1

S Gio Dc V o To Qung Nam Trng THPT L Qu n BI TP TH TCH KHI A DIN GIO VIN : TRNG QUANG THNH

T : Ton - Tin

Trng THPT L Qu n Trong chng trnh gio dc ph thng th mn ton c nhiu hc sinh yu thch v say m, nhng ni n phn mn hnh hc th li mang nhiu kh khn v tr ngi cho khng t hc sinh, thm tr ta c th dng t S hc.c bit l hnh hc khng gian tng hp. y l phn c trong cu trc thi cao ng v i hc v thng xuyn xut hin trong cc thi tuyn chn hc sinh gii v kin thc phn ny yu cu hc sinh phi t duy cao,kh nng phn tch tng hp v tng tng m mt ch im ca quan trng ca hnh hc khng gian tng hp l tnh th tch khi a din. Nhm gip hc sinh vt qua kh khn v tr ngi v ngy cng yu thch v hc ton hn yu cu cc thy c chng ta phi c nhiu tm huyt ging dy v nghin cu .Qua thc t ging dy ti c cht kinh nghim ging dy phn ny mong c chia s cng cc thy c ng nghip v nhng ngi yu thch mn ton.

I )TNH TH TCH KHI A DIN THEO CNG THC

Vic p dng cng thc thng thng yu cu a) xc nh ng cao

b) tnh di ng cao v din tch mt y

xc nh ng cao ta lu

Hnh chp u c chn ng cao trng vi tm ca y.

Hnh chp c cc cnh bn bng nhau th chn ng cao trng vi tm ng trn ngoi tip mt y.

Hnh chp c cc mt bn cng to vi y nhng gc bng nhau th chn ng cao chnh l tm ng trn ni tip mt y.

Hnh chp c mt mt bn vung gc vi y th chn ng cao nm trn giao tuyn ca mt phng v y.

Hnh chp c hai mt bn cng vung gc vi y th ng cao nm trn giao tuyn ca hai mp tnh di ng cao v din tch mt y cn lu

Cc h thc lng trong tam gic c bit l h thc lng trong tam gic vung.

Cc khi nim v gc, khong cch v cch xc nh.

Sau y l cc bi tp Bi1 Chp tam gic u SABC c y l tam gic u cnh bng a, cc cnh bn to vi y mt gc 600.Hy tnh th tch ca khi chp .

Bi gii

Gi D l trung im ca BC v E l tm y

Khi AE=AD=

Ta c SAD=600 nn SE=AE.tan600=a

SABC= Do VSABC=SE.SABC=

BI 2: Cho hnh chp tam gic SABC c SA=5a,BC=6a,CA=7a. Cc mt bn SAB,SBC,SCA cng to vi y mt gc 600.Tnh th tch ca khi chp Bi gii Ta c hnh chiu ca nh S trng tm D ng trn ni tip y Ta c p==9a Nn SABC==6a2.

mt khc SABC=pr r==

trong SDK c SD=KDtan600 = r.tan600= 2a.

Do VSABC=SD.SABC=8a3.

Bi 3 Cho hnh chp SABC c cc cnh bn bng nhau cng hp vi y gc 600, y l tam gic cn AB=AC=a vBAC=1200 .Tnh th tch khi chp . Bi gii

Gi D l trung BC v O l tm ng trn ngoi tip tam gic ABC

C SO chnh l ng cao SABC=1/2.AB.AC.sin1200= v BC=2BD=2.ABsin600=a.

OA=R==a SO=OA.tan600=a.

Do vy VSABC=SO.SABC=1/4a3. Bi 4

Cho hnh chp SABCD c y ABCD l hnh vung cnh 2a,SA=a, SB=a v mpSAB vung gc vi mt y. Gi M,N ln lt l trung im ca AB,BC. Hy tnh th tch khi chp SBMDN.Bi gii

H SH AB ti H th SH chnh l ng cao SADM=1/2AD.AM=a2

SCDN=1/2.CD.CN=.a2 Nn SBMDN=SABCD-SADM-SCDN=4a2 -2a2=2a2.

mt khc

EMBED Equation.3 SH==

do VSBMDN=.SH.SBMDN=

Bi 5

Cho hnh chp S.ABCD, y ABCD l hnh thang vung ti A,D; AB=AD=2a,CD=a. Gc gia hai mpSBC v ABCD bng 600. Gi I l trung im ca AD, Bit hai mp SBI,SCI cng vung gc vi mpABCD. Tnh th tch khi chp S.ABCD.

Bi gii Gi H trung im l ca I ln BC, J l trung im AB. Ta c SImpABCD

IC==a

IB= =a v BC==a

SABCD=1/2AD(AB+CD)=3a2 SIBA=1/2.IA.AB=a2 v SCDI=1/2.DC.DI=1/2 SIBC=SABCD-SIAB-SDIC= mt khc SIBC=.IH.BC nn IH =

SI=IH.tan600=.

Do VABCD=SI.SABCD=a3 Bi 6 Cho chp SABC c SA=SB=SC=a, ASB= 600,CSB=900, CSA=1200 CMR tam gic ABC vung ri tnh th tch chp.

Bi gii

Gi E,D ln lt l AC,BC

SAB u AB=a, SBC Vung BC=a.

SAC c AE=SA.sin600=

EMBED Equation.3 AC=av SE=SAcos600=a.

EMBED Equation.3 ABC c AC2=BA2+BC2 =3a2 vy ABC vung ti B

C SABC=.BA.BC=

SBE c BE=AC=

SB2=BE2+SE2=a2 nn BE SE AC SE Do SE chnh l ng cao

VSABC= SE.SABC=

Bi 7

Cho khi lng tr ng ABC.A1B1C1 c y l tam gic vung ti A,AC=a,ACB=600ng thng BC1 to vi mp(A1ACC1)mt gc 300.Tnh th tch khi lng tr.

Bi gii

Trong tam gic ABC c AB=AC.tan600=a

ABAC v ABA1A

Nn AB mp(ACC1A) do AC1B=300 v AC1=AB.cot300=3a. .D pitago cho tam gic ACC1 : CC1==2a

Do vy VLT=CC1.SABC= 2a..a.a=a3.

Bi 8 Cho khi tr tam gic ABCA1B1C1 c y l tam gic u cnh a, im A1 cch u ba

im A,B.C,cnh bn A1A to vi mp y mt gc 600.Hy tnh th tch khi tr .

Bi gii

Ta c tam gic ABC u cnh a nn SABC=

mt khc A1A= A1B= A1C A1ABC l t din u

gi G l trng tm tam gic ABC c A1G l ng cao

Trong tam gic A1AG c AG=2/3AH=vA1AG=600 A1G=AG.tan600=a. vy VLT=A1G.SABC=

Bi 9

Cho khi tr tam gic ABCA1B1C1 c y l ABC l tam gic vung cn vi cnh huyn AB=.Cho bit mpABB1vung gc vi y,A1A=,Gc A1AB nhn, gc gia mpA1AC v y bng 600. hy tnh th tch tr. Bi gii

Tam gic ABC c cnh huyn AB=v cn nn CA=CB=1;

SABC=1/2.CA.CA=1/2.

. MpABB1vung gc vi ABC t A1 h A1G AB ti G.

A1G chnh l ng cao T G h GH AC ti H

Gt gc A1HG=600 t AH=x(x>0)

Do AHG vung cn ti H nn HG=x v AG=x

HGA1 c A1G=HG.tan600=x.

A1AG c A1A2=AG2+A1G2 3=2x2+3x2 hay x=

Do A1G= vy VLT=A1G.SABC=

Bi 10

Cho khi hp ABCD.A1B1C1D1 c y l hcn vi AB= v AD=. Cc mt bn ABB1A1 v A1D1DA ln lt to vi y nhng gc 450 v 600. Hy tnh th tch khi hp bit cnh bn bng 1.gii

Gi H l hnh chiu ca A1 ln mpABCD T H h HMAD ti M v HNAB ti N

Theo gt A1MH=600 v A1NH=450 t A1H=x(x>0) ta c A1M==

t gic AMHN l hcn( gc A,M,N vung)

Nn HN=AM m AM==

Mt khc trong tam gic A1HN c HN=x.cot450

Suy ra x = hay x= vy VHH=AB.AD.x= 3.

II ) TNH GIN TIP Ngha l ta s dng phn chia lp ghp khi a din, a v bi ton p dng tnh th tch theo cng thc hoc dng bi ton tnh t l hai khi t din(chp tam gic) Cho hnh chp SABC. Trn cc on thng SA,SB,SC ly ln lt ba im A1,B1,C1 khc vi S th Chng minh bi ton T s th tch hai khi t din(chp tam gic) Gi H,E ln lt l hnh chiu ca A,A1 trn mpSBC

AH / / A1E nn SAH v SA1E ng dng

Khi VSABC=AH.SSBC=AH.SB.SC.sinBSC. VSABC = A1E.SSBC = A1E.SB1.SC1.sinBSC. Do vy

Nn Bi 1

Cho hnh chp SABC c SA=a,SB=2a,SC=3a v BSA=600, ASC=1200, CSB=900. Hy tnh th tch chp

Bi gii

Nhn xt cc mt y khng c cc lu nn vic xc nh ng cao l kh nhng ta thy cc gc nh S l rt quen thuc. Ta lin tng n bi 6 phn I Vy ta c li gii sau

Trn SB ly B1 Sao cho SB1=a, Trn SC ly C1 sao cho SC1=a,

Ta c (theo bi 6)

M =

Bi 2 : Cho khi tr tam gic ABCA1B1C1 c y l tam gic u cnh a. A1A =2a v A1A to vi mpABC mt gc 600. Tnh th tch khi t din A1B1CA.gii

Gi H l hnh chiu ca A1 trn mpABC

Khi A1H=A1A.sinA1AH=2a.sin600=a.

M VLT=A1H.SABC=

nhn thy khi lng tr c chia lm ba khi chp

khi chp CA1B1C1 c =VLT khi chp B1ABC c =VLT

Khi chp A1B1CA do =VLT =

Bi 3 :Cho khi hp ch nht ABCD.A1B1C1D1 c AB=a,A1A=c,BC=b. Gi E,F ln lt l trung im ca B1C1 v C1D1. Mt phng FEA chia khi hp thnh hai phn. hy tnh t s th tch hai khi a din Bi gii

DDF

Mp(FEA) ct cc on thng A1D1,A1B1,B1B,D1D ln lt ti J,I,H,K(hv)

Gi V1,V2 ln lt l th tch phn trn v phn di mp

Ta nhn thy rng hai phn khi a din cha phi khi hnh quen thuc nhng khi ghp thm hai phn chp HIEB1 v chp KFJD1 th phn di l hnh chp AIJA1 Ba tam gic IEB1,EFC1,FJD1 bng nhau c.g.c Theo TA-LET V

V1= -2.=

V2= Vhh-V1= do vy

III) BI TON N TP

Sau khi trang b phn phng php nh vy ta cng gip hc sinh a ra cch gii mt bi ton linh hot bng c hai phng php hc sinh so snh i chiu la chn v a ra bi tp mc tng hp

Bi 1

Cho khi lng tr ng ABC.A1B1C1 c tt c cc cnh u bng a.

a) hy tnh th tch khi t din A1BB1C.

b) Mp i qua A1B1v trng tm tamgic ABC ct AC,BC ln lt ti E,F. Hy tnh th tch chp C.A1B1FE.

Gii a) Cch 1 tnh trc tip

gi H l trung im B1C1 suy ra Vtd=

Tng t gi K l trung im AB

Cch 2

Nn

b) cch 1 Tnh trc tip gi Q l trung im ca A1B1,G l trng tm tam gic ABC

Khi qua G k d // vi AB th E=ACd v F=BC d

MpCKQ chnh l mp trung trc ca AB,FE Nn khong cch t C n QG chnh l khong cch t C n mpA1B1FE Ta c

Mt khc

Cch 2 dng gin tip (s dng bi ton t l th tch )

Bi 2 :Cho hnh chp S.ABCD c y ABCD l hcn,AB=a,AD=a,SA=2a v SA ABCD, Mt mp i qua A v vung gc vi SC,ct SB,SC,SD ln lt ti H,I,K. Hy tnh th tch khi chp S.AHIK theo a Bi gii

Cch 1 tnh trc tip

Ta c

Nn cn ti A m AI SC nn I l trung im SC AI=SI=

M cho nn Trong tam gic vung HAI c

Tng t ta c AK=

Cch 2 tnh gin tip Tng t nh cc 1 ta ch lp lun AH SB, AKSD

Tng t Do VSAHIK=

Bi 3

Cho hai ng thng cho nhau x v y. ly on thng AB c di a trt trn x, on thng CD c di b trt trn y. CMR VABCD khng i

gii

nhn xt cc yu t khng i a,b,gc v khong cch gia hai ng thng x v y

t (x,y)= v d(x,y)=d Ta dng hnh lng tr ABF.CED nh (hv)

Khi d=d(x,y)=d(AB,CD)=d(AB,CDE)=d(B,CDE) hay d chnh l chiu cao lng tr

VLT= d.SCDE=d.CD .CE.sin=d.b.a.sin

mt khc Khi lng tr c ghp t 3 khi t din gm

T din BCDE c VBCDE=.d(B,CDE).SCDE=.VLT T din BACD v BAFD c th tch bng nhau

Do vy VABCD=.VLT=.d.a.b.sin= hng s

lCch 2 Dng hnh hp, cch 3 dng hbh Nh hai hv sau

Bi 4 Bi ton th tch lin quan n cc tr Cho hnh chp S.ABCD,SA l ng cao,y l hcn vi SA=a,AB=b, AD=c. Trong mpSDB ly G l trng tm tam gic SDB qua G k ng thng d ct cnh BS ti M, ct cnh SD ti N,mpAMN ct SC ti K . Xc nh M thuc SB sao cho VSAMKN t gi tr ln nht v nh nht, Hy tm gi tr ln nht v nh nht

Bi gii

Gi O L tm hcn ABCD Ta c SG=.SO v K=A GSC v K l trung im SC

Tng t

Do

Trong mpSBD

Do M,N ln lt nm trn cnh SB,SD nn

t t=( ) th

Nhn thy VSAMKN t GTLN,GTNN nu f(t)= vi

Ta c

Nn (loi)

f(1/2)=3/2 , f(1)=3/2 f(2/3)=4/3

do vy VSAMKN =l GTLN khi M l trung im SB hoc M trng vi B

VSAMKN =l GTNN khi MB chim 1 phn SB

III. BI TP T GII Bi 1 Cho tam gic ABC vung cn A v AB=a. Trn ng thngqua C v vung gc vi mp(ABC) ly im D sao cho CD=a. Mt phng qua C vung gc vi BD,ct BD ti F v ct AD ti E. tnh th tch khi t din CDEF.

Bi 2 cho hnh chp S.ABC c tam gic ABC vung ti C,AC=a,AB=2a,SA vung gc vi y.Gc gia mpSAB v mpSBC bng 600. Gi H,K ln lt l hnh chiu ca A ln SB,SC. Chng minh rng SA vung KH v tnh th tch khi chp S.ABC

Bi 3

Cho hnh chp u S.ABC cnh y bng a, Hy tnh th tch khi chp S.ABC bit

a) MpSBA vung gc vi mpSCA

b) Gi M,N ln lt l trung im SA,SC v mpBMN vung gc mpSAC

Bi 4 Cho khi lng tr ABC.A1B1C1 c BB1=a. Gc gia ng thng BB1v mpABC bng 600. Tam gic ABC vung ti C v gc BAC bng 600. Hnh chiu vung gc ca im B1 ln mpABC trng vi trng tm tam gic ABC, tnh th tch khi t din A1ABC theo a

Bi 5 Cho khi lng tr u ABC.A1B1C1 c cnh y bng a,khong cch t tm O ca tam gic ABC n mpA1BC bng .hy tnh th tch khi tr Bi 6 Cho khi lng tr ng ABC.A1B1C1 c y ABC l tam gic cn ti A,gc gia A1A v BC1 bng 300, khong cch gia chng bng a. Gc gia hai mt bn qua A1A bng 600. hy tnh th tch khi tr Bi 7 Cho lng tr xin ABC.A1B1C1 c y ABC l tam gic vung ti A,AB=a,BC=2a. Mt bnABB1A1 l hnh thoi nm trong mp vung gc vi y v hp vi mt bn mt gc . hy tnh th tch khi lng tr. Bi 8 cho hnh chp t gic u S.ABCD c cnh y bng a, cnh bn hp vi y gc 600, gi M l im i xng vi C qua D. N l trung im SC.mpBMN chia khi S.ABCD thnh hai phn. Hy tnh t s th tch ca hai phn Bi 9 cho hnh hp ch nht ABCD.A1B1C1D1 c AB=a,BC=2a,A1A=a,M thuc on AD sao cho AM=3MD.Hy tnh th tch khi t din MAB1C1,

Bi 10 Cho hlp ABCD.A1B1C1D1 c cnh a, im K thuc CC1 sao cho CK=2/3.a.Mt phng (P) qua A,K v song song vi BD chia khi lp phng thnh hai phn. Tnh t s th tch hai phn .

Bi 11 Cho hnh chp S.ABCD c y l hnh thang vung ti Av D. Tam gic SAD l tam gic u cnh 2a, cnh BC =3a. Cc mt bn to vi y cc gc bng nhau. Hy tnh th tch khi chp.

Bi 12 Cho hnh chp S.ABCD c y l hnh thang vi cc cnh AB=BC=CD=1/2.AD

Tam gic SBD vung nm trong mp vung gc vi y v c cc cnh gc vung l SB=8a,SD=15a. hy tnh th tch khi chp

Bi 13 Cho t din ABCD c tam gic ABC,ABD l hai tam gic u cnh a,mpADC vung gc mpBCD. Tnh VABCD.

Bi 14

Cho t din ABCD, cc im M,N,P ln lt BC,BD,AC sao cho BC=4BM, BD=2BN,AC=3AP. MpMNP chia t din lm hai phn tnh t s th tch hai phn .

Bi 15 Cho khi lng tr ABC.A1B1C1 c cc mt bn (A1AB),(A1BC),(A1CA) hp vi y (ABC) gc 600,gcACB=600,AB=a,AC=2a. tnh VLT. Bi 16 Cho hlp ABCD.A1B1C1D1 c cnh a.Gi M,N,P ln lt thuc cc on A1A,BC,CD sao cho A1A=3A1M,BC=3BN,CD=3DP.MpMNP chia khi lp phng lm hai phn. tnh th tch tng phn.

Bi 17 Cho t din ABCD.Gi M l trung im DA.Cc im N,P thuc BD sao cho BN=NP=PD.Hy tnh t s th tch ca hai phn t din ct bi a) mp qua MN v song song vi trung tuyn AI ca tam gic ABC

b) mp qua MP v song song vi AI

c) mp qua MN song song vi trung tuyn CE ca tam gic ABC

Bi 18 Cho t din ABCD c AB=BD=AC=CD=, Cnh BC=x, khong cch gia BC v AD bng y.Tnh VABCD theo x v y,tm x,y VABCD t gi tr Max,min. Ba 19 Trong mp(P) cho hnh vung ABCD c cnh AB=a, tia Ax v tia Cy cng vung gc vi mp(P) v cng thuc na mp b AC. Ly im M bt k thuc tia Ax v chn im N thuc tia Cy sao cho mpBDM vung gc vi mpBDNa) Tnh AM.CN theo a.

b) Xc nh v tr ca im M th tch khi t din BDMN t min.

Bi 20 Hai na ng thng Am,Bn vung gc vi nhau v nhn AB=a lm on vung gc chung. Cc im M,N ln lt chuyn ng trn Am,Bn sao cho MN=AM+BN.a) CMR VABMN khng i, tnh gi tr

b) Goi O l trung im AB,H l hnh chiu ca O trn MN. CMR. ....HT...H

K

A

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