.com Bi 2 Cho t din ABCD c AB = CD, AC = BD, AD =
BC, qua mi cnh ca t din k mt phng song song vi cnh i din, cc mt
phng nhn c xc nh mt hnh hp: 1)Chng minh hnh hp ni trn l hnh hp ch
nht? 2)CMR hhcn ABCDV 3V =3)Gi IJ, EF, MN l cc dng trung bnh ca t
din. CMR: ABCD1V3= IJ.MN.EF Bi 3 Cho hnh hp ch nht ABCD.EFGH, M l
trung im ca AD, mt phng (ABM) ct ng cho AG ti I, tnh t s th tch ca
hai khi a din c to bi mt phng (EBM) ct hp? 4. Bi ton cc tr th tch
Bi 1 Cho hnh chp S.ABC c SA = x, SB = y, cc cnh cn li bng 1,vi gi
tr no ca x, y th th tch ca khi chp l ln nht, tm gi tr ln nht ?
ABCSNM -Trnh by li gii Gi M, N l trung im ca SA, BC, ta c: S.ABCV=
2.S.MBCV , cc tam gic ABS, ACS c: BA = BS, CA = CSA ABS =A ACS v l
cc tam gic cn Ta c:BM SA,CM SA SA (MBC) SM (MBC) , Generated by
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evaluation only.www.vnmath.com SM l ng cao, SM = x2 Tnh din tch y:
MB = MC = 2x14 , MN = 22BCBM4= 2 2x y14+
MBCSA = 12MN.BC = 2 2y x y12 4+Th tch: S.MBCV= 1 x y3 2 2 2 2x
y14+= xy12, S.ABCV= 2 2xy x y16 4+Ta c: ( x-y)2> 0 x2 + y2>
2xy 2 2x y xy4 2+>
S.ABCV= 2 2xy x y16 4+ s xy6xy12 s 1622 xy(xy)2 s 16xy xy2 (2
xy)2 2p dng BT Cauchy cho ba s xy2, xy2, (2-xy) ta c: xy2xy2(2-xy)s
33xy xy(2 xy)2 23| |+ + |\ . = 1627 Vs 161627 = 2 327, du bng xy ra
khi 2 2x y 2xyxy2 xy2 + == x = y = 23 Bi 2 Cho hnh chp t gic u
S.ABCD c khong cch t nh A n mt phng (SBC) bng 2a, gi l gc gia mt bn
vi mt y, vi gi tr no caoth th tch ca khi chp l ln nht? Generated by
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evaluation only.www.vnmath.com A BCSHMIND -Trnh by li gii M, N l
trung dim ca BC v AD nn
SMN = o, v AD // BC suy ra AD // (SBC)d(A,SBC) = d(N,SBC) (1) Mt
khc:MN BC,SM BC BC (SMN) (SBC) (SMN) Do SM =(SBC) (SMN) , kNH SM NH
(SBC) ,d(N,SBC)= NH(2) t (1) v (2) ta c NH = 2a H thc lng trong tam
gic vung MNH ta c: MN = NHsino = 2asino Din tch y : 22 2ABCD4aS AB
MNsin= = =o Gi I l tm y th ta c SI = MI.tano= a atansin coso =o o
Th tch: th tch V = 13SI.ABCDS= 324a3sin .cos o o 2minV sin .cos o o
t GTLN cos o(1 - 2cos o) t GTLNt x = cos o , xt hm s y = x - x3 trn
(0,1), xt du hm y ta cGenerated by Foxit PDF Creator Foxit
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only.www.vnmath.com maxy= 3 2 3y( )3 9=khi v ch khi x = 33cos o= 33
Vy 3minV 2a 3 =( vtt ) Bi 3 Cho tam gic u OAB c AB = a, trn ng thng
i qua O v vung gc vi mt phng(OAB) ly im M, t OM = x,Gi E, F ln lt l
cc hnh chiu vung gc ca A ln MB v OB.ng thng EF ct d ti N. Xc nh x
th tch khi chp ABMN l nh nht? AOBMEFN -Trnh by li gii Gi V l th tch
khi t din ABMN ta c V = M.OAB N.OABV V += OAB OAB1 1OM.S ON.S3 3A
A+= OAB1(OM ON).S3A+Do th tch V nh nht ( OM + ON ) t GTNNHai tam
gicAOMB AOFN suy ra: OM.ON = OF.OB = hng s v O, F, B c nh, ta c: OM
+ ON2 OM.ON sdu = xy ra OM = ON nn ( OM + ON ) t GTNNOM = ON = x V
OM.ON = OF.OB 22ax2= a 2x2=( OF = a2, OB = a ) Generated by Foxit
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evaluation only.www.vnmath.com Vy M thuc d sao cho OM = a 2x2=th th
tch l nh nht , khi : V = OAB1(OM ON).S3A+= 2a 612 ( vtt ) Bi 4 Cho
hnh chp S.ABCD c 7 cnh bng 1, cnh bn SC = x. Tinh sth tch ca khi
chp, vi gi tr no ca x th th tch l ln nht? ABCSDOH -Trnh by li gii
SH l ng cao trong tam gic C nn ta c: SH.AC = SA.SC SH = 2SA.SC xACx
1=+ Ta c: OB2 = AB2 - AD2 = 2(3 x )4 nn OB = 23 x2 ( 0s xs3) Din
tch y ABCDS= AC.OB = 213 x2 .2x 1 +Th tch V = ABCD1S .SH3 = 2 221
x(3 x )(x 1)6x 1 ++ = 21x 3 x6Ta c: V2 = 2 21x (3 x )36v 2 2x (3 x
) + = 3 khng i nn Max2 2x (3 x )( = 94 2 2x (3 x ) = 6x2=Generated
by Foxit PDF Creator Foxit Softwarehttp://www.foxitsoftware.com For
evaluation only.www.vnmath.com V2 t gi tr ln nht l 94.36 6x2=
max1V4=6x2=Bi 5 Cho hnh chp S.ABCD c y l hnh vung cnh a, hai mt bn
(SAB) v (SAD) cng vung gc vi y, mt gc
0xAy 45 =chuyn ng trn y quay quanh im A cc cnh Ax, Ay ct CB v CD
ti M, N, t BM = x, CN = y, tm x, y th tch caAMCNVt GTLN? BCDSMNA
-Trnh by li gii Tr ht ta chng minh ng thc: x + y =2a xya
Ta c
0BAM NAD MAN 90 + + = , 0BAM NAD 45 + =t 0BAM NAD 45 = o = | o
+| = tan( ) o + |= 1Ta c tan tan11 tan tano + |= o |, m x ytan ,
tan2 2o = | = suy ra 2x y21 x yxy1a+= + = 2a xya,Generated by Foxit
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evaluation only.www.vnmath.com ta c 2AMCN ABCD ABM ADNax ayS S S S
a2 2= =
2AMCN AMCN1 a ax ayV SA.S (a )3 3 2 2= = = 2a(a xy)6+ ta c xy
2(x y)4+s suy ra Max (xy) = 2(x y)4+
maxV (xy) t GTLN khi (xy) = 2(x y)4+ t GTLN suyra22(x y)a(x y)
a4++ + = suy ra x y 2a( 2 1) + = 2maxaV (2 2)3= khi x = y =a( 2 1)
Bi tp ngh Bi 1 Cho hnh chp S.ABC trong SA (ABC) , ABC l tam gic
vung cn ti C. Gi s SC = a. Hy tm gc gia hai mt phng (SCB) v (ABC)
th tch khi chp l ln nht, tm gi gi tr ln nht ? Bi 2 ( s 21- Chuyn
luyn thi vo H - Trn Vn Ho) Cho ba tia Ox, Oy, Oz vung gc vi nhau
tng i mt. Xt tam din Oxyz. im M c nh nm trong gc tam din. Mt mt
phng qua M ct Ox, Oy, Oz ln lt ti A, B, C gi khong cch t M n cc mt
phng (OBC), (OCA), (OAB) ln lt l a, b, c. Tnh OA, OB,OC theo a, b,
c th tch khi t din l nh nht?5. Chng minh h thc hnh hc chng minh cc
h thc trong khi adin ta c th s dng cc kin thc v th tch gii nh sau:
-Gn bi ton cn chng minh vo mt h thc no v th tch, cc h thc ny thng
l: Th tch ca mt khino c th biu din thnh tng hoc hiu cc th tch ca
khi a din c bn ( nh khi chp, khi lng tr ) - Vi cc h thc v th tich y
sau cc php bin i tng ng n gin ta nhn c iu phi chng minh.Bi 1 Cho t
din ABCD, im O nm trong t din v cch u cc mt ca t din mt khong r. Gi
A B C Dh , h , h , hln lt l khong cch t cc im A, B, C, D n cc mt i
din. Generated by Foxit PDF Creator Foxit
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only.www.vnmath.com CMRA B C D1 1 1 1 1r h h h h= + + +ABCDH Khi t
din ABCD c chia thnh 4 khi t din OBCD, OCAD, OABD, OABC. Ta c:
OBCDABCD AV rV h= OCADABCD BV rV h= OABDABCD CV rV h= OABCABCD DV
rV h=Cng v vi v ca cc ng thc trn ta c: OBCD OCAD OABD OABCABCD A B
C DV V V V 1 1 1 1rV h h h h ( + + += + + + ( ABCDABCD A B C DV 1 1
1 1rV h h h h (= + + + ( A B C D1 1 1 1 1r h h h h= + + +( pcm )Bi
2 Generated by Foxit PDF Creator Foxit
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only.www.vnmath.com Chng minh rng tng khong cch t mt im nm trong t
din n cc mt i din ca n khng ph thuc vo v tr ca im nm trong t din ?
AB COFGHK Gi s M l im ty thuc min trong ca t din u ABCDGi 1 2 3 4d
,d , d , dl khong cch t im M n cc mt (BCD), (ACD), (ABD), (ABC) Gi
1 2 3 4V, V , V , Vl th tch ca 4 khi t din chung nh M, V l th tch
ca t din ABCD ta c: V = 1 2 3 4V V V V + + + 1 2 3 4V V V V1V V V
V= + + +V ABCD l t din u nn khong cch t nh xung mt i din bng nhau.
Ta gi s khong cch ny l h, hai t din ABCD v MBCD c chung ynn: 1 1d
Vh V=Hon ton tng t ta c kt qu: i iV dV h=(i= 2,3,4) Do : 1 2 3 4d d
d d1h h h h= + + + h = 1 2 3 4d d d d + + +( pcm) Bi 3 Cho gc tam
din vung Oxyz nh O trn Ox, Oy, Ox ln lt ly cc im A, B, C sao cho OA
+ OB + OC + AB + AC + BC = L, gi V l th tch ca t din ABCD. CMR: 3L
( 2 1)V162sGenerated by Foxit PDF Creator Foxit
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only.www.vnmath.com Hng dn gii AC OB t OA = a, OB = b, OC = c p dng
BT Bunhiacpxki: a + b 2 22(a b ) s + , a + c 2 22(a c ) s + , b + c
2 22(c b ) s +cng v vi v cc BT trn: 2 2 2 2 2 22(a b c) (a b ) (a c
) (c b ) + + s + + + + +(a + b + c)( 1 +2 ) s2 2 2 2 2 2(a b ) (a c
) (c b ) (a b c) + + + + + + + + (a + b + c)( 1 +2 )s L (1)du =
trong (1) xy ra khi a = b = c p dng BT Cauchy cho a, b, c ta c: a +
b + c 3abc > (2) Ta c V =abc6, BT (2)a + b + c 33 6V > (3) Du
= trong (3) xy ra khi a = b = c T (1), (3) ta c3L 3.(1 2).3. 6V
> +hay3L ( 2 1)V162s(4) Du = (4) xy ra khi a = b = c = L( 2 1)3
Bi 4 Cho OABC l t din vung ti nh O, vi OA = a, OB = b, OC = c, gi r
l bn knh hnh cu ni tip t din.Generated by Foxit PDF Creator Foxit
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only.www.vnmath.com CMR: 1 1 1 1 3 3r a b c a b c= + + ++ + Hng dn
gii A BCOH K OH (ABCD) v gi s OH = h Do OABC l t din vung nn a, b,
c, h l 4 ng cao ca t din ln lt k t A, B, C, O theo kt qu ca bi tp 2
ta c: 1 1 1 1 1r a b c h= + + +(1)T (1)suy ra BT cn chng minh c
dng: 1 3 3h a b c=+ +(2) V OABC l t din vung nn ta c kt qu sau: 2 2
2 21 1 1 1h a b c= + +(3) Theo (3) v p dng BT Cauchy ta c: 32 2 2 2
2 2 21 1 1 1 1 1 13h a b c a b c= + + >(4) Ta li c: ( a + b + c
)2 2 2 2 39 a b c > (5) T (3), (4), (5) ta c: 2 2 2 21 93h a b
c>+ + Generated by Foxit PDF Creator Foxit
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only.www.vnmath.com 1 3 3h a b c>+ + Vy (2) ng suy ra pcm, du =
xy ra khi a = b = c T bi ton trn ta c kt qu: h1 3r s +Tht vy: Theo
(1)1 1 1 1 1r a b c h= + + +Do:22 2 2 21 1 1 1 1 1 13 3a b c a b c
h| | | |+ + s + + = ||\ . \ . hay 1 1 1 3a b c h+ + s 1 1 3 h1 3r h
h rs + s +( pcm )Bi tp ngh Bi 1 Chng minh khong cch t mt dim nm
trong hnh lng tr n cc mt ca n khng ph thc vo v tr ca im nm trong
lng tr ? Bi 2 Cho hnh chp tam gic c a b csin sin sin= =o | , trong
a, b, c l ba cnh ca tam gic y. Ccgc o,|,tng ng l cc gc nh din cankj
a, b, c. Chng minh tng khong cch t mt im O trn mt y n cc mt xung
quanh ca hnh chp l mt hng s Generated by Foxit PDF Creator Foxit
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