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8/10/2019 PHN DNG V PHNG PHP GII TON HNH HC 12 - NGUYN VN PHC
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ThS Nguyen Vn Phco.03e
PHN DNG V
PHNG PHP GII TONH N H H O C
1 2
. TM TT L THU YT
. PH N DNG V PHNG PHP GII TON
. TON TRC NGHIM - T LUN
. BI TP NGH
NH XUT BN I HC QUC GIA TP. H CH MINH
2008
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PHN DNG V PHNG PHP GII TON HNH HC 12Nguyn Vn Phc
NH XUT BN
I HC QUC GIA TP H CH MINHKhu ph 6, phng Linh Trang, qun Th c, TP HCM
T .-72 42 18 1,7 242 60 + (1421, 1422, .1423, 1425, 1426)
Fax: 7 242 194 - Em ail: [email protected]
* * *
Chu trch nhim xu t bn
TS HUNH B LN
Bin t p
THN TH HNG
Sa bn in
TRN VN THNG
Trnh by ba
n v/Ngi lin kt
Cty TNHH MTV Sch Vit
iK.Ol.T(T)BHQG HCM 08 130'2008/ CB/296*04 T.TK.474-08(T)
In 3.000 c u n , kh 16 X 24 cm ti Cng ty TNHH mt thnh vin in Ngi Lao ng, s 131 Cng Qunh, Q.l, TP.HCM - T: 8374604.
S ng k k hoch xut bn: 130-2008/CXB/296-04/HQGTPHCM Quyt nh xut bn s 294/Q-HQGTPHCM ngy 20/5/2008.
In xong v np lu chiu thng 5 nm 2008.
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Nhm gip cc enc sinh lp 12 c ti liu t hc mn ton, chng
ti bin son cun Phlng v phng ph p g i i ton Hnh hc 12.
Sch gm ba ch, Ni dung mi chng gm c:
A. Tm t t l thuy
B. Phn dng v png php gii ton
c. Ton trc nghi- T lun
D. p n.
Sau mi chng clig ti a ra mt s bi tp n chng di dng
ton trc nghim (c hiig dn gii) v ton t lun.
c bit phn tn trc nghim chng ti xp theo th t cc bi
hc cc em d dng tp.
Hy vng cun scy s gip cc em hc tp tin b, c th t rn
uyn hc tt mn to.
Chc cc em thnhng.
Cui cng chng t( mong nhn c nhng kin chn rhnh, xy
dng ca qu ng nghip, qu c gi v cc em hc sinh cun sch
ngy mt hon thin hn
Tc gi
ThS Ngun Vn Phic
------- ------------------------------------------:--------------------------------------------------------------- .
Sch Vit lununnnhnmi kirt nggp, ph b)nhv6chcaquc ti, xingi v ach:
PHNG XUT BNCTTNHH MTV SCH VIT
39U15A Hu nh Tn Ph t, P .T n Thun ng, Q.7 - TPHCM
T: 08.8720897 - Fax: 0S.8720897 - Email: [email protected]
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Chng I
KHI A DIN
1. KHI NIM V KHI A DIN
A . T M T T L TH IT Y T
I. KHI NIM V HNH A DIN V KHOI A DIN
1. Khi nim v hnh a din
Hnh a din (gi tt l .a din) l hnh c to bi mt s hu
hn cc min a gic tha mn hai tnh cht :
a) Hai min a gic phn bit ch c th khng giao nhau hoc ch
c mt nh chung hoc ch c mt cnh chung.b) Mi cnh ca min a gic no cng l cnh chung ca ng
hai min a gic.
V du:Hnh lp phng
ABCD.ABCD l mt
hnh a din. 'A
A, B, c,. .. gi l cc nh
AB, BC, CD,... gi l cc cnh
ABCD, AB BA,... gi l
cc mt. A
* Nhn xt
Hai min a gic ABCD v ABCD khng giao nhau.
Hai min a gic ABCD v ABBA c chung nhau cnh AB.
2. Khi nim v khi a din
* Khi a din l phn khng gian c gii hn bi mt hnh a
din, k c hnh a din .* Nhng im khng thuc khi a din c gi l im ngoi
ca khi a din. Tp hp cc im ngoi c gi l liin
ngoi ca khi a din.
* Nhng im thuc khi a din nhng khng thuc hnh a din
tng ng c gi l im trong ca khi a din. Tp hp cc
im trong c gi l min trong ca khi a din.
B"
111111
D
/ B '
D
5
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II. HAI A DIN BNG NHAU
1. Php di hnh trong khng gian
* Php bin hnh v php di hnh trong khng gian c nh ngha
tng t nh trong mt phng.
* Trong khng gian quy tc t tng ng mi im M vi im M
xc nh duy nht c gi l mtphp bn hnhtrong khng gian.Php bin hnh trong khng gian c goi l php d i hnh nu n
bo ton khong cch gia hai im ty .
V d: Cc php bin hnh trong khng gn sau y l nhng php
di hnh:
a) Php tnh tin theo vect V, l php bin hnh bin im M
thnh im Msao cho M M ' - V .
V
M --------------------M
b) Php i xng qua mt phng (P), l php bin hnh bin cc
im thuc (P) thnh chnh n, bin im M khng thuc (P)
ithnh im M sao cho (P) l mt phng trung trc ca M M \
Nu php i xng qua mt phng (P) bin hnh (H) thnh
chnh n th (P) c gi l mt phng i xngca (H)
c) Php i xng tm o, l php bin hnh bin im o thnhchnh n, bin im M khc o thnh im M sao cho o l
tming im ca MM\
Nu php i xng tm o bin hnh (H) thnh chnh n th
o c gi l tm i xngca (H).
6
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M
d) Php i xng qua ng thng l php bin hnh bin mi
im thuc A thnh chnh n, bin im M khng thuc A thnh
im M sao cho A l ng trung trc ca MM.
* Nhn xt
Thc hin lin tip cc php di hnh s c mt php di
hnh.
Php di hnh bin a din (H) thnh a din (H) v bin
nh, cnh, mt ca (H) thnh nh, cnh, mt tng ng ca
(H).
2. Hai hnh bng nhau
Hai hnh c gi l bng nhau nu c mt php di hnh bin
hnh ny thnh hnh kia.
c bit, hai a din c gi l bng nhau nu c mt php di
hnh bin a din ny thnh a din kia.
III. PHN CHIA V LP GHP CC KHOI A DIN
Ta ni khi a in (H) c phn chia thnh hai khi a din (Hi), (H2) nu tha mn hai tnh cht sau:
1. Hai khi a din (Hi), (H2) khng c im trong chung.
2. Hp ca hai khi a din (Hi), (H2) chnh l khi a din (H).
Khi ta cng ni hai khi a din (Hi), (H 2) c lp ghp li
thnh khi a din (H).
7
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B . G I I T O N
* VAN t 1
Phn chia v lp ghp cc khi a din
* Phng Php
Chn mt phng thch hp phn chia hoc lp ghp cc khdin.
Bi 1.1. Hy ch ra mt cch phn chia khi chp S.ABCD c y
a gic li thnh hai khi chp tam gic.
Hng dn gi i
Khi chp t gic S.ABCD cth c phn chia thnh hai
khi chp tam gic: S.ABC v
S.ACD.
^ Ch : C th phn chia
khi chp S.ABCD thnh
hai khi chp: S.ABD v
S.BCD. c
Bi 1.2. Hy ch ra mt cch phn chia khi thp S.ABCDE c
mt a gic li thnh cc khi chp tam gic.
Hng dn gi is
Khi chp ng gic
S.ABCDE c th cphn chia thnh ba khi
chop tam gic:
S.ABC;
S.ACD;
v S.ADE.c
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Bi 1.3. Hy ch ra mt cch phn chia khi lng tr ABC.ABC thnh
ba khi t din.
Khi lng tr
ABC.ABC c th c
chia thnh b
khi t din:
AABC;
BABC
v CABC'
is.VN t 2
Hng dn gi i
Chng minh hai a din b ng nhau
Phng Php
chng minh hai a din bng nhau ta chng minh c mt php
di hnh bin a din ny thnh a din kia .
Bi 1.4. Cho khi chp t gic u S.ABCD. Gi K l trung im ca s c . Mt mt phng (a ) i qua AK v song song vi BD ct SB ti .VI
v SD ti N.
Chng minh hai khi chp A.BCKM v A.DCKN bng nhuu.
B D / / (a )
Ta c:
Hng dn gi i
MN // BDB D c ( S B D )
(a) MN 1 (SAC)
Ngoi ra: OB = OD v OM = 0 N
nn B, D i xng nhau qua (SAC)
v M, N cng i xng nhau qua (SAC).
9
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t: p = (SAC), Ta c: .
p() = A
;(B) = D
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Bi 1,6. Cho khi chp t gie F.ABCD c y l hnh vung, cnh FC
vung gc vi y v c di bng AB. Chng minh rng dng ba
khi chp bng khi chp trn c th ghp li thnh khi lp phng.
H
D
Hng dn gi i
T khi chp cho ta dng khi
lp phng HEFG.ABCD.
Ta thy hai khi chp F.ABCD v
F.ABEH i xng nhau qua mt
phng (ABF) (hc sinh t chng
minh), hai khi chp F.ABCD v
F.AHGD i xng nhau qua mt
phng (ADF) (hc sinh t chng
minh).Do ba khi chp F.ABCD, F.ABEH v F.AHGD bng nhau.
Nh vy, c th phn chia khi lp phng HEFG.ABCD thnh ba
khi chp bng khi chp F.ABCD. Hay ni cch khc, c th dng ba
khi chp bng khi chp F.ABCD ghp li thnh mt khi lp
phng.
^ VN 3Chng minh mt s'tnh cht lin quan n cc nh,
cc cnh v cc mt ca khi a din
* Phng Php
Dng cc tnh cht trong nh ngha hnh a din.
c bit l tnh c h t : Mi cnh ca min a gic no cng l
cnh chung ca ng hai min a gic.
Bi 1.7. Chng minh rng mt khi a din c cc mt l nhng tam
gic th s cc mt ca n phi l s chn.
Hng dn gi i
Gi mv cln lt l s" mt v scnh ca khi a din (H).
11
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V mi mt ca (H) l mt tam gic nn m mt c 3mcnh. -
Mt khc, mi cnh ca (H) l cnh chung ca ng hai min a g
nn s cnh ca (H) l: c = .2
T suy ra m phi l s" chn.
Bi 1.8. Chng minh rng nu khi a din m mi nh l nh chu
ca 5 cnh th snh phi ] s chn.
Hng dn gi i
Gi dv c Hnh v s cnh ca khi a din (H).
V mi nh ca (H) l nh chung ca 5 cnh nn dnh c 5/ cnh
Trong cch tnh ny mi cnh c tnh ng 2 ln .5d
Do : 2c = 5d=> c =.2
T suy ra d phi l s" chn.
Bi 1.9. Gi d\ c; m ln lt l s" nh, s cnh v s" mt ca mt
din th:
) c > m b) c > d .
Hng dn gi i
a) V s" cnh ca mt mt ln hn hoc bng 3 nn s cnh ca m
ln hn hoc bng 3m.
Ngi ra mt cnh l cnh chung ca ng 2 mt nn s" cnh
m mt bng 2c.
Vy: 2 c > 3 m=> c > m .
b) V s cnh qua 1 nh ln hn hoc bng 3 nn s cnh qua cl
ln hn hoc bng 3d .
Trong cch tnh s cnh qua d nh, mi cnh c tnh ng 2
do scnh qua dnh l 2c.
Vy: 2c > d => c> d .
12
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c. TON TRC NGHIM
11. Cho khi chp S.ABCD c y ABCD l hnh bnh hnh.
Xt cc mnh :
(I) Khi chp S.ABCD c th phn chia thnh hai khi chp S.ABC
v S.ADC.(II) Khi chp S.ABCD c th phn chia thnh hai khi chp S.ABC
v S.ABD. Mnh no ng?
A. (I) ng, (II) sai B. (I) sai, (II) ng
c . C (I) v (II) u ng D. C (I) v (II) u sai.
2. Cho khi chp S.ABC c y ABCD l hnh bnh hnh tm o .
Xt cc mnh :
(I) Kh'i chp S.ABCD c th phn chia thnh bn khi chp S.AOB,S.BOC, S.COD v S.DOA.
(II) Khi chp S.ABCD c th phn chia thnh hai khi chp S.AOB
v S.AOBCD.
Mnh no ng?
A. (I) ng, (II) sai B. (I) sai, (II) ng
c . C (I) v (II) u ng D. C (I) v (II) u sai.
3. in vo ch trng trong mnh sau n tr thnh mnh nsi:S cnh ca khi a din lun ... s mt ca khi a din y.
A. bng B. nh hn
c . nh hn hoc bng D. ln hn.
4. in vo ch trng trong mnh sau n tr thnh mnh ng:
Scnh ca khi a din lun ... s"nh ca khi a din y.
A.b ng B.nh hn
c . nh hn hoc bng D. ln hn.5. Cho khi a din. Trong cc mnh sau, mnh no sai?
A. Mi nh l nh chung ca t nht ba cnh.
B. Mi nh l nh chung ca t nht ba mt.
c. Mi cnh l cnh chung ca t nht ba mt.
D. Mi mt c t nh't ba cnh.
13
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Trong cc mnh sau mnh no ng?
A. Snh v so' mt ca mt hnh a din lun lun bng nhau
B. Tn ti hnh a din c s" nh v s" mt bng nhau;
c . Tn ti mt hnh a din cscnh bng s nh;
D. Tn ti mt hnh a din cs cnh v mt bng nhau;
Trong cc mnh sau, mnh no ng?
S cc nh hoc cc mt ca b't k hnh a din no cng:
A. Ln hn hoc bng 4 B. Ln hn 4
c . Ln hn hoc bng 5 D. Ln hn 5.
Trong cc mnh sau, mnh no ng?
S cc cnh ca hnh a din lun lun:
A. Ln hn hoc bng 6 B. Ln hn 6
c. Ln hn 7 D. Ln hn hoc bng 8.
Xt cc mnh :
(I) Tn ti mt hnh a din c s nh bng smt.
(II) Tn ti mt hnh a din c s nh ln hn s" mt.
(HI) Tn ti mt hnh a din c s nh nh hn s" mt.
Mnh no ng ?
A. Ch (I) v (II) B. Ch (I) v (III)
c . Ch (l ) v (III) D. C (I ) , (II) v (III).
Xt cc mnh :
(I) Tn ti mt hnh a din sao cho: d - c + m schn.
(II) Tn ti mt hnh a din sao cho: d - c + m l s l.
Trong d, c, mln lt l s nh, s cnh v s mt ca hnh a din.
Mnh no ng ?
A. (I) ng, (II) sai B. (I) sai, (II) ng
c . C (I) v (II) u ng D. C (I) v (II) u sai.
P N
1A 2C 3D 4D 5C 6B 7A 8 9D 10C
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2. KHI A DIN Li V
KHI A DIN U
. TM TT L THUYT
I. KHI A DI N LI
Khi a .din (H) c gi l khi a din li nu on thng ni hai
im bt k ca (H) lun thuc (H). Khi hnh a din xc nh (H)
c gi l hnh a din li.
* Nhn xt:Cc khi lng tr tam gic, khi hp, khi t din l cc
khi a din li.
* nh l: Gi d, c, m theo th t l s nh, s cnh v s mt ca
mt khi a din li. Khi ta c mi lin h sau y gi l cng,thc -le:
\ d - c + m = 2
Ch :Nu (H) l mt khi a din bt k th s:
(H ) = d - c + m
c gi l c s -le (cn gi l c s) ca khi a din (H).
Ch Hy Lp 2 -c l khi.
I. KH A DIN U
1. h ngha: Khi a din u l khi a din li c hai tnh cht
sau y:
a) Cc mt l cc a gic u v c cng s' cnh.
b) Mi nh l nh chung ca cng mt scnh.
2. Ch
a) Ch c 5 loi khi a din u:- Khi t din u.
- Khi lp phng.
- Khi 8mt u.
- Khi 12 mt u.
- Khi 20 mt u.
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b) Khi a din u m mi mt l a gic u n cnh v
nh l nh chung ca p cnh c gi l khi a din u l
{;/>}
Nh vy:
- Khi t din u l loi {3; 3}.
- Khu lp phng l loi {4; 3}.
- Khi 8 mt u l loi 3; 4}.
- Khi 12 mt u l loi {5; 3}.
Khi 20 mt u l loi {3; 5}.
c) Hai khi a din u cng loi th ng dng vi nhau.
d) Vi khi a din u loi {n \ p } ta c h thc:
pd = 2c = nm
B . G I I T O N
* VN 1
Chng minh mt s tnh cht lin qiian n cc nh,
cc cnh v cc mt ca khi a din li
* Phng Php
- Dng cc tnh cht trong nh ngha hnh a din.
- i vi khi a din li c th s dng cng thc - le:
d - c+ m= 2
Trong d, c, m ln lt l s nh, s" cnh v s" mt ca m
khi a din li.
Bi 1.10. Tnh s cnh ca mt khi a din li c 6nh v 8mt.
Hng dn gi i
Theo gi thit ta c: d =6; m - 8.
Mt khc, theo cng thc -Ie:
d - c + m - 2 < ^ c = d + m - 2 = 6 + S - 2 = 12
Vy khi a din c 12 cnh.
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Bi 1.11. Chng minh rng khng tn ti mt hnh a din li c s nh, s' cnh, s mt u l .
Hng dn gi i
Gi d, c, m ln lt l s nh, s cnh v smt ca mt khi a din
li. Theo cng thc -le , ta c: d c+ m= 2
Nu d, c, mu l th d - c + m l. iu ny v l.
Vy khng tn ti mt hnh a din li c s nh, s cnh, s mt u l.
Bi 1.12. Tnh s nh, s cnh v s mt ca mt khi a din u loi{3; 4}. __
Hng dn gi i
Khi a din u oi {3; 4} l khi a din m mi mt ca n l mt tam gic u, mi nh ca n c ng 4 cnh i qua. Do , ta c:
, d c m d - c + m 2
= _ = _ = _ = 1 ^1 1 = _ =4 2 3 4 2 + 3 12
=> d= 6; c= 12 ; m = 8
y l khi tm mt u.
c. TON TRC NGHIM
21. Trong cc mnh sau, mnh no sai?A. Hnh t in l hnh a din li.
B. Hnh hp l hnh a din li.
c. Hnh chp l hnh a din li.
D. Hnh lng tr tam gic l hnh a din li.
2. Trong cc mnh sau, mnh no ng?
A. S nh v s" mt ca mt khi a din li lun bng nhau.
B. Tn ti mt khi a din li c snh v s cnh bng nhau,
c . Khng c khi a din li no c s cnh bng snh hoc s mt.
D. C mt khi a din li c s cnh v mt bng nhau.
3. Trong cc mnh sau, mnh no ng?
C khi a din li m:
A. S" nh, scnh v s mt u l.
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B. S nh v s cnh chn, cn s mt l.
c. S nh v smt chn, cn s" cnh l.D. S' nh v s cnh l, cn s" mt chn.
4. Khi a din li c 7 nh, 7 mt th s' cnh bng:
A. 12 B. 14
. 10 D. Mt p s khc.
5. Khi 12 mt u c my nh?A. 12 B. 20 c. 24 . 30.
6. Khi 20 mt u c my nh?
A. 8 B. 10 c . D. 20.
7. Khi 20 mt u c my cnh?
A. 16 B. 20 c. 24 D. 30.
8. Xt cc mnh :
(I) Hnh chp t gic l mt a din li.(II) Hnh lng tr t gic l mt a din li.
Mnh no ng?
A. (I) iig, (II) sai B. (I) sai, (II) ng
c . C (I) v (II) u ng ' D. C (I) v (II) u sai.
9. Xt cc mnh :
(I) C mt a din li m : d - c + m l s l.
(II) C mt a din li m : d - c + m b ng 4.
Mnh no ng ?
A. (I) ng, (II) sai B. (I) sa, (II) ng
c . C (1) v (II) u ng D. c (I) v (II) u sai.
10. Xt cc mnh :
{]) C mt a din li m s"nh bng s"mt.
(II) C mt a din li m s' nh ln hn smt.
Mnh no ng ?
A. (I) ng, (II) sai B. (I) sai, (II) ngc. C (I) v (II) u ng D. C (I) v (II) u sai.
D . P N
1C 2C 3D 4A 5B 6C 7D 8D 9D 10C
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3. TH TCH CA KHI A DIN
A . T M T T L T H U Y T
I. TH TCH CA KHI LNG TR
1. Th tch ca khi lp phng cnh al:V = a3
2. Th tch ca khi hp ch nht c ba kch thc a, b, cl:
V = a.b.c
3. Th tch ca khi lng tr c din tch y B v chiu cao hl:
II. TH TCH CA KHI CHP V KHI CH P CT
1. Th tch ca khi chp c din tch y B v chiu cao hl:
V = - B./z3
2. Th tch ca khi chp ct c din tch hai y bng B v B \ chiu
cao bng hl:
V;^ - ( b + b '+ V b b 7)^
III. MT S TNH CH T
1. T s th tch ca hai khi a din ngdng bng lp phng t s ng dng.
2. Cho khi chp S.ABC. Trn cc onSA, SB, s c ln lt ly 3 im A \ B \ Ckhc vi s. Khi :
V(S.AB'C} SA' SB' SC'
V(S.ABC) SA SB s c
B . G I I T O N
^ VN /
Tnh th tch ca m t khi a din
* Phng Php
1. Nu khi a din l khi lng tr, khi chp hoc khi chp ctth ta dng cc cng thc tng ng tnh.
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Tv/1 IUIU1uu uiyn UI Jh.1, ia \*\Jme lctll CiliU UdUl CclCK10 tr hoc khi chp n gin hn tnh.
3. C th tm t s th tch gia khi a din cho v khdin bit th tch suy ra th tch cn tm.
Bi 1.13. Cho khi lp phng ABCD.ABCD cnh a. Tnh th ca khi lp phng, khi chp A.ABCD v khi chp A.ABD.
Hng dn gi i
Gi V|, V v V3 ln lt l th tch cakhi lp phng, khi chp A.ABCDv khi A'.ABD. Ta c:
V, =/*
V2= r S ,
B
.AA'
a= - a 2.a = 3 3
V, = - Sabd.AA' = - AB.AD.AA' = 3 3 2 6
A
' A
Bi 1.14. Tnh th tch ca khi t din u cnh a.
Hng dn gi i
Xem t din u ABCD (cnh a)nh l kh'i chp nh A v y ltam gic BCD. Khi , ta c din tch y:
lsn) - 4
Gi AH l ng cao ca hnh chp A.BCD th H l tm ca tam gic
u BCD. Ta c:
AH2 = A B 2- B H2 = a 2 -
' u _ 0V2=> AH =
'2 2/3
l 3 ' 2
A
D
/
a 2 _ 2a 2~a_T _~T
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Vy: VABCD SgC.AH l a 2S gy2 ^ 4 2
3' 4 73 12
Bi 1.15. Tnh th tch ca khi 8mt u cnh a.
Hng dn g i i
Trong hnh bn, ta c khi 8 mtu (H). vi cc nh A, B, c , D, M,N. Ta c th phn chia khi 8 mtu (H) thnh hai khi chp t gicu M.ABCD v N.ABCD.
D thy M.ABCD v N.ABCD i
xng nhau qua mt phng (ABCD)
nn bng nhau. V do , th tchca chng cng bng nhau.
Gi o l tm ca hnh vung ABCD.
V MANC cng l hnh vung cnh a nn:
m o = M N = W2
D
M/l
/ h j . . ./ / l '
/ *' ' "' \ c / /
\ . /
N
1 2 , a\f a3V2Do : V(HI =2Vmabcd = 2 S ABCD-MO = ^ fl = - y -
Bi 1.16. Cho khi lng tr ABC.ABC c th tch V. Tnh th tch khi
chp C.ABBA theo V.
Hng dn g ii
Gi h l chiu cao ca khi lng tr
ABC.AB'C. Khi h cng l chiu
cao ca khi chp C.ABC.Ta c th tch khi lng tr:
V - SA-B-C-.
Gi V| l th tch khi chp C.ABC,
ta c:
v , = ^ j s a.b.c ..a = ^ v
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Gi v 2 l th tch khi chp C.AB BA, ta c:
V = v - v , = v - i v = - v .2 ' 3 3
Bi 1.17. Cho khi lng tr t gic u ABCD.AiBiCiDi c khong cch
gia hai ng thng AB v A[D bng 2 v di ng cho ca
mt bn bng 5.
a) H AK 1 A tD (K e AD). Chng minh rng: AK = 2.
b) Tinh th tch ca khi lng tr ABCD.A]BiC|D|._____________
Hng dn gi i
a) Ta c: AB -L (A ADiD) => AK JLAB
Mt khc: AK 1 A,D
T ( I) v (2) => AK l ng vung
gc chung ca hai ng thng AB
v A]D.
=> AK = z(AB, A,D) => AK = 2.
b ) t : A i K - X = > K D = 5 - X
Tam gic AiAD vung ti A,
c ng cao AK nn:
AK = A]K.KD
~ X 1
( 1)
(2)
Bi
Ci> 4 = 4 5 - x ) X' - 5 x + 4 = 0x = 4
Vi X = 1 => AiK = 1 v KD = 4
Ta c: AD2 = K2 + KD2 = 4 + 16 = 20 => AD = 2V5
AA-7= AK2 + A,K 2 = 4 + 1= 5 ^ AA, = V5
Khi: ABCD.A,B]C|D| = Sabcd.AA -Vs = 20V5
Vi X = 4, tng t ta c: VABCD.A,B,C|Dt=lO-v/5.
Bi 1.18. Cho kh'i chp t gic u S.ABCD c cnh y bng a. Tn
th tch khi chp.
a) Bit gc gia mt bn v y bng a.
b) Bit gc gia cnh bn v y bng [3.________________________
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Hng dn gi i
a) Gi M l trung im ca BC, ta c:
SM JLBC
OM 1 BC=> SMO = a
Tam gic SOM vung ti o nn:SO
tan a = OM
SO= OMtan a =7 tana2 A
Vy: Vs ABCD - |JABCD= - S Anr-n.SO = - . a 2.tanar = a itanr.- S ABCD.SO = - .3 ABC 3 2 6
b) Ta c: s o (ABCD) => OB l hnh chiu ca SB trn (ABCD).
=> SBO = p
Tam gic SOB vung ti o nn:
tan p = ~ =} SO = OB tan p = tan pOB 2
Vy: Vs abCd -abcd1
= S .nr,n.SO =.a .l y l
tanp =a3V2
tanp.
i 1.19. Tnh th tch ca khi chp ct tam gic u c cnh y ln l
2a,y nh av gc ca mt bn v mt y bng 60.
Hting dn g ii
Xt khi chp ct tam gic u ABC.ABC c ct ra t khi chp
u S.ABC.
Gi o , O l tm ca hai y v M, M l trung im ca BC v B C.
O M 1B C - 0R rng: < =>OMS = O
s m -LBC
1 A 1 2ayf yTa C: OM = - AM = =
3 3 2 3
0'M' = - A'M' =3 3 2 6
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Tam gic SOM vung ti o nn:
SOtan 60 = => SO = OM tan 60 = a
OM *
Tam gic SOM c 0 M// OM nn:
, : _ , . U s - s o - SO QM 2 2 2
Do : 0 0 '= S O - S O ' = a A2 2
Gi V th tch khi chp ct ABC.ABC.
Ta c: V = - ( b +B ' + 7 b 7)
4a2V3Vi
3'
B = S a
,/
/t
ttA/
'V '> \ c
- 1/ \
B\
60Ao
"
B
4
W 3
4
v h= OO = -2
T ta c: V =7v5.a3
24
7/n t s th tch ca hi khi a din
* Phng Php
1. Tm th tch ca tng khi a din ri suy ra t scn tm.
2. S dng cng thc:
v (uw r) SA' SB' SC'
Xs.ABC) SA SB s c
Bi 1.20. Cho khi lng tr ABC.ABC. Tm t s th tch ca kh
din CABC v khi lng tr cho.
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Hng dn gii
H CH (ABC) th CH l ng
cao ca khi lng tr ABC.ABC
v cng l ng cao ca khi t
din CABC.
Gi V] v v 2 ln lt l th tch cakh'i lng tr v khi t din. Ta c:
V ,= S ABC.H
V2 = y S ABt.C'H
= > v , . v 1^ .2 3 V,
C
A
Bi 1.21. Cho khi lng tr ABC.ABC. Gi M, N ln lt i trun
im ca hi cnh AA v B B \ Mt phng (CMN) chia khi lnc ( r
cho thnh hai phn. Tnh t s th tch hai phn .
Hng dn gi i
Gi V l th tch ca khi lng tr.
Khi th tch khi t din CABC
Vl . Ta c:
3
V,.. = V , - V ..(" ' '
= V - X = 2 V
3 3
=> Th tch khi chop C.MNBA l:
1 2 2 3 3
=> Th tch khi a din ABCMNC l:
v , = v - ^ =
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Bi 1.22. Cho khi chp tarri gic S.ABC. Trn cc on SA, SB, s c ln
lt ly ba im A \ B \ C khc vi s. Chng minh rng:
V,(S.ABC )
V,(S.ABC)
SA' SB' SC'
SA SB s c
Ghi ch:C s dng kt qu ny lm bi tp (xem phn tm tt
l thuyt).
Hng dn gi i
t: BSC = a
H AH 1 (SBC) v AH (SBC)
Khi : s, H \ H thng hng.
Ta c: V(SA.B.C) - - . S sb.c .AH
= SB'.S. sin a.A'H'6
V = AH( ) 2
= -SB.SC.sinaAH 6
Do :
Mt khc:
Vy:
V(S VB.C) SB S C A'H'
SB s c AHV( )
A'H' SA'
AH SA
V ,S.AB-C, SA' SB' s c
SA S B s cV,(S.ABC)
I 1.23. Cho khi chp tam gic S.ABC. Trn ba cnh SA, SB, s c ln lt
ly ba im A \ B \ C sao cho: SA' - -S A , SB' = -S B v SC' = s c .4 5 2
Tnh t s th tch ca hai khi chp S.ABC v S.ABC.
Ta c:V..........
( ")
V( )
Hng dn gi i
S A SB' SC' _ 1 1 1 1
S A ' S B ' S C _ 4 5 2 _ 40
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Bi 1.24. Cho khi chp S.ABCD c y l hnh bnh hnh. Gi B \ D 'ln lt l trung im ca SB, SD. Mt phng (ABD ) ct s c ti C.Tm t s" th tch ca hai khi chp S.ABCD v S.ABCD.
Hng dn gi i
Gi o l giao im ca AC v BD.
D thy AC, BD v s o ng quiti I v I cng l trung im ca s o .
K o c // AC
R rng: SC = C C = CC
SC' _ 1
s c _ 3
VS,AB.C. = SA SB' S C
VcaDr. S A 'SB s c
Do :
Ta c:
2 3 6
Ngoi ra: V S .ABCD = 2V s . ABC
Nn:V,S.AB'C'
S.ABCD
12
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Hng dn gi i
a) Ta c: VSAUC^SSAB.SC
= SA.SB.SC3 2
SA13C c
= - .3 .4 .4 = 86
b) Tam gic SAC vung ti s nn:
AC2 - SA2+ s c 2= 9 + 16 = 25
=> AC = 5
Tng t: AB = 5 v BC = 4V2 A
Gi p l na chu vi ca tam gic ABC, ta c:
AB + BC + AC 5 + 4 \ f +5 r-p = -------- --------= ------ -----= 5 + 2V2
2 2p dng cng thc H-rng, ta c:
SABC = Vp(p - BC)(p - AB)(p - AC)
= 5 +2V2)(5 -2V2).2V2.2V2
= V(25-8).8 = V
Mt khc nu gi SH l khong cch t s n mt phng (ABC) th:
Vs a b c = ^ S sa b.SH=>8 = ^.2V34.SH
OTT 24 _ IV _ VI* o H f:. . .
2V34 34 17
Bi 1.26. Cho khi chp S.ABC c y ABC l tam gic vung ti
cnh SA xung gc vi y. Bit rng AB = a,BC = bv SA = c.
Tnh khong cch t A n mt phng (SBC).
nn theo nh l ba ng vuna gc th SB 1 BC.
Hng dn gi i
SA 1 (ABC)
28
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Ta c: V,1
S.ABC 3
= -.-A B .B C S A = (1)3 2 6
Mt khc nu gi h l khong cch
t A n m tphng (SBC) th:
Vsabc= ^ S sbc.A = .S B .B C ./i
I= .\la2 + c 2h.h (2)
6
T (1) v (2)abc 1 r ~ L
- = .s a + c .b.h-6 6 + c
Bi 1.27. Cho hnh hp ch nht ABCD.ABCD c AB = a, BC = 2a
v AA = a. L'y im M trn cnh AD sao cho MA = 3MD.
a) Tnh th tch khi chp M.ABC.
b) Tnh khong cch t M n mt phng (ABC).__________________
Hng dn gi i
a) Th tch khi chp M.ABC bng th tch khi chp B.ACM.
Theo gi thit:
MA = 3MD => MA = -A D = 4 2
SACM=-.M A .CD = - . .a = 2 2 2 4
Vy: Vmab.c = -SACM.BB
= 1 3 l = .(1)3' 4 'a 4
b) Gi h l khong cch t M n mt
phng (ABC).
D c
C
Khi : Vmab.c = Sabx-h
Tam gic ABC c: AB' = ay ; AC = CB' = a \ [
29
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Do na chu vi: p = -AB' + AC + CB'
Ta c: SAB.C= Vp ( p - AB' )(p - AC)(p - CB')
rz CL'.' \l . ^2.CV5
/
1 _ 1 3a2V y 1 Y vi.A B 'c ~ 2 AB'C 2 ' 2 2
2 '
(2)
a . .T(I) v (2) suy ra: = .h=i>h = -
4 2
c. X)N TRC NGHIM 3
1. Th tch ca khi chp t gic u c tt c cc cnh bng al:
A. B. . D.
Cho t din ABCD. Gi B v C ln lt l trung im ca AB v AC.
Khi t s" th tch ca khi t din ABCD v khi t din ABCDhng bao nhiu?
A.2
4
.1
D.1
3. Cho khi chp S.ABC, trn ba cnh SA, SB, s c ln lt l'y ba im
A \ B v C sao cho: SA' = SA; SB' = SB v SC' = s c . T s5 6 7
th tch gia khi t din S.ABC v S.ABC bng:,
A.420
B.630
.210
_107
210
4. Cho khi lng tr tam gic u ABC.ABC c tt c cc cnh u
bng a.Th tch khi t din ABBC bng bao nhiu?
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5. Cho khi chp S.ABC c ba cnh SA, SB, s c vng gc nhau tng i
mt v SA = SB = s c = a.Th tch khi chp S.ABC bng bao nhiu?
A. B. c . 3 D.6 12 2
6. Cho khi lp phng ABCD .AB CD. T s th tch gia khi chop
A.ABCD v khi lp phng bng bao nhiu?
A. B. 6 3
c . D. Mt p s' khc.2 9
7. Cho khi lp phng ABCD.AB CD. T s th tch gia khi chp
A .ABD v khi lp phng bng bao nhiu?
A. - B. - c . - D. .
3 2 6 12
8. Cho khi lng tr ABC.ABC c din tch y bng B v chiu cao
bng h. Xt cc mnh :
(I) Th tch khi lng tr: V] =B.h
(II) Th tch khi chp A.ABC: v 2 = -B .fr3
Mnh no ng ?
A. (I) ng, (II) sai B. (I) sai, (II) ng
c . C (I) v (II) u ng D. c (I) v (II) u sai.
9. Cho khi chp S.ABCD c y l hnh vung cnh a v c tm o , h l
chiu cao ca khi chp. Xt cc mnh :
(I) Th tch khi chp S.ABCD: Vi = a2.h
(II) Th tch khi chp S.AOB: v 2= a 2.h
Mnh no ng ?
A. (I) ng, (II) sai B. (I) sai, (II) ng
c . C (I) v (II) u ng D. C (I) v (II) u sai.
10. Cho khi chp lc gic u S.ABCDEF c cnh y bng a v chiu
cao bng h.Gi o l tm ca y. Xt cc mnh :2P
(I) Th tch khi chp S.ABCDEF l: V, = .h2
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, , a2V3(II) Th tch khi chp S.AOB l: v 2= .h
12
Mnh no ng ?
A. (I) ng, (II) sai B. (I) sai, (II) ng
c . C (I) v (II) u ng D. C (I) v (II) u sai.
D. P N
1A 2B 3C 4D 5A 6B 7C 8G 9B 10C
4. TON TNG HP
Bi 1.28. Hnh lng tr ng ABC.ABC c y ABC l mt tam
vung ti A, AC = b, C= 60. ng cho BC ca mt bn BB
to vi mt phng (AACC) mt gc 30.
a) Tnh d di on AC.
b) Tnh th tch ca khi lng tr.
Hng dn gi i
BA AC(A BACvung)__*=> BA ( A A ' C ' C )
B A l A A v i A A ' l ( A B C )B C
Suy ra: BC'A l gc gia BC v
mp(AACC).
ABAC v ABAC' vung ti A nn:AB = Ac tan c = btan 60 = byj3
A C ' = A B c o t C ' = 3. cot 30
Vy: AC = 3b A
32
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M S ^ = A B . A C = ^ -
h - AA' - VAC'2 - A C 2 = J % 2 - b 2 = 2bs
Do : V = AA'.Siy = ^ - .2 b y 2 =3V.
Vy: V= 3V6.
Bi 1.29. Cho lng tr tam gic ABC.ABC c y ABC l mt tam
gic u cnh a v im A cch u cc im A, B, c. Cnh bn
AA to vi mt phng y mt gc 60.
a) Tnh th tch ca khi lng tr.
b) Chng minh mt bn BCCB l mt hnh ch nht.
c) Tnh tng din tch cc mt bn ca khi lng tr (tng ny thng
gi l din tch xung quanh ca hnh lng tr cho).
Hng dn gi i
a) Gi o l tm ca tam gic y ABC
V A cch u A, B, c nn A 'O 1 ( A B C ) .A ' AO gc gia AA v (ABC) nn A ' AO = 60.
AABC u nn trung tuyn AM cng l ng cao:
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Vy th tch khi lng tr l:
V . S ^ O - ^ - t
aVy: V =
4
B C I A ' 0I ^ BC 1 A A '=> B C 1 BB' (v BB'II AA' ) \ B C L A M
Vy BBCC l hnh ch nht.
l a y c) V BB ' = AA' = - nn:
3
2aV3 2a2\3SBCC.B. = BC.B B' = a . - = ^ - = - ^ L-
3 3
Gi N l trung im ca AB th AB _LCN
\ AB LO N _ => AB ( A' NO )= > AB A N
[ A B I A ' 0 v
13a 2
12
V3 _ g y f . j 2 _ as l396
a 2J39
A ' N = - , -V2 12 6
SB.A. = S ACC.r =AB. A' N: 6
Vy: + =
v = 4 ( V 5 5 + 2V ) .
Bi 1.30. Cho khi lng tr ABC.ABC, y ABC l tam gic vung
cn nh A. Mt bn ABBA l hnh thoi cnh , nm trong mt
phng vung gc vi y. Mt bn ACCA hp vi y mt gc a .
Tnh th tch ca lng tr.
Hng dn gi i
Gi H |. hnh chiu ca A xung AB.
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Ta c: mp(A'B'B)-L mp(ABC) nn A ' H L ( AB C)
A / l A C = > A A l A C = > ' A H = a
Ta c: h= A' H = AA 's in a = as i n a,
S A B C = A B A C = a
Th tch lng tr ABC.ABC l:
V =R h = - a 2, asin a2 A'
V - -- a 3sin or.2 c
Bi 1.31. Cho hnh chp t gic u S.ABCD.a) Bit AB = av gc gia mt bn v y bng a ,tnh th tch khi
chp.
b) Bit trung on bng dv gc gia cnh bn v y bng , tnh
th tch ca khi chp.___________________ _____________________
Hng dn gi i
a) Gi I l trung im AB, o l
tm cua hnh vung ABCD.
=>SOl gc gia mt bn (SAB)
v mt y.
Theo gi thit: SO = a
ASO I vung ti o v s o (ABCD )
A B 1 S
A B 1 0 1
SO - IOtan a = tan a2
B
; = - A B 2. SO = - . a 2. - t a n a = 3 3 2 6
a tan a
y - 3tan6
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b) Vi SO 1 (BC D ) nn SAO l gc gia SA v (ABCD).
Theo gi thit SAO = , trung on SI = d.
Gi a l s o cnh hnh vung y.
SO - A o tanSO - tan
IO = ~2
ASO vung ti o nn:
SO2 + O2 = SI2 => tan2 + = d 2 ^ a = 2d2 4 ^ I t a n 1 + 1
Ta c: F( / . =--y52.5 = a3V2 tan?3 6
4V2J3 tan
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b) Gi bl cnh . Gc gia mt bn v y l.SIO= a .
l r v - 1 b S _ b S 6
OI = - C I = - : - 3 3 2
C O r= -C I = = AO 3 3
__ h
SO= 10 tanSIO = tan a
Tam gic SAO vung ti o nn :
SO2 + O A 2 = SA2 => tan2a + ^ = l 2=> = -t L12
1 2V3 V 3tan a 1------- tan a -
+ tan a
cho'p ^
Hay VchB>
3' 4 6 24
V/2tan a(4+ tan2a ) V 4 + tan2 a
Bi 1.33. Cho khi hp ABCD.ABCD c th tch V. Tnh th tch
khi t din ACBD theo V.
Hng dn gi i
Ta c: VCB-D-) V(ABCD.A'B'CD) ~{ B'ABC)+ (ACD) + \cB'C') )
(h l chiu cao ca khi hp)
Tng t: V(DACD} V(AA'BD') ~ CB'C'D)
V VVy: 3 -
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Bi 1.34. Cho khi lng tr ABC.ABC. Mt mt phng i qua AB vtrung im ca AC, chia khi lng tr thnh hai phn. Tnh t s' th
tch ca hai phn ._____________________
Hng dn gii
Gi s, h, V l din tch y, chiu cao, th tch ca khi lng tr
ABC.ABC.
Gi I l trung im ca cnh AC.Mt phng (AB1) song song vi AB nn ct
mt phng (ABC) theo giao tuyn
IJ//AB e BC) . .
Ta V| lth tch khi chp ct A BC.IJC
v v 2 l th tch phn cn li. .
Ta c: vx= - h ( B + y[BJ p+ B' }.
Trong : B = s.
ACIJ ~ACAB :} (CU)
(CAB)CA
__
4
(CU)
Do : V. - h
= - hay B' = - 4 4
75S + J S . - + -
4 4V
7 7V, = - S h - V
12 2
7 5VSuy ra: K, = v - V =
12 12
= h.3 4
V.Vy:
V,
Bi 1.35. Cho khi t din ABCD. Gi a l gc gia AB v GD, d l
khong cch gia AB v CD. Chng minh rng th tch ca khi t
din ABCD c tnh bng cng thc:
V - AB.CD.d.sin a .6
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Hng dn gi i
V hnh bnh hnh ABED.
DE//AB => B^CD^ = CD= a .
ABD - AEBDv hai tam gic ny
cng nm trong mt mt phng.
Do : VABCD V c .A B D = 'c.EBD
hay V.4BCD = V'b.CDEABCD
ABCD
A
C
M - Vb .c d e ~
D E
SCDE - DE.CD sin a = AB.CDsin a
Mt phng (CDE) cha CD v song song vi AB, do :
d(B,(CDE)) = C D ) = d
Vy: VABCD = - . AB.CD.sin a .d
Hay V4BCD = AB.CD.d.s \na .6
i 1.36. Cho hnh chp S.ABC c cc cnh bn SA, SB, s c i mt
vung gc. Gi M l trung im cnh AB. Tnh th tch khi chp
S.BCM, bit AB = AC = 3; BC = 4.
Do : V s . b c m - - Vs.ABC2
t: SA = a\SB = b\s c = c
Cc tam gic SAB, SAC v SBC
vung ti s nn theo nh l Pi-ta-gota c:
Hng dn gii
V M l trung im ca AB nn: S a s a b = 2 S a s m b
1 , , c
a2+b2- 9 a2 =1s
c
B
A
39
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Ngoi ra: VSABC= - .SA.SB.SC = - a b c = - . \:
Vy: V,S.BCM
Bi 1.37. Cho hnh chp S.ABCD c tt c cc cnh u bng nhau. Bi
V th tch ca khi chp l V = fl3. Tnh di cnh ca kh
chp.
Hng dn gi i
T gi thit ta suy ra S.ABCD l khi chp u.
Gi o l tm ca y v X l di cnh. Ta c:
SO 2 = x 2 - = => s o = - i- '2 2 V2
Mt khc:
=> X3 = 21 a3
=> JC= 3a
Vy cnh ca khi chp bng 3a.
Bi 1.38. Cho khi chp S.ABC c y l tam gic vung ti B. Cnh SA
vung gc vi y, gc ACB = 60, BC = a v SA = ay3 . Gi M l
trng im ca cnh SB.
a) Chng minh: (SAB) 1 (SBC).
b) Tnh th tch khi t din MABC.
Hng dn gi i
BClAB(gt )a) Ta c: \ = B C ( S A B >
[BC _LSA (do SA J_( ABC))
. Mt khc: BC c (SBC)
Nn: (SAB) (SBC)
40
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b) Tam gic ABC vung ti B nn:
tanC = 4 ^ => AB = BCtanO0 = 3BC
Vi MS = MB nn: ^
V = VMABC 2 s a b c
= 4 j s AABC.SA
= . AB. BC.SA6 2
= .a>/3,a.a>/3 = .12 4
Bi 1.39. Cho khi chp S.ABCD c y ABCD l hnh vung cnh a ,
SA = av vung gc vi y. Gi M l trung im ca SD
a) Tnh khong cch gia hai ng thng AB v sc .
b) Tnh th tch khi t din MACD._______________________________
a) Ta c:AB//CD
Hng dn gi i
AB//(SCD)[CDc(SCD)
M SCc(SCD)=> d(AB, SC) = d(AB, (SCD))
= d(A,(SCD))
Tam gic SA vung cn ti A, c
M l trung im ca SD nn:
A M 1 S D B
Mt khc: AM 1 C (v CD 1 (SAD))
Do-: AM 1 (SCD) => d(A, (SCD)> = AM =a 4
Vy: MH // S A ^ > M H 1 (ACD)
. . . . . T SA Ngoi ra: MH = =
41
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Vy: VMACD = i v ACD.MH = i A D . C D . M H
1 a 3:,a.a.=6 2 12
Hi 1.40. Cho khi chp t gic u S.ABCD c cnh y bng a.Gi Gl trng tm ca tam gic s AC v khong cch t G n mt bn
SCD bng2 \ 3
Tnh khong cch t tm o ca y n mt bn6
SCD v th tch khi chp S.ABCD.
Hng dn g i i
Gi I l trung im ca CD.
Khi : OI _LCD
Mt khc: SI -LCD
Nn: CD-KSO)
Ta c: (SOI) n (SCD) = SI
K OK v GH vung gc vi SI
=> OK 1 (SCD) v GH 1 (SCD)
:=> d(0 , (SCD)) = OK
Ta c: OK = GH
2
Vy: d(o, (SCD)) =
3 aV
2 ' 6
1 ^
V3
Tam gic SOI vung ti o c OK l ng cao nn:
aSOK
Vy:
1 J _
S F
1
>so=-
V = - S : 2 * '
- . 2 .a 4 _ fl3V3
2 ~
6 '
Bi 1.41. Tnh th tch ca hnh chp S.ABC, bit rng y ABC l mt
lam gic u c cnh bng a, mt bn SAB vung gc vi mtphng y, hai mt bn cn li cng tc vi mt phng y mt gc
bng .
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Hng dn gi i
Dng ng cao SH ca hnh chp.
V (SAB) 1 (ABC) nn H e AB .
Dng H -LBC=>S 1 BC
(nh l ba ng vung gc)
=>SIH l gc gia mt bn A'
(SBC) v y (ABC)
= > s = a
Tng t SK H = a => ASHI = ASHK => HI= HK
AB a
2 ~ 2
aV3
T : AH = BH
H = HK =rr, \ 3
SH= HI.tan a = tan a
Vy: KS.ABC1 c ct / _ 1 a 2J a j 3 t a3- S , Rr.SH = . ;. tan a = ta na .3 3 4 4 16
Bi 1.42. Cho hnh chp t gic u S.ABCD c cnh y bng a , gc
gia cnh bn v mt y bng )(o0 < < 90)! Tnh tang ca gc
gia hai mt phng (SAB) v (ABCD) theo . Tnh th tch khi
chp S.ABCD theo av . *
Hng dn gi i
Gi giao im ca AC v BD l o th
S O l ( A B C D ) .
Suy ra: SAO =
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. 5 r~Do : tanSM O = = V2 tan .
OM
1 2 V2 3
Bi 1.43. Cho khi chp S.ABC c y ABC l tam gic u cn
Cnh bn SA vung gc vi y SA = .
a) Tnh khong cch t im A ti mt phng (SBC).
b) Tnh th tch khi chp S.ABC v din tch tam gic SBC. __
Hng dn gi i
a) Gi H trung im ca BC.
H AK1SH
mt phng (SBC).
Vy AK l khong cch t A ti
=> AK _L(SBC) A
B
ASAH vung ti A c AK l ng cao nn:
AK2 SA2 AH2 6a2 3a 2 a2
1 1 1 1 1 2TTT ~ T+ +~ T
4 4
2 2
b) Ta c: S.ABC abc-S-A- 2 '
SBC AK 8 ' 2
4 ' 2 8
_ 3 V s a b c _ . a & _ 3= 4
44
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Bi 1.44. Cho tam gic vung cn ABC c cnh huyn BC = a. Trn
ng thng vung gc vi mt phng (ABC) ti A ly im s saocho gc gia hai mt phng (ABC) v (SBC) brig 60. Tnh th tch
khi chp SABC.
Hng dn gi iGi M l trung im ca BC.
Khi do tam gic ABC cn ti A nn: AM -L BC
SAl (ABC)Ta c:
A M 1 B C
=> SM _LBC (nh l 3 ng vung gc)
AM1BC ^Nh vy : r =>SMA = 60
S M 1 B C
AABC vung ti A => AM -
ASAM vung ti => tan 60
=> SA = AM. tan 60
BC a
2 ~ 2
0_ SA
M
a j3
w*... X, l c l a 2 a j3 3Vy: VSABC S^gj-.SA 3' 4 ' 2 24
c
Bi 1.45. Tnh th tch khi t din ABCD bit AB = a, AC = b, AD = c
v cc gc B A C , CAD , DAB u bng 60.
Hng dn gi i
Gi s a= min{a, b, c}.
Trn AC, AD ln lt ly im C v D sao cho: C = AD = a.
Khi ABCD' l khi t din u cnh a.
Gi H l hnh chiu vung gc ca B trn mt phng (ACD) th H l
tm ca y ACD \
45
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Ta c:2 . . . 2 B'C' = aV3
Tam gic BC1 vung ti C nn:
BI2 = BC2+ C I22 fl2 _ 132
= 3a + - =
= 3a
zV3
Tai gic ABB vung ti B nn:
AB' = ABV = a 4 2
46
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Tam gic ACI vung ti c nn:
AI2 = AC2 + CI2 = a 2 + = ^ A I = 4 4 2
Nh vy, ta c:
AB'2+ AI2 = 2a +5 _ l j
4 4 =>AB'2+ AI2 = B'I2B =
=> Tam gic AB1vung ti A.
b) Gi cpl gc gia hai mt phng (ABC) v (ABI).
Ta c: s . . nr =^AB.AC.sinl20 =-2V5
AAB'1 AB'.AI =2V
2 4 .
D thy tam gic ABC l hnh chiu ca tam gic AB1 trn mt
phng (ABC), do :
o __o y V ABC = >AAB I.C( = > = ,C (
4 4
. . 7 3 - 7 3 0=> COS (p = = =
V 10
T ' \ / - c _ a 3y3c) Ta co: VABCA.B.C. S.^gc-AA .a
Bi 1.47. Cho hnh chp S.ABC c y ABC vung ti B; AB = a,BC = 2a.
Cnh SA _L(ABC) v SA = 2a.Gi M l trung im ca sc.
a) Chng minh rng tam gic AMB cn ti M.
b) Tnh din tch tam gic AMB.c) Tnh th tch khi chp S.AMB, suy ra khong cch t s n mt
phng (AMB).
Hng dn g i i
SA _L(ABC)a) Ta c: = > S B B C (inh l 3 ng vung gc)
AB JLBC
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Tam gic SBC vung ti B v BM
s cl trung tuyn nn: BM =
scTng t: M =
Do : AM = BM
Vy tam gic AMB cn ti M.
b) Tam gic ABC vung ti B nn:
AC2= AB2+ BC2 = a2 +4a2= 5a2
=> AC = a 5
Tam gic SAC vung ti A nn:
, SC2= SA2 + AC2 = 4a2 + 5a2= 9a2
=> s c = 3a => AM = = 2 2
Gi H l trung im ca AB. Khi : MH AB
Tam gic AHM vung ti H nn:
MH2 = a m 2- a h 2 =9a
= 2 a
=> MH = aV2
Saamb = AB.MH = - .a .a S = 2 2 2
c) Gi Vi v v 2 ln lt l th tch khi chop S.AMB v
Ta c: v2= sABC.sA = .AB.BC.SA= .a.2a.2a3 3 2 6
V, _ SA SB SM
S.ABC.
2a3
3
,-u, VI SA SB SM 1Mt khc: = = -r
v2 SA SB sc 2
V, V,
a
T
Vy:
Gi h l khong cch t s n mt phng (AMB)
T ' V - U - 1Ta c: V, - s AMB.h= > = - . - -.h 4 .
48
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Bi 1.48. Cho hnh chp S.ABC c SA = 3av SA vung gc vi mt phng
(ABC). Tam gic ABC c AB = BC = 2a,gc ABC = 120.
a) Tnh khong cch t A n mt phng (SBC).
b) Tnh th tch khi chp S.ABC, suy ra din tch tam gic SBC.
Hng dn gii
a) Gi I l hnh chiu vung gc ca Atrn BC.
BC-LAITa c: < => BC -L (SAI)
IBC -L SA
H
Ta c:
AH 1 SI
AHJ-SI
AH B C(doB C 1 (SAI)
=> AH 1 (SBC)Vy AH l khong cch t A n
mt phng (SBC).
= 180 -12 0 = 60
Tam gic ABI vung ti I nn:
sin 60 = => AI = AB.sin 60 = a VAB
Tam gic SAI vung ti A v c ng cao AH nn:
1 1 1 1 1 4
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Bi 1.49. Cho lng tr ABC.ABC c A.ABC l hnh chp tam gic
u, cnh y AB = a, cnh bn AA = b. Gi a l gc gia hai mt
phng (ABC) v (ABC) .Tnh tana v th tch khi chp ABBCC.
Hng dn gii
Gi H l tm ca tam gic ABC v
M l trung im ca BC.
V A.ABC l hnh chp tam gic
u nn AH l ng cao ca hnh
chp A.ABC v cng l ng cao
ca lng tr ABC.ABC.
AM 1 BC ------- ,Ta c: < __=>AMA = a
AM 1 BC '
\ \
/ B \ ( /
60/
- ~ T '
N U " H / M
ABC l tam gic u cnh ac AM
A * Q'fel ng cao nn: AM = ;2
Do : AH = -A M = v HM = -AM3 3 3 6
a
Tam gic AHA vung ti H nn:
A'H2 = AA'2 - AH2 = b 2 -2>a 9b - 3 a
-> A'H = -.V92-3
3a L
Gi V| , v 2 v V ln lt l th tch khi lng tr ABC.AB C, khi
chp A.ABC v khi chp ABBCC.
T ' \ / _ c A ' u _ a2^ 1 7 T _ b 2- a 2l a c: V , = S aabc.A H = - . y l 9 b - 3 a - -------------4 '3'
1-A'H = -V , = al
3 1
V:
3 . 3 12
a2\l3b2 - a 2v = v , - v 2
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Bi 1.50. Cho hnh chp S.ABCD c y ABCD l hnh vung cnh bng
a, cnh SA (ABCD ) v c di SA = a. Mt mt phng i qua
CD ct cc cnh SA, SB ln lt M, N. t AM = X.
a) T gic MNCD l hnh g ? Tnh din tch t gic MNCD theo a, X .
b) Xc nh gi tr ca X the tch ca hnh chp S.MNCD bng
ln th tch hnh chp S.ABCD.
Hng dn gi i
a) Ba mt phng (MNCD), (SAB) v
(ABCD) giao nhau i mt m AB//CD ^
nn MN//CD//AB.
Suy ra: SMNC0 = j a 2 + x 2 (a + a - x ) D
Mt khc: AB -LD M nn MN_LDMVy MNCD l hnh thang vung.
V SA = AB = a nn MN = SM = a - X
DM =yj'a2 + x 2
B
c
= yj a2 +X2 ( 2 a - x )
b) K SH _LM D ; H e MD\
ASHM ~ DAM =>SH = AD.SM _ a ( a - x )
=>KSMNCD _
D M yja 2_ x2
( 2 - x ) ( a - A : ) 2
a - X'
V.SABCD 2 a 9
2a
X =
x = - -a loi3
Vy: X = 3
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Bi 1.51. Cho t din u ABCD cnh a. Gi H l hnh chiu vung
ca A xung mt phng (BCD) v o l trung im ca AH.
a) Tnh th tch V ca t din theo a.
b) Chng minh rng AB _LC D . Tnh khong cch gia hai
thng AB, CD theo a.
c) Chng minh rng cc ng thng OB, o c , OD tng
vung gc nhau.
d) Xc nh im M trong khng gian sao cho: MA1+MB2 +MC2 +
_____t gi tr nh nht.__________________ ____________________
Hng dn gi i
a) Gi I, J l trung im ca AB v CD.
Ta c: H l tm tam gic DBC
A
3 2
= > a h = ^ a b 2 - b h 2
^ V A BCD - SBCD.AH 3 3 4
1 a2yj3 \6 _ a y2ABCD 4 ' 3 12
b) Ta C: CD 1 AJ;C D 1 BJ *> CD _L(ABJ)
=> CD L A B . ,
Khong cch gia AB v CD chnh l IJ.
Ta c: J2 = JA2 - A 2 = - = > I J =2 2
c) Ta
Trong tam gac vung OBH, ta c:
36 9 2
Tng t: O B 1 + O C 2 = a2 = BC2 => AOB C vung.
Chng minh tng t: OB1 OD\ o c J_ OD.
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d) Gi G l trng tm ca t din, ta c:
MA2 +MB2 +M C 2 +M D 1
= (7 + i)2 + { MG + G )2 + (m + c ) 2 +^MG + GD ) 2
= m - + G4*+gF + GD1+IM & G + GB+ GC+ G)
= 4MG1 + G + g + G + G D1( v G+ GB+ GC+ GD = )
Vy: MA2 +MB2 + M C 2 + M D 2 nh nht khi M trng G.
(Ta thay G chnh l trung im ca IJ)
N TP CHNG IA. TON TRCNGHIM
4
1. Cho hnh chp S.ABC c SA_L(ABC). Tam gic ABC vung ti A v
SA = a,AB = b,AC = c.Khi th tch hnh chp bng:
A. abc B, \ a b c c . \ a b c D. ab c.2 3 6
2. Cho khi chp S.ABC c y ABC l tam gic u cnh a. Cc cnh
bn to vi y mt gc 60. Khi chiu cao SH ca kh'i chp
bng:
A. aV 3 B. a c. a 4 D.a
2 .
3. Cho hnh chp t gic u S.ABCD c cnh y bng a , gc hp bi
cc cnh bn v y bng 60. Tnh chiu cao SH.
A V6 /7 ' V3 ^ A. _ B. a v 6 c. D. a v 3 .
4. Cho hnh chp t gic u S.ABCD. Bit AB = a v gc gia mt bn
v mt y bng a. Khi , th tch ca khi chp bng:
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a h a n a 3taimA. B.
3tan a ^ a3 cot ac. D.
2 6
5. Cho hnh chp t gic u S.ABCD c cnh y bng a v gc gia mt
bn v mt y bng 60. Tnh din tch xung quanh ca hnh chp.A. a2 B. 2a2 c. 4a2 D. 8a2.
6. Cho hnh chp S.ABCD. Gi A, B , c \ D ln lt l trung im ca
SA, SB, SC, SD. Trong cc kt qu sau, kt qu no ng? T s th
tch ca hai khi chp S.ABCD v S.ABCD bng:
A. B. c . - D. .2 4 6 8
7. jCIio hnh hp ABCD.ABCD. Trong cc mnh sau, mnh nong? T s th tch ca khi t din ACBB v khi hp
ABCD.ABCD bng:
A. - B. c . - D. .2 3 4 6 .
8. Cho hnh chp S.ABCD c th tch V. Ly im A trn cnh SA sao
cho SA' = SA. Mt phng qua A v song song vi y ca hnh chp
ct cc cnh SB, s c , SD ln lt ti B \ c \ D \ Khi th tch ca
hnh chp S.ABCD bng:
V V V _ V.A. B. c . D. .
5 25 125 625
9. Th tch khi lng tr c chiu cao bng h, y l ng gc u ni
tip trong mt ng trn bn knh rbng:
A. - hr 2 B. - h r 12 . 4
c . - h r 2 n l l D . - h r 2s m l 2 \4 2
10. Th tch Vca khi t din u cnh a bng:
A . B. C . H D. .12 12 12 4
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5
C my loi khi din r ...
A. 3 B. 4 c. 5 D. 6
Cho khi lng tr tam gic u c tt c cc cnh bng a. Khi th
tch khi lng tr bng:
A 0^/3 C . ^ - D . c f S .. ^
Cho hnh chp t gic u S.ABCD c cnh y bng a,gc gia cnh
bn v mt y bng cp. Khi , th tch khi chp S.ABCD bng:
A. tancp ' B. ^r-tan2 0
c . a tan D. ^ 2^-cotcp.6
Cho khi chp tam gic u S.ABC c cnh y bng a v gc ASB
bng 2
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Mnh no ng?
A. (I) ng, (II) sai B. (I) sai, (II) ng
c . C (I) v (II) u dng D. c (I) v (II) u sai.
7. Cho hnh chp tam gic u c cnh y bng a v mt bn c g
y bng a. Khi chiu cao ca hnh chp bng:
A. aV9tan2a - 3 B. ^V9tan2a - 36
c . ^V 9tan2a + 3 D. ay9tan2a + 3 .6
8. Cho hnh lng tr ng ABC.AB C c y ABC l tam gic vu
ti A, gc c = 60(>, AC = a v AC = 3a. Khi th tch khi lng
bng:
A. 3V B. c . 3V3 D.| 3V3 .
9. Cho khi lng tr tam gic ABC.ABC. Gi M l trung im
A A \
Xt cc mnh :
(I) VA.A'B'C = 2Vm.a-B'C- (II) Va .bb'c c = Va.bbc c
Mnh no ng?
A. (I) ng, (II) sai B. (I) sai, (II) ng
c . C (I) v (II) u ng D. c (I) v (II) u sai.
10. Cho khi chp S.ABC . Gi M, N ln lt l trung im ca SA v
T sth tch ca hai khi chp S.ACN v S.BCM bng:
1A. B. 1
2
c . 2 . D. Khng xc nh c.
P N
4 1D 2B 3A 4A 5B 6D 7D 8C 9D 10A
5 1C 2B 3C 4A 5D 6A 7B 8A 9C 10B
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B. HNG DN GII
4
Ta c : V = ^ S mbc.SA
- ..bca3 2
= abc6
p s:Chn D.
Gi s AH ct BC ti I, th AI l
trung tuyn ca tam gic ABC.
'T ' M _ a '_ AUTa c: AI - - => AH = AI =-2 3 3
Tam gic SAH vung ti H v
c A = 60 nn:
SH = AH.tan60 = 3
p s:Chn B.
Ta c: AC = AB^= a 4 .
= ,C H = .2
Tam gic SHC vung ti H nn:
tan60 =SH
HC
=>S t = H C.tan 60
>SH_ aV2 ^ _ a\6
p s:Chn A.
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4. Tam gic SOI vung ti o nn:
SOtan a -
O l
Vy: V SABCD
> SO = O tan a = tan a 2
1 2.SO= - .a ..tana3 2
a3tan a
/? .T: Chn A.
5. Tam gic SOI vung ti o nn:
--- A
p s: Chn B.
6. p dng tnh cht:
* T sth tch ca hai khi a din ng dng bng lp phng t s"
ng dng.* Hai khi chp S.ABCD v S.ABCD l ng dng vi t s' ng
dng bng nn t s th tch ca chng bng .2 8
p s:Chn D.
7, Dng ng cao BH ca hnh hp ABCD.ABCD .
* Ta c: Th tch hnh hp AB CD.ABCD l:
Vi = Sa b c d .B'H = 2Sa b c.B'H a . b ,
* Th tch khi t din ACBB:
V2 = X- Sab c.B'H
Vy: = ~ .K 6
p s: Chn D.
B
58
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8. S.ABCD v S.ABCD l hai khi chp ng dng vi t ng dng
V ' l "bng . Do :
5 V(vi V l th tch ca S.ABCD)
V' V
125
p s: Chn c .
9. Xt ng gic u ABCDE ni tip ng trn tm o bn knh r.
Ta c: AOB = 72
S0AB = -O A.O B.sin 72 = - r 1sin 72
s a b c d e ~ 2 r S l n
Vy th tch khi lng tr:
V = SABCDE.h = h r 2s m l 2 .
p s: Chn D.
10. Ta c: B =
3 3
!V3
Tam gic AHB vung ti H nn:
A H 2 = AB2 - B H 1
a2 2a2
AH --x4
V T
v , f / _ l l a y / a \ [ a 4 Vy. V - - S ^ C. AH= .-= - =
3 3 4 73 12
p s:Chn A.
59
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Theo l thuyt c 5 loi khi a din u: khi t din u, khi
phng, khi 8 mt u, khi 12 mt u v khi 20 mt u.
p s: Chn c.
Th tch kh-! lng tr: V = Sy Xchiu cao =:2V3 _ a 3V3-.a=
p s:Chn B.
Ta c: AC = a-Jz=> OA -V2
Tam gic SOA vung ti o nn:
v otancp =- => S = Atarxp
tancp
a* \2tancp
p s: Chn c.
Ta c: ASK = , AK = O K = ~ 6
* Tam gic SAKvung ti Knn:
tan(p =AK
S K = -a
SK 2tancp
* Tam gic s oKvung ti o nn:
SO2 - S K2 - OK2= - - V - - 4tan cp 12
a 2 ( 3 - t a n 2 cp
12tan2(p
=>S =tan (p
J ' tan(p
p s:Chn A.
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5 Ta c: . S/' SB' S C ' = 1 i 1 _ 1VSAHC S A S B S C 234 24
p : Chn D.
6. * Tam gic ABD cn ti A v A =60 nn l tam gic u r=>BD = a
S b d d -b - = BD.OO = .2. = 2a2
Vy (I) ng.
D
= 2 S a b d = 2,-4 2
Vy: V = S a b c d - O O
= ^ ^ - . 2 a = a2S
Vy (II) sai.
p s:Chn A.
7. Gi I l trung im AB v s o l ng cao.
T ' 4 / * n , l r , a S Ta c: AI = v OI = C =2 3 6
Tam gic SIA vung ti I, nn:
/B
D
^60" C1 /
tan a =SI
S= ~rtanaAI 2
Tam gic SOI vung ti o nn:
SO2 = S 2 - OI2
4 36
=>SO =V9tan2a - 3D
p s:Chn B,
8. Tam gic ABC vung ti A
=> tan60 = -^1
=>AB= a>/3
Tam gic ACC vung ti c
=> CC2 = AC2 - AC2
= 9a2- a2 = 8a2
=>CC = Idy
c
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Th tch khi lng tr:
V = S.CC' = AB .AC .CC ' = .a j3 .2 a -j 2 =3V
p s: Chn A.
9. C (I) v (II) u ng.
p s: Chn c.
, a - r , Vs .ACN _ SA SN sc_ 1 / _ 1\ / \r 0 : i u M ' M 2 V ( I ) -
\!>- u E E S M S B s c = s V = - L \ / ( n \
v "AfC SA SB SC 2 sliCM 2 S'AIC
VT( n v (2) ta c : -SJL= 1 .
^ S.BCM
p s:Chn B.
c. TON T LUN
Bi . Hy ch ra mt cch phn chia khi chp lc gic u S.ABCDEF
thnh 4 khi chp tam gic.
ii 2. Hy ch ra mt cch phn chia khi chp t gic u S.ABCD thnh
4 khi chp tam gic.
Bi 3. Cho hnh lp phng ABCD.ABCD. Gi K l trung im ca
c c . Mt mt phng (a ) i qua AK v song song vi BD ct BB ti M
v DD ti N. Chng.minh hai khi chp A.BMKC v A.DNKC bng
nhau.
Bi 4. Tnh s cnh ca mt khi a din li c 12 nh v 20 mt.
p s: 3:0 cnh.
Bi 5. Tnh s nh ca mt khi a din li c 30 cnh v 12 mt.
p s: 20 nh.
Bi 6. Tnh s nh, s cnh v s mt ca mt khi a din u loi {3; 5}.
p s: 1.2 nh, 30 cnh v 20 mt.
>2
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Bi 7. Cho hnh lng tr ng ABC.ABC c y ABC l mt tam gic
vung ti A, AC = a, c - 60. ng cho BC ca mt bn BBCC
to vi (AA,CC) mt gc 30.
a) Tnh AB.
b) Chng minh AB -L ( ' , C C ) .
c) Tnh AC.d) Tnh th tch khi lng tr.
e) Tnh tng din tch cc mt bn ca hnh lng tr (tng ny c gi
l din tch xung quanh ca hnh lng tr. K hiu s .)
f) Tnh tng din tch cc mt ca hnh lng tr (tng ny c gi l
din tch ton phn ca hnh ng tr. K hiu Sn )
Bi 8. Cho lng tr tam gic ABC.ABC c y ABC l mt tam gic u tm o, cnh a v im A cch u cc im A, B, c. Cnh bn AA
lo vi mt phng y mt gc 60.a) Chng minh hnh chp A.ABC l hnh chp tam gic u. Suy ra
A ' O l ( A B C ) .
b) Tnh th tch khi chp A.ABC.
c) Tnh tng din tch cc mt bn ca hnh chp A.ABC (tng ny
c gi l din tch xung quanh ca hnh chp. K hiu S x ).
d) Tnh tng din tch cc mt ca hnh chop A.ABC (tng ny c
gi l din tch ton phn ca hnh lng tr. K hiu 5, ).
e) Tnh th tch kh lng tr.
f) Chng minh mt bn BCCB ca hnh lng tr l mt hnh ch nht.
g) Tnh Sx ca hnh lng tr.
h) Tnh s ca hnh lng tr.
Bi 9. Cho hnh chp S.ABC, c SA ( A B C ) .y ABC l tam gic
vung ti B c AB = a, BC= 2a. Gi I ltrung im ca BC. Bit
S =45.a) Tnh VSABI.
b) Tnh Slp ca hnh chp S.ABC.
c) Tnh vs ABC.Suy ra khong cch t A n (SBC) v khong cch t
c n (SAB).
63
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Bi 10. Cho hnh chp t gic u S.ABCD, cnh dy bng a, ng
SO v gc hp bi mt bn v y bng 60.
a) Gi ( l gc hp bi cnh bn SA v y. Tnh tan?.
b) Tnh Sxq v th tch V ca hnh chp.
Bi 11. Cho hnh chp t gic u S.ABCD cnh y bnga ,
ng cav gc hp bi cnh bn v y bng 45.
a) Gi (p l gc hp bi mt bn (SBC) v y. Tnh tan .
b) Tnh s v th tch V ca hnh chp.
c) Tnh khong cch t o n (SBC).
d) Gi a l gc hp bi SA v (SBC). Tnh sin a .
Bi 12. Cho hnh chp t gic u S.ABCD, c cnh y bng a. Gc bng 60. ng cao so.
a) Tnh so.
b) Tnh SX/, Slp v th tch V ca khi chp.
c) Tnh gc hp bi SB v y (ABCD).
d) Tnh khong cch t o n (SBC).
e) Tnh khong cch t D n (SBC).
64
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B C H
PHP V T V PHP NG DNG
TRONG KHNG GIAN
I. PH P V T
1. nh ngha
Cho s" k 0 v mt im o . Php bin hnh bin mi im M
trong khng gian thnh im M sao cho OM' = :OM gi l php
v t tm o t s k.
* Nhn xt: Php v t trong khng gian c nh ngha hon
ton nh php v t trong hnh hc phng. Do php v t
trong khng gian cng c cc tnh chtnh php v t trong hnh
hc phng.
2. V d
Cho t din ABCD. Gi A \ B \ c \ D ln lt l trng tm ca cc
tam gic BCD, ACD, ABD, ABC. Chng minh rng c php v t
bin t din ABCD thnh t din ABCD.
Hng dn gii
Gi G l trng tm ca t din A
ABCD. Khi ta bit rng:
G' = - g 3
GB' = - - G B 3
OI2 = 225 => OI = 15 (cm)
- Tam gic OIB vung ti I nn:
IB2 = OB2 - OI2 = 625 - 225 = 400 => IB = 20 (cm)
=> AB = 40 (cm)
- Tam gic SOI vung ti o nn:SI2 = SO2 + OI2 = 400 + 225 = 625 => SI = 25 (cm)
Vy din tch thit din SAB l:
SasAB = - AB.SI = - .40.25 = 500 (cm2).2 2
Bi 2.4. Mt mt phng (a) i qua hai ng sinh ca hnh nn ct my hnh nn theo mt dy cung c di gp 4 ln ng cao c
hnh nn. Tnh gc cp gia mt phng (a ) v mt y ca hnh n
nu (p bng na gc nh ca thit din ca hnh nn khi ct bi m
phng ().___________________________________________________
Hng d n gi i
Gi l tm ca y v mt phng (a) qua hai ng sinh A, SB
Gi I l trung im ca AB-
\ O I L A B -Khi : =>SO= ^ - = = - ^ = -= > co s< p = - = x p = 60.
tan A 2so 2 . 272
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Bi 2.5. Cho hnh nn nh s, ng cao so, A v B i 2 im thuc
ng trn y sao cho khong cch t o n AB bng av SAO= 30,
SAB= 60 Tnh di ng sinh ca hnh nn theo a.____________
Hng dn g iis
Gi I l trung im ca AB.
Khi : OI AB; SI J. AB v OI = at: SA =
- Tam gic SAO vung ti o nn:OA 0 3
cos30 = => OA = -cos30 =SA 2
- Tam gic SAB cn ti s c gc SAB =60
nn tam gic u =>SA = AB =>AB = t IA =
- Tam gic AIO vung ti I nn:
'XP2 p2 r-OA2=AI2 + OI2 = > = + a 2 => t 2 = 2 a2 => = a^ .
4 4
'& .VN93
Din tch hnh nn - th tch khi nn
* Phng php
Cho hnh nn N c chiu cao h,ng sinh tv bn knh y R.
Gi Sxq l din tch xung quanh ca hnh nn v V l th lch ca
khi nn N.
Ta c: Sxq= n R t
V = - 7 1R2h3
Bi 2.6. Ct mt hnh nn N bng mt mt phng i qua trc ca n, ta
c thit din l mt tam gic u cnh 2a.Tnh din tch xung quanh,
din tch ton phn ca hnh nn N (din tch ton phn l tcng din
tch xung quanh v din tch y) v th tch ca khi nn N._________
73
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