BAI TAP TONG HOP
J,). Tinh ban kinh R cua mat cau ngoai tiep tu difn SABC
L O I G I A I
a)- d u n g d i n h l i h a m so'cosin t r o n g ASAB:: j girfir*]
= SA^ + SB^ - 2.SA.SB.cosl20° , j k*:
^ B^ = a 2+ a^ - 2 a 2 — = 3 a 2= > A B = aV3
{2)
^SAB deu nen BC = a
A S A C v u o n g can tai S nen A C = ayfl .
Suy ra A B ^ = A C ^ + B C ^ ' Vay t a m giac A B C v u o n g tai C. * '
b). Gpi H la h i n h chie'u v u o n g goc ciia S tren m p ( A B C ) .
Do SA = SB = SC nen H A = H B = S C , do d o H la t a m d u o n g t r o n ngoai tiep cua A A B C . M a A A B C v u o n g tai C nen H la t r u n g d i e m ciia A B .
Goi r la ban k i n h ciia m a t cau n o i tiep h i n h chop S.ABC t h i ta c6: r = —
T i n h V : V = i s ^ 3 C . S H = i ia.a72 2
a a^72 12
Tinh S , „ : S , „ = S ^ B C + SgAC + ^SBC + S 'SAB
a^^f2 a^ a 2 2 aV2
•73 a^Vs _2
4 4 2 • V 'UaSm'
Suyra: r = j= = .
2(V3 + V 2 + 1 )
Goi O la t a m m a t cau ngoai tiep h i n h chop S.ABC va K la t r u n g d i e m ciia SA. T u cau b) suy ra S H la true d u o n g t r o n ngoai tiep A A B C n e n O e S H . Vi O A = OS nen O n a m tren t r u n g true ciia SA. •>'* t.ô,( j , a -
Hai t a m giac v u o n g S K O va S H A d o n g d a n g nen ta c6:
SO SK SK.SA SA^
SA S H S H 2SH
2 - 2
- = a ( S H = j )
JO;- >
^ ^ { g ^ a n k i n h m a t cau ngoai tiep h i n h chop S.ABC la R = a'.
hinh chop S.ABCD c6 day A B C D la h i n h thoi, goc A B C = 60, SA = SB = SC, 'eu cao ciia h i n h chop la a, goc g i i i a (SCD) va ( A B C D ) b a n g 60. T i n h the I g*^*^ S.ABCD. Xac d i n h t a m va ban k i n h m a t cau ngoai tiep t u d i $ n S.ACB,
^^^CP. T i n h k h o a n g each t u D deh m a t p h g n g (SBC).
L O I G I A I
361
Vi A B C = 60° nen A A B C v a A A C D deu.
Gpi G la trQng tam AABC tW: G A = GB = G C . Ma SA = SB = S C => G la hinh chieu vuong goc ciia S tren mat phSng ( A B C D ) . . l O i ? (w^m De dang chungminh G C l C D . ' 'i;i j Co C D 1 G C va C D 1 G C
C D 1S C (djnh ly ba duong vuong goc ).
Vay goc giua hai mat phSng (SCD) va (ABCD) la goc giua S C va C G la goc
S C G = 6 0 ° .
Trong ASCG c6: C G - SG a 73 tan 60°
SC = S G 2aV3 sin 60 ,0
A H n-m 'X He
AB.S aS . AB = a . Trong AABC c6: C G =
1 1 /- a ^Vs J. ^ u
Dien tich A B C D : S ^ B C D = - A C . B D = - a . a V 3 = '
1 1 a^yjs a^-Js
The tich khoi chop S.ABCD V = - S G . S A B C D = T - ^ - ^ = • Vi G la tarn duong tron ngoai tiep tam giac A B C va S G 1 ( A B C D ) . Suy ra S G la true duong tron ngoai tiep tam giac A B C .
Trong mat ph^ng (SGC) dung duong trung true cua doan thSng S C cat duong S G tai diem H . H chinh la tam mat cau ngoai tiep khoi chop S.ABC C O ban kinh R = S H .
ASMH ~ ASGC S H S M
• S H = S C '
2aV3 N2
.Ay.
2a
SC S G 2SG 2a 3
Gpi J la tam duong tron ngoai tiep A A C D , thi J cung la tam duong ngoai tiep tu giac A G C D . T u J dung duong thing a vuong goc vai ( A G C D ) thi a la true duong tron ngo^i tiep tu giac A G C D va a // S G . Trong mp (SGD) dung duong trung true ciia S G cat duong thSng a ''^
Vay: I la tam mat cau ngoai tiep khoi chop S . A G C D , ban kinh R = G I a a\f3 Ta c6: GJIN la hinh chCr nhat c6 IJ = G N = ^ , G J = 20J = ^ .
ay: R = G I = V G J ^ T J I ^ = a^ aV2T - +
I .
Vay: K = ằ^i = V<-J-+Ji- =-y 2 ^ ^
Vi A D / / B C : ^ d ( D , ( S B C ) ) = d ( A , ( S B C ) ) . ,, „^
Muon tinh khoang each tu A den mp (SBC), ta phai tinh khoang each tu G (hinh chieu cm dinh) den mp(SBC) truoc, sau do su dung cong thuc ti le khoang each thi tinh dugc khoang each tu A.
Gpi E la trung diem ciia BC. f' - . u ? ,
Co B C1G E va B C1 S G => B C1 ( S G E ) (SBC) 1 ( S G E ) [vi B C c (SBC) ].
Hai mat phang nay vuong goc voi nhau theo giao tuyen SE.
Dvrng G F 1 SE ( F e S E ) =^ G F 1 ( S B C ) . Vay d ( G , ( S B C ) ) = G F . ' ' Trong A S G E co: — - = — - + — - = _ + _ _ = = > G F = ^ ^ .
G F ^ G S ^ G E ^ 13 ^
T a c o A E = 3 G E d ( A , ( S B C ) ) = 3.d(G.(SBC)) = ^ '^^ ^ ^ f .
13
Cho hinh chop S A B C D c6 day A B C D la hinh thoi canh a, goc A B C = 60.
SA vuong goc voi A B C D , goc giiia (SCD) va (ABCD) bang 60.
Tinh the tich khoi chop S.ABCD.
Xac dinh tam va tinh ban kinh mat cau ngoai tiep \xx di^n SABC, SABD, SBCD.
L O I G I A I
Vi A B C = 60° nen A A B C v a A A C D deu. ' ^m.^^lAl,^ M'-'^'L'^
Gpi E la trung diem cua C D , c6 C D 1A E va C D 1 S A , C D 1 S E (dinh ly ba duong vuong goc).
T u do suy ra: goc giCra hai mat phSng (SCD) va (ABCD) la goc S E A = 60°.
Trong A S A E c o SA = AE.tan60° = ^ . ^ 3 = y T i n h the tich khoi chop S . A B C D :
V _ l c A c _ 1 3 a a ^ ^ / 3 a^S
^ S . A B C D - 3 ^ ^ - ^ A B C D - 3- Y - - y — = —^ ^ .. 0 A 3 A - (14,?/. . V ; (Voi S A B C D = f A C . B D = l a . a V 3 = ^ ) . ' Xac djnh tam va tinh ban kinh m$t cau ngoai tiep tu di^n S.ABC
363
GQI G la tam duang tron ngoai tiep AABC. Tu G ke duang thing d vuong goc voi mat phing (ABC) thi d la tryc ciia duong tron ngoai tiep tam giac i( -
^BCvad//SA.
Tronc, mat phang chua d va SA, ke duong thang a la duong trung tryc doan SA, duong thSng a cat duong thang d tai I thi I la tam mat cau ngdai tiep khoi chop S.ABC, ban kinh R = AI.
Ta c6: AGIM la hinh chu nhat c6: GI = A M =
Trong AAIG c6: A I = N/AG
3 16 12 Xac dinh tam va tinh ban kinh mat cau ngoai tiep tu di^n S.ABD
Vi CO CA = CB = CD => C la tam duong tron ngoai tiep tam giac ABD.
Tu C ke duang thSng b vuong goc voi (ABCD) thi b la true ciia duong tron ngoai tiep tam giac ABD va b // SA.
Trong mat ph5ng chiia b va SA, ke duang thSng c la duong trung true doan SA, duong thSng c cat duong thSng b tai H thi H la tam mat cau ngo^i tiep khoi chop S.ABD, ban kinh R = AH. '' ,
S A _ 3 a Ta c6: A C H M la hinh chu nhat c6: CH = A M = — = — . ~ 4
2 9a^
a +• 5a
Trong AAIG c6: AH - N/AC^TCH^ o AH = I ^
Xac dinh tam va tinh ban kinh mat cau ngoai tiep tu di?n SBCD.
Vi CO A B = A C = A D => A la tam duong tron ngoai tiep tam giac B C D . Vi S A 1 ( A B C D ) SA la true duong tron ngoai tiep tam giac B C D .
Trong A S A D ke duong trung true doan S D eat S A tai diem J, thi J la tam mat cau ngoai tiep khoi chop S . B C D c6 ban kinh R = JS.
Co: ASNJ ~ ASAD SJ SN
SD SA SA 2SA
13a^
2 . ^ 2
13a 12
Cho hinh chop S.ABCD c6 day ABCD la hinh vuong tam O, SA vuong go^
voi day ABCD, SA = a , goe giua SC va SAB bang 30°, mat phSng (P) qua A va vuong goc vai SC, eat SB, SC, SD Ian lugt tai B', C, D'.
364
a) Chung minh nam diem A , S , B', C , D' ciing nam tren mgt m|it cau Tim tam va tinh ban kinh mat cau do . , i,a •
b) Chung minh rang A, B, C, D, B', C, D' nam tren mot mat cau eo dinh.
Tim tam va tinh ban kinh ciia mat cau do.
c) Ti'nh the tieh cua S.AB'C'D'.
L O I G I A I
Vi CB 1 (SAB) nen SB la hinh chieu vuong goc cua SC tren mat phJing (SAB).
Vay: goc giua SC va (SAB) la CSB = 30° . Trong ASBC c6: SB = BC: tan 30° = BC.S . Trong ASAB vuong tai A c6:
SB^ = SA^ + AB^ ô 3BC^ =a^+ AB^
^ AB = — (vi AB = BC).
2 , t. , , . . . > • ; • A
Ma ABCD la hinh vuong AC = BC = AB.V2 = a Vi S C l ( P ) = > S C 1 A C ' . . .
Ta c6: BD 1 AC va BD 1 SA BD 1 ( S A C ) =^ BD 1 SC . " ' '
Tu do suy ra BD // ( p ) . ' ' ' "-'^ ^^'Wl ort:v
Goi O = AC n BD . Ta CO : SO = (SAC) n (SBD) . ' l , ; l ' t J r r 2 r D Goi I = AC'nSO. I la diem ehung eiia hai mat phSng (SBD) va (P) va
BD // (P). Suy ra giao tuyeh ciia ehung qua I va song song vai BD, giao tuyen nay cat SB va S D Ian lugt tai B' va D'.
Vi BC 1 AB va BC 1 SA BC 1 ( S A B ) ^ BC 1 AB' (vi AB' c ( S A B ) ).
A B ' I B C , ,
r u - , . / X • " " ^ ^ ^ ô ^ ' " •
'-nung mmh tuong tu eo: A D ' 1 (SCD).
Vi AB' 1 (SBC) AB' 1 SB'. Vay tam giac SAB' vuong tai B'. ' Tuong tu: ASAD' vuong tai D'. ' f ' >'SW> ^r'ằn wno uJ •••:
Vi ba diem B', C, D' cijng nhin doan SA dual mgt goc vuong nen ehung nam tren mat cau duang kinh SA, c6 tam trung diem SA va ban kinh
SA _ a , Ta eo:
Theo Chung minh tren c6: A B ' C = A C ' C = A D ' C = A D C = A B C = 90° . T u do suy ra bay diem A , B , C , D , B', C, D ' ciing nkm tren mat cau duorig kinh A C ban kinh R = V2 A C = V2 a. • ,
A p dyng h? thiic lugng trong A vuong cho hai A S A B va S A D c6:
. . . ^ T . 0 * 2 S D ' . S D S A ^ S D - S A ^ 2
S D ' . S D = S A ^ o — = — - ô — = — 5 7 = r - S D ^ S D ^ S D S A ^ + A D ^ 3
S B ; ^ 2 ^ '
SB 3'
Co: S A = A C = a => A S A C can tai A suy ra C trung diem ciia S C . Tuong tu ta tinh dugc: ^„ - , • , ,„-^ .
V i c o : VsABCD = 3 -S A . S A B C D = a 3 a^! .0-;. :)fi<r!/
V s . A B C _ S A SB' S C ' _ 2 T _ l _ 1 - - S C - 3 • 2 -1 ^ ^s-^^'^' - 3 ^^-^^^ " 6 ^S-^^^D •
Tuong tu Vg A D C = 3 Vs.ADC = ^VsABCD
1 1 1 a-^
^ S . A B ' C ' D ' ='Vs.AB'C' + ^ S . A D ' C ' ^ ^ S . A B C D + ^ ^ S . A B C D = 3' ^ S . A B C D = ^ Cho hinh chop S . A B C D day A B C D c6 hai duang cheo vuong goc voi nhau tai H , S H vuong goc voi day A B C D .
Chung minh rang bon tam mat cau ngoai tiep cac hinh chop S.HAB, S.HBC, S H C D , S . H D A tao thanh mpt hinh chu nhat.
Gpi H i , H 2 , H\, la hinh chieu vuong goc cua H Ian lupt tren AB, BC, C D , D A . Chung minh rang hinh chop S H 1 H 2 H 3 H 4 c6 mat cau ngoai tiep Tinh di^n tich cua thiet dif n mat cau ay khi cat boi mp(ABCD), neu biet
H , H 3 = a , B A C = a , B D C = p.
L O I G I A I
a). Gpi Ij la trung diem cua A B va O , la tam mat cau ngoai tiep hinh choi S.ABH thi I j O i // S H va ip^ = - S H .
Tuong tu nhu tren, neu I2/I3/I4 thu tu la trung diem ciia BC, C D , D A . V a © 2 , 0 3 , O 4 thu ty la tam cac mat cau ngoai tiep hinh chop S.HBC, S.H^P S.HDA thi: I2O2 = | s H ; I3O3 = | s H ; I4O4 = ^ S H .
Va I2O2,13O3, I4O4 cung song song voi S H .
' ^ thay: I^Is // Op^ Ijls // A C ;
^ 1 3 / 7 0 2 0 3 v a l 2 l 3 / / B D . I3I4 / / O 3 O 4 va I3I4 / / A C ; I ^ I , / / O 4 O 1 va I 4 I 1 / / B D . Ket hgp voi A C 1 B D , ta c6:
O J O 2 O 3 O 4 la hinh chCr nhat.
I,). De thay: H H^ 2 = ™ H 2 = H B C
H H; H 4 = H A H 4= r i A b _ ^
. I'Oi, J
H H 3 H 2 = H C H 2 = H C B H H ^ = H C H 4 = H D A
Ttrdo: H H , H 2 + H H 1 H 4 + H H 3 H 2 + H H 3 H 4
= H B C + H C B + H A D + H D A = 180"
V?y: H J H 2 H 3 H 4 la tii giac npi tiep duong tron.
Tu do hinh chop S. H J H 2 H 3 H 4 c6 mat cau ngoai t i ^ . D
Di?n tich thiet dien cua hinh cau ^ do va mat phSng (ABCD) la di^n
tich hinh tron ngogi tiep tu giac HJH2H3H4. • - -
Vi B A C = a , B D C = p nen H, H 4 H 3 = a + p .
^ gj„/^ = 2^ '^"h ditang tron ngoai tiep tu giac HJH2H3H4).
( a + p) Tu do: R =
2 s i n ( a + p ) ' . H ^ > J l a r : , V^y: di?n tich hinh thu dupe la 47tR^ =
H i
sm ' ( a . p )
Cho hinh chop luc giac deu S . A B C D E F c6 canh day bSng a va mat ben
I hpp voi day m^t goc a .
Xac djnh tam va tinh ban kinh ciia mat cau ngoai tiep hinh chop theo a va a . T ^ac dinh tam va tinh ban kinh ciia mat cau noi tiep hinh chop theo a va a .
a de tam mat cau ngoai tiep va npi tiep cua hinh chop trung nhau.
L O I G I A I
a) Gpi H la tam luc giac deu ABCDEF thi SH la duang cao ciia hinh chop.
Vi S ABCDEF la hinh chop deu nen SH la true aia duong tron ngoai tiep ABCD^
:^ GQ'I K la trung diem ciia SA.
, Dung mat trung true ciia SA cat SH tai diem O.
' Ta dupe diem O la tam ciia mat cau ' ''^
: ! ngo^i tiep hinh chop S.ABCDEF. .- ./-ur
Gpi J la trung diem cua AB. A~7r-/-,;o , Ta c6: AB 1 SH va AB 1 HJ nen AB 1 (SHJ). / A'X _ i _ . 1.1 £
Suy ra: SJ1 AB . , , ^ '^^
-J Suy ra goc SJH la goc phang tao boi A
V , hai mat (SAB) va (ABCDEF), va SJH = a . j Hai tam giac vuong SHA va SKO dong dang nen: B
so SK
SA " S H SO = SA.SK SA^
SH 2SH
Tu ASHJ vuong tai H ta CO SH = SJ.tana = .tana.
Tam giac SHA vuong tai H cho:
SA^ =SH^ +HA^ = 3a ^
tan a + a 2 a= —f3tan^a + 4 ^ , _ 2 4 V
Suy ra: SO = •
U 3 t a n 2 a + 4) 37^(3^3^2 ^ ^
2 ^ t a n a 12 tan a
Vay ban kinh ciia mat cau ngoai tiep hinh chop S.ABCDEF la:
a\/3(3tan^ a + 4) f .-y., , :.
R =
12 tan a
b) Trong ASHJ ke phan giac goc SJH cM SH tai diem I . Goi P la hinh chieu vuong goc cua I tren SJ.
Ta c6: IP 1 SJ. Suy ra AB I I P .
Nhu the IP 1 (SAB), hie la dp dai doan IK la khoang each tu I den mp (SAB) Mat khac: I nam tren phan giac cua goc SJH nen I H IP .
Do do diem I la diem each deu hai mat (ABCDEF) va (SAB).
Suy ra I each deu tat ca cac mat cua hinh chop.
Vay: I la tam mat cau npi tiep hinh chop S.ABCDEF.
Tu tam giac IHJ vuong tai H ta c6 I H = HJ.tan IJH = a^ tan—
2 2
Tam m|t cau ngoai tiep va npi tiep trung nhau khi va chi khi:
c u a V 3 ( 3 t a n 2 a + 4) aVs^ a aVs
R + r = SH<ằ \ i + t a n - = —!^tana ., 12tana 2 2 2
3tan'^a + 4 ^ a ^ ^ 2 . a
<=>—; + tan—= tana<::>3tan a - 6 t a n a . t a n 4 = 0
6tana 2 2 Dat t = tan— ta dupe phuong trinh 3 a 2t \
1-t^ - 6 2t
l - t ^ . - 4 = 0 ' >
Hay 2t'' + 2t2 - 1 = 0 . Giai dupe = o t = N/ 3 - 1 (t>0)
Vay tam mat cau ngoai tiep va npi tiep cua hinh chop S.ABCDEF trung nhau khi goc a dupe xac djnh boi t a n ^ =
Cho hinh chop S.ABCD c6 day ABCD la hinh vuong canh a, hai mat phMng (SAB) va (SAD) cung vuong goc voi mat phJing (ABCD). Gpi I, J, K Ian lupt la hinh chieu vuong goc ciia A tren SB, SC,SD.
a) Chung minh bay diem A, B, C, D, I, J, K ciing n3m tren mpt mat cau. Tinh di^n tich cua mat cau do.
b) Gpi (T) la duong tron c6 duong kinh la AC nam tren mat phSng (SAC).
Tren (T) lay diem M bat ky (khac A va C). Chung minh rang MBD la mpt tam giac vuong.
LOI GIAI Tu gia thiet ABC = ADC = AJC = 90° (1)
(SAB) 1 (ABCD) Cung tu gia thiet:
B C I A B Ta co: < B C I S A Do d o B C l A I .
J A I 1 BC A l l SB
(SAD) 1 (ANBCD) (SAB)n(SAD) = SA B C l ( S A B ) .
>;ằ<jeii i M' x I l i i^\"t%iH^
. S A l j A B C D )
A I 1 (SBC) Suy ra A I 1 I C hay AIC = 90°
Tuongtv: AKC = 90° (3)
369
Tu (1), (2), (3) suy ra: B^y diem A, B, C, D, I, J, K ciing nam tren mpt mat ca\
(S) duong kinh AC. Ban kinh ciia (S) la R = ^ = • i f .
Di?n tich cua m|it cau (S) la S = 47i = 27ta%:nfil + - • - 'on.i!.:?',.
Diem M n l m tren duong tron (T) c6 duong kinh la AC, nen M nam tre^
mat cau (S). De y r^ng BD la duorig kinh cua (S), nen voi diem M e (s) ta phai CO BMP = 90° . Vay tam giac MBD vuong tai M . ' ' '
De thi d?i hpc KHOI D 2003
Cho hai mat phSng (P) va (Q) vuong goc voi nhau, c6 giao tuyen la duong thing. Tren giao hiyen lay hai diem A, B voi AB = a. Trong mat phSng (P) lay diem C, trong mat phing (Q) lay diemO sao cho AC, BD vuong goc voi nhau va AC = BD = AB. Tinh ban kinh mat cau ngoai tiep tii di^n ABCD va tinh khoang each tu A den mat phJng (BCD) theo a.
LOI GIAI finTaco: ( P ) 1 ( Q ) va A = ( P ) n ( Q )
^ ^ ^ ^ , = ^ A C 1 ( Q ) = > A C 1 A D .
A C e P ) ^
(:: Ma:
•jom ^'
Hay CAD = 90". i^, . 4f
Chung minh hoan toan tuong ty: BD 1 (P^
r : > B D l B C hay CBb = 90°. "Ur
Vay: bon diem A, B, C, D thuoc mat cau tarn 1 trung diem cua CD ban kinh R =
C D = V B D ^ + B C ^ = V A B ^ + A C ^ + B D ^ = aS • Vay ban kinh R = ^ Dung A H l B C ( H e B C ) (1). 18
Vi BD 1 ( P ) => (BCD) 1 (ABC) c6 giao tuyen AB
Tu (1) va (2) suy ra: A H 1 ( B C D ) ^ d(A,(BCD)) = A H 86.1
\ a a^/2 l U i ,>lx.lA ôn.
AH^
J L i - = _
a^\^^a^
M A T T R U 1. Mat try tron xoay
Trong mp(P) cho hai duong thang A va I song song nhau, each nhau mpt khoang r. Khi quay mp(p) quanh true CO dinh A thi duong thang 1 sinh ra mot mat tron xoay dugc gpi la mat try tron xoay hay goi tat la mat tru. ^ ., - Duong thang A dupe gpi la true.
- Duong thang 1 dupe gpi la duong sinh.
- Khoang each r dupe gpi la ban kinh cua mat try^
2. Hinh try tron xoay
Khi quay hinh chu nhat ABCD xung quanh duong thSng chiia mpt canh, chang han canh AB thi duong gap khiic ABCD tao thanh mpt hinh, hinh do dupe gpi la hinh try tron xoay hay gpi tMt la hinh try.
- Duong thang AB dupe gpi la tryc. p n ? * i w i t i o O - Doan thang CD dupe gpi la duong sinh.
- Dp dai doan thang AB = CD = h dupe gpi la chieu cao ciia hinh try.
- Hinh tron tam A , ban kinhr = ADva hinh tron tamB, ban kinh r = BC dupe gpi la 2 day cua hinh try.
- Khoi try tron xoay, gpi tat la khoi try, la phan khong gian gioi han boi hinh try tron xoay keca hinh try. ^
3. Cong thuc tinh dien ti'ch va the rich cua hinh try M'Wff^-^^.
Cho hinh try c6 chieu cao la h va ban kinh day bang r, khi do: riniT - Dif n tich xung quanh ciia hinh try: S^q = 27trh
- Di^n tich toan phan ciia hinh try:
- The tich khoi tru: /,
Stp = + 2.Sj)ay-= 27trh + 27tr^
V = B.h = Tcr^h
rbi.' n%\h rim
4. Tinhcha't:
- Ne'u cat mat try tron xoay (c6 ban kinh lar) boi mptmp(a) vuong goc voi tryc A thi ta dupe duong tron c6 tam tren A va c6 ban kinh bang r voi
r ciing chinh la ban kinh ciia mat try do.
- Ne'u cat mat try tron xoay (c6 ban kinh la r ) boi mpt mp(a) khong vuong goc voi tryc A nhung cat tat ca cac duong sinh, ta dupe giao tuyen la mpt 37!
d u o n g e l i p c6 t r u n h o b a n g 2 r va t r y c I a n b a n g - ^ ^ , t r o n g d o (p la goc giug true A v a m p ( a ) v a i 0° < cp < 9 0 ° .
- C h o m p ( a ) song song v o l t r u c A c u a mat t r u t r o n xoay va each A mot k h o a n g k .
+ Ne'u k < r t h i m p ( a ) cat m a t t r y theo hai d u o n g sinh=> t h i e t d i # n la hinh c h i i nhat.
+ Ne'u k = r t h i m p ( a ) tiep xiic v o i m a t t r y theo m p t d u a n g sinh.
+ N e u k > r t h i m p ( a ) k h o n g cat m a t t r u .
M p t h i n h t r u c6 ban k i n h day bang R va thiet d i ^ n qua tmc la m p t h i n h vuong.
a) . T i n h d i e n t i c h x u n g q u a n h va di?n t i c h toan p h a n ciia h i n h t r y b) . T i n h the t i c h cua k h o i t r y
fir.
L O I G I A I G p i thiet d i ^ n qua t r y c la h i n h v u o n g A B B ' A ' . N e n CO A B = A A ' = O O ' = 2R
S^q = 27tRl = 27r.OA.OO' = 271.R.2R = 4nR^
= S , „ + 2 Sd.iy = 47iR^ + 27tR^ = 6nR2 V = Sd.iy. h = 7t.R^2R = 27iR^
M o t h i n h t r y c6 ban k i n h day r = 5cm va khoang each g i i i a hai day bang 7cm.
a) T i n h d i ^ n t i c h x u n g q u a n h va d i ^ n tich toan p h a n ciia h i n h t r y b) T i n h the t i c h ciia k h o i t r y
c) Cat k h o i t r y b o i m p t m a t phSng song song v o i t r y c va each t r y 3cm. Hay t i n h d i ? n tich ciia thiet d i ^ n d u p e tao nen.
Up
L O I G I A I a) . T i r J i d i e n t i c h x u n g q u a n h : i
S^q = 27irh = 27t.OA.OO' - 2:1.5.7 = VOn ( c m ^ ) . D i ? n t i c h t o a n p h a n ciia h i n h t r y :
Stp = Sxq + 2Sday = 70 K + 50 71 = 120 71 (cm^) b) . T i n h the tich ciia k h o i t r y :
V = TiR^h = 7 i. O A ^ O O ' = 71.52.7 = 17571 (em3)
V i O O ' v u o n g goc v o i hai day, m p t mp(P) / / O O ' va cat h i n h t r y , t h i thiet dien la m p t h i n h chCr nhat A B B ' A ' .
G p i I t r u n g d i e m ciia day A B , theo de bai eo O I J. A B va O I = 3cm .
T r o n g A O A I eo: A I = VoA^ - O I ^ =4 => A B = 2 A I = 8 , c6 A A ' = O O ' = 7 . D i ? n t i c h thiet d i e n A B B ' A ' S = A B . A A ' = 8.7 = 56^cm^ j .
M p t h i n h t r y c6 ban k i n h r va chieu eao h = r \/3 .
a) . T i n h d i e n t i c h x u n g q u a n h va d i f n tich toan p h a n ciia h i n h t r y . b) . T i n h the t i c h ciia k h o i t r y tao nen b a i h i n h t r y da cho .
c) . Cho h a i d i e m A va B Ian l u p t n a m tren hai d u a n g t r o n d a y sao cho goc g i i i a d u o n g t h a n g A B va t r y c ciia h i n h t r y bang 30°. T i n h k h o a n g each g i i i a d u o n g t h a n g A B va true ciia h i n h t r y .
L O I G I A I a) . T i n h d i ^ n t i c h x u n g q u a n h :
S^q =27rrh = 27:.r.rV3 = 2V37ir2. [ D i ^ n t i c h toan p h a n ciia h i n h t r y :
Stp = Svq + 2Sd.iy = 2V37rr2 + 27:r^ = 2(V3 + l)7tr^
b) . T i n h the tich ciia k h o i t r y : ^ V = T i r^ h = ni^rS = yf3ni^
D u n g A C / / O O ' .
Goc g i u a A B va O O ' b a n g goc g i u a A B va A C , c h i n h la goc B A C = 30° . T u d o s u y ra: d ( A B , 0 0 ' ) - d ( 0 0 ' , ( A B C ) ) = d ( 0 , ( A B C ) ) .
D u n g O H l B C ( H e B C ) t h i O H l ( A B C ) . Vay d ( 0 , ( A B C ) ) = O H . T r o n g A A B C v u o n g tai C eo BC = A C . t a n 30° = rVs. = r . •i-n njn "
Suy ra A O B C n e n eo O H = . J t^m mv.Sax/^'SH {4*. n%v .
K e t l u a n : d ( A B , 0 0 ' ) =
M p t h i n h t r y c6 cac d a y la hai h i n h t r o n t a m O va O ' b a n k i n h r va c6 d u a n g cao h = rv'2 . G p i A la m p t d i e m tren d u o n g t r o n t a m O va B la m p t d i e m tren d u a n g t r o n t a m O ' sao cho O A v u o n g goc v o i O ' B . a) C h u n g m i n h r a n g cac m a t ben ciia t u d i f n O A B O ' la n h i i n g t a m giac
^ v u o n g . T i n h the t i c h ciia h i d i ^ n nay.
171
b) Gpi (a) la mat phang qua AB va song song voi OO'. Tinh khoang each giua tryc OO' va mat phang (a) .
c) . Chung minh rang (a) tiep xuc voi m$t try tryc OO' c6 ban kinh bSng doc theo mpt duong sinh.
' . £ v i - riLOIGIAI a) . Vi true OO' vuong goe voi eac day nen
O O ' l O A va O O ' l O ' B .
Vay eae tam giac AOO' va BO'O vuong tgi O va O'.