Môn Toán - Học và ôn luyện theo cấu trúc đề thi: Phần 2

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Nối tiếp nội dung phần 1 tài liệu Học và ôn luyện theo cấu trúc đề thi môn Toán, phần 2 giới thiệu các nội dung phần Hình học bao gồm: Đường thẳng và mặt phẳng trong không gian, thể tích các khối, đường thẳng trong mặt phẳng, đường tròn và elip,... Mời các bạn cùng tham khảo nội dung chi tiết.

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Nội dung Text: Môn Toán - Học và ôn luyện theo cấu trúc đề thi: Phần 2

?H4N II HIJVHHOC 1 9 2 k'': TS, Vu The Huu - Nguygn Vinh Can
mm i I. tiJHG THIG VA HJT PHIB TRDNE KHONG emii B A I T O A N 1 . QUAN H E SONG SONG I. KIEN THLfC QUAN TRQNG 1. Dvfcfng t h ^ n g s o n g s o n g vofi diiofng t h a n g a) Dinh nghia : - H a i dudng t h a n g goi l a cheo nhau neu chung k h o n g cung n a m t r o n g mot m a t phang. - H a i di/ofng t h a n g goi l a song song neu chung dong phSng va k h o n g c6 diem chung. b) Cdc tinh chat : Tinh chat 1 Trong k h o n g gian, qua mot diem ngoai mot diicfng t h a n g c6 m o t va chi mot diiorng t h a n g song song vdi dudng thSng do. Tinh chat 2 Hai diTofng t h a n g p h a n b i e t cung song song vdfi m o t di/dng t h a n g thuf ba t h i song song v d i nhau. Dinh li (ve giao tuyen ciia ba m a t phang) Neu ba m a t phang cSt nhau theo ba giao tuyen p h a n b i e t t h i ba giao tuyen ay hoac dong quy hoac doi mot song song. He qua Neu hai m a t phang p h a n b i e t I a n lirgt d i qua h a i du'&ng t h a n g song song t h i giao tuyen ciia chiing (neu c6) song song v d i h a i diTdng t h a n g do (hoac t r i j n g vdi mot t r o n g hai diTdng thSng doj. 2. Dvfcfng t h ^ n g s o n g s o n g vdfi m a t p h ^ n g a) Dinh nghia : M o t diidng thSng va mot m a t phSng goi l a song song v d i nhau neu chiing k h o n g c6 d i e m chung. b) Dieu kien de mot dudng thang song song vdi mot mat phang : Dinh li 1 Neu diidng thang a song song vdi mot dudng thang b nao do n a m t r e n mot mat phang (P) khong chufa a t h i a song song vdi m a t phang (P;. Dinh li 2 Neu mot dddng t h a n g song song vdi mot m a t phang t h i no song song vdi mot di/dng t h a n g nao do n a m t r o n g m a t phang. HQC va on luyen theo CTDT mon Toan THPT 193
He qua 1 N e u diTcfng t h a n g a s o n g s o n g v d i m o t m a t p h d n g ( P ) t h i m o i m a t p h a n g ( Q ) chiifa a m a c a t ( P ) t h i cSt ( P ) t h e o g i a o t u y e n s o n g s o n g v d i a. He qua 2 N e u h a i m a t p h S n g p h a n b i e t c u n g s o n g s o n g v d i m o t difofng t h a n g t h i g i a o t u y e n c i i a c h i i n g ( n e u c6) c u n g s o n g s o n g v d i diTdng t h a n g do. 3. H a i m a t p h a n g song song a) Dinh nghia : H a i m a t p h a n g g o i l a s o n g s o n g k h i c h u n g k h o n g c6 d i e m c h u n g . b ) Dieu kien de hai mat phdng song song : Dinh li : N e u m o t m a t p h a n g ( P ) chijfa h a i d i i d n g t h a n g a v a b c a t n h a u va cung song song v d i m a t p h a n g (Q) t h i (P) song song v d i (Q). c) Tinh chat : Tinh chat 1 Q u a m o t d i e m n g o a i m o t m a t p h a n g c6 m o t v a c h i m o t m a t p h a n g song song v d i m a t p h a n g do. He qua 1 N e u d i i d n g t h a n g a s o n g s o n g v d i m a t p h a n g ( Q ) t h i q u a a cd m o t v a chi m o t m a t p h a n g (P) song song v d i m a t p h a n g (Q). He qua 2 H a i m a t p h d n g p h a n b i e t c i i n g s o n g s o n g v d i m o t m a t p h a n g t h d ba t h i song song v d i nhau. Tinh chat 2 N e u h a i m a t p h a n g ( P ) v a ( Q ) s o n g s o n g t h i m o i m a t p h a n g ( R ) d a cat (P) t h i p h a i c a t ( Q ) v a cac giao t u y e n ciia c h u n g s o n g song. d) Dinh li Ta-let trong khong gian : Dinh li thuan Ba m a t p h a n g d o i m o t song song chan r a t r e n h a i cat tuyen b a t k i cac d o a n t h a n g ti/cfng t i n g t i l e . Dinh li ddo G i a sijf t r e n h a i d i i d n g t h a n g a v a a' I a n l i i d t l a y h a i bo d i e m ( A , B , C) . ^, , AB BC CA v a ( A , B , C ) sao c h o = = A'B' B'C CA' K h i do b a d u d n g t h a n g A A ' , B B ' , C C c t i n g s o n g s o n g v d i m o t m a t phang. e) Hinh Idng tru va hinh hop : Binh nghla hinh Idng tru H i n h l a n g t r u l a h i n h c6 h a i d a y l a h a i d a g i a c n a m t r o n g h a i m a t 1 9 4 j - ' l TS. Vu Th6' Hau - NguySn VTnh CJn
phSng song song. T a t ca cac canh k h o n g thuoc h a i day deu song song vdri nhau. H i n h l a n g t r u c6 : - Cac canh ben b&ng nhau. - Cac m a t ben l a cac h i n h h i n h h a n h . - H a i day l a h a i da giac bang nhau. Dinh nghia hinh hop H i n h l a n g t r u c6 day l a h i n h b i n h h a n h diTcfc goi l a h i n h hop. H i n h hop c6 sau m a t deu l a h i n h b i n h h a n h , cac m a t d o i dien t h i song song vdri nhau. Cac diJcfng cheo ciia h i n h hop cdt nhau t a i t r u n g d i e m ciia m o i di/dfng. II. CAC DANG BAI TAP VA PHl/dNG PHAP GIAt 1. C a c d a n g b a i t^p a) Cac bai t a p t r o n g muc nay, chii yeu l a chijfng m i n h h a i dudng t h a n g cheo nhau, h a i di/dng t h a n g song song, diicJng t h a n g va m a t phSng song song, h a i m a t phang song song. De g i a i cac b a i t a p n a y t a thifdng suf dung cac t i n h chat ciia quan he song song va t r o n g n h i e u t r i i d n g hgfp t a suf dung phiCang phdp chiing minh phdn chiJcng. b) Ngoai r a cac t i n h chat ciia h i n h l a n g t r u , h i n h hop cung giup t a g i a i cac b a i t a p ve quan he song song (cac canh ciia h i n h l a n g t r u , h i n h hop, cac m a t ben ciia h i n h hop v.v...). ^ Bait^ip 1. Cho bon d i e m k h o n g dong phang A , B, C, D. Chufng m i n h r a n g cac du'cfng t h a n g sau l a cac cap dudng t h a n g cheo nhau : A B va C D A D va BC AC va B D . Chi dan : Suf dung phi/cfng phap p h a n chufng. 2. Cho h i n h chop tuf giac S.ABCD. Goi M , N , P theo thuf t\i l a t r u n g diem ciia cac canh SA, SB, SC. a) Chufng m i n h mp ( M N P ) // m p (ABCD). b) Chufng m i n h giao d i e m ciia m p ( M N P ) v d i SD l a t r u n g d i e m Q cua doan t h a n g SD. c) Goi O l a giao diem ciia A C va B D , O' l a giao d i e m cua MQ va N P . Chufng m i n h ba d i e m S, O, O' thSng hang. Chi dan : Doc gia tir g i a i . 3. Cho h i n h chop tuf giac S.ABCD; day A B C D l a h i n h vuong. G o i M , N theo thuf tir la t r u n g d i e m ciia h a i canh SB, SD. Hpc va on luyen theo CTDT mon Toan THPT 195
^^^^^^^^^ a) Chirng m i n h M N // (ABCD). b) Neu each xac d i n h giao d i e m P cua m p ( A M N ) vdfi canh SC. e) Xac d i n h giao t u y e n eiia m p (ABCD) \6i mp ( A M N ) . C h i dSn : Doc gia tii g i a i . 4. Cho h i n h hop ABCD.A'B'C'D'; O la giao diem eiia AC va B D ; O' la giao d i e m eiia A ' C va B D ' . a) Chufng m i n h m p (AB'D') // mp(C'BD). b) Chijfng m i n h A O / / C O ' . C h i dan : Doc gia tiT g i a i . 5 . Cho h i n h hop ABCD.A'B'C'D': goi O, O' theo thiif tiT la giao diem cua AC v d i B D va ciia A ' C v d i B'D' a) Chufng m i n h mp(AB'D') // mp(C'BD) b) Chijfng m i n h A'O // CO' C h i dan : Doc gia t\i g i a i . BAITOAN 2. QUAN HE VUONG GOC NH0NG BAI TOAN VE KHOANG CACH 1. K h o a n g e a c h tijf m p t d i e m d e n mpt m a t p h a n g Cho m a t p h a n g (P) va d i e m O; H la h i n h chieu vuong goc ciia O t r e n (P). K h o a n g each tii d i e m O den m a t p h a n g (P) l a dp d a i doan t h a n g O H . /p\ 2. d(d, (P)) = d ( 0 e d, (P)). 196 tJ.; TS, VQ The Huu - Nguygn VTnh Can
3. K h o a n g e a c h givia h a i dxXdng t h a n g c h e o n h a u Khoang each giCifa h a i difcfng thSng cheo nhau l a do d a i doan vuong goe ehung eiia h a i di/ofng t h ^ n g . Tinh chat : K h o a n g each giijfa hai diidng t h a n g cheo nhau a; b t h i bSng khoang each tCr m o t d i e m A e a den mot m a t p h a n g (P) qua b va song song vdri difcfng t h a n g a. Chuy : a) Ngoai ba t r U c f n g hcrp t r e n day, t a con c6 k h o a n g each giiJa h a i diTcrng t h a n g song song, k h o a n g each giiJa h a i m a t phang song song. Cac kien thufc n a y dofn g i a n va eung i t gap t r o n g cac b a i t o a n n e n t a k h o n g nhae l a i of day. b) De dung mat phang (P) qua b va song song vdfi a, ta l a m n h i i sau : Lay mot diem M e b. Q u a M diing difofng t h a n g a' // a. M a t phang xae dinh bofi hai dUcfng t h i n g b va a ' chinh la mat phang (P). II. CAC BAI TAP VA PHl/dNG PHAP GlAi Lj C a c dang bai tap a) Bai tap ve tim khoang each tit mot diem den mot mat phang De t i m khoang each ttr diem O den mat phang (P), triTdrc het ta t i m h i n h chieu H eiia O t r e n (P), sau do t i n h do dai doan thang O H . Bai toan nay lien quan den v a n de difofng t h a n g vuong goe vdfi m a t phang va k h i t i n h O H , thiTcfng t a phai sijf dung den cac k i e n thiife ve he thufc Itfofng trong t a m giac, n h a t la he thijfe lUgfng trong t a m giae vuong. b) Bai tap ve khoang each til mot diCang thang den mot mat phang song song Ta t i n h khoang each gi-iifa ducfng t h a n g A den m o t m a t phang (P) song song vdi A bang each t i m k h o a n g each tiT m o t diem M G A d e n m a t phang (P). Dieu quan t r o n g t r o n g viee g i a i b a i toan nay l a of cho chon diem M e A m o t each t h i c h hgp, giup t a suf dung het cac gia t h i e t ciia bai toan t i n h duoe d ( M , (P)). e) Bai toan ve khoang each giita hai duang thdng cheo nhau De t i m k h o a n g each giOfa h a i diTcfng t h a n g cheo nhau a; b , t a thiidfng l a m nhif sau : Hoc va on luyen theo CTDT mon Toan THPT , 1 9 7
- T i m m o t m a t p h a n g (P) chufa a va song song v d i b (hoac chijfa b va song song v d i a). M a t p h a n g (P) thiicfng diTgrc xac d i n h b o i a va mot dirofng t h a n g cAt a va song song v 6 i b. - T i m mot diem M thich hop thupc a va t i m khoang each tir M den (P). Diem M can chon la diem c6 Hen he vcfi cac gia thiet ciia bai toan de giup t a van dung dirge cac c o n g thijfc, he thijfc liTpfng trong cac tam giac. 2. C a c b a i tap 6. Cho h i n h chop t a m giac S.ABC, h a i m a t phang (SAB) va (SAC) vuong goc vdfi m a t phang day (ABC). a) N e u each xac d i n h h i n h chieu H cua dinh A t r e n mat phang (SBC). b) B i e t A B = 13cm, B C = 14em, CA = 15cm va SA = 16em. T i n h khoang each tiT d i n h A den m a t phang (SBC). CHI DAN (SAB) 1 (ABC) a) Ta CO : (SAC) 1 (ABC) ^ SA ± (ABC) (SAB) n (SAC) = SA T r o n g t a m giac A B C , k e diiomg cao A M A M ± BC. SA 1 (ABC) Tir S M 1 BC AM ±BC ( D i n h li 3 diTotng v u o n g g o c ) . T r o n g t a m g i a c S A M ke A H 1 S M (1) BC 1 S M Tir • BC 1 (SAM) BC ± A M BC ± ( S A M ) ' • A H ± BC (2) A H cz (SAM)J AHXBC TCr (1) va (2) t a diroc A H 1 (SBC) AH I S M => H l a h i n h chieu cua A t r e n m a t phang (SBC), b) Trirdtc h e t , t a t i n h d i e n t i c h t a m giac A B C theo cong thuTc H e r o n , t a c6 : 2p = 42 => p = 2 1 suy r a p - a = 8; p - b = 7; p - c = 6 SABC = V21.8.7.6 = 84 2S. ^ABC _ 19 Tir day t a eo A M = BC 198 ei TS. Vu The Hi;u - Nguyen VTnh CJn
Tarn giac S A M vuong t a i A v a A H l a di/dng cao thuoc canh huyen. Theo he thiic li/gfng t r o n g tarn giac vuong t a c6 : 1 1 1 1 1 1 1 1 + TT = + AH^ SA^ M A ^ A H ^ 16^ 12^ 256 144 => A H ^ = ^ A H = 9,6 (cm). 400 Chiiy : 1. Ta CO the t i n h A H theo each khac. Trirdrc het, t a t i n h S M tiT t a m giac vuong S A M . S M ' = S A ' + A M ' = > S M = 20 T i n h A H t\X hai t a m giac vuong dong dang : .c^TTA A T A H A M .^^ AM.SA A S H A CO A S A M = > = =^AH = SA S M S M AH =1^=9,6 20 T i n h A H tiT dien t i c h t a m giac S A M . 2SsAM = A H . S M OA A M 2SsAM = S A . A M => A H . S M = S A . A M =^ A H = = 9,6. SM 2. Ta cung c6 t h e dirng d i e m H theo each l i luan sau : Do SA 1 (ABC) =^ SA 1 BC. Dirng qua SA m o t m a t p h a n g (P) vuong goc vdri BC; mSt p h a n g nay cat BC t a i M . T r o n g m a t p h a n g (P), t a ke AH 1 SM t h i v i (SAM) ± (SBC) va (SAM) n (SBC) = SM ma A H 1 A M nen A H 1 (SBC) hay H l a h i n h chieu ciia A t r e n m a t p h a n g (SBC). 7. Cho h i n h chop S.ABC; h a i m a t p h a n g (SAC) va (SAB) vuong goc v6i m a t day (ABC) va SA = a V 2 . T i n h k h o a n g each tCr d i n h A den m a t phang (SBC) t r o n g cac t r u d n g hop : a) Day ABC l a t a m giac deu canh a. b) Day ABC l a t a m giac can d i n h A, goc A - 1 2 0 ° va A B = A C = a. c) Day ABC l a t a m giac vuong t a i B; A C = 5a, BC = 4 a . CHI D A N a) T a CO SA 1 (ABC). Goi M l a t r u n g d i e m ciia canh BC. T r o n g t a m giac S A M ke A H ± S M t h i A H 1 (SBC) T r o n g t a m giac S A M t h i AH' AS' A M ' Hgc va on luyen theo CTDT mon Toan THPT 199
AH'' (aV2)' • + iV66 AH = 11 b) Goi M la t r u n g d i e m ciia BC. T r o n g t a m giac S A M ke A H ± S M =^ A H 1 (SBC) T r o n g t a m giac S A M t h i 1 1 1 AH^ AS' AM' vdi AS - aV2 va A M = AB.coseO" = - 2 l^f2 ta t i n h duoc A H = c) Ta CO : SA ± (ABC) BC ± (SAB) T r o n g t a m giac SAB, ke A H 1 SB ^ A H ± (SBC) va t a cung c6 1 1 1 AH' AS' + AB' De t h a y A B ' = 9 a ' 6aVl3 AH = 13 8. Cho h i n h chop S.ABCD, chieu cao bSng a\/3, hai m a t phang (SAC) va (SBD) vuong goc v d i m a t p h a n g day (ABCD); day A B C D la h i n h thoi, canh a, goc n h o n A = 60". T i n h khoang each giOfa h a i dUcfng t h a n g A D va SB. CHI DAN Ta CO : Goi O la giao diem ciia hai diTdng cheo AC va BC ciia h i n h thoi A B C D : 200 ,.'. TS. Vu The Hi/u - Nguy§n VTnh Can
(SAC) 1 (ABCD) (SBD) ± (ABCD) (SAC) n (SBD) = SO o SO 1 (ABCD) =:> SO = aVs BC // A D => m a t p h a n g (SBC) la m a t p h a n g chijfa (SB) va song song vdfi A D . ^ ^ Qua SO ta difng mot mat phang vuong goc vofi canh BC, mat phang nay cat AD d P va cat BC d Q. Ttr P ke P K 1 SQ t h i P K 1 (SBC) do d6d(AD, SB) = d(AD, (SBC)) = d(P, (SBC)) = P K TCr O ke O H 1 SQ =o O H 1 (SBC) => O H // P K IN/3 De t h a y PQ = AB.sin60° = OQ- Trong t a m giac vuong SOQ t h i + • OH^ OQ*-^ OS' 3a^ OH^ = OH = OH' ^aV3^' (aVS)^ 17 17 Trong t a m giac P Q K t h i O H l a diTcfng t r u n g b i n h nen aVsi PK = 2 0 H = 17 Chii y : Co t h e t i n h O H t i ^ h a i t a m giac vuong dong d a n g SOQ va SHO. OH SO SO.OQ ASOQ ASHO OH = OQ SQ SQ vdi SQ-" = SO' + O Q l 9. Cho tuf dien A B C D . H a i m a t b e n A B C va DBC n a m t r o n g h a i m a t phang hop \6i nhau m o t goc 60°. M a t ben A B C l a m o t t a m giac deu con m a t ben DBC l a m o t t a m giac vuong can, d i n h D. B i e t D B = a. Goi M l a t r u n g d i e m cua canh BC. 1. T i n h canh A D . 2. T i n h khoang each tCr d i n h A den m a t phang (DBC) va k h o a n g each tCf d i n h D den m a t phang (ABC). 3. T i n h khoang each giufa difdng t h a n g A D va diTcfng t h a n g BC, k h o a n g each giuTa h a i diidng t h a n g AC va D M . H Q C va on luyen theo CTDT mon Toan THPT.'.'; 201
CHI DAN 1. M l a t r u n g d i e m cua BC t h i A M 1 BC => A M D - 60° D M ± BC T a m giac D B C can vuong t a i D #D va D B = a, cho t a BC = a^f2 ^ D M = ^ 2 T a m giac A B C deu, canh aV2 => A M = A p dung d i n h l i cosin vao t a m giac A M D : AD^ = A M ' + D M ' 2AM.DM.C0SAMD AD = -^8-2^. 2 2. BC 1 ( A M D ) (DBC) ± (AMD). Trong t a m giac A M D , ke A H 1 M D ==> A H 1 (DBC) hay d(A, (DBC)) = A H T r o n g t a m giac vuong A H M , A M = ; A M H = 60° 3a V2 A H = AM.sin60° => A H = T a cung c6 BC 1 ( A M D ) ^ (ABC) 1 ( A M D ) . T r o n g t a m giac A M D ke DK 1 A M ^ D K 1 (ABC) D K = d(D, (ABC)) T r o n g t a m giac vuong D K M , D M = 1V2 va A M D = 60° D K = DM.sin60° ^ DK = 3. Goi J la giao diem cua A H va D K t h i M J ± A D t a i I hay M I 1 A D . Ta l a i c6 BC ± ( A M D ) ma M I c: (AMD) M I ± BC => M I la doan vuong goc chung ciia A D va BC. IM AD AM.AD A A I M c/5 A A K D IM = AM KD KD => I M = a V 8 - 2 V 3 . T r o n g m a t p h a n g (DBC), tir C t a ke Cx // D M t h i m a t phang (A, Cx) l a m a t p h a n g chufa A C va song song vdi D M , do do t a c6 d(AC, D M ) = d ( D M , (A, Cx)) Ta lay m o t d i e m t u y y t r e n D M , diem H chang h a n t h i v i H G D M nen d(AC, D M ) = d ( D M , (A, Cx)) = d ( H , (A, Cx)) 202 IS Vu The Hiju - Nguyin VTnh Can
Tii H t a ke H K 1 Cx, do A H 1 (DBC) => theo d i n h h' 3 difofng vuong goc t a suy r a A K 1 Cx va Cx 1 ( A H K ) , cho ta (AHK) 1 (A, Cx). Til day, suy ra r k n g neu ta ke H P 1 A K t h i H P ± (A, Cx) hay H P la k h o a n g each ti^ H de'n m a t phang (A, Cx). Tong hop cac t h o n g t i n t r e n , ta c6 : d ( D M , AC) = d ( D M , (A, Cx)) = d ( H , (A, Cx)) = H P 1 1 1 T r o n g t a m giac vuong A H K t h i HP2 H A ' HK^ a-j2 3a V2 D i thay H K = M C = v-a H A = 3a V5 Suy r a H P = 10 Chii y : V i A H i. (DBC) =^ A H 1 Cx, do do t a c6 the xac d i n h m a t p h a n g ( A H K ) bSng each diing qua A H m o t m a t phSng vuong goc v d i Cx, m a t phang n a y cat Cx t a i K va t a c6 A K _L Cx, H K 1. Cx. 10. Cho h i n h l a p phuofng ABCD.A'B'C'D', canh a. a) T i m khoang each tii d i n h C d e n m a t phang (C'BD). b) T i m khoang each giiifa h a i difcirng t h a n g A D ' va CB'. c) * T i m khoang each giiifa h a i dLforng t h a n g A B ' va B C . CHI DAN a) Ti^ CB = C C = CD va A ' C = A'B' = A D t a suy r a CA' 1 (C'BD). Goi H la giao d i e m ciia CA' vdti m p (C'BD) t h i H l a h i n h chieu eiia C t r e n mp (C'BD) va C H l a k h o a n g each t i f C den mp (C'BD). H i n h chop C.C'BD l a h i n h chop t a m giac deu n e n H l a t a m ciia t a m giac deu C'BD. T a m giac deu C'BD c6 canh la aV2 nen chieu cao a^.Vs aVe CO = Ta l a i c6 C'H = - C O n e n C H = 2 2 3 T a m giac C H C vuong t a i H cho t a V3 CH^ = CC'2 - C'H^ ^ CH^ = a' - >CH = a Hoc va on luy?n theo CTDT mon Toan THPT L'i 203
Chii y : Co t h e t i n h C H theo each khac. 1 a^ a'^ Ta CO : Vc.c BD = - • y . a = — (aV2)^V3 a^Vs T a m giac C'BD deu, c6 canh aV2 n e n SCBU = = 4 2 M a t khac Vc.cBD = ^ . C H . S C B D C H = ^iXtoe. = _ 6 ^ £ ^ 3 bcBD a v3 3 2 b) D a p so': a. e) T a CO A B ' // C D n e n m a t phSng (C'BD) la m a t phang chufa B C va song song v d i A B ' . Goi I la giao d i e m cua A B ' va BA' t h i I thuoe A B ' va k h o a n g each giOfa h a i diicfng thSng A B ' va B C t h i b^ng khoang each giCfa difcfng t h a n g A B ' vdri m a t phSng ( C B D ) va cung bSng khoang each i\i d i e m I den m p (C'BD) : d(AB', B C ) = d(AB', (C'BD)) = d ( I , (C'BD)) Ta xac d i n h h i n h chieu J ciia d i e m I t r e n m p ( C B D ) . V i CA' i . ( C B D ) m a (A'BC) 3 CA' n e n t a eo ( A B C ) L (C'BD) Ta l a i CO (A'BC) n (C'BD) = B H V i I e A ' B m a A ' B c (A'BC) n e n h i n h chieu J ciia I t r e n m p (C'BD) p h a i n a m t r e n diicfng t h a n g B H . A'C L ( C B D ) Ta cung c6 I J // A'C. IJKC'BD) J T r o n g t a m giac A ' B H t h i I J l a diicrng t r u n g b i n h n e n I J = — A H 2 V i A'C = aV3; C H = ^ nen A ' H = 3 3 1 2aV3 aV3 aVs => I J = => I J = ; d(AB , BC ) = . 2 3 3 3 11. Cho tuf d i e n A B C D c6 h a i m a t ben A B C va D B C n a m t r o n g h a i m a t phang tao vdri nhau m o t goc 60°. M a t ben ABC la t a m giac deu, con mat ben DBC l a t a m giac vuong can t a i dinh D va D B = a. a) T i n h k h o a n g each tiT d i n h A den m a t phSng (DBC) va khoang each tir d i n h D den m a t phSng (ABC). b) T i n h k h o a n g each giuTa h a i diTcfng t h a n g A D va BC, khoang each giCfa h a i diTcfng t h a n g A C va D M ; M l a t r u n g d i e m ciia canh BC. 204 . . ' ; TS, Vu The Hifu - NguySn Vinh Can
CHI D A N a) Goi M la t r u n g d i e m ciia canh BC t h i A M 1 BC; D M 1 BC =:> A M D = 60" Ta cung c6 BC ± ( A M D ) ^ ( A M D ) ± (DBC) Trong t a m giac A M D , ke A H 1 M D x> A H 1 (DBC) => A H l a khoang each tii diem A den m p < D B C ) . Trong t a m giac DBC t h i D B = a => BC = aV2 Tam giac ABC l a t a m giac deu canh aVe aV^ n e n A M - Trong t a m giac vuong A H M , AM = A M D = 60° n e n 3aV2 A H = AM.sin60° ^ A H = Trong t a m giac A M D , ke D K _L A M . Ta da c6 BC ± ( A M D ) i:^ B C l DK, suy ra D K 1 (ABC) D K la khoang each tU D den mp (ABC). aV2 Trong t a m giac vuong D M K t h i D M = va K M D = 60° D K = DM.sin60° DK = b) Ta CO, t r o n g t a m giac A M D ke M I 1 A D BC 1 ( A M D ) BC ± M I => M I l a doan vuong goc chung ciia h a i diiorng t h d n g A D , B C h a y chinh la khoang each giCfa hai di/cfng thSng A D , BC. H a i t a m giac vuong M I D va A H D dong dang : MI MD MD.AH AMID ^ AAHD MI = AH AD AD 3a V2 Ta da c6 M D - AH = 2 2 T i n h A D tCr t a m giac A M D vdri iV2 A M D = 60", A M - MD bang each ap dung d i n h l i cosin : AD^ = A M ' + M D ' - 2 . A M . M D . C 0 S A M D =0 A D = -Vs - 2^3. 2 Hoc va on luyen theo CTDT mon Toan THPT ' 2 0 5
3a Til day ta t i n h dirge MI V8-2V3 Chu y : Co the t i n h M I theo each M I . A D = A H . M D = 2SAMD- TCr C ta ke Cx // M D t h i m a t phang (A, Cx) la m a t ph^ng chufa AC va song song vdfi M D , va v i H e M D n e n ta c6 d ( M D , AC) = d ( M D , (A, Cx)) = d ( H , (A, Cx)) Ke H F ± Cx => A F ± Cx T r o n g t a m giae A H F ke H E ± A F ^ H E J. (A, Cx) K e t hop v6i cac k e t qua t r e n , ta eo d ( D M , AC) = d ( H , (A, Cx)) = H E 1 1 1 T r o n g t a m giae vuong A H F , ta eo : HE*-^ HA' HF' ^ , „ . 3aV2 . -.^ aV2 _ 2V5a Ta eo : H A = va H F = M C = , suy r a H E = . BAITOAN 3. CAC BAITOAN VE G O C I- C A C KIEN THLTC C d BAN 1. X a c dinh c a c loai goc a) Goc giUa hai duang thdng De xae d i n h goe giOfa h a i diTcfng thSng a va b, ta lay m o t d i e m O thuoc a va qua O, ve diidng t h a n g b' // b. Goc tao bofi hai di/cfng t h a n g a va b' la goc giiifa h a i difdng t h a n g a, b. b) Goc giUa duang thdng va mat phang - Goc giOfa di/crng t h d n g a va m a t phang (P) la goc giOfa difcfng t h a n g a va h i n h chieu a' eiia a t r e n m a t phSng (P). - Ta xae d i n h goc giOfa diTorng thang a va mat phang (P) nhir sau : + L a y m o t d i e m A thuoc a va ha diTdng vuong goc A H xuong mat p h a n g (P) : A H ± (P) + Gia sijf a cat (P) t a i d i e m O t h i goe A O H la goc giiifa di/cfng t h i n g a va m a t phSng (P). c) Goc giQa hai mat phang Gia SLf h a i m a t p h a n g (P), ( Q ) giao nhau theo giao tuyen d. Ta xac 206 ; TS. Vu Thg Huu - Nguyin Vinh CJn
dinh goc giufa hai mat phang (P), (Q) nhu sau : - Lay mot diem O thuoc giao tuyen d. - Trong mat p h i n g (P), ke tia Ox vuong goc vdfi giao tuyen d va trong mat phSng (Q), ke tia Oy vuong goc vdi giao tuyen d. Goc 3cOy 6 [0°; 90°] la goc giaa hai mat phang (P), (Q). Doi khi ta cung noi goc xOy la goc mat phang (P) tao vdi mat phang (Q). T a cung CO the phat bieu each xac dinh goc xOy nhif sau : - Qua mot diem O thuoc giao tuyen d cua hai mat phang (P), (Q) ta difng mat phang (R) vuong goc vdi d. Mat phang (R) cat (P) theo giao tuyen Ox va cSt mat phang (Q) theo giao tuyen Oy. Goc xOy e [0°; 90°] CO di/Oc la goc giUa hai mat phSng (P), (Q). 2. C h i i y. Cac bai toan ve goc, trong phan Idn trtfcfng htfp, c6 lien he den quan he vuong goc giOfa diXofng thang va mat phang. II- C A C DANG BAI TAP VA PHLfdNG PHAP OIAI l] C a c d a n g b a i t|lp v a phifofng p h a p c h u n g de g i a i - Viec xac dinh cac goc, nhat la xac dinh goc giOfa difdng thang va mat phang, goc giOra hai mat phang, thifcfng la mot trong nhufng cong viec dau tien khi bat dau giai bai toan hinh, khong gian va cong viec nay nhieu khi c6 tinh chat quyet dinh vi neu khong xac dinh dMc cac goc hoac xac dinh sai cac goc thi ta khong the tinh difcfc hoac tinh sai cac ket qua khac. - Viec tinh toan cac yeu to c6 lien quan den cac goc, cac do dai thtfcfng lien quan den : + Cac he thufc Itfqng trong tam giac vuong. + Cac ti so ItfOng giac cua goc nhon. + Dinh li sin, cosin. Do vay, co gang tim ra tren hinh ve cac tam giac vuong c6 chufa cac goc da cho va nhieu khi ta phai ve them cac diTcfng phu de tao ra cac tam giac vuong ay. 2. C a c b a i tap 12. Cho hinh chop tuf giac deu S . A B C D ; O la giao diem cua hai difdng cheo AC, BD. 1. Hay chi r5 : HQC 6n luyOn theo C T D T mOn Jo&n THPT £3 207
a) Chieu cao cua hinh chop. b) Goc giiJfa canh ben va mat phang day. c) Goc giOfa mat ben va mat phSng day. d) Goc giufa chieu cao va mat ben. e) Khoang each tCr tam O cua day den mat ben. 2. Cung v6i cac cau hoi tren, giai bai toan trong trifcfng hop hinh chop dacho la hinh chop tam giac deu (O la tam cua day). CHI D A N 1. a) Hinh chop tijf giac deu c6 day la hinh vuong va O la tam ciia day. SO 1 (ABCD) => SO la di/orng cao ciia hinh chop. S b) SO 1 (ABCD) => OC la hinh chieu ciia SC tren day : Goc SCO la goc giOfa canh ben SC vdri day. Tiiong tu, ta c6 cac goc S A O , S B O , S D O theo thijf tiT la cac goc giiJa cac canh ben SA, S B , S D vdfi mat phang day. ^ c) Goi M la trung diem ciia canh B C t h i S M 1 B C , O M ± B C => goc S ' M O la goc giiJfa mat ben ( S B C ) voi mat phang day ( A B C D ) . d) Goc O S M la goc giCifa chieu cao S O va mat ben ( S B C ) . e) Trong tam giac S O M , ke O H ± S M =^ O H 1 ( S B C ) => H la hinh chieu cua O tren mp ( S B C ) => O H la khoang each t i i tam O den mat ben ( S B C ) . 2. Trong trirdng hop hinh chop S . A B C la hinh chop tam giac deu t h i day A B C la tam giac deu, tam O la trong tam cua day. Goi M la trung diem canh B C t h i A M la trung tuyen. A M = ^5:^;A0= ^AM; 2 3 A M 1 BC;S M ± BC. a) S O J- ( A B C ) ^ S O la chieu cao hinh chop. b) S A M la goc giOfa canh ben S A vdi mat phang day. c) S M O la goc giOfa mat ben ( S B C ) v(Ji mat phang day. d) O S M la goc giOra diTofng cao S O va mat ben ( S B C ) . e) Trong tam giac S O M ke O H ± S M t h i O H ± ( S B C ) OH la khoang each tir diem O den mp ( S B C ) . Chii y : K h i can t i n h toan cac yeu to' tren t h i ta nen diTa vao cac he thiirc liigrng trong tam giac, dac biet la he thufc liitfng trong tam giac vuong. 2 0 8 ; T S . Vu The Hiju - Nguygn Vfnh Can
YtNfini.TH'E'TiCHCIlCKil B A I TOAN 4. T I N H THE T I C H CAC K H O I DA DIEN I- CAC KIEN THlfC C6 BAN 1. C o n g thiJc t i n h V v a Sxq c a c k h o i d a d i ^ n Hinh The t i c h V D i e n t i c h xung quanh Sxq Chop V= -Bh Sxq=^pd(*) 3 L a n g t r y dufng V = Bh (**) Sxq = 2p.h H i n h hop diifng V = Bh (**) Sxq = 2p.h B l a dien t i c h day, h l a chieu cao, 2p l a chu v i day. Chiiy Cong thufc (*), Sxq = — pd chi diing cho h i n h chop deu, t r o n g do d l a 2 t r u n g doan ciia h i n h chop (chieu cao ciia m a t ben). The t i c h h i n h hop chOf n h a t V = abc. Cong thufc (**) cung diing v d i triTorng hop k h o i l a n g t r u x i e n . II- CAC BAI TAP VA CACH GIAI 1. C a c l o a i b a i t^p a) Cac bai tap chu yeu l a t i n h the t i c h va dien t i c h xung quanh cac k h o i da dien, doi k h i k e t hop v d i viec t i n h mot so yeu to ve k h o a n g each. Cung can biet t h e m cong thijfc : Stp = S d + Sxq D i e n t i c h toan p h a n = dien t i c h day + dien t i c h xung quanh. b) De t i n h the t i c h cac k h o i da dien, t a thiTcfng p h a i l a m h a i viec : - Xac d i n h v a t i n h chieu cao ciia k h o i da dien. - T i n h dien t i c h day. Viec xac d i n h va t i n h di/orng cao ciia k h o i da dien l i e n quan den cac quan he vuong goc t r o n g k h o n g gian. Viec t i n h dien t i c h day l i e n quan den viec t i n h dien t i c h cac t a m giac, tuf giac, cac da giac. Can ghi nhdf mot so cong thufc t i n h cho mot so h i n h dac biet de t i e t k i e m thcfi gian k h i l a m toan. HQC va on luyen theo CTOT mon Toan THPT 209
+ Dien tich tarn giac deu canh a : S = (chieu cao tarn giac deu canh a la h = + Dien tich tarn giac c6 hai canh a, b va goc xen giOfa hai canh ay la a : S = —ab.sina 2 + Dien tich tarn giac biet ba canh a, b, c la : S = .^pCp - a)(p - b)(p - c) (cong thufc Heron) vdri a + b + c = 2p (2p la chu vi tarn giac). + Dien tich tarn giac biet chu v i 2p va ban kinh diicfng tron noi tiep r : S = pr. + Lien he giuTa dien tich tarn giac vdfi ban kinh diicfng tron ngoai tiep : abc S= 4R' c) De tinh the tich khoi chop tarn giac, ta con c6 the suf dung cong thiic ti so the tich. ^ Cac bai t^p 13. Cho hinh chop S.ABC, cac mat ben SAB, SAC la cac tarn giac vuong can tai A, con mat ben SBC c6 dien tich la —a'Vs. Chieu cao hinh 2 chop la a. a) Tinh the tich khoi chop. b) Tinh khoang each t\i dinh A den mat phang (SBC). CHI DAN SA 1 AB a) SA L (ABC) SA 1 AC => SA la chieu cao hinh chop, SA = a. De thay SB = SC = aV2 SsBc = -SB.SC.sinS 2 ^ -aV2.aV2.sinS = => sinS = — ^ S = 6 0 ° ^ B C = a>^ 2 2 2 Goi M la trung diem cua BC : AM^ = AB^ - MB^ 1V2 2a^ A M ' = a' - >AM- 4 Vs.ABC = ^SA.S,3c = \.^\ ^ . a V ^ ^ V, S.ABC = a" (dvtt) 210 ta TS. Vu The Huu - Nguyin VTnh C0n
b) Goi H l a h i n h chieu cua A t r e n m a t phSng (SBC) => A H l a chieu cao ke tif d i n h A t r o n g h i n h chop A.SBC. VA.SBC = — A H . S S B C = VS.ABC >AH = o 3 2 6 Chii y : Ta CO the t i n h A H theo each khac. Goi M la t r u n g d i e m ciaa canh BC BC 1 (SAM) SMIBC => (SAM) 1 (SBC) Trong m a t p h a n g (SAM) ke A H 1 S M A H ± (SBC) Trong t a m giac vuong S A M t h i 1 1 • + • 3 =>AH^ = — AH = AH' SA' A M ' raV2 14. Cho l a n g t r u duTng A B C D . A ' B ' C ' D ' , chieu cao a; day A B C D l a h i n h vuong. DLfdng t h a n g A D ' tao vdri m a t phang ( A A ' C ) m o t goc 45°. T i n h the tich cua k h o i l a n g t r u va t i n h goc giufa h a i di/cfng t h a n g A I va BD', t r o n g do I la t a m cua h i n h vuong A'B'C'D'. CHI DAN Goi I la giao d i e m cua A ' C va B D ' . D'l L A'C'l De t h a y : D ' l l A'A D' => D ' l 1 ( A A ' C ) ^ D ' A I = 45° B Goi canh day l a n g t r u la x, A'D' = x ^ D ' I = i ^ D D'l T r o n g t a m giac vuong D ' l A : A D ' = = 2x sin45° T r•ong o n g t a m giac vuong D ' D A : D'D^ D'D' = D'A^ - DA^ aVs a' = ( 2 x ) 2 - x ^ ^ a ' = 3 x 2 ^ x = ABCD.ABCD' = (dvtt) V i B B ' // AA' nen goc giufa h a i difcfng t h S n g A I va B B ' cung bSng goc giufa h a i di/cfng t h a n g A I va AA'. T r o n g t a m giac vuong A A ' I t a c6 : A I cosA'AI = cosA'AI = — >A'AI = 60°. lA' 2 Hoc va 6n luyen theo CTDT mSn To4n THPT £0 211