1 Chuyn luyn thi i hc PHNG PHP GII CC BI TP HNHKHNG GIAN TRONG K THI TSH Bin son: GV Nguyn Trung KinTrong k thi TSH bi ton hnh khng gian lun l dng bi tp gy kh khn cho hc sinh. Nguyn nhn c bn l do hc sinh cha bit phn bit r rng dng bi tp la chn cng c, phng php gii cho ph hp. Bi vit ny s gip hc sinh gii quyt nhng vng mc . Phn 1: Nhng vn cn nm chc khi tnh ton - Trong tam gic vung ABC (vung ti A) ng cao AH th ta lun c:

b=ctanB, c=btanC;2 2 21 1 1AH AB AC= +-Trong tam gic thng ABC ta c: 2 2 22 2 22 cos ; cos2b c aa b c bc A Abc+ = + = . Tng t ta c h thc cho cng b, c v gc B, C: - 1 1 1sin sin sin2 2 2ABCS ab C bc A ac BA= = =-V(khi chp)=1.3 B h (B l din tch y, h l chiu cao) -V(khi lng tr)=B.h -V(chp S(ABCD)=13(S(ABCD).dt(ABCD)) -S=p.r (Trong p l na chu vi, r l bn knh vng trn ni tip tam gic) Phng php xc nh ng cao cc loi khi chp: -Loi 1: Khi chp c 1 cnh gc vung vi y chnh l chiu cao. -Loi 2: Khi chp c 1 mt bn vung gc vi y th ng cao chnh l ng k t mt bn n giao tuyn. -Loi 3: Khi chp c 2 mt k nhau cng vung gc vi y th ng cao chnh l giao tuyn ca 2 mt k nhau . -Loi 4: Khi chp c cc cnh bn bng nhau hoc cc cnh bn cng to vi y 1 gc bng nhau th chn ng cao chnh l tm vng trn ngoi tip y. CB H A 2 -Loi 5: Khi chp c cc mt bn u to vi y 1 gc bng nhau th chn ng cao chnh l tm vng trn ni tip y. S dng cc gi thit m: -Hnh chp c 2 mt bn k nhau cng to vi y gc th chn ng cao h t nh s ri vo ng phn gic gc to bi 2 cnh nm trn mt y ca 2 mt bn (V d: Hnh chp SABCD c mt phng (SAB) v (SAC) cng to vi y gcth chn ng cao h t nh S thuc phn gic gc BAC) -Hnh chp c 2 cnh bn bng nhau hoc hai cnh bn u to vi y mt gc th chn ng cao h t nh ri vo ng trung trc ca on thng ni 2 im cn li ca cnh bn thuc mt y. (V d: Hnh chp SABCD c SB=SC hoc SB v SC cng to vi y mt gc th chn ng cao h t S ri vo ng trung trc ca BC) Vic xc nh c chn ng cao cng l yu t quan trng tm gc to bi ng thng v mt phng hocgc to bi 2 mt phng. V d: Cho khi chp SABCD c mt bn SAD vung gc (ABCD), gc to bi SC v (ABCD) l 600, gc to bi (SCD) v (ABCD) l 450, y l hnh thang cn c 2 cnh y l a, 2a; cnh bn bng a. Gi P,Q ln lt l trung im ca SD,BC.Tm gc to bi PQ v mt phng (ABCD).Tnh V khi chp? R rng y l khi chp thuc dng 2. T ta d dng tm c ng cao v xc nh cc gc nh sau:-K SH vung gc vi AD th SH l ng cao(SC,(ABCD))= ; ( , ( )) ) SCH SM ABCD HMS = , vi M l chn ng cao k t H ln CD -T P h PK vung gc vi AD ta c ( , ( )) PQ ABCD PQK = Phn 3: Cc bi ton v tnh th tchA.Tnh th tch trc tip bng cch tm ng cao:V d 1) (TSH A 2009) Cho hnh chp SABCD c y ABCD l hnh thang vung ti A v D., c AB=AD=2a; CD=a. Gc gia 2 mt phng (SCB) v (ABCD) bng 600. Gi I l trung im DA BC M H S P Q K 3 AD bit 2 mt phng (SBI) v (SCI) cng vung gc vi (ABCD). Tnh th tch khi chp SABCD? TEL: 0988844088 HD gii: V 2 mt phng (SBC) v (SBI) cng vung gc vi (ABCD) m (SBI) v (SCI) c giao tuyn l SI nn SI l ng cao. K IH vung gc vi BC ta c gc to bi mt phng (SBC) v (ABCD) l 060 SHI = . T ta tnh c: 212; 5; ( ) ( ) 32IC a IB BC a S ABCD ADAB CD a = = = = + =2 22 21 3. ( ) ( ) ( ) ( ) 32 2 2a aIH BC SIBC SABCD S ABI SCDI a a = = = =nn 2 ( ) SIBCIHBC= =3 35a . T V(SABCD)=33 155a . V d 2) (TSH D 2009) Cho lng tr ng ABCABC c y ABC l tam gic vung ti B, AB=a; AA=2a; AC=3a. Gi M l trung im ca on AC, I l trung im ca AM v AC. Tnh V chp IABC theo a? HD gii:- ABC ABC l lng tr ng nn cc mt bn u vung gc vi y.V Ie(ACC) (ABC), t I ta k IHAC th IH l ng cao v I chnh l trng tm tam gic AAC2 43 3IH CI aIHAA CA = = =' ' C 22 2 2 2 2AA 9 4 5 2 AC A C a a a BC AC AB a ' ' = = = = = =V(IABC)=31 1 4 1 4. ( ) . . .2 .3 3 3 2 9aIH dt ABC a a a = = ( vtt) S IA B H D C 4 B.Tnh th tch bng cch s dng cng thc t s th tch hoc phn chia khi a din thnh cc khi a din n gin hn Khi gp cc bi ton m vic tnh ton gp kh khn th ta phi tm cch phn chia khi a din thnh cc khi chp n gin hn m c th tnh trc tip th tch ca n hoc s dng cng thc tnh t sth tch tm th tch khi a din cn tnh thng qua 1 khi a din trung gian n gin hn. Cc em hc sinh cn nm vng cc cng thc sau:( ) . .( ) . .VSA B C SASBSCVSABC SASB SC' '' ' ' '= (1) ( AABC) AA( ) SAVSVSABC' '= (2). Cng thc (2) c th m rng cho khi chp bt k. CBAC'B'A'S BCM A B A I H C 5 V d 3) Cho hnh chp SABCD c y ABCD l hnh thoi cnh a, 060 BAD = , SA vung gc vi y(ABCD), SA=a. Gi C l trung im SC, mt phng (P) i qua AC song song vi BD ct cc cnh SB, SD ca hnh chp ti B, D. Tnh th tch khi chp HD gii: Gi O l giao 2 ng cho ta suy ra AC v SO ct nhau ti trng tm I ca tam gic SAC. T I thuc mt phng (P)(SDB) k ng thng song song vi BD ct SB, SD ti B, D l 2 giao im cn tm. Ta c: 1 2;2 3SC SD SB SISC SD SB SO' ' '= = = =D thy ( ) ( ) ( ) ( )2 ; 2SAB C D SAB C SAB C SABCV V V V''' '' ''= =( ) ( ) . . 1( ) ( ) . . 3VSAB C D VSAB C SASBSCVABCD VSABC SASB SC''' '' ' ' = = = Ta c 3( )1 1 1 3 3. ( ) . . . . . .3 3 3 2 6SABCDV SAdt ABCD SA AD AB sinDAB a a a a = = = =3( )318SAB C DV a''' = (vtt) V d 4) (D b A 2007) Cho hnh chp SABCD l hnh ch nht AB=a, AD=2a, cng SA vung gc vi y, cnh SB hp vi y mt gc 600. Trn cnh SA ly M sao cho AM=33a. Mt phng BCM ct DS ti N. Tnh th tch khi chp SBCMN.HD gii:T M k ng thng song song vi AD ct SD ti N l giao im cn tm, gc to bi SB v (ABCD) l 060 SBA = . Ta c SA=SBtan600=a 3 . S B C D O BC DA 6 T suy ra SM=SA-AM=3 2 3 233 3 3SM SNa a aSA SD = = =D thy ( ) ( ) ( ) ( ) ( )2 2SABCD SABC SACD SABC SACDV V V V V = + = =

( ) ( ) ( ) SBCMN SMBC SMCNV V V = +( ) ( ) ( ) ( ) ( ) 1. . . 1. . .( ) ( ) 2 ( ) 2 ( ) 2. . . 2. . .1 2 53 9 9VSMBCN VSMBC VSMCN VSMCN VSMCN SM SB SC SM SC SNVSABCD VSABCD VSABC VSACD SASB SC SASC SD+ = = + = += + =M 3 3( ) ( )1 1 2 3 10 3. ( ) 3 .23 3 3 27SABCD SMBCNV SAdt ABCD a a a a V a = = = = Phn 4: Cc bi ton v khong cch trong khng gian A.Khong cch t 1 im n 1 mt phng gii quyt nhanh gn bi ton khong cch t mt im n mt mt phng hc sinh cn nm chc bi ton c bn v cc tnh cht sau * Bi ton c bn: Cho khi chp SABC c SA vung gc vi y. Tnh khong cch t A n (SBC) - H AM vung gc vi BC , AH vung gc vi SM suy ra AH vung gc vi (SBC). Vy khong cch t A n (SBC) l AH. Ta c 2 2 21 1 1AS AH AM= +S M N AD CB 7 HMCBAS * Tnh cht quan trng cn nm: - Nu ng thng (d) song song vi mt phng (P) th khong cch t mi im trn (d) n mt phng (P) l nh nhau - NuAM kBM =th /( ) /( ) A P B Pd kd =trong (P) l mt phng i qua M Trn c s cc tnh cht trn ta lun quy c khong cch t mt im bt k v bi ton c bn. Tuy nhin 1 s trng hp vic tm hnh chiu tr nn v cng kh khn, khi vic s dng cng thc tnh th tch tr nn rt hiu qu. Ta c V(khi chp)=1 3.3VB h hB= V d 1) Cho hnh chp SABCD c y ABCD l hnh vung cnh a. Hnh chiu ca S trng vi trng tm tam gic ABD. Mt bn (SAB) to vi y mt gc 600. Tnh theo a th tch ca khi chp SABCD v khong cch t B n mt phng (SAD). Li gii:Gi G l trng tm ca tam gic ABD, E l hnh chiu ca G ln AB Ta c:( ) ; SG ABGE AB AB SGE 060 SAG =. tan 3 SG GE SEG GE = =Mt khc G l trng tm ca tam gic ABD 13 3aGE BC = =31 3.3 9SABCD ABCDaV SG S = = 8 H GN vung gc vi AD, GH vung gc vi SN. Ta c /( ) / ( )2 2 2233 .3 . 33 33 3233 3B SAD G SADa aGN GS ad d GHGN GSa a= = = = =+| || | + | |\ .\ . HNEGDCABS V d 2) Cho hnh lng tr ng. ABCD A B C D ''''c y ABCDl hnh thoi ,3 AB a = , 0120 BAD Z = . Bit gc gia ng thng AC' v mt phng ( ) ADD A ' ' bng 030 .Tnh th tch khi lng tr trn theo a. v khong cch t trung im N ca BB n mt phng (CMA).Bit M l trung im ca AD Ta c . ' ' ' ''.ABCDA B CD ABCDV AAS =(1). y ABCD l hnh thoi gm 2 tam gic u ABC, ACD nn: ( )223 33 32 2.4 2ABCD ABCaaS SA= = =(2) Gi CM l ng cao ca tam gic u CAD th( ) ' ' ' CM ADAD nn 0' 30 CAM =Ta c 0 2 23 3 3' ' .cot 30 ' ' 62 2a aCM AM CM AA AM AM a = = = = =(3) Thay (2),(3) vo (1) ta c:2 3. ' ' ' '3 3 9 2. 62 2ABCDA B CDa aV a = = . 9 Ta c /( ' ) / ( ' ) N CMA K CMAd d =vi K l trung im ca DD (V K v N i xng nhau qua trung im O ca AC) T K h KH vung gc vi AM th /( ' )1( ' ) ; . ( ' ' ) ( ' ) ( ' ) ( )2K CMAKH ACM d KH KHAM dt AADD dt AAM dt MDK dt AKD = = 3 3 1 3 1 6 3 1 6 6. 6. 3 6. . . . . 34 2 2 2 2 2 2 2 2a a a a aKH a a a a KH a = =Vy /( ' )62N CMAd a =HKMB'C'A'D'DCBAN V d 3) Cho hnh chp SABC c gc to bi 2 mt phng (SBC) v (ABC) l 600, ABC,SBC l cc tam gic u cnh a. Tnh khong cch t nh B n mp(SAC).( d b khi A 2007) HD: Cch 1: Coi B l nh khi chp BSAC t gi thit ta suy ra BS=BA=BC=a. Gi O l chn ng cao h t B xung mp(SAC). O chnh l tm vng trn ngoi tip tam gic SAC. Gi M l trung im BC ta c; SM BCAM BC . Nn gc to bi (SBC) v (ABC) l 0a 360 AS=2SMA SM AM = = = . By gi ta tm v tr tm vng ngoi tip tam gic SAC. Tam gic SAC cn ti C nn tm vng trn ngoi tip nm trn trung trc ca SA v CN (N l trung dim ca SA). K trung trc ca SC ct trung trc ca SA ti O l im cn tm 222232 1316cos4SAaSCaNCSNCSC SC a| ||\ .= = = = 10 22 2 22 4 32;13 cos 13 13SCa a aOC BO BC OC aSCN = = = = = . Cch 2: 0( ) ( )1 22 2 . ( ) . .sin 603 3.2SABCD SABMaV V BM dt SAM AM MS = = =33( )16a dt SAC ==21 1 13 3 39 3 ( ) 3.AS= . . ( , ( )2 2 4 2 16 ( ) 13a VSABC aCN a a dB SACdt SAC= = =V d 4) Cho hnh chp SABCD c y ABCD l hnh thang0 90 ABC BAD = = , BA=BC=a, AD=2a. Cnh bn SA vung gc vi y v SA= 2 a , gi H l hnh chiu ca A ln SB. Chng minh tam gic SCD vung v tnh theo a khong cch t H n mp(SCD) (TSH D 2007) HD gii: Ta c 2 2 2 22; 6; 2 AC a SD SA AD a SC SA AC a = = + = = + = . Ta cng d dng tnh c2 CD a = . Ta c 2 2 2SD SC CD = + nn tam gic SCD vung ti C. 2 2 22 2 2 22 21 1 1 .AS . 2 2AS 3AB AS 222 233 3 3AB a aAH aAH ABa aaSHSH SA AH aSB a= + = = =+ + = = = = O S P C M B A N 11 21. .( ) 1( ) ( ) ( ) . ;2 2 2AB BC AD adt BCD dt ABCD dt ABD AB AD+= = =2231( ) . 22( ) . . 2 1 1. 2. 2; ( ) . ( )( ) . . 3 3 3.2 6dt SCD SC CD aVSHCD SH SC SD a aVSBCD SAdt BCD aVSBCD SB SC SD= == = = = = 32( )9VSHCD a = .Ta c 323 ( ) 2 1( /( )) .3( ) 9 3 2VSHCD adH SCD adt SCD a= = = B.Khong cch gia 2 ng thng cho nhau trong khng gian Khi tnh khong cch gia 2 ng thng cho nhau a v b trong khng gian ta tin hnh theo trnh t sau: - Dng (tm) mt phng trung gian (P) cha a song song vi b sau tnh khong cch t 1 im bt k trn b n mp(P)- Khi tnh khong cch t 1 im n mt phng ta c th vn dng 1 trong 2 phng php trnh by mc A. V d 1) Cho lng tr ng ABCABC c y ABC l tam gic vung AB=BC=a, cnh bn 2 AA a ' = . Gi M l trung im ca BC. Tnh theo a th tch khi lng tr ABCA B C '''v khong cch gia 2 ng thng AM, BC.(TSH D2008) HD gii: 32( ) .2VABCA B C S h a ' '' = = . Gi N l trung im ca BB ta c BC song song vi mp(AMN). T ta c:( , ) ( , ( )) ( , ( )) dB CAM dB AMN dB AMN ' ' = = v N l trung im ca BB. Gi H l hnh chiu vung gc ca B ln (AMN), v t din BAMN l t din vung ti B nn ta c 2 2 2 21 1 1 17aBHBH BA BN BM= + + =chnh l khong cch gia AM v BC. B C DA H S 12 Ch 1) Trong bi ton ny ta dng mt phng trung gian l mp(AMN) tn dng iu kin BC song song vi (AMN). Ti sao khng tm mt phng cha BC ccem hc sinh t suy ngh iu ny Ch 2) Nu mt phng (P) i qua trung im M ca on AB th khong cch t A n (P) cng bng khong cch t B n (P)) V d 2) Cho hnh chp t gic u SABCD c y l hnh vung cnh a. Gi E l im i xng ca D qua trung im ca SA, M l trung im ca AE, N l trung im ca BC. Chng minh MN vung gc vi BD v tnh khong cch gia 2 ng thng MN v AC.(TSH B 2007) HD gii: Gi P l trung im ca SA, ta c t gic MPNC l hnh bnh hnh.Nn MN// PC. T suy ra MN//(SAC). Mt khc BDmp(SAC) nn BDPCBD MN . Ta c: d(MN, AC)=d(N,(SAC))=1 1 1( , ( )) 22 4 2dB SAC BD a = = S MP E A N C D B B C A N BH M CA K 13 ( Ch vic chuyn tnh khong cch t N n (SAC) sang tnh khong cch t B n (SAC) gip ta n gin hobi ton i rt nhiu. Cc em hc sinh cn nghin cu k dng ton ny vn dng) V d 3) Cho hnh chp SABC c y ABC l tam gic vung cn ti B,2 , AB BC a = =hai mt phng (SAC) v (SBC) cng vung gc vi y (ABC). Gi M l trung im AB, mt phng qua SM song song vi BC ct AC ti N. Bit gc to bi (SBC) v (ABC) bng 600. Tnh th tch khi chp SBCNM v khong cch gia hai ng thng AB v SN (TSH A 2011) Gii: - Ta c 0 0 ( ); 90 60 2 3 SA ABC ABC SBA SA a = = =Mt phng qua SM song song vi BC ct AC ti N suy ra N l trung im AC T tnh c 33 V a =- K ng thng (d) qua N song song vi AB th AB song song vi mt phng (P) cha SN v (d) nn khong cch t AB n SN cng bng khong cch t A n(P). Dng AD vung gc vi (d) th/ /( ) AB SND , dng AH vung gc vi SD th / /( )2 2. 2 39( )13ABSN A SNDSA AD aAH SND d d AHSA AD = = = =+ MNDHCBAS Phn 5: Cc bi ton tnh gc gia 2 ng thng cho nhau trong khng gian. Khi cn tnh gc gia 2 ng thng cho nhau a v b trong khng gian ta phi tm 1 ng thng trung gian l c song song vi a v c ct b. Khi gc to bi a v b cng chnh l gc to bi b v c. Hoc ta dng lin tip 2 ng thng c v d ct nhau ln lt song song vi a v b. Sau ta tnh gc gia c v d theo nh l hm s csin hoc theo h thc lng trong tam gic vung. V d 1) Cho lng tr ABCABC c di cnh bn bng 2a , y ABC l tam gic vung tiA. AB = a , AC = a v hnh chiu vung gc ca A ln mp (ABC) l trung im ca cnh BC , Tnh theo a th tch khi chp AABC v tnh csin gc to bi AA v BC . (TSH A 2008) HD gii :Gi H l trung im caBC. Suy raAH(ABC) v2 21 132 2AH BC a a a = = + =Do AH =2 2' 3. AA AH a = 14 V(AABC) =13AH.dt (ABC) =32aTrong tam gic vung ABH ta cHB=2 2' ' 2 AB AH a + = nn tam gic BBH cn ti B. t l gc to bi AA v BC th 1' cos2.2 4aBBHa = = =Tel 0988844088 V d 2) Cho hnh chp SABCD c y ABCD l hnh vung cnh 2a , SA = a, SB = a 3mp (SAB) vung gc vi mt phng y . Gi M,N ln lt l trung im ca cc cnh AB,BC. Tnh theo a th tch khi chp SBMDN v tnh cosin gc to bi SM v DN. Hd gii: T S h SH vung gc AB th SH vung gc vi mp (ABCD). SH cng chnh l ng cao khi chp SBMDN . Ta c SA2 + SB2 = 4a2 = AB2SAB Avung ti S2ABSM a SAM = = A l tam gic u 32aABCH = D thy ng thng(BMDN)=1/2dt(ABCD)=2a2 . Do V(SBMDN)=31 3. ( )3 3aSH dt BMDN =K ME song song vi DN ( E thuc AD) suy ra AE =2a gi s(SM,DN)= ( , ). SMME = Ta c SA vung gc vi AD (nh l 3 ng vung gc ) suy B H C A B C A 15 ra 2 2 2 25 5,2 2a aSA AE SE SA AE ME AM ME = + = = + =Tam gic SME cn ti E nn cos525SMME = = PHN 4) CC DNG BI TP V MT CU NGOI TIP KHI A DIN gii quyt tt dng bi tp ny hc sinh cn nm vng kin thc c bn sau:** Nu I l tm mt cu ngoi tip khi chp 1 2..nSA A Ath tm I cch u cc nh 1 2; ; .....nS A A A - V vy tmI thuc trc ng trn y l ng thng qua tm vng trn ngoi tip y v vung gc vi y 1 2...nAA A(ng thng ny song song vi ng cao khi chp) (Phi ch vic chn mt y cn linh hot sao cho khi xc nh trc ng trn y l n gin nht) - Tm I phi cch u nh S v cc nh 1 2; .....nA A A nn I thuc mt phng trung trc ca iSAy l vn kh i hi hc sinh cn kho lo chn cnh bn sao cho trc ng trn xc nh v cnh bn ng phng vi nhau vic tm I c d dng ** Trong mt s trng hp c bit khi khi chp c cc mt bn l tam gic cn, vung, u ta c th xc nh 2 trc ng trn ca mt bn v y . Khi tm I l giao im ca 2 trc ng trn. Nu hnh chp c cc nh u nhn cnh a di mt gc vung th tm mt cu l trung im ca cnh a. ** Khi tnh ton cn lu cc cng thc: 4 4abc abcS RR S= = ;2 sin ,... a R A =Ta xt cc v d sau:S AE M BN C D 16 V d 1) Cho hnh chp SABCD c y ABCD l hnh thang vung ti A v B a AD a BC AB 2 ; = = = .Cnh bn SA vung gc vi y (ABCD) v SA=a. Gi E l trung im ca AD.Tnh th tch khi chp SCDE v tm tm bn knh mt cu ngoi tip khi chp . HD gii: 63aV = Gi M, N ln lt l trung im ca SE v SC ta c mt phng (ABNM) l mt phng trung trc ca SE. Vy tm O ca mt cu ngoi tip hnh chp SCDE l giao im ca mt phng (ABMN) v trc ng trn ngoi tipy CDE. GiA l ng thng qua I l trung im ca CD v song song vi SA.Gi K l trung im ca AB thKN //AM. KN vA ng phng suy ra O KN = A l im cn tm Tam gic OIK vung cn nn OI=IK=232a AD BC=+; Ta c 21141142492 2 22 2 2aOC Ra a aIC OI OC = = = + = + =(0,25 im) jOCEIMNKABS Trong v d ny ta dng mt phng trung trc ca SE tn dng iu kin tam gic SAEvung cn A V d 2) Cho hnh chp SABCD c y ABCD l hnh ch nht cnh; 2 AB aAD a = =gc gia hai mt phng (SAC) v ABCD bng 600. Gi H l trung im ca AB. Bit mt bn SAB l tam gic cn ti nh S v thuc mt phng vung gc vi y. Tnh th tch khi chp SABCD v xc nh tm bn knh mt cu ngoi tip khi chp SAHC 17 - Ta c( ) SH AB SH ABCD .K HM vung gc vi AC th gc to bi (SAC) v (ABCD) l 060 SMH =C 02 6 2sin ; tan 602 6 2 3BC a a a aHM AH HAM AH SH HMAC a= = = = = =31( )3 3SABCDaV SHdt ABCD = =QPEMN KIDOHCBAS - Gi E, Kln lt l trung im ca SA, HA . K ng thng qua K song song vi AD ct CD F th KF ( ) SAH . Dng Ex song song vi KF th Ex l trc ng trn ngoi tip tam gic SHA. Dng ng thng qua tm O ca mt y vung gc vi AC ct KF, AD ti N, Pth N l tm vng trn ngoi tip tam gic AHC. Trong mt phng cha Ex v KF k ng thng Ny vung gc vi y (ABCD) (ng thng song song vi EK) th Ny l trc ng trn ca tam gic AHC.Giao imI Ny Ex = l tm mt cu ngoi tip hnh chp SAHC.Ta c 2 2 2 2 2 2R IH IN NH KE NH = = + = + .222 2223 3 3 1 5 3 3. ; ( )2 2 4 cos 2 2 2 4 2 4 22 3 3 314 32 4 2AO a a a AHAP aKN HO AP HN KN aCAD aa aR a= = = = + = = + =| | | | = + = || ||\ . \ . Vy 3132R a = Cach2) Gi J, r ln lt l tm v bn knh ng trn ngoi tip tam gic AHC. Ta c.2 43 32. .4. . aSAC HC AHSAC HC AHrABC AHC= = = 18 K ng thngA qua J v. // SH AKhi tm I ca mt cu ngoi tip hnh chpAHC S.l giao im ca ng trung trc on SH vA trong mt phng (SHJ). Ta c.4222 2rSHJH IJ IH + = + =Suy ra bn knh mt cu l.3231a R =V d 3) Cho t din ABCD c ABC l tam gic u cnh a, 3aDA DB = = , CD vung gc vi AD.Trn cnh CD ko di ly im E sao cho 090 AEB = .Tnh gc to bi mt phng (ABC) v mt phng (ABD).Xc nh tm v tnh th tch khi cu ngoi tip khi t din ABCE. Gii:-Gi I l trung im ca AB th CI vung gc vi AB v DI vung gc vi AB. Nn gc to bi (ACD) v (ABD) l CID.Do hai tam gic ACD v BCD bng nhau nn 2 2 20 2 2 23 90 ( ) ; ;2 3 4 12a a a aBDC ADC CD ABD CD DI CI DI DA AI = = = = = =3 1cos :2 3 2DI a aCIDCI= = =- Tam gic vung ACD c 2 2 223CD CA DA a = = . Tam gic ABE vung cn, do 2 22;2 6a aAE DE AE DA ACE = = = Ac AD l ng cao v 22.3aCD DE DA ACE = = Avung ti A.Tng t ta c tam gic BCE vung ti B. Vy mt cu ngoi tip t din ABCE c CE l ng knh tm I ca mt cu l trung im ca CE. Bn knh 3331 1 2 6 4 4 6 6( )2 2 3 4 3 3 4 8 6a a a aR CD DE a V R | | | |= + = + = = = = || ||\ . \ . IBACDE 19 MT S BI TP CHN LC V HNH KHNG GIAN THNG DNG TRONG K THI TSH BIN SON GV NGUYN TRUNG KIN 0988844088 Cu 1) Khi chp SABCD c y l hnh bnh hnh, M l trung im ca SC. Mt phng (P) i qua AM, song song vi BD chia khi chp lm 2 phn. Tnh t s th tch hai phn . Cu 2) Cho hnh chp t gic u SABCD c cc cnh bng a. a)Tnh th tch khi chp. b)Tnh khong cch t tm mt y n cc mt ca hnh chp. Cu 3) Khi chp SABCD c y l hnh vung cnh a. SA(ABCD); SA=2a. Gi E, F l hnh chiu ca A trn SB v SD. I l giao im ca SC v (AEF). Tnh th tch khi chp SAEIF. Cu 4) Cho lng tr ng ABCA1B1C1 y l tam gic u. Mt phng (A1BC) to vi y 1 gc 300 v tam gic A1BC c din tch bng 8. Tnh th tch khi lng tr. Cu 5) Khi lng tr ABCA1B1C1 c y l tam gic vung cn, cnh huyn AB= 2 . Mt phng (AA1B) vung gc vi mt phng (ABC), AA1= 3 ; gc A1AB nhn, gc to bi (A1AC) v mt phng (ABC) bng 600. Tnh th tch khi lng tr. Cu 6) Khi lng tr t gic u ABCDA1B1C1D1 c khong cch gia 2 ng thng AB v A1D bng 2, di ng cho mt bn bng 5. a)H AHA1D (KeA1D). chng minh rng AK=2. b)Tnh th tch khi lng tr ABCDA1B1C1D1. Cu 7) Cho hnh t din ABCD c cnh AD vung gc vi mt phng (ABC), AC=AD=4cm; AB=3cm; BC=5cm. Tnh khong cch t im A ti mt phng (BCD). Cu 8) Cho hnh chp tam gic u SABC nh S, di cnh y bng a. Gi M, N ln lt l trung im ca cc cnh SB v SC. Tnh theo a din tch tam gic AMN, bit rng mt phng (AMN) vung gc vi mt phng (SBC). Cu 9) Cho hnh chp SABC c SA=3a v SA vung gc vi mt phng (ABC). Tam gic ABC c AB=BC=2a, gc ABC=1200. Tnh khong cch t nh A n mt phng (SBC). Cu 10) Cho hnh chp SABCD c y ABCD l hnh vung cnh a, tam gic SAB u v nm trong mt phng vung gc vi y. Tnh gc gia 2 mt phng (SAB) v (SCD). Cu 11) Cho hnh chp tam gic SABC c y ABC l tam gic u cnh a, SA=2a v SA vung gc vi mt phng (ABC). Gi M v N ln lt l hnh chiu vung gc ca A trn cc ng thng SB v SC a)Tnh khong cch t A n mt phng (SBC)b)Tnh th tch ca khi chp ABCMN. Cu 12) Hnh chp tam gic SABC c cc cnh bn SA=SB=SC=a, gc ASB=1200, gc BSC=600, gc ASC=900. Chng minh rng tam gic ABC vung v tnh th tch hnh chp SABC theo a. Cu 13) Cho hnh chp t gic u SABCD. Khong cch t A n mt phng (SBC) bng 2a. Gc gia cc mt bn v mt y l . a)Tnh th tch khi chp theo a vb)Xc nh th tch khi chp nh nht. Cu 14) Cho hnh chp SABCD c y ABCD l hnh ch nht vi AB=a, AD=2 a , SA=a v SA vung gc vi mt phng (ABCD). Gi M v N ln lt l trung im ca AD v SC, I l giao im ca BM v AC. a)Chng minh rng mt phng (SAC) vung gc vi mt phng (SMB). b)Tnh th tch ca khi t din ANIB. 20 Cu 15) Cho lng tr ng ABCABC c y ABC l tam gic vung ti B, AB=a, AA=2a, AC=3a. Gi M l trung im ca on thng AC, I l giao im ca AM v AC a)Tnh theo a th tch khi t din IABC b)Tnh khong cch t im A n mt phng (IBC)Cu 16) Cho hnh chp SABCD c y ABCD l hnh thang vung ti A v D, AB=AD=2a, CD=a, gc gia 2 mt phng (SBC) v (ABCD) bng 600. Gi I l trung im ca cnh AD. Bit 2 mt phng (SBI) v (SCI) cng vung gc vi mt phng (ABCD), tnh th tch khi chp SABCD theo a. Cu 17) Cho hnh lng tr tam gic ABCABC c BB=a, gc to bi BB v mt phng (ABC) l 600, tam gic ABC vung ti C v gc BAC=600. Hnh chiu vung gc ca im B ln mt phng (ABC) trng vi trng tm ca tam gic ABC. Tnh th tch khi t din AABC theo a. Cu 18) Trong khng gian cho hnh chp tam gic u SABC c7 SC a = . Gc to bi (ABC) v (SAB) =600. Tnh th tch khi chp SABC theo a. Cu 19) Trong khng gian cho hnh chp SABCD vi ABCD l hnh thoi cnh a, gc ABC=600, SO vung gc vi y ( O l tm mt y), 32aSO = . M l trung im ca AD. (P) l mt phng qua BM v song song vi SA, ct SC ti K. Tnh th tch khi chp KABCD. Cu 20) Cho hnh chp SABC c y ABC l tam gic u cnh a, cnh bn SA vung gc vi y (ABC). Tnh khong cch t A n mt phng (SBC) theo a bit 6.2aSA =Cu 21) Cho hnh chp SABCD c y l hnh ch nht,2, 2 . AD a CD a = = Cnh SA vung gc vi y v3 2 . SA a = Gi K l trung im AB. a)Chng minh rng (SAC) vung gc vi (SDK) b)Tnh th tch khi chp CSDK theo a; tnh khong cch t K n (SDC). Cu 22) Cho hnh chp SABCD c y ABCD l hnh vung cnh a. Mt phng (SAC) vung gc vi y, gc ASC=900, SA to vi y 1 gc 600. Tnh th tch khi chp. Cu 23) Cho lng tr ABCABC c y ABC l tam gic u cnh a, hnh chiu vung gc ca A ln mt phng (ABC) trng vi tm O ca tam gic ABC. Mt mt phng (P) cha BC v vung gc vi AA ct lng tr theo 1 thit din c din tch 238a. Tnh th tch khi lng tr Cu 24) Cho hnh chp SABC c AB=AC=a;; 32aBC SA a = =; gc SAB bng gc SAC v bng 300. Tnh th tch ca khi chp theo a. Cu 25) Cho hnh chp t gic u SABCD cnh y bng a. Gi G l trng tm tam gic SAC v khong cch t G n mt bn (SCD) bng 3.6a a)Tnh khong cch t tm ca mt y n mt bn (SCD) b)Tnh th tch ca khi chopSABCD. Cu 26) Cho hnh chp SABC c ng cao AB=BC=a; AD=2a. y l tam gic vung cn ti B. Gi B l trung im ca SB, C l chn ng cao h t A xung SC.Tnh th tch khi chp SABC. 21 Cu 27) Cho lng tr ng ABCABC c y ABC l tam gic vung, AB=BC=a, cnh bn AA= 2 a . Gi M l trung im ca cnh BC a)Tnh theo a th tch ca khi lng tr ABCABC b)Tnh khong cch gia 2 ng thng AM v BC. Cu 28) Cho hnh chp SABCD c y ABCD l hnh vung cnh 2a; SA=a; SB= 3 av mt phng (SAB) vung gc vi mt phng y. M v N ln lt l trung im ca cnh AB v BC. Tnh th tch khi chp SBMDN v gc gia (SM;ND). Cu 29) Cho hnh chp SABCD c y ABCD l hnh thang, gc BAD bng gc ABC v bng 900; AB=BC=a; AD=2a. SA vung gc vi y v SA=2a. Gi M, N ln lt l trung im ca SA; SD. Tnh th tch khi chp SABCD v khi chp SBCMN. Cu 30) Cho lng tr ABCABC c di cnh bn bng 2a, y ABC l tam gic vung ti A, AB=a; AC= 3. av hnh chiu vung gc ca A trn (ABC) l trung im ca cnh BC. Tnh theo a th tch khi chp AABC v cosin ca gc gia 2 ng thng AA v BC. Cu 31) Cho hnh chp SABCD c y ABCD l hnh vung cnh a, mt bn SAD l tam gic u v nm trong mt phng vung gc vi y. Gi M, N, P ln lt l trung im ca cc cnh SB, BC, CD. Chng minh AM vung gc vi BP v tnh th tch khi t din CMNP. Cu 32) Cho lng tr ng ABCA1B1C1 c AB=a; AC=2a; AA1= 2 5 av gc BAC=1200. Gi M l trung im ca cnh CC1. Chng minh rng MBMA1 v tnh khong cch d t im A n mt phng (A1MB) Cu 33) Cho hnh chp SABC c gc gia 2 mt phng (SBC) v (ABC) bng 600 . Cc tam gic ABC v SBC l cc tam gic u cnh a. Tnh theo a khong cch t nh B n mt phng (SAC). Cu 34) Cho hnh chp SABCD c y ABCD l hnh vung tm O, SA vung gc vi y. Cho AB=a; SA= 2 a . Gi H v K ln lt l hnh chiu ca A ln SB; SC. Chng minh SC(AHK) v tnh th tch khi chp OAHK. Cu 35) Trong mt phng (P) cho na ng trn ng knh AB=2R v im C thuc na vng (SAB;SBC)=600. Gi H, K ln lt l hnh chiu ca A trn SB, SC. Chng minh tam gic AHK vung v tnh VSABC Cu 36) Lng tr ng ABCA1B1C1 c y l tam gic vung AB=AC=a; AA1= 2 a . Gi M, N ln lt l trung im ca AA1 v BC1. Chng minh rng MN l on vung gc chung ca AA1 v BC1. Tnh th tch khi chp MA1BC1 Cu 37) Cho lng tr ng ABCA1B1C1 c tt c cc cnh u bng a. M l trung im ca on AA1. Chng minh BMB1C v tnh ( )1; BMB CdCu 38) Cho hnh chp t gic u SABCD c y l hnh vung cnh a. E l im i xng ca D qua trung im SA, M l trung im ca AE, N l trung im ca BC. Chng minh MN vung gc vi BD v tnh khong cch gia MN v AC theo a. Cu 39) Cho hnh chp SABCD c y l hnh thang, gc ABC= gc BAD= 900; AD=2a; BA=BC=a. Cnh bn SA vung gc vi y v SA= 2 a . Gi H l hnh chiu vung gc ca A trn SB. a)Chng minh rng tam gic SCD vung b)Tnh khong cch t H n mt phng (SCD) 22 Cu 40) Cho hnh chp SABC m mi mt bn l 1 tam gic vung. SA=SB=BS=a. Gi M, N, E ln lt l trung im ca cc cnh AB, AC, BC. D l im i xng ca S qua E, I l giao im ca AD v (SMN) a)Chng minh rng AD vung gc vi SI b)Tnh theo a th tch khi t din MBSICu 41) Cho hnh hp ng ABCDABCD c cc cnh AB=AD=a; AA=32av gc BAD=600. Gi M v N ln lt l trung im ca AD v AB. Chng minh AC vung gc vi mt phng (BDMN) v tnh th tch khi chp ABDMN. Cu 42) Hnh chp SABCD c y ABCD l hnh ch nht vi AB=a, AD=2a, cnh SA vung gc vi y, cnh SB to vi mt phng y gc 600. Trn cnh SA ly M sao cho 33aAM = , mt phng (BCM) ct SD ti N. Tnh th tch khi chp SBCNM. Cu 43) Cho hnh chp SABCD c y ABCD l hnh thoi cnh a. Gc BAD=600. SA vung gc vi mt phng (ABCD), SA=a. Gi C l trung im ca SC, mt phng (P) i qua AC v song song vi BD, ct cc cnh SB, SD ca hnh chp ln lt ti B, D. Tnh th tch ca khi chp SABCD. Cu 44) Cho lng tr ABCABC c AABC l hnh chp tam gic u, cnh y AB=a, cnh bn AA=b. Gi l gc gia 2 mt phng (ABC) v (ABC). Tnhtan v th tch khi chp ABBCC. Cu 45) Cho hnh chp t gic u SABCD c cnh y =a. Gi SH l ng cao ca hnh chp. Khong cch t trung im I ca SH n mt phng (SBC) bng b. Tnh th tch khi chp SABCD. Cu 46) Cho hnh lp phng ABCDABCD c cnh =a v im K thuc cnh CC sao cho:23aCK = . Mt phng i qua A, K v song song vi BD chia khi lp phng thnh 2 khi a din. Tnh th tch ca 2 khi a din . Cu 47) Cho 1 hnh tr trn xoay v hnh vung ABCD cnh a c 2 nh lin tip A; B nm trn ng trn y th nht, 2 nh cn li nm trn ng trn y th 2 ca hnh tr. Mt phng (ABCD)to vi y hnh tr gc 450. Tnh din tch xung quanh v th tch ca hnh tr. Cu 48) Cho hnh nn nh S, y l ng trn tm O, SA v SB l 2 ng sinh. Bit SO=3a, khong cch t O n mt phng (SAB) bng a, din tch tam gic SAB=18a2. Tnh th tch v din tch xung quanh. Cu 49) Cho hnh tr c 2 y l 2 hnh trn tm O v O. Bn knh y bng chiu cao v bng a. Trn ng trn y tm O ly im A, trn ng trn y tm O lyim B sao cho AB=2a. a)Tnh din tch ton phn ca hnh tr v th tch ca khi tr b)Tnh th tch t din OOAB. Cu 50) Cho hnh chp ct tam gic u ngoi tip 1 hnh cu bn knh r cho trc. Tnh th tch khi chp ct bit rng cnh y ln gp i cnh nh. (Hnh chp ngoi tip hnh cu nu hnh cu tip xc vi tt c cc mt ca hnh chp). Cu 51) Cho hnh chp tam gic u SABC c di cnh bn bng a. Cc mt bn hp vi mt phng y mt gc . Tnh th tch khi cu ni tip hnh chp. 23 Cu 52) Cho hnh chp SABCD. Hai mt bn (SAB) v (SAD) cng vung gc vi mt y. y ABCD l t gic ni tip trong ng trn tm O, bn knh R. Xc nh tm v tnh th tch khi cu ngoi tip hnh chp SABCD bit SA=h. Cu 53) Hnh cu ng knh AB=2R. Ly H trn AB sao cho AH=x ( 0