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{"id":22195,"date":"2018-03-17T11:21:08","date_gmt":"2018-03-17T04:21:08","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=22195"},"modified":"2018-03-17T11:21:08","modified_gmt":"2018-03-17T04:21:08","slug":"hinh-hoc-chuong-2-on-tap-chuong-2-bai-tap","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/hinh-hoc-chuong-2-on-tap-chuong-2-bai-tap\/","title":{"rendered":"H\u00ecnh h\u1ecdc – Ch\u01b0\u01a1ng 2 – \u00d4n t\u1eadp ch\u01b0\u01a1ng 2 – B\u00e0i t\u1eadp"},"content":{"rendered":"

B\u00e0i 1 (trang 63 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

Cho mp (P) v\u00e0 \u0111i\u1ec3m A kh\u00f4ng thu\u1ed9c m\u1eb7t ph\u1eb3ng (P). ch\u1ee9ng minh r\u1eb1ng m\u1ecdi m\u1eb7t c\u1ea7u \u0111i qua A c\u00f3 t\u00e2m n\u1eb1m tr\u00ean (P) lu\u00f4n lu\u00f4n \u0111i qua hai \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh.<\/span><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

\"\"<\/p>\n

Gi\u1ea3 s\u1eed S l\u00e0 m\u1eb7t c\u1ea7u \u0111i qua A v\u00e0 c\u00f3 t\u00e2m O \u2208 (P). (h\u00ecnh v\u1ebd b\u00ean). G\u1ecdi A\u2019 l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi A qua (P) th\u00ec OA\u2019 = OA n\u00ean m\u1eb7t c\u1ea7u S c\u0169ng \u0111i qua A\u2019.<\/p>\n

V\u1eady c\u00e1c m\u1eb7t c\u1ea7u S \u0111i qua hai \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh A v\u00e0 A\u2019.<\/p>\n

B\u00e0i 2 (trang 63 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

X\u00e1c \u0111\u1ecbnh t\u00e2m v\u00e0 b\u00e1n k\u00ednh m\u1eb7t c\u1ea7u ngo\u1ea1i ti\u1ebfp h\u00ecnh ch\u00f3p S.ABC, bi\u1ebft<\/span><\/p>\n

\"\"<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

\"\"<\/p>\n

Theo (gt) => AB = a, BC = a \u221a2 v\u00e0 AC = a \u221a3<\/p>\n

V\u1eady \u0394ABC vu\u00f4ng t\u1ea1i B.<\/p>\n

G\u1ecdi SH l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a h\u00ecnh ch\u00f3p th\u00ec do SA = SB = SC n\u00ean HA = HB = HC, do \u0111\u00f3 H l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AC. G\u1ecdi O l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi \u0111i\u1ec3m S qua \u0111i\u1ec3m H.<\/p>\n

Khi \u0111\u00f3 OS = OA = OC = OB = a. suy ra m\u1eb7t c\u1ea7u ngo\u1ea1i ti\u1ebfp h\u00ecnh ch\u00f3p c\u00f3 t\u00e2m O v\u00e0 b\u00e1n k\u00ednh R = a.<\/p>\n

B\u00e0i 3 (trang 63 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

Cho hai \u0111\u01b0\u1eddng tr\u00f2n (O, r) v\u00e0 (o\u2019, r\u2019) c\u1eaft nhau t\u1ea1i hai \u0111i\u1ec3m A, B v\u00e0 l\u1ea7n l\u01b0\u1ee3t n\u1eb1m tr\u00ean hai m\u1eb7t ph\u1eb3ng ph\u00e2n bi\u1ec7t (P) v\u00e0 (P\u2019).<\/span><\/p>\n

a) Ch\u1ee9ng minh r\u1eb1ng c\u00f3 m\u1eb7t c\u1ea7u (S) \u0111i qua \u0111\u01b0\u1eddng tr\u00f2n \u0111\u00f3<\/span><\/p>\n

b) T\u00ednh b\u00e1n k\u00ednh c\u1ee7a R c\u1ee7a m\u1eb7t c\u1ea7u (S) khi r = 5, r\u2019 = \u221a10, AB = 6, OO\u2019 = \u221a21<\/span><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

\"\"<\/p>\n

a) G\u1ecdi M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AB th\u00ec OM \u22a5AB,O’ M\u22a5AB. Do (P) v\u00e0 (P\u2019) ph\u00e2n bi\u1ec7t n\u00ean ba \u0111i\u1ec3m O, M, O\u2019 kh\u00f4ng th\u1eb3ng h\u00e0ng.<\/p>\n

T\u1eeb \u0111\u00f3 AB \u22a5 m\u1eb7t ph\u1eb3ng (OMO\u2019).<\/p>\n

G\u1ecdi \u0394 v\u00e0 \u0394’ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung tr\u1ef1c c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n (O, r) v\u00e0 (O\u2019, r\u2019) th\u00ec \u0394 v\u00e0 \u0394’ cung vu\u00f4ng g\u00f3c v\u1edbi AB.<\/p>\n

T\u1eeb \u0111\u00f3 suy ra \u0394 v\u00e0 \u0394’ c\u00f9ng n\u1eb1m trong m\u1eb7t ph\u1eb3ng (OMO\u2019). \u0394v\u00e0 \u0394’ c\u1eaft nhau t\u1ea1i \u0111i\u1ec3m I. Khi \u0111\u1ea5y m\u1eb7t c\u1ea7u (C ) c\u00f3 t\u00e2m I v\u00e0 b\u00e1n k\u00ednh R = IB l\u00e0 m\u1eb7t c\u1ea7u c\u1ea7n t\u00ecm.<\/p>\n

b)Ta c\u00f3:<\/p>\n

\"\"<\/p>\n

T\u01b0\u01a1ng t\u1ef1: O\u2019M = 1<\/p>\n

X\u00e9t \u0394OMO’ ta c\u00f3:<\/p>\n

\"\"<\/p>\n

Nh\u01b0 v\u1eady R2<\/sup>=IB2<\/sup>+IO2<\/sup>=25+12=37 t\u1ee9c R = \u221a37<\/p>\n

V\u1eady R=\u221a37<\/p>\n

B\u00e0i 4 (trang 63 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

Cho m\u1ed9t h\u00ecnh n\u00f3n H sinh b\u1edfi m\u1ed9t tam gi\u00e1c \u0111\u1ec1u c\u1ea1nh a khi quay quanh m\u1ed9t \u0111\u01b0\u1eddng cao.<\/span><\/p>\n

a) M\u1ed5 m\u1eb7t c\u1ea7u c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n c\u1ee7a h\u00ecnh n\u00f3 H th\u00ec c\u00f3 b\u00e1n k\u00ednh b\u1eb1ng bao nhi\u00eau?<\/span><\/p>\n

b) M\u1ed9t kh\u1ed1i c\u1ea7u c\u00f3 th\u1ec3 t\u00edch b\u1eb1ng th\u1ec3 t\u00edch c\u1ee7a kh\u1ed1i n\u00f3n H th\u00ec c\u00f3 b\u00e1n k\u00ednh b\u1eb1ng bao nhi\u00eau?<\/span><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

Theo gt h\u00ecnh n\u00f3n H c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y r=a\/2, c\u00f3 chi\u1ec1u cao h=a\u221a3\/2 v\u00e0 c\u00f3 \u0111\u01b0\u1eddng sinh l=a. v\u1eady n\u00f3 c\u00f3 di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n l\u00e0 S1<\/sub>\u00a0v\u00e0 V th\u1ec3 t\u00edch<\/p>\n

\"\"<\/p>\n

a) N\u1ebfu m\u1eb7t c\u1ea7u c\u00f3 b\u00e1n k\u00ednh R th\u00ec c\u00f3 di\u1ec7n t\u00edch l\u00e0 S2<\/sub>=4 \u03c0R2<\/sup><\/p>\n

Theo b\u00e0i ra<\/p>\n

\"\"<\/p>\n

b) N\u1ebfu m\u1eb7t c\u1ea7u b\u00e1n k\u00ednh R th\u00ec c\u00f3 th\u1ec3 t\u00edch V c\u1ee7a h\u00ecnh n\u00f3n th\u00ec<\/p>\n

\"\"<\/p>\n

B\u00e0i 5 (trang 63 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

Cho tam gi\u00e1c vu\u00f4ng ABC vu\u00f4ng t\u1ea1o A, AB = c, AC = b, g\u1ecdi V_1,V_2,V_3 l\u00e0 th\u1ec3 t\u00edch c\u00e1c kh\u1ed1i tr\u00f2n xoay sinh b\u1edfi tam gi\u00e1c \u0111\u00f3 (k\u1ec3 c\u1ea3 c\u00e1c \u0111i\u1ec3m trong) khi l\u1ea7n l\u01b0\u1ee3t quay quanh AB, AC, BC.<\/span><\/p>\n

\"\"<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

<\/div>\n

\"\"<\/p>\n

a) Khi quay tam gi\u00e1c ABC quanh AB ta \u0111\u01b0\u1ee3c kh\u1ed1i n\u00f3n c\u00f3 \u0111\u01b0\u1eddng cao c v\u00e0 b\u00e1n k\u00ednh \u0111\u00e1y b, n\u00ean n\u00f3 c\u00f3 th\u1ec3 t\u00edch<\/p>\n

\"\"<\/p>\n

Khi quay tam gi\u00e1c ABC quanh AC ta \u0111\u01b0\u1ee3c kh\u1ed1i n\u00f3n c\u00f3 \u0111\u01b0\u1eddng cao bv\u00e0 b\u00e1n k\u00ednh \u0111\u00e1y c, n\u00ean n\u00f3 c\u00f3 th\u1ec3 t\u00edch<\/p>\n

 <\/p>\n

G\u1ecdi AH l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c ABC. Khi quay tam gi\u00e1c ABC quanh BC (h\u00ecnh v\u1ebd) ta \u0111\u01b0\u1ee3c kh\u1ed1i tr\u00f2n xoay h\u1ee3p th\u00e0nh c\u1ee7a hai kh\u1ed1i n\u00f3n sinh ra b\u1edfi tam gi\u00e1c ABH v\u00e0 ACH khi quay quanh BC.<\/p>\n

V\u00ec v\u1eady:<\/p>\n

\"\"<\/p>\n

b) Ta c\u00f3:<\/p>\n

\"\"<\/p>\n

B\u00e0i 6 (trang 63 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao):<\/b><\/p>\n

M\u1ed9t h\u00ecnh thang c\u00e2n ABCD c\u00f3 d\u1ea1ng \u0111\u00e1y AB = 2a, DC = 4a. C\u1ea1nh b\u00ean AD = BC =3a. cho h\u00ecnh thang \u0111\u00f3 (k\u1ec3 c\u1ea3 \u0111i\u1ec3m trong) quay quanh tr\u1ee5c \u0111\u1ed1i x\u1ee9ng c\u1ee7a n\u00f3. H\u00e3y t\u00ednh th\u1ec3 t\u00edch v\u00e0 di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n c\u1ee7a kh\u1ed1i tr\u00f2n xoay \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh.<\/span><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

\"\"<\/p>\n

G\u1ecdi S l\u00e0 giao \u0111i\u1ec3m c\u1ee7a hai c\u1ea1nh b\u00ean AD v\u00e0 BC c\u1ee7a h\u00ecnh thang (h\u00ecnh v\u1ebd b\u00ean). \u0110\u01b0\u1eddng cao SO c\u1ee7a tam gi\u00e1c c\u00e2n SCD l\u00e0 tr\u1ee5c \u0111\u1ed1i x\u1ee9ng c\u1ee7a h\u00ecnh thang d\u00f3 \u0111\u00f3 SO c\u1eaft AB t\u1ea1i trung \u0111i\u1ec3m O\u2019 c\u1ee7a AB.<\/p>\n

Khi quay quanh SO, tam gi\u00e1c SCD sinh kh\u1ed1i n\u00f3n H1<\/sub>\u00a0c\u00f3 th\u1ec3 t\u00edch V1<\/sub>\u00a0tam gi\u00e1c SAB sinh ra kh\u1ed1i n\u00f3n H2<\/sub>\u00a0c\u00f3 th\u1ec3 t\u00edch V2<\/sub>\u00a0c\u00f2n h\u00ecnh thang ABCD sinh ra m\u1ed9t kh\u1ed1i tr\u00f2n xoay H c\u00f3 th\u1ec3 t\u00edch V = V1<\/sub>-V2<\/sub><\/p>\n

\"\"<\/p>\n

Ch\u00fa \u00fd r\u1eb1ng AB l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c SCD n\u00ean SB = 3a v\u00e0 do \u0111\u00f3:<\/p>\n

\"\"<\/p>\n

G\u1ecdi S1<\/sub>,S2<\/sub>\u00a0l\u1ea7n l\u01b0\u1ee3t l\u00e0 di\u1ec7n t\u00edch xung quanh c\u1ee7a kh\u1ed1i n\u00f3n H1<\/sub>,H2<\/sub>\u00a0th\u00ec di\u1ec7n t\u00edch xung quanh c\u1ee7a kh\u1ed1i tr\u00f2n xoay H l\u00e0 Sxq<\/sub>=S1<\/sub>-S2<\/sub>=\u03c0.OC.SC-PI O’ B.SB=9 \u03c0a2<\/sup><\/p>\n

=> Di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n STP<\/sub>=9 \u03c0a2<\/sup>+\u03c0a2<\/sup>+4 \u03c0a2<\/sup>=14 \u03c0a2<\/sup><\/p>\n

\n
<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

B\u00e0i 1 (trang 63 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao): Cho mp (P) v\u00e0 \u0111i\u1ec3m A kh\u00f4ng thu\u1ed9c m\u1eb7t ph\u1eb3ng (P). ch\u1ee9ng minh r\u1eb1ng m\u1ecdi m\u1eb7t c\u1ea7u \u0111i qua A c\u00f3 t\u00e2m n\u1eb1m tr\u00ean (P) lu\u00f4n lu\u00f4n \u0111i qua hai \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh. L\u1eddi gi\u1ea3i: Gi\u1ea3 s\u1eed S l\u00e0 m\u1eb7t c\u1ea7u \u0111i qua A v\u00e0 […]<\/p>\n","protected":false},"author":3,"featured_media":22203,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1298],"tags":[1395,1393],"yoast_head":"\nH\u00ecnh h\u1ecdc - Ch\u01b0\u01a1ng 2 - \u00d4n t\u1eadp ch\u01b0\u01a1ng 2 - B\u00e0i t\u1eadp<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/lop12.edu.vn\/hinh-hoc-chuong-2-on-tap-chuong-2-bai-tap\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"H\u00ecnh h\u1ecdc - Ch\u01b0\u01a1ng 2 - \u00d4n t\u1eadp ch\u01b0\u01a1ng 2 - B\u00e0i t\u1eadp\" \/>\n<meta property=\"og:description\" content=\"B\u00e0i 1 (trang 63 sgk H\u00ecnh H\u1ecdc 12 n\u00e2ng cao): Cho mp (P) v\u00e0 \u0111i\u1ec3m A kh\u00f4ng thu\u1ed9c m\u1eb7t ph\u1eb3ng (P). ch\u1ee9ng minh r\u1eb1ng m\u1ecdi m\u1eb7t c\u1ea7u \u0111i qua A c\u00f3 t\u00e2m n\u1eb1m tr\u00ean (P) lu\u00f4n lu\u00f4n \u0111i qua hai \u0111i\u1ec3m c\u1ed1 \u0111\u1ecbnh. 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