cos(-) =cossin(-) = - sintan(-) = - tancot(-) = - cot sin( - )
=sincos( - ) = - costan( - ) = - tancot( - ) = - cot sin(2- ) =
cos, cos(2- ) = sintan(2- ) = cot, cot(2- ) = tan ++ ++ tan tantan(
)1 tan .tantan tantan( )1 tan .tana baba ba baba bTrng THPT Ni
ThnhCHUYN PHNG TRNH LNG GIC TRONG CC K THI TUYN SINH H - CTrong thi
H C cc nm gn y, bi ton gii phng trnh lng gic (PTLG) l mt trong cc
bi ton thng xuyn xut hin trong cc thi. Bi ton ny khng thuc trong
nhm cc bi ton kh trong cc thi nn ch cn phng php hc khoa hc hc sinh
c th t im ti a i vi bi ton ny. Trong vic gii PTLG vic tp cho hc
sinh nhn xt mi quan h cc gc ca cc hm s trong PTLG rt quan trng v iu
ny s gip hc sinh p dng cng thc lng gic hp l.Trong bi vit ny ti xin
gii thiu cc phng php gii, mt s php bin i v mt s k nng c bn gip hc
sinh nhn dng v vn dng cc cng thc lng gic hp l gii quyt tt bi ton
gii phng trnh lng gic trong cc thi H C. PHN A C S L THUYTI CNG THC
LNG GIC:1. Cc h thc lng gic c bn: Nh: Cng gc2 2sin cossin cos 1;
tan , cot ; 1 sin , cos 1cos sinx xx x x x x xx x+ Suy ra: 2 22 21
11 tan ,1 cot ; tan .cot 1.cos sinx x x xx x+ + 2. Cung c lin quan
c bit: Nh: Cos i Sin b - Ph choc bit: khikhi t t 't t 'sin kchnsin(
k ) ;tan( k ) tansinkhi klcos kchncos( k ) ;cot( k ) cotcos khi
kl3. Cng thc cng: Nh: Sin th sin cos, cos sinCos th cos cos, sin
sin du i
sin(a + b) = sina.cosb + cosa.sinb sin(a - b) = sina.cosb -
cosa.sinb cos(a + b) = cosa.cosb - sina.sinb cos(a - b) = cosa.cosb
+ sina.sinb 4. Cng thc nhn i:Nh: Suy ra t cng thc cng bng cch thay
b bng a5. Cng thc h bc:Nh: c suy ra t cng thc nhn i.1sin2a =
2sina.cosa22.tantan21 tanaaacos2a = 2.cos2a 1= 1 2.sin2a= cos2a
sin2aTrng THPT Ni Thnh+ 2 21 cos2 1 cos2cos ,sin2 2x xx x6. Cng thc
bin i tng thnh tch:Nh: Sin cng sin bng hai ln sin cos.Sin tr sin
bng hai ln cos sin Cos cng cos bng hai ln cos cos.Cos tr cos bng
hai ln cos sin+ ++ + + + + + sin( )sin sin 2.sin .cos , tan tan2 2
cos .cossin( )sin sin 2.cos .sin , tan tan2 2 cos .coscos cos 2.cos
.cos , cos cos 2.sin .sin2 2 2 2ab ab aba b a ba bab ab aba b a ba
bab ab ab aba b a b7. Cng thc bin i tch thnh tng:Nh: Suy ra t cng
thc tng thnh tch + + 1 ] + + 1 ] + 1 ]1sin .cos sin( ) sin( )21cos
.cos cos( ) cos( )21sin .sin cos( ) cos( )2a b ab aba b ab aba b ab
ab8. Cng thc tnh theo t = tan a: + + 22 2 22 1 2sin2 ,cos2 ,tan21 1
1t t ta a at t tCng thc ny a s hc sinh khng nh c nhng hay dng trong
vic gii PTLG nn cn lu cho hc sinh9. Cng thc nhn ba: 3 332sin3 3.sin
4.sin , cos3 4.cos 3.cos3tan tantan31 3tana a a a a aa aaaII CC
PHNG PHP GII PTLG: gii bi ton ny phng php thng gp l thc hin mt s
php bin i hp l (v cc cng thc lng gic rt a dng) a bi ton v:+ PTLG c
bn.+ PTLG thng gp: 1. Phng trnh bc nht, bc hai, bc ba, i vi hslg:2.
Phng trnh bc nht i vi sinu, cosu: 3. PT thun nht bc hai i vi sinu,
cosu:4. Phng trnh i xng i vi sinu v cosu:+ Phng trnh tch cc PTLG c
bn, cc PTLG thng gp.+ H cc PTLG: phn ny ta thung s dng: a v tng cc
bnh phng, nh gi hoc dng bt ng thc . Cc nm gn y t thy ra dng ny nn
ti khng gii thiu trong chuyn ny.Ngoi ra, ta cn s dng cch t n s ph
hp l a v phng trnh theo n ph v gii tm nghim.PHN B PHN BI TPI - MT S
PHP BIN I THNG DNG: a v PT tch hay rt gn1) 21 sin 2 (sin cos ) x x
x + + ; 21 sin 2 (sin cos ) x x x 2) sin cos1 tancosx xxxtt ,cos
sin1 cotsinx xxxtt , 3) sin 2sin cos2xx x 2Trng THPT Ni Thnh4)( ) (
)2 2cos 2 cos sin cos sin . cos sin + x x x x x x x5)( ) ( )2 2cos
x 1 sin x 1 sin x . 1 sin x + ( ) ( )2 2sin x 1 cos x 1 cosx . 1
cos x +6)2 2sin x cos x 2t anx+cot xsin x.cosx sin 2x+ , 2 2sin x
cos x 2cos2xt anx cot xsin x.cosx sin 2x 7)3 3sin cos (sin cos )(1
sin .cos ) x x x x x x + + , 3 3sin cos (sin cos )(1 sin .cos ) x x
x x x x +8) 4 4 2 2cos sin cos sin cos 2 x x x x x
4 4 2 2 21 1 1 cos 4 3 1sin cos 1 2sin .cos 1 .sin 2 1 . .cos 42
2 2 4 4xx x x x x x _+ + ,
6 6 2 2 23 3 1 cos 4 5 3sin cos 1 3sin .cos 1 .sin 2 1 . .cos 44
4 2 8 8xx x x x x x _+ + ,9) 3 3 3sin sin .cos cos .sin cos2 2 2x x
x x _+ + ,
7 7 7cos cos .cos sin .sin sin2 2 2x x x x _+ ,( )2sin sin .cos
cos .sin . sin cos4 4 4 2x x x x x _+ + + , .II - MT S K NNG NHN
DNG THNG DNG: vn dng cng thc lng gic hp l gii bi ton gii PTLGKhi gp
PTLG c cha:- Bnh phng, khc gc ta thng s sng cng thc h bc.- Tch cc
hm s lng gic sin v cos ta thng bin i v tng.- Tng cc hm s lng gic
sin v cos ta thng bin i v tch.- Gc gp i nhau ta thng s dng cng thc
nhn i.- Cc gc c bit, VD nh: x4+ ,32x ,74x ta thng s dng cng thc cng
bin i trc. Lu cc cp gp ph nhau.III - MT S BI TON P DNG:Bi ton 1:
Gii PTLG sau: 2cos 2 1cot 1 sin sin1 tan 2xx x xx + +Nhn xt : bi
ton ny ta vn dng cc php bin i trn a v PT tchHD gii: iu kin: sin
.cos 0 x x va tanx 1 ( ) ( )( )2cos 2 1cot 1 sin sin1 tan 2cos sin
cos sincos sinsin . sin coscos sinsincosxx x xxx x x xx xPT x x xx
xxx + + + + +( )1sin cos sin cos 0sinx x x xx _ + ,3Trng THPT Ni
Thnhsin cos 0 (1)1sin cos 0 (2)sinx xx xx
+
S: 4x k +Bi ton 2: Gii PTLG sau:(1 sin x cos 2x) sin x1 4cos x1
tan x 2 _+ + + ,+ (A 2010)Nhn xt : bi ton ny ta thy c cha( )2sin .
sin cos4 2x x x _+ + , v mu c cha sin cos1 tancosx xxx++ nn ta phn
tch rt gn t v mu cho (sinx + cosx)HD gii: iu kin:cos 0 x va tanx 1
2(1 sin cos 2 ). .(sin cos )12.cos .cossin cos 2x x x xPT x xx x+ +
+ +(1 sin cos 2 ).(sin cos ).cos cossin cosx x x xx xx x+ + + +(1
sin cos 2 ) 1 sin cos 2 0 x x x x + + + Gc 2x v 1x: nn s dng CThc
nhn i212sin sin 1 0 sin 1( ) sin27S: 2 2 ( )6 6x x x loai hay x x k
hay x k k + + Bi ton 3: Gii PTLG sau:1 1 74sin3 sin 4sin2xxx _+ _ ,
, (A 2008)Nhn xt : bi ton ny ta thy c cha 3sin2x _ ,v 7sin4x _ ,nn
ta s dng cng thc cng bin iHD gii: 3 3 3sin sin .cos cos .sin cos2 2
2x x x x _ ,( )7 7 7 2sin sin .cos cos .sin . sin cos4 4 4 2x x x x
x _ + ,iu kin: sin 0, cos 0 x x PT tr thnh: 1 1 12 2(sin cos ) 0
(sin cos ) 2 2 0sin cos sin .cossin 04 5S: , ,4 8 8 1sin 22x x x xx
x x xx x k x k x kx _+ + + + + ,
_+
,
+ + +
Bi ton 4: Gii PTLG sau: ( ) 2 cos sin1tan cot 2 cot 1x xx x x+
4Trng THPT Ni ThnhNhn xt : bi ton ny ta vn dng cc php bin i trn rt
gn v phi, v tric cha tanx + cot2x ta bin i trcHD gii: Ta c: ( ) cos
2sin .sin 2 cos .cos 2 1tan cot 2cos .sin 2 cos .sin 2 sin 2x xx x
x xx xx x x x x++ iu kin: sin2x.(tanx + cot2x) 0 v cotx 1( ) 2 cos
sin1 2sin 2 2.sin cos1 cos sin2sin 2 sinx xPT x x xx xx x Tm nghim
v kt hp iu kin ta c:( ) 24x k k + Bi ton 5: Gii PTLG sau: cot sin 1
tan . tan 42xx x x _+ + ,Nhn xt : bi ton ny ta v tri c cha 1 tan .
tan2xx _+ , ta bin i trcHD gii: Ta c: coscos .cos sin .sin1 22 21
tan . tan2 coscos .cos cos .cos2 2xx xxx xxxx xxx x _+ ,+ iu kin:
sinx 0, cosx 02cot tan 4 tan 4. tan 1 0 tan 2 3 PT x x x x x + + t
. S:( )arctan 2 3 x k t +Bi ton 6: Gii PTLG sau: (1 2sin x) cos
x3(1 2sin x)(1 sin x)+ (A 2009)Nhn xt : Bin i v s dng cch gii PT:
a.sinx + b.cosx = c HD gii: ( )2cos sin 2 3. 1 sin 2.sin PT x x x x
( ) cos sin 2 3. cos 2 sin x x x x Ta chuyn cng gc qua mt v a v dng
a.sinx + b.cosx( ) cos sin 2 3. cos 2 sin x x x x 3.sin cos sin 2
3.cos x x x x + + Chia hai v ca PT cho 2 sin sin 26 3x x _ _ + + ,
,S: 2 3, 218 3 2x k x l + +Bi ton 7: Gii PTLG sau:3sin x cos xsin2x
3cos3x 2(cos 4x sin x) + + + (B 2009)Nhn xt: Bin i v s dng cch gii
PT: a.sinx + b.cosx = c bi ton ny ta thy c cha tch: cosx.sin2x nn
ta bin i v tng v c sin3x nn ta s dng cng thc nhn ba h bc 3HD gii: (
)1 3 1sin sin3 sin 3 cos3 2(cos 4 sin sin3 )2 4 41 3 3 1sin3 sin 3
cos3 2cos 4 sin sin32 2 2 2PT x x x x x x xx x x x x x + + + + + +
+ sin3 3 cos3 2cos 4 x x x + Ta chuyn cng gc qua mt v a v 5Trng
THPT Ni Thnhdng a.sinx + b.cosx1 3sin3x cos3x cos4x2 2 + Chia hai v
ca PT cho 2cos 3x cos 4x6 _ ,S: 2x k , x k242 7 6 + + Bi ton 8: Gii
PTLG sau: 22.sin 2 sin 7 1 sin x x x + (B 2007)Nhn xt: Bnh phng,
khc gc ta thng s dng cng thc h bc HD gii: 1 cos 42. sin 7 1 sin2xPT
x x + sin 7 sin cos 4 0 x x x Tng ta thng bin i v tch t nhn t
chung2cos 4 .sin3x cos 4 0 x x cos 4 0cos 4 (2sin3x 1) 0sin3
sin6xxx
2 5 2: , ,8 4 18 3 6 3KL x k x k x k + + +Bi ton 9: Gii PTLG
sau: 2 2cos 3x cos2x cos x 0 (A 2005)Nhn xt: Bnh phng, khc gc ta
thng s dng cng thc h bc HD gii: 2 2cos 3x.cos2x cos x 0 ( ) 1+cos6x
cos 21+cos2x02 2x cos6x.cos 2 1 x Tch ta thng bin i v tng( )1cos8
cos 4 12x x + Gc 8x v 4x: nn s dng cng thc nhn i22.cos 4 cos 4 3 0.
S: .2x x x k + Bi ton 10: Gii PTLG sau: sin 2x cos2x 3sinx cos x 1
0 + (D 2010)Nhn xt : Gc 2x v 1x: nn s dng cng thc nhn i bin i HD
gii: 2sin .cos cos 2 3sin cos 1 0 PT x x x x x + y ta nhm
2.sinx.cosx vi cosx do khi nhm vi 3.sinx ta khng gii tip c( )2cos .
2sin 1 2sin 3sin 2 0 x x x x + +
( ) ( ) ( ) cos . 2sin 1 2sin 1 . sin 2 0 x x x x + + ( ) ( )
2sin 1 cos sin 2 0 x x x + + ( )( )2sin 1 0 15S: 2 , 26 6 cos sin 2
0 2 ,x x k x kx x PTVN + +
+ +
Bi ton 11: Gii PTLG sau: 25.sin 2 3.(1 sin ). tan x x x (B
2007)Nhn xt : a v cng mt hm s lng gic bi ton ny ta nhn thy cng gc
nn s dng cc h thc lng gic c bn v a PT v cng mt hm s sinxHD gii: 2 2
25sinx(1 sin x) 2(1 sin x) = 3(1 sinx).sin x PT 3 22sin x+sin x
5sinx+2=0 2( 1)(2 3 2) 0 t t t + (t = sinx)11, , 22 t t t6Trng THPT
Ni ThnhS: 2 , 22 6 x k x k + +Bi ton 12: Gii PTLG sau:cos3x cos2x
cosx 1 0 + (D 2006)Nhn xt:a v cng mt hm s lng gic bi ton ny ta nhn
thy cos3x v cos2x ta u chuyn c v cosx nn s dng cng thc nhn ba v cng
thc nhn i a PT v cng mt hm s sinxHD gii: 3 2cos3x cos2x cosx 1
04.cos 3.cos 2cos 1 cos 1 0 x x x x+ + 3 22cos x cos x 2cosx 1 0 +
2(2cos 1)(cos x 1) 0 x + 1cos , sin 02 x x2S: 2 ,3 x k x k t + Bi
ton 13: Gii PTLG sau: 3sin .sin 2 sin3 6cos x x x x + Nhn xt: Bin i
a v PT dng: 3 2 2 3a.sin x +b.cos x.sinx +c.cosx.sin x +d.sinx
+e.cosx +f.cos x = 0 HD gii: 2 3 32.sin .cos 3sin 4sin 6.cos 0 PT x
x x x x + Khi cosx = 0 2sin 1 x (khng tha phng trnh). Khi cosx
0:Chia 2 v cho cos3x, t t = tanx ta c:( ) ( )3 2 22 3 6 0 2 . 3 0 t
t t t t + Bi ton 14: Gii PTLG sau: 3 3 2 2sin 3 cos sin .cos 3.sin
.cos x x x x x x(B 2008)Nhn xt: Bin i a v PT dng: 3 2 2 3a.sin x
+b.cos x.sinx +c.cosx.sin x +d.sinx +e.cosx +f.cos x = 0 HD gii:
Khi cosx = 0 2sin 1 x (khng tha phng trnh). Khi cosx 0:Chia 2 v cho
cos3x, t t = tanx ta c:3 2t 3 3 0 t t + 2( 3)( 1) 0 t t + 3, 1 t t
t S: ,3 4 x k x k + t +Bi ton 15: Gii PTLG sau: cos cos cos3 6 4x x
x _ _ _+ + + + , , ,Nhn xt: bi ny ta nhn xt gc:3 62 4x xx _ _+ + +
_ , , + , nn ta p dng cng thc bin i tng thnh tch bin i PT HD gii:
2.cos .cos cos cos 04 12 4 4PT x x x _ _ _ + + + , , ,S: 2x k +Bi
ton 16: Gii PTLG sau: 4 42sin coscos 4tan . tan4 4x xxx x + _ _+ ,
,7Trng THPT Ni ThnhNhn xt: bi ny ta nhn xt gc 4x+v 4xph nhau, t ta
s dng cc php bin i thng gp HD gii: iu kin: Ta c: 4 4 2x x _ _+ + ,
, nntan cot4 4x x _ _+ , ,Khi : 2 23 1.cos 4 cos 4 4cos 4 cos 4 3
04 4PT x x x x + cos 4 13cos 44xx
CC PHNG TRNH LNG GIC TRONG CC THI I HC T 2002 N NAYI. Bin i a v
phng trnh bc 2, bc 3 i vi mt hslg1. (KA2002) Tm cc nghim thuc khong
(0; 2) ca phng trnh cos3x + sin3x5(sinx +) os2x + 31+sin2xc S 5;3 3
2. (D b2002) 4 4sin cos 1 1cot 25sin 2 2 8sin 2x xg xx x+ S: 6x k t
+3. (D b2002)tgx + cosx - cos2x = sinx(1 + tgxtg2x) S:2 x k 4.
(KB2003) ) cotgx - tgx + 4sin2x = 2sin 2x S x = 3k t +5. (D b2003)
2cos4xcotx = tanx + sin2xS 3x k t +6. (KB2004) 5sinx 2 = 3(1
sinx)tan2x. S 52 ; 26 6x k k + +7. (KA2005) cos23xcos2x-cos2x = 0 S
2x k 8. (KD2005) sin4x + cos4x + cos(x-4)sin(3x-4) - 32= 0 S 4x k
+9. (KA2006) ( )6 62 os sin sinxcosx02 2sinxc x x + S 524x k +10.
(KB2006) xcotx + sinx(1 + tanxtan ) 42 S 5;12 12x k k + +11.
(KD2006 ) Cos3x + cos2x cosx 1 = 0 S 22 ;3x k k t +12. (KA2010) (1
sin x cos 2x)sin x1 4cos x1 tan x 2 _+ + + ,+II. Bin i a v phng
trnh bc nht i vi ) ( cos ), ( sin x u x u8Trng THPT Ni
Thnh1.KA2009(1 2sin x) cos x3(1 2sin x)(1 sin x)+ 218 3 + x k 2.KB
20093sin x cos x sin 2x 3 cos 3x 2(cos 4x sin x) + + + x= 7242; 26
k k + + 3.KD2009 3cos5x 2sin3xcos 2x sin x 0 x k18 3 ;x k6 2
4.(KB2008) 3 3 2 2sin 3 os sinxcos 3sin osx x c x x xc S; ;4 4 3x k
x k k + + +5.(KD) 2xsin os 3 osx = 22 2xc c _+ + ,6. (D b2005) Tm
nghim trn khong( ) 0;ca pt 2 234sin 3 os2x = 1 + 2cos x - 2 4xc _
,S 5 17 5; ;18 18 6 7. (D b2002) Cho pt 2sinx + cosx + 1sinx -
2cosx + 3a (*)a. Gii pt (*) khi a = 13b. Tm a pt (*) c nghim. S 1;
24 2x k a + III. Bin i, nhm, t nhn t chung a v phng trnh tch 1.
(KB2002) sin23x - cos24x = sin25x - cos26xS;9 2k kx x
2.(KA2003)cotgx - 1 = 21cos xtgx + + sin2x - 12sin2x S 4x k +3:
(KD2003) 2 2 2sin tan os 02 4 2x xx c _ ,S 2 ;4x k k + +4.(D b2003)
3 - tgx(tgx + 2sinx) + 6cosx = 0 S 3x k t +5.(D b2003) cos2x +
cosx(2tg2x - 1) = 2S2 ; 23x k k + t +6.( KD2004) (2cosx 1)(2sinx +
cosx) = sin2x sinx S2 ;3 4x k k t + +7. (KB2005) 1 + sinx + cosx +
sin2x + cos2x = 0 S 22 ;3 4x k k t + +8.(KA2007) (1 + sin2x)cosx +
(1 + cos2x)sinx = 1 + sin2x S; 2 ; 24 2x k x k k + +9. (KB2007)
2sin22x + sin7x 1 = sinx10.(KA2008) 1 1 74sin( )3sinx 4sin( )2xx S
5; ;4 8 8x k x k k + + +11.(KB2008) 3 3 2 2sin 3 os sinxcos 3sin
osx x c x x xc S; ;4 4 3x k x k k + + +9Trng THPT Ni Thnh12.
(KD2008) 2sinx(1 + cos2x) + sin2x = 1 + 2cosx S 22 ;3 4x k k t +
+13. KB2010 (sin 2x + cos 2x) cosx + 2cos2x sin x = 0x = 4 2k +(k
Z)14. KD2010 sin 2 cos 2 3sin cos 1 0 x x x x + S26k x + 265k x +
15. KA201121 sin 2 cos 22 sin sin 21 cotx xx xx+ ++/sk x + 2, 24k x
+ 16. KB2011 sin 2 cos sin cos cos 2 sin cos x x x x x x x + + + /S
22k x + 323 kx + 17. KD2011sin2x 2cos x sin x 10tan x 3+ + /S 23k x
+ Trn y l mt s phng php, mt s php bin i v mt s k nng thng s dng
trong vic gii PTLG, khng c k nng hay phng php no l tuyt i. Mun gii
tt cc bi tp dng ny hc sinh phi nm vng l thuyt lng gic v gii nhiu bi
tp t rt ra kinh nghim ring cho bn thn mnh. 10Trng THPT Ni ThnhCHUYN
HNH HC KHNG GIANTRONG K THI TUYN SINH I HCA. Bi ton tnh th tch khi
a din:I/ C s l thuyt cn nm:+ Th tch khi chp:V =h S.31(S: din tch y,
h: chiu cao)+ Th tch khi hp: V = a.b.c (a,b,c: di ba cnh)+ Th tch
khi lng tr: V = S.h (S: din tch y, h: chiu cao)II/ Cc dng ton v tnh
th tch:Loi 1: Tnh th tch bng cch s dng trc tip cc cng thc ton + xc
nh chiu cao ca khi a din cn tnh th tch (da vo cc nh l quan h vung
gc bit: nh l 3 ng vung gc, nh l k ng thng vung gc mt phng )+ tm din
tch y bng cc cng thc quen bit.V d:Cho hnh chp S.ABCD c y ABCD l hnh
vung ti A v D, c AB=AD=2a; CD=a. gc gia 2 mt phng (SCB) v (ABCD)
bng 060. Gi I l trung im AD bit 2 mt phng (SBI) v (SCI) cng vung gc
vi (ABCD). Tnh th tch khi chp SABCD?Gii:V 2 mt phng (SBI) v (SCI)
cng vung gc vi (ABCD) m (SBI) v (SCI) c giao tuyn l SI l ng cao. K
IH vung gc vi BC ta c gc to bi mt phng (SBC) v (ABCD) l 0SHI 60 =.
T ta tnh c:IC a 2; IB BC a 5 = = =2ABCD1S AD(AB CD) 3a2= + =nn
IBC2S 3 3IH aBC 5= = . T 3SABCD3 15V a5=Cc bi ton cng dng: H
A-2009; H B-2009; H D-2009; H A-2007; H B-2006Loi 2: 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.+ phn chia khi a din thnh tng hoc hiu cc khi c bn (
hnh chp hoc hnh lng tr) m cc khi ny d tnh hn.+ Hoc so snh th tch
khi cn tnh vi mt khi a din khc bit trc th tch.Vi loi ny ta hay s
dng kt qu sau y:Cho hnh chp S.ABC. ly A', B', C' tng ng trn cnh sau
y SA, SB, SC. Khi :11Trng THPT Ni Thnh SCSCSBSBSASAVVABC SC B A
S'.'.'' ' ' .>V d p dng:Cho hnh chp S.ABCD c y l hnh thoi cnh a,
0BAD 60 =, SAvung 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, SC ca hnh chp ti
B', D' . Tnh th tch khi chp S.AB'C'D'.Hng dn 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 (SDB) k ng thng song song vi BD ct SD ti B', C' l 2
giao im cn tm.Ta c: 32 ' ';21 ' SBSBSDSDSCSCD thyVS.AB'C'D' =
2VS.AB'C'; VS.ABCD = 2VS.ABC 31 '.'..' ' ..' ' ' .
SCSCSBSBSASAVVVVABC SC AB SABCD SD C AB STa c 3SABCD ABCD1 1 1 3 3V
SA.S SA.AD.AB.sin DAB a.a.a. a .3 3 3 2 6= = = =3' ' ' .183a VD C
AB S Dng ton tng t:Cho hnh chp S.ABCD c yABCDl hnh ch nht AB=a,
AD=2a, cnh SA vung gc vi y, cnh SB hp vi y mt gc 060. Trn cnh SA ly
M sao cho a 3AM3= . Mt phng BCM ct DS ti N. tnh th tch khi chp
SBCMN.Cc bi ton cng dng: H A-2004; H D-2006; H A-2003Loi 3: Tnh th
tch khi a din bng php tnh ta trong khng gianV d: Cho hnh chp S.ABCD
c yABCDl hnh ch nht AB=a, AD a 2 = , 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. Tm th tch khi t din ANIB.Gii: dng h trc ta Axyz vi gc A12Trng
THPT Ni ThnhTrong h trc ta ny, ta cA(0; 0; 0); D(a 2; 0; 0)B(0; a;
0); C(a 2; a; 0);S(0;0; a)Khi ta cIB MI IB MI2121 Nh vy a 3 aI ( ;
; 0)2 3=Ta c:
,_
,_
,_
2;6;62;2;2;22;2;2;23 a a aNIa a aNBa a aNAT : [ ]
,_
22; 0 ;2,2 2a aNB NAV vy:[ ]362. ,613aNI NB NA VANIB Cc bi ton
cng dng: H A-2003; H A-2004; H B-2006; H D-2009Cc dng ton khc: Ngoi
cc dng thng gp nu trn, cn c dng ton S dng phng phpth tch tm khong
cchCc bi ton v th tch khi a din c kt hp vi vic tm GTLN, NNCc bi ton
v so snh th tch.V d:Cho hnh chp u S.ABC c cnh y bng a, ng cao hnh
chp bng a 3 . Mt phng (P) qua cnh y BC v vung gc vi cnh bn SA chia
khi chp S.ABC thnh hai phn. Tnh t s th tch ca hai phn .Hng dn
gii:VS.ABClhnhchpunnchn ng cao H ca ca hnh chp l tm tam gicu ABC.
Ta c AH ct BC ti trung im M ca BCv BC SA. H BN vung gc vi SA suy ra
SA (BCN), suy ra tam gic BCN l thit din m mp(P)ct hnh chp S.ABC.V
thit din chia khi chp S.ABCthnh hai khi t din c chung y (BCN) nn t
sth tchbngt s hai ng cao AN/SN.13SNABCM HTrng THPT Ni ThnhVSAH MAN
nn: 20331023.33. .22 2 2 + aa aAH SHAM AHSAAM AHSAANVy t s th tch
l: 173SNBCANBCVV hoc 317ANBCSNBCVVIII/ Mt s bi tp cng dng:Cu 1) Cho
khi chp S.ABCD 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?HD: 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)) = PQKCu 2) Cho hnh chp
S.ABCD c y ABCD l hnh thoi cnh a, BAD = 600 , 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
S.ABCD S (833' ' 'a VD C SAB vtt)Cu 3) 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.S: V = 8 3Cu 4) 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.S: V = 105 3Cu 5) Khi lng tr t gic u
ABCD.A1B1C1D1 c khong cch gia 2 ng thng AB v A1D bng 2, di ng cho
mt bn bng 5.a) H AK A1D (K thuc A1D). Chng minh rng AK=2.b) Tnh th
tch khi lng tr. ABCD.A1B1C1D1S: b) V = 20 5Cu 6) Cho hnh chp tam
gic S.ABC 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
SCa) Tnh khong cch t A n mt phng (SBC)b) Tnh th tch ca khi chp
A.BCMN.S: a) d = 1957 2 a; b) V = 503 33aB. Bi ton v Khi nn, khi
trI. Kin thc c bn:1/ Mt nn, hnh nn v khi nn+ Th tch: V =h R231 +
Din tch xung quanh: Sxq =Rl + Din tch ton phn: Stp = 2R Rl +(R: bn
knh y, h: chiu cao, l: ng sinh)14Trng THPT Ni Thnh2/ Mt tr, hnh tr
v khi tr+ Th tch: V =h R2 + Din tch xung quanh: Sxq =Rh 2+ Din tch
ton phn: Stp = 22 R Rh +(R: Bn knh y, h: chiu cao)3/ Ch :+ Ct mt nn
bi mt mt phng i qua nh ta c thit din l mt tam gic cn.+ Ct mt nn bi
mt mt phng vung gc vi trc ta c thit din l mt hnh trn.+ Ct mt tr bi
mt mt phng song song hoc cha trc ta c thit din l mt hnh ch nht.+ Ct
mt tr bi mt mt phng vung gc vi trc ta c thit din l mt hnh trn.II/
Bi tp vn dng:1/ Cho hnh lng tr lc gic u ABCDEF.A'B'C'D'E'F' cnh y
bng a, chiu cao bng h. Tnh th tch khi tr ni tip hnh lng tr.S: V =h
a2432/ Cho hnh trc c trc O1O2. Mt mt phng song song vi trc ct hnh
tr theo thit din dn l hnh ch nht ABCD. Gi O l tm ca thit din , bn
knh ng trn ngoi tip hnh ch nht ABCD bng bn knh ng trn y ca hnh tr.
Tnh s o gc O1OO2.S: 02 190 OO O3/ Cho hnh tr c chiu cao v bn knh y
u bng aa) M, N l hai im ly trn hai ng trn y sao cho MN to vi trc ca
hnh tr mt gc . Tnh khong cch t trc ca hnh trc n ng thng MN.b) Mt mt
phng( ) song song vi trc ca hnh tr v ct hnh tr theo thit din l hnh
vung. Tnh khong cch t trc ca hnh tr n mt phng( ) c) Mt mt phng( )
khng song song vi trc ca hnh tr v ct hnh trc theo mt thit din l hnh
vung. Tnh gc to bi mt phng( ) vi trc ca hnh tr.S:a) d =2tan 42a; b)
d' = 23a ; c) 510cos 4/ Mt hnh tr c bn knh R v chiu cao3 R . A v B
l hai im trn hai ng trn y sao cho gc to bi AB v trc ca hnh tr bng
300. a) Tnh din tch ca thit din qua A v song song vi trc ca hnh
tr.b) Tnh gc gia hai bn knh qua A v B.c) Dng v tnh di on vung gc
chung ca AB v trc ca hnh tr.S: a) S =32R , b) 060 c) Dng ng thng
qua H v song song OO' ct AB ti I- Dng IJ//OH (J thuc OO'), IJ chnh
l on thng vung gc chung phi dng, IJ = 23 R5/ Cho mt hnh tr trn xoay
y l ng trn (O) v (O') c bn knh bng 3 n v, chiu cao ca hnh tr l 4 n
v. Gi AB l mt ng knh c nh ca (O). M l mt im lu ng trn (O'). Gi MC l
ng sinh qua C, C trn ng trn (O). K HC vung gc vi AB v th Ah = x.a)
1. Chng minh rng tng s bnh phng cc cnh ca hnh chp MABC l mt hng
s.2. Tnh MH theo x.3. nh v tr ca M din tch S ca tam gic MAB t cc
i.4. Tnh th tihcs V ca hnh chp MABC. Chng minh rng V cc i khi S cc
i.b) nh x V = 4k (k l s cho trc)S:a) 1. T = 156; 2. MH = ( ) 6 0 ,
16 62 + + x x x; 3. S = 3MH, S t cc i khi x = 3, H trng vi O, M l
im m ng sinh MC i qua im chnh gia C ca cung AB.(dng phng php th);
4. V =) 6 ( 4 x x , V cc i khi x = 3, khi S cc i.15Trng THPT Ni
Thnhb) ( ) 3 0 , 9 3 , 9 32 2 < + k k x k x6/ Mt hnh nn c ng
sinh l v gc gia ng sinh v y l .a) Tnh din tch xung quanh v th tch
khi nn.b) Gi I l im trn ng cao SO ca hnh nn sao cho( ) 1 0 , <
< k kSOSI. Tnh din tch ca thit din qua I v vung gc vi trc.S: a)
Sxq = cos2l ; b) Sthit din = 2 2 2cos l kC. Bi ton v khong cchI/ Cc
dng ton v khong cch1/Khong cch t 1 im M n 1 mt phng( ) : +Bc1: Chon
mp() cha ( qua ) M v vung gc vi( ) +Bc2: Tm giao tuyn d ca mp( ) v
mp() +Bc3: Dng MHdti HMH( ) MH= [M;( )]dHnh v minh ha:dHM2 /Khong
cch gia ng thng v mt phng song song vi ng thng :Bng khong cch t mt
im bt k trn ng thng n mt phng
3 /Khong cch gia hai mt phng song song Bng khong cch t mt im bt
k trong mt phng ny n mt phng kia(hoc ngc li)
4/Khong cch gia hai ng thng cho nhau: *Phng php1:Nn dng cho 2 ng
thng cho nhau m vung gc vi nhau Dng on vung gc chung
baHO[a;b]d OH 16Trng THPT Ni Thnh*Phng php2: tnh khong cch gia
hai ng thng a v b cho nhau lm cc bc sau +Bc1: Tm mp( ) cha bvmp( )
// a+Bc2: [ ] [ ] [ ] ; ;( ) ;( ) a b a Md d d MH HMba*Phng php3:
tnh khong cch gia hai ng thng a v b cho nhau lm cc bc sau +Bc1: Tm
mp( ) cha a v mp( ) cha b mmp( ) // mp( ) +Bc2: [ ] [ ] [ ] ; ( );(
) ;( ) a b Md d d MN NMba*Phng php4: tnh khong cch gia hai ng thng
a v b cho nhau lm cc bc sau +Bc1: Tm mp( ) vung gc a v ct a ti O
+Bc2: Tm hnh chiu b ca b ln mp( ) ; r rng a//mp(b,b)Suy ra:[ ] [ ]
[ ] ; ; ( , ') ; ( , ') a b a mp b b O mp b bd d d OH *Ni thm: MN l
on vung gc chung ca a v b
NMHO b'baLu cn thit:1/ tnh khong cch t M n mp( ) ta c th lm nh
sau : + Tm mt ng thng a qua M m a // mp( ) + Chn mt im N trn a
(thch hp vi gi thit bi ton) , tnh khong cch t N n mp( ) + Khi ; [ ]
[ ] [ ] ; ( ) ; ( ) ; ( ) M mp N mp a mpd d d 2/ tnh khong cch t M
n mp( ) ta c th lm nh sau : + Tm mt ng thng a qua M m a ct mp( ) ti
I + Chn mt im O trn a (thch hp vi gi thit bi ton) , tnh khong cch t
O n mp( ) 17Trng THPT Ni Thnh + Khi ; tnh t s:IOkIM , suy ra c
:''OOkMM '1' OO MMk
[ ] [ ] ; ( ) ; ( )1.M mp O mpd dk M'O'IMOII/ Bi tp:BI1: Cho hnh
chp u S.ABC c cnh y bng a v cnh bn bng a2, ng cao l SO.Gi M v N ln
lt l trung im ca AB v BCa/Chng minh rng (SBC) (SAN) v tnh di SOb/
Tnh khong cch t O n (SBC)c/Tnh khong cch gia 2 ng thng AB v SCd/Tnh
khong cch t M n (SAN)e/Tnh khong cch gia 2 ng thng MC v
SAxPJKIHONMECBAS GI a/Chn ng BC chng minh vung gc vi (SAN) suy ra
(SBC) vung gc vi (SAN)*Tnh SO : Xt tam gic vung SOC ti O v lu : tam
gic ABC u nn ta c3 2 3;2 3 3 3aMC a OC MC a b/Ta chia lm 3 bc cho d
hiu: + Chn mp(SAN) cha O , ta c:(SBC) (SAN) (chng minh trn) +Ta
c:(SBC)(SAN) =SN + Dng OH vung gc vi SN ti H OH (SBC)OH l khong cch
t O n (SBC)Xt tam gic vung SON ti O. c OH l ng cao 2 2 21 1 1OS OH
ON + ..c/Chng minh c AB SC ,dng MK vung gc vi SC ti K, suy ra MK l
on vung gc chung+ Xt tam gic SMC c 2 ng cao: SO v MK , suy
ra:MK.SC=SO.MC MK = ?d/ + Chn mp(ABC)cha M, ta c: (ABC) (SAN) ( vSO
(ABC)18Trng THPT Ni Thnh+(ABC)(SAN) =AN+Dng: MI AN ti I (MI // BC),
suy ra:MI (SAN) ..( Nh: MI=2 4BN a )e/ * Dng Ax//MC (khi :Ax nm
trong (ABC) v Ax AB,gi s Ax ct BC ti E)Suy ra: MC //(SAE) [ ] [ ] [
]; ;( ) ;( ) MC SA MC SAE O SAEd d d (im O rt quan trng)*DngOJ AE
ti J, d dng chng minh c (SOJ) (SAE)( vAESO; AEOJ) +Chn (SOJ) cha O
v vung gc vi (SAE)+(SOJ)(SAE) = SJ +Dng OP vung gc vi SJ ti P , suy
ra :OP(SAE) OP l khong cch t O n (SAE)- TnhOP? Xt tam gic vung SOJ
ti O c OP l ng cao2 2 21 1 1OS OP OJ +Bi2:Cho hnh chp S.ABCD ; y
ABCD l hnh thang vung ti A v B,c: AB=BC=a;AD = 2a; SA= a .E l trung
im ca y ln AD; SA vung gc vi mt ya/Chng minhBESC v (SAB) (SBC)b/Tnh
khong cch gia 2 ng thng : BC v SD , AC vi SDc/Tnh khong cch t O n
(SCD) . Tnh khong cch t D n (SCE)HOJQKxPE DC BSAGi : +AQ chnh l
khong cch gia AC v SD+DP chnh l khong cch t D n (SCE); OH chnh l
khong cch t O n (SCE)Bi3:Cho hnh chp S.ABCD ; y ABCD l hnh thoi cnh
a tm O. Mt phng (SAB) vung gc vi mt phng (ABCD). Tam gic SAB cn ti
S, H l trung im ca AB v SH=a, gc BAD = 600.a/Tnh gc gia hai mt phng
(SCD) v (ABCD).b/Tnh khong cch t H n (SCD),tnh khong cch t O n
(SCD),c/Tnh khong cch gia 2 ng thngBC v SD .d/Tnh gia ng thng SO v
(SAB).Bi4:Cho hnh chp S.ABCD , y ABCD l hnh thang c y ln AD=2a, y b
BC=a,AB=a, gc BAD bng 1200. SA vung gc vi mt y v3 SA a . Gi H v K
ln lt l trung im ca AB v AD.a/ Chng minh BK vung gc vi SC, tnh
khong cch gia BK v SC.b/ Tnh khong cch t A n (SCD)c/ Tnh gc giang
thng SC v (SAB).d/ Tnh gc gia hai ng thng AD v SC.Bi 5: Cho hnh chp
S.ABCD c y ABCD l hnh vung cnh a , tm O . SA vunggc vi mt y v SA= a
6 . Gi M l hnh chiu vung gc ca A ln SB . a/Chng minh rng CB(SAB) v
AMSC .19Trng THPT Ni Thnh b/Tnh gc gia ng thng SB v mt phng (SAC).
c/Gi G l trng tm ca tam gic SCD .Tnh gc gia hai ng thng AG v BD D.
Bi ton v gc gia hai ng thng trong khng gianI/ Phng php gii ton:Khi
cn tnh gc gia 2 ng thng cho nhau a v b trong khng gian ta c 2 cch
sau:Cch 1:Dng gcta phi tm 1 ngthng trung gian l c song song vi a v
c ct b. Khi gc to bi a v b cng chnh l gcto 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. Cch 2: Gi l s o ca gc hp bi a v b Gib a,ln lt l cc vect ch
phng ca a v b ta c b ab ab a..) , cos( cos .Ch :Khi tnh gc gia 2 ng
thng thng gp cc cng thc sau:1/nh l hm s csinA bc c b a cos . 22 2 2
+ (a,b,c l cc cnh i ca gc A,B,C) V bca c bA2cos2 2 +2/Cng thc tnh
di ng trung tuyn4 2 22 2 22a c bma + MT S BI TON LIN QUAN N GC:BI
1: Cho hnh chp S.ABCD c y ABCD l hnh ch nht tm O, SA vung gc vi mt
y .Bit3 SA a , BC=a v tam gic OBC u.a/ Gi M l hnh chiu vung gc ca A
ln SB. Chng minh rng AM vung gc vi SC.b/ Tnh gc gia SB v mt phng
(SAC).c/ Gi G l trng tm ca tam gic ACD. Tnh gc gia MG v SC.cIIaa
33aa 33aaaaaaaaLLKK GGOOHHMMSS\\ \\\ \DDCCBBAA\\ \//HNG DNb/ ( )(
), ( ) , SB SAC SB SI BSI c/Ta c: ( ).cos ,.MG SCMG SCMG SCuuuur
uuuruuuur uuur.20Trng THPT Ni Thnh2 222 2 2 2 23 3 2 232 6 3 18a a
a aMG MK GK MH HK GK _ _ _ + + + + + , , ,uuuur( )( )222 23 2 7 SC
SA AC a a a + + uuur( ) ( )5. . . . . . . .6MG SC AG AM SC AG SC AG
SA AC AG AC AG AC AL AC AO AC + uuuur uuur uuur uuuuruuur uuur uuur
uuuruur uuur uuur uuur uuur uuur225 5 5. .412 12 3aAC AC a ( )(
)2055 23cos , , 5623 161. 718aMG SC MG SCaa BI 2:Cho hnh chp S.ABCD
c y ABCD l hnh vung cnh a , tm O . SA vunggc vi mt y v SA= a 6 . Gi
M l hnh chiu vung gc ca A ln SB . a/Chng minh rng AMSC . b/Tnh gc
gia ng thng SB v mt phng (SAC). c/Gi G l trng tm ca tam gic SCD
.Tnh gc gia hai ng thng AG v BD . HNG DN GIO NDCBMASc/Gi I l trng
tm caACD GI (ABCD)+Ta c :( ) ( )os , os , c AG BD c AG BD uuur uuur
..AG BDAG BDuuur uuuruuur uuur
-Trong tam gic vung AGI ti I ,c :
22 2 2 223AG GI IA GI AN _ + + ,( )2 2 2 2 2 24 4 119 9 9GI AN
GI AD DN a + + + 119AG a +( ). . . . AG BD AI IG BD AI BD IG BD +
+uuur uuur uur uur uuur uur uuur uuuruuur. . . AI BD OI BD OI BD
uur uuur uur uuur 2.3aBDOI Suy ra:( )1os ,22c AG BD.21Trng THPT Ni
Thnh* Nhn xt:( )21 1 1 1 1 2. . . . . . . 23 3 3 3 3 2 3a aAG BD AS
AC AD BD AD BD AO BD AO BD a + + uuur uuur uuur uuur uuur uuur
uuuuruuur uuuuruu urBI 3: H 2008 BCho hnh lng tr ABC.A'B'C' c cnh
bn bng 2a. Tam gic ABC vung ti A, AB=2a,AC a 3 . Hnh chiu vung gc
ca A ln (ABC) trng vi trung im M ca BC.a/ Tnh th tch khi lng tr
ABC.A'B'C' .b/ Tnh gc gia hai ng thng AA v BC.
HNG DNHaaNNaa 33aa2aaaaaaHHddMMC''B''A''CCBBAb/Cch1: Qua A dng
d//BC.Suy ra ( )( )', ' ' ', ' AA B C AA d A AH Cch2: ( )( ) (
)'.os ', ' ' os ', os ',' .AA BCc AA B C c AA BC c AA BCAA BC uuur
uuuruuur uuuruuur uuur( )( )( )02' ' 212 os ', ' ' ', ' ' 754'. ' .
.AA AA aBC BC a c AA B C AA B CAA BC AM MA BC AM BC a ' +
uuuruuuruuur uuur uuuur uuuuruuur uuuur uuurII/ Bi tp:1) Cho lng tr
ABCABC c di cnh bn bng 2a , y ABC l tam gic vung ti A. AB = a ,AC =
a 3 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.S: 41cos 2)
Cho hnh chp SABCD c y ABCD l hnh vung cnh 2a , SA = a, SB = a 3
mp(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.22Trng THPT Ni ThnhS: V = 333a,55cos E. Bi ton v mt cu ngoi tip
hnh chp v lng tr:I/ Kin thc c bn: 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
SA1A2..An th tm I cch u cc nhS; A1; A2.....An- V vy tm I thuc trc
ng trn y l ng thng qua tm vng trn ngoi tip y v vung gc vi y
A1A2...An (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 A1; A2.....An nn I thuc mt phng trung trc ca
SAi y 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.II/ Bi tp:1) Cho hnh chp S.ABCD c y ABCD l hnh thang vung ti
A v B, AB = BC = a; AD = 2a .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:+ V = 63a+ 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
tip y CDE. Gi l ng thng qua I l trung im ca CD v song song vi SA.Gi
K l trung im ca AB th KN //AM. KN v ng phng suy ra KN = O l im cn
tm. R = 211 a2) Cho hnh chp SABCD c y ABCD l hnh ch nht cnh AB = a;
AD = a2 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.HD:+ V = 33a+ Gi E, K ln 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, P th 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 tamgic AHC.Giao im
I = Ny Ex l tm mt cu ngoi tip hnh chp SAHC. R =a32313) Cho t din
ABCD c ABC l tam gic u cnh a, DA = DB = 3a , 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
ABCEHD:23Trng THPT Ni Thnh+ 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. 31cos CID+
Chng minh tam gic ACE vung ti A (AD l ng cao v CD.DE = AD2 = 32a)
.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. V = 863a 4) Cho hnh chp
SABCD c y ABCD l hnh vung cnh bng a v ng cao l SH.vi H tha mn HM HN
3 trong M, N l trung im AB, CD. Mt phng (SAB) to vi y ABCD gc 600.
Tnh khong cch t N n mt phng (SAC) v xc nh th tch khi cu ngoi tip
hnh chp SABCD.HD:+ VSNAC = 1421)) ( , (483.313aSAC N d a dt SHNAC +
Trc ng trn y l ng thng d qua O v //SH d(SMN) . V tam gic SAB vung
cn ti S nn trc d ca mp(SAB) qua M v vung gc vi SAB. Theo trn ta c
(SAB) vung gc vi (SMH) nn k HE vung gc vi SM th HE(SAB) nn (d)
//HE. Ta c d 'd = I l tm mt cu ngoi tip hnh chp SABCD. R = IA = 621
a, V = 5421 73a E. Gii bi ton hnh khng gian bng phng php ta :I/
Phng php gii ton:Vn quan trng nht trong vic gii bi ton hnh khng
gian bng phng php ta l thit lp h ta cho ph hp. Sau y chng ti xin
gii thiu mt s phng php thit lp h ta .1/ Thit lp h ta i vi tam
din:Vi gc tam din Oabc vic ta ha thng c thc hin kh n gin, c bit
vi:+ Tam din vung th h trc ta vung gc c thit lp ngay trn tam din .+
Tam din c mt gc phng vung, khi ta thit lp mt mt ca h trc ta cha gc
phng .2/ Thit lp h ta cho hnh chp:Vi hnh chp, vic ta ha thng c thc
hin da trn c tnh hnh hc ca chng. Ta c cc trng hp thng gp sau:* Hnh
chp u th h ta c thit lp da trn gc O trng vi tm ca y v trc Oz trng
vi ng cao ca hnh chp. C th:* Hnh chp c mt cnh bn (SA) vung gc vi y
th ta thng chn trc Oz l cnh bn vung gc vi y (SA), gc ta trng vi chn
ng vung gc (A).Trong cc trng hp khc ta da vo ng cao ca hnh chp v
tnh cht a gic y chn h ta ph hp.3/ Thit lp h trc ta cho hnh hp ch
nht:Vi hnh hp ch nht th vic thit lp h ta kh n gin, thng c hai cch:+
Chn mt nh lm gc ta v ba trc trng vi ba cnh ca hnh hp.+ Chn tm ca y
lm gc ta v ba trc song song vi ba cnh ca hnh hp.4/ Thit lp h ta cho
hnh lng tr:+ Vi lng tr ng th ta chn trc Oz thng ng, gc ta l mt nh
no ca y hoc tm ca y. Cc trc Oy, Ox th da vo tnh cht ca a gic y m
chn cho ph hp.+ Vi lng tr nghing, ta da trn ng cao v tnh cht ca y
chn h ta cho thch hp.Ngoi cc trng hp trn, trong cc trng hp khc ta
da vo quan h song song, vung gc v cc tnh cht ca ng cao, y,... thit
lp h ta cho thch hp.24Trng THPT Ni ThnhII. Cc dng bi tp:* Phng php
chung: Ta thc hin theo hai bc:+ Thit lp h trc ta thch hp, t suy ra
ta ca cc im cn thit.+ Thit lp biu thc gii tch cho cc gi tr cn xc
nh.* V d:1) Cho gc tam din Oxyz, trn Ox, Oy, Oz ly cc im A, B, C
sao cho OA = a, OB = b, OC = c. Trong t din OABC v ni tip mt hnh lp
phng sao cho mt nh trng vi O cn nh i din thuc mt phng (ABC). Tnh di
cnh ca hnh lp phng.HD:+ Chn h ta Oxyz vi A thuc Ox, B thuc Oy, C
thuc Oz. Ta c: A(a; 0; 0), B(0; b; 0), C(0; 0; c).+ mp(ABC):1 +
+czbyax+ Gi t l cnh ca hnh lp phng v A' l nh i din vi O, khi A'(t;
t; t).+ A' thuc (ABC) suy ra t = ca bc ababc+ +2) Trong mt phng (P)
cho tam gic u ABC cnh bng a. Dng on SA = a vung gc vi mp(P). Tnh
tan ca gc nhn gia hai cnh AB v SC.HD: Chn h ta nh hnh v:Ta c: A(0;
0; 0), B
,_
0 ;23;2a aC(a; 0; 0), S(0; 0; a)S: 7 tan42..cos SC ABSC AB3) Cho
hnh lp phng ABCD.A1B1C1D1 cnh bng a. Gi M, N theo th t l trung im
cc cnh AD v CD. Ly P thuc BB1sao cho BP = 3BP1. Tnh din tch thit
din do (MNP) ct hnh lp phng.HD:+ Chn h ta Axyz vi B thuc Ax, D thuc
Ay v A1 thuc Az. Ta cA(0; 0; 0), B(a; 0; 0), C(a; a; 0), D(0; a;
0), A1(0; 0; a), B1(a; 0; a), C1(a; a; a), D1(0; a; a), M ,_
0 ;2; 0a, N ,_
0 ; ;2aa, P ,_
43; 0 ;aa.+ Gi l gc to bi (MNP) v (ABCD), ta c: 62cos + Gi S1 v
S ln lt l din tch thit din v hnh chiu ca n ln mt phng (ABCD). Ta
c:( )166 7..cos1cos cos21aS SS SSDMN ABCDABCMN 4) Cho hnh lng tr
tam gic u ABC.A1B1C1c cc cnh bng a. Tnh gc gi hai mt phng (ABC1) v
(BCA1).+ Chn h ta nh hnh v. Ta c:25ABCSzxyTrng THPT Ni ThnhA(0; 0;
0), B(a; 0; 0), C
,_
0 ;23;2a aA1(0; 0; a), B1(a; 0; a), C1
,_
aa a;23;2+ Gi l gc gia (ABC1) v (BCA1). Ta c:71.cos2 12 1 n nn
n* V d:1) Cho tam gic vung cn ABC c AB = AC = a, M l trung im cnh
BC. Trn cc na ng thng AA1, MM1 vung gc vi mp(ABC) v cng mt pha, ly
tng ng cc im N, I sao cho 2 MI = NA = a. Gi H l chn ng vung gc h t
A xung NM. Chng minh rngNI AH HD:+ Chn h trc ta Axyz vi B thuc Ax,
C thuc Ay v N thuc Az. Ta c:A(0; 0; 0), B(a; 0; 0), C(0; a; 0), M
,_
0 ;2;2a a, N(0; 0; a), I ,_
2;2;2a a a, H ,_
2; 0 ;2a a+ Tnh c NI AH NI AH 0 . 2) Cho hnh chp S.ABCD c y ABCD
l hnh vung cnh bng a, SA vung gc vi y. Gi M, N l hai im theo th t
thuc BC, DC sao cho BM = 2a, DN = 43a. Chng minh rng hai mt phng
(SAM) v (SMN) vung gc vi nhau.HD:+ Chn h trc ta Axyz vi B thuc Ax,
D thuc Ay, S thuc Az, khi :A(0; 0; 0), B(a; 0; 0), C(a; a; 0), D(0;
a; 0), S(0; 0; a), M ,_
0 ;2;aa, N ,_
0 ; ;43aa+ Tnh c AM MN AM MN 0 ..+ Mt khc SAMN, suy ra MN ) ( )
( ) ( SAM SMN SAM 3) Cho hnh chp S.ABC c y ABC l tam gic u cnh a,
SA = SB = SC, khong cch t S n mp(ABC) bng h. Tm iu kin ca a v h hai
mt phng (SAB) v (SAC) vung gc vi nhau.HD:+ Chn h trc ta Oxyz vi O l
trng tm tam gic ABC, BC song song Ox, A thuc Oy, S thuc Oz, khi
:A
,_
0 ;33; 0a, B
,_
0 ;63;2a a, C
,_
0 ;63;2a a, S(0; 0; h)+6 0 ' . ) ( ) ( h a n n SAC SAB 4) Cho
hnh lng tr ABC. A1B1C1 c y l tam gic u cnh bng a, AA1 = h v vung gc
vi mp(ABC). Bit rng khong cch gia A1B1 v BC1 bng d. Chng minh rng:
a = ) ( 322 2d hdhHD:+ Chn h trc ta Axyz vi B thuc Ax, A1 thuc Az,
khi :26BACA1zxyC1B1Trng THPT Ni ThnhA(0; 0; 0), B(a; 0; 0), C
,_
0 ;23;2a aA1(0; 0; h), B1(a; 0; h), C1
,_
ha a;23;2+ Gib a,theo th t l VTCP ca A1B1 v BC1. Ta c:a cng
phng) 0 ; 0 ; 1 ( ) 0 ; 0 ; (1 1 a a B A , bcng phng
,_
ha aBC ;23;21 ( ) h a a b 2 ; 3 ; .Khi : [ ][ ] ) ( 323 43,,2 2
2 21d hdhaa hahb aBB b ad + 5) Cho gc tam din Oxyz, trn Ox, Oy, Oz
ly cc im A, B, C.a/ Tnh khong cch t O n mp(ABC) theo OA = a, OB =
b, OC = c.b/ Gi s A c dnh cn B, C thay i nhng lun tha mn OA = OB +
OC. Hy xc nh v tr ca B v C sao cho th tch t din OABC l ln nht.HD:+
Chn h trc Oxyz, khi : A(a; 0; 0), B(0; b; 0), C(0; 0; c)+ PT
(ABC):1 + +czbyax, d(O, (ABC)) = 2 2 2 2 2 2b a a c c babc+ ++
VOABC = 24 2.616132a c ba abc
,_
+(Theo bt csi)Do Max(VOABC) = 243a t c khi b = c = 2a6) Cho t
din SABC c SC = CA = AB = 2 a, SC vung gc vi mp(ABC), tam gic ABC
vung ti A, cc im M thuc SA, N thuc BC sao cho AM = CN = a (0 < t
< 2a).a/ Tnh di on MN. Tm gi tr ca t on MN ngn nht.b/ Khi on MN
ngn nht, chng minh MN l ng vung gc chung ca BC v SA.HD:+ Chn h trc
ta Cxxyz vi B thuc Cx, S thuc Cz, khi :A(a; a; 0), B(2a; 0; 0),
C(0; 0; 0), S(0; 0; 2 a).+ Phng trnh SA: [ ] a uu zu a yu a x; 0
,2' suy ra M( ) 2 ; ; u u a u a + V AM = t suy ra u = 2t. Khi M
,_
22;21;21 ta aa/ MN2 = 3236322 4 322 2ataMinMNaa at t + b/ Khi on
MN ngn nht th ,_
,_
0 ; 0 ;32,32;32;32 aNa a aM. Lc '0 .0 .BC MNSA MN, tc l Mn l on
vung gc chung ca SA v BC.7) Cho hnh lp phng ABCD.A1B1C1D1, cnh bng
a. Trn cnh AA1 ko di v pha A1 ly im M v trn cnh BC ko di v pha C ly
im N sao cho MN ct cnh C1D1. Tnh gi tr nh nht ca di on MN.27Trng
THPT Ni ThnhHD: + Chn h trc ta Axyz trong B thuc Ax, D thuc Ay v A1
thuc Az, khi : A(0; 0; 0), B(a; 0; 0), A1(0; 0; a), C1(a; a; a).+
AA1: ( ) ( ) ) 0 ; ; ( ; ,0: ); ; 0 ; 0 ( ; , 00v a N a vzv ya xBC
u M a uu zyx + ' + '+ V MN ct C1D1 nn MD1 // NC1 a vavuau av aa
.Khi MN = u + v - a = a va av v+ 2 2, suy ra MinMN = 3a khi v = 2aa
u 2 v khi MN i qua trung im I ca C1D1 (Dng phng php o hm)F. Bi ton
hnh khng gian trong d thi i hc, Cao ng cc nm va qua:Bi 1)H 2002
K.ACho hnh chp tam gic u S.ABC nh S, c di cnh y bng a. Gi M,N ln lt
l cc trung imca cc cnh SB v SC . Tnh theo a din tch tam gic
Agiacsbieets rng mt phng (AMNphawngrvuoong gc vi mt phng (SBC).Bi
2)H 2002 K.BCho hnh lp phngABCDA1B1C1D1c cnh bng a.a) Tnh theo a
khong cch gia hai ng thng A1B v B1D.b) Gi M,N,P ln lt l trung im ca
cc cnh BB1, CD, A1D1. Tnh gc gia hai ng thng MP, C1N.Bi 3)H 2002
K.DCho hnh t din ABCD c cnh ADvung gc vi mt phng (ABC) ; AC = AD =
4cm; AB = 3cm; BC = 5cm. Tnhkhong cch t im A n mt phng (BCD).Bi 4)H
2003 K.ACho hnh lp phng ABCD.ABCD. Tnhs o gc phng nh din
[B,AC,D].Bi 5) H 2003 K.BCho hnh lng tr ng ABCD.ABCD c y ABCD l mt
hnh thoi cnh a, gc BAD= 600. Gi M l trung im ca cnh AA v N l trung
im ca cnh CC. Chng minh rng 4 im B, M, D, N cng thuc mt mt phng. Hy
tnh di cnh AA theo a t gicBMDN l hnh vung.Bi 6)H 2003 K.DCho hai mt
phng (P) v (Q) vung gc vi nhau, c giao tuyn l ng thng. Trn giao
tuyn ly hai im A, B vi AB = a . Trong mt phng (P)ly im C , trong mt
phng (Q)ly imD sao cho AC, BD vung gc vi nhau v AC = BD = AB.Tnh bn
knh mt cu ngoi tip t din ABCDv tnh khong cch t A n mt phng (BCD)
theo a.Bi 7)H 2004 K.BCho hnh chp t gic u S.ABCD c cnh y bng a,gc
gia cnh bn v mt y bng (00 < < 900). Tnh tangca gc gia hai mt
phng (SAB) v (ABCD) theo . Tnh th tch khi chp S.ABCD theo a v .Bi
8)H 2006 ACho 2 hnh tr c y ln lt l 2 ng trn (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 ly im
Bsao cho AB=2a.Tnh th tch khi t din OOAB /S1233.aVBMDN SBi 9)H 2007
A28Trng THPT Ni ThnhCho hnh chp S.ABCD c y 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 SB,BC,CD .Chng minh AM vung gc BP v tnh th tch ca t din
CMNP /S9633.aVBMDN SBi 10)H 2007 BCho hnh chp t gic u SABCD c y
ABCD l hnh vung cnh a Gi 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 BD v tnh
theo a khong cch gia 2 ng thng MN,AC S:42) ; (aAC MN d Bi 11)H 2007
Khi DCho hnh chp S.ABCDc y l hnh thang Gc DAB=ABC=900
,BA=BC=a,AD=2a.cnh bn SA vung gc vi y v SA=a2.Gi H l hnh chiu vung
gc ca A trn SB.Chng minh tam gic SCD vung v tnh theo a khong cch t
H n mp (SCD) S:3) ( ; (aSCD H d Bi 12)H 2008 ACho lng tr ABC.ABC c
di cnh bn bng 2a, y ABC l tam gic vung ti A,AB=a,AC=3a v hnh chiu
vung gc ca nh A trn mt phng ABC l trung im cnh BC.Tnh theo a th tch
khi chop AABC v tnh cosin ca gc gia 2 ng thng AA,BCS:33'.aVABC A
,41cos Bi 13)H 2008 B Cho hnh chp S.ABCD c y ABCD l hnh vung cnh
2a, SA=a, SB =a 3v mt phng (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 ca khi chp S.BMDN
v tnh cosin ca gc gia hai ng thng SM, DN.S: 333.aVBMDN S55cos Bi
14)H 2008 DCho lng tr ng ABC.A'B'C' c y ABC l tam gic vung, AB = BC
= a, cnh bn AA' a 2 . Gi M l trung im ca cnh BC. Tnh theo a th tch
ca khi lng tr ABC.A'B'C' v khong cch gia hai ng thng AM, B'C.
S:223' ' ' .aVC B A ABC77) ' ; (aC B AM d Bi 15)H 2009 ACho hnh chp
S.ABCD c y ABCD l hnh thang vung ti A v D; AB = AD = 2a, CD = a; gc
gia hai mt phng (SBC) v (ABCD) bng 60. Gi I l trung im ca cnh AD.
Bit hai mt phng (SBI) v (SCI) cng vung gc vi mt phng (ABCD). Tnh th
tch khi chp S.ABCD theo a.S:515 33.aVABCD SBi 16)H 2009 B Cho hnh
lng trtam gic ABC.ABC c BB = a, gc gia ng thng BB v mt phng (ABC)
bng 600; tam gic ABC vung ti C v 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 S:20893'aVABC A29Trng THPT Ni ThnhBi 17) H 2010
ACho hnh chp S.ABCD c y ABCD l hnh vung cnh a. Gi M v N ln lt l
trung im ca cc cnh AB v AD; H l giao im ca CN vi DM. Bit SH vung gc
vi mt phng (ABCD) v SH = a 3 . Tnh th tch khi chp S.CDNM v tnh
khong cch gia hai ng thng DM v SC theo a.S:243 53.aVCDNM S193 2) ;
(aSC DM d Bi 18) H 2010 B Cho hnh lng tr tam gic u ABC.ABC c AB =
a, gc gia hai mt phng (ABC) v (ABC)bng 600. Gi G l trng tm tam gic
. Tnh th tch khi lng tr cho v tnh bn knh mt cu ngoi tip t din GABC
theo a.S: 83 33' ' ' .aVC B A ABC127aR Bi 19) H 2010 D Cho hi nh
cho p S.ABCD coay ABCD lahi nh vung ca nh a, canh bn SA = a; hi nh
chiu vung goc cua inh S trnmt ph ng (ABCD) laim H thuc oan AC,
4ACAH . Go i CM la ng cao cua tam giac SAC. Ch ng minh M latrung im
cua SA va ti nh thtich kh i tdin SMBC theo a. S: 48143.aVBCM SBi
20) H 2011 A Cho hnh chp S.ABC c y ABC l tam gic vung cn ti B, AB =
BC = 2a; hai mt phng (SAB) v (SAC) cng vung gc vi mt phng (ABC). Gi
M l trung im ca AB; mt phng SM v song song vi BC, ct AC ti N. Bit
gc gia hai mt phng (SBC) v (ABC) bng 60o. Tnh th tch khi chp S.BCNM
v khong cch gia hai ng thng AB v SN theo a.S:33.a VBCNM S1339 2) ;
(aSN AB d Bi 21) H 2011 BCho lng tr ABCD.A1B1C1D1c y ABCD l hnh ch
nht. AB = a, AD =3 a . Hnh chiu vung gc ca im A1 trn mt phng (ABCD)
trng vi giao im AC v BD. Gc gia hai mt phng (ADD1A1) v (ABCD) bng
600. Tnh th tch khi lng tr cho v khong cch t im B1 n mt phng (A1BD)
theo a. S: 233.1 1 1aVD C B A ABCD23) A ( ; (1 1aBD mp B d Bi 22) H
2011 DCho hnh chp S.ABCc y ABCl tam gic vung tiB, BA = 3a, BC =
4a;mt phng (SBC) vung gc vi mt phng (ABC). Bit SB =2 3 avSBC= 300 .
Tnh th tch khi chp S.ABC v khong cch t im B n mt phng (SAC) theo
a.S: 3.3 2 a VABC S76)) ( ; (aSAC mp B d BI TP B SUNGBI: Cho hnh tr
c chiu cao OO' = 8 , bn knh y bng 5. M v N l hai im ln lt nm trn 2
ng trn y sao cho MN cch trc OO' mt khong cch bng 3. Tnh th tch khi
t din OO'MN. 30Trng THPT Ni ThnhBI:Cho hnh chp S.ABCD c y ABCD l
hnh vung cnh a , tm O.Bit SA vunggc vi mt y v SB hp vi mt y mt gc
bng 600 . Gi G l trng tm ca tam gic SCD. Tnh th tch khi a din
S.ABCG v tnh gc gia hai ng thng AG v BDBI:Cho hnh chp S.ABCD c y
ABCD l hnh vung cnh a tm O. Gi H l trung im ca AB, SH vung gc vi
(ABCD). SC to vi y mt gc bng 600. Gi M l trung im ca SC.1)Tnh th
tch khi t din MCHD v din tch tam gic MHD2)Tnh khong cch gia hai ng
thng BC v HM.BI:ChohnhlngtrngABCD.A'B'C'D' cyABCDlhnhthoi
cnhavgcnhn 060 BAD . Bit AB' BD' .Tnh th tch khi lng tr
ABCD.A'B'C'D' theo a.BI: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 ABCD. Gi
M,N,P ln lt l trung im ca cc cnh SB,BC,CD . Tnh th tch khi t din
CMNPBI:Trong khng gian, cho tam gic ABC vung cn ti C, cnh huyn
AB=2a. Trn ng thng vung gc vi (ABC) ti A ly im S sao cho(SBC) to vi
(ABC) mt gc bng 600. Tnh din tch mt cu ngoi tip t din SABCBI:Cho
hnh chp SABCD c y ABCD l hnh thoi cnh a , gc ABC bng 600. SO^(ABCD)
ti O ( vi O l giao im ca hai ng cho ca hnh thoi) v SO =a 32.Gi M l
trung im ca AD.Mt phng ( )acha BM v song song vi SA, ct SC ti K.
Tnh th tch ca khi a din K.BCDMBI:Cho hnh chp t gic u SABCD c cc cnh
bn to vi mt y mt gc bng 600 . Mt phng(P) cha AC v vung gc vi (SAD)
. Tnh t s ca th tch hai phn ca hnh chp chia bi mp(P).BI:Cho lng tr
ng ABC.A'B'C' c hai y l hai tam gic vung ti B v B'. Bit AB=a,BC=a v
din tch tam gic B'AC bng 22a3.Tnh th tch khi lng tr ABC.A'B'C'
BI:Cho hnh hp ng ABCD.A'B'C'D' c y l hnh thoi cnh a, cnh bn bng3 a
, 060 BAD . Gi M l trung im ca BB' .Tnh th tch khi t din MD'AC.
BI:Cho hnh tr c hai tm y l O v O'. Bn knh y bng3 a , chiu cao bng
4a . Ly hai im M v N ln lt trn haing trn hai y sao cho MN =5a .
Chng minh rng MN v OO' cho nhau v tnh th tch khi t din MOO'N.
BI:Cho hnh hp ng ABCD.A'B'C'D' c AB=AD=a,3AA'2av 060 BAD .Gi M,N ln
lt l trung im ca A'D' v A'B'.Tnh th tch khi chp A.BDMN theo
a.BI:Cho hnh chp SABCD c y ABCD l hnh thoi cnh a , gc ABC bng 600.
SO^(ABCD) ti O ( vi O l giao im ca hai ng cho ca hnh thoi) ,SO =a
32.Gi M l trung im ca AD.Mt phng ( )acha BM v song song vi SA, ct
SC ti K.Tnh th tch ca khi chp K.BCDM.BI:Cho mt hnh tr trn xoay v
hnh vung ABCD cnh a c hai nh lin tip A, B nm trn ng trn y th nht ca
hnh tr, hai nh cn li nm trn ng trn y th hai ca hnh tr. Mt phng
(ABCD) to vi y hnh tr gc 450. Tnh din tch xung quanh v th tch ca
hnh tr.BI:Cho hnh nn c nh S, y l ng trn tm O, SA v SB l hai ng
sinh, bit SO = 3, khong cch t O n mt phng SAB bng 1, din tch tam
gic SAB bng 18. Tnh th tch v din tch xung quanh ca hnh nn
cho.--------------------------------------------------------------------------------------------------------------------
HT 31Trng THPT Ni Thnh32
