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{"id":28634,"date":"2018-04-26T00:16:39","date_gmt":"2018-04-25T17:16:39","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=28634"},"modified":"2018-04-26T00:16:39","modified_gmt":"2018-04-25T17:16:39","slug":"bai-tap-trac-nghiem-hinh-hoc-12-chuong-2-bai-2-mat-cau-phan-5","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/bai-tap-trac-nghiem-hinh-hoc-12-chuong-2-bai-2-mat-cau-phan-5\/","title":{"rendered":"B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m H\u00ecnh h\u1ecdc 12 \u2013 Ch\u01b0\u01a1ng 2 – B\u00e0i 2: M\u1eb7t c\u1ea7u (Ph\u1ea7n 5)"},"content":{"rendered":"

C\u00e2u 11:<\/b>\u00a0Cho h\u00ecnh ch\u00f3p S.ABC c\u00f3 SA = SB = SC = a v\u00e0 tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A. G\u00f3c gi\u1eefa SB v\u1edbi \u0111\u00e1y l\u00e0 60o<\/sup>\u00a0. Th\u1ec3 t\u00edch kh\u1ed1i c\u1ea7u ngo\u1ea1i ti\u1ebfp h\u00ecnh ch\u00f3p l\u00e0 :<\/p>\n

\"\"<\/p>\n

C\u00e2u 12:<\/b>\u00a0Cho l\u0103ng tr\u1ee5 \u0111\u1ee9ng ABC.A’B’C’ c\u00f3 \u0111\u00e1y ABC l\u00e0 tam gi\u00e1c vu\u00f4ng c\u00e2n t\u1ea1i A, BC = a\u221a2 v\u00e0 g\u00f3c gi\u1eefa A\u2019B v\u00e0 m\u1eb7t ph\u1eb3ng (ABC) l\u00e0 60o<\/sup>\u00a0. B\u00e1n k\u00ednh c\u1ee7a m\u1eb7t c\u1ea7u ngo\u1ea1i ti\u1ebfp l\u0103ng tr\u1ee5 l\u00e0 :<\/p>\n

\"\"<\/p>\n

C\u00e2u 13:<\/b>\u00a0Cho h\u00ecnh ch\u00f3p S.ABCD c\u00f3 \u0111\u00e1y ABCD l\u00e0 h\u00ecnh vu\u00f4ng c\u1ea1nh b\u1eb1ng 2a, \u0394SAB l\u00e0 tam gi\u00e1c vu\u00f4ng c\u00e2n \u0111\u1ec9nh S v\u00e0 m\u1eb7t ph\u1eb3ng (SAB) vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng \u0111\u00e1y h\u00ecnh ch\u00f3p. B\u00e1n k\u00ednh m\u1eb7t c\u1ea7u ngo\u1ea1i ti\u1ebfp h\u00ecnh ch\u00f3p l\u00e0 :<\/p>\n

\"\"<\/p>\n

C\u00e2u 14:<\/b>\u00a0Cho h\u00ecnh n\u00f3n c\u00f3 \u0111\u01b0\u1eddng cao b\u1eb1ng \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y. X\u00e9t m\u1eb7t c\u1ea7u (S) n\u1eb1m trong h\u00ecnh n\u00f3n ti\u1ebfp x\u00fac v\u1edbi \u0111\u00e1y v\u00e0 t\u1ea5t c\u1ea3 \u0111\u01b0\u1eddng sinh c\u1ee7a h\u00ecnh n\u00f3n. T\u1ec9 s\u1ed1 th\u1ec3 t\u00edch c\u1ee7a kh\u1ed1i c\u1ea7u v\u00e0 kh\u1ed1i n\u00f3n l\u00e0 :<\/p>\n

\"\"<\/p>\n

C\u00e2u 15:<\/b>\u00a0Cho h\u00ecnh ch\u00f3p S.ABC c\u00f3 \u0111\u00e1y ABC l\u00e0 tam gi\u00e1c vu\u00f4ng c\u00e2n \u0111\u1ec9nh A v\u00e0 AB = SB = a , SB vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng (ABC). B\u00e1n k\u00ednh nh\u1ecf nh\u1ea5t c\u1ee7a m\u1eb7t c\u1ea7u ti\u1ebfp x\u00fac v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng SC v\u00e0 AB l\u00e0 :<\/p>\n

\"\"<\/p>\n

H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n\n\n\n
11-A<\/td>\n12-B<\/td>\n13-B<\/td>\n14-A<\/td>\n15-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

C\u00e2u 11:<\/b><\/p>\n

H\u1ea1 SH \u22a5 (ABC) t\u1ea1i H, khi \u0111\u00f3 ta c\u00f3 \u0394SHA = \u0394SHB = \u0394SHC n\u00ean HA = HB = HC hay H l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c ABC. Ta c\u00f3 :<\/p>\n

\"\"<\/p>\n

Trong m\u1eb7t ph\u1eb3ng (SAH), \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a SA c\u1eaft SH t\u1ea1i I. Khi \u0111\u00f3 I c\u00e1ch \u0111\u1ec1u c\u00e1c \u0111\u1ec9nh c\u1ee7a h\u00ecnh ch\u00f3p n\u00ean I l\u00e0 t\u00e2m m\u1eb7t c\u1ea7u ngo\u1ea1i ti\u1ebfp h\u00ecnh ch\u00f3p. G\u1ecdi M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a SA, khi \u0111\u00f3 ta c\u00f3 :<\/p>\n

\"\"<\/p>\n

Khi \u0111\u00f3 th\u1ec3 t\u00edch kh\u1ed1i c\u1ea7u l\u00e0 :<\/p>\n

\"\"<\/p>\n

C\u00e2u 12:<\/b><\/p>\n

Trong tam gi\u00e1c vu\u00f4ng ABC ta c\u00f3<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

=> AA’ = AB.tan60o<\/sup>\u00a0= a\u221a3.<\/p>\n

G\u1ecdi I l\u00e0 t\u00e2m c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt BCC\u2019B\u2019 v\u00e0 M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a BC. Do tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A n\u00ean M l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c ABC v\u00e0 do \u0111\u00f3 IM l\u00e0 tr\u1ee5c c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp \u0111\u00e1y ABC v\u00e0 I c\u00e1ch \u0111\u1ec1u B, B\u2019 n\u00ean I l\u00e0 t\u00e2m c\u1ee7a m\u1eb7t c\u1ea7u ngo\u1ea1i ti\u1ebfp l\u0103ng tr\u1ee5. Khi \u0111\u00f3 ta c\u00f3 :<\/p>\n

\"\"<\/p>\n

C\u00e2u 13:<\/b><\/p>\n

G\u1ecdi H l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AB, suy ra SH \u22a5 AB. Khi \u0111\u00f3 ta c\u00f3 :<\/p>\n

\"\"<\/p>\n

G\u1ecdi O l\u00e0 t\u00e2m c\u1ee7a h\u00ecnh vu\u00f4ng ABCD. X\u00e9t tam gi\u00e1c vu\u00f4ng SAB c\u00f3 H l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c SAB. Khi \u0111\u00f3 ta c\u00f3 HO l\u00e0 tr\u1ee5c c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c SAB. V\u1eady \u0111i\u1ec3m O c\u00e1ch \u0111\u1ec1u c\u00e1c \u0111\u1ec9nh c\u1ee7a h\u00ecnh ch\u00f3p v\u00e0 O l\u00e0 t\u00e2m m\u1eb7t c\u1ea7u ngo\u1ea1i ti\u1ebfp h\u00ecnh ch\u00f3p S.ABCD. B\u00e1n k\u00ednh m\u1eb7t c\u1ea7u ngo\u1ea1i ti\u1ebfp l\u0103ng tr\u1ee5 l\u00e0 :<\/p>\n

r = OA = AC\/2 = a\u221a2<\/p>\n

C\u00e2u 14:<\/b><\/p>\n

G\u1ecdiV1<\/sub>, V2<\/sub>\u00a0l\u1ea7n l\u01b0\u1ee3t l\u00e0 th\u1ec3 t\u00edch c\u1ee7a kh\u1ed1i c\u1ea7u v\u00e0 kh\u1ed1i n\u00f3n. T\u1eeb gi\u1ea3 thi\u1ebft ta c\u00f3 h = 2r => l = r\u221a5.<\/p>\n

G\u1ecdi S, O l\u1ea7n l\u01b0\u1ee3t l\u00e0 \u0111\u1ec9nh v\u00e0 t\u00e2m c\u1ee7a h\u00ecnh n\u00f3n. G\u1ecdi I l\u00e0 t\u00e2m c\u1ee7a m\u1eb7t c\u1ea7u (S), khi \u0111\u00f3 I thu\u1ed9c \u0111o\u1ea1n SO. L\u1ea5y m\u1ed9t \u0111i\u1ec3m A thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n \u0111\u00e1y c\u1ee7a h\u00ecnh n\u00f3n. G\u1ecdi H l\u00e0 ch\u00e2n \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c h\u1ea1 t\u1eeb I l\u00ean SA, g\u1ecdi r\u2019 l\u00e0 b\u00e1n k\u00ednh c\u1ee7a m\u1eb7t c\u1ea7u (S). Khi \u0111\u00f3 ta c\u00f3 IO = IH = r\u2019<\/p>\n

\"\"<\/p>\n

C\u00e2u 15:<\/b><\/p>\n

M\u1eb7t c\u1ea7u S(I,r) ti\u1ebfp x\u00fac v\u1edbi AB, SC l\u1ea7n l\u01b0\u1ee3t t\u1ea1i T, K. Khi \u0111\u00f3 ta c\u00f3:<\/p>\n

2r = IT + IK \u2265 d(AB; SC) => r \u2265 d(AB, SC)\/2<\/p>\n

D\u1ef1ng h\u00ecnh b\u00ecnh h\u00e0nh ABDC, khi \u0111\u00f3 ta c\u00f3 ABDC l\u00e0 h\u00ecnh vu\u00f4ng c\u1ea1nh a. H\u1ea1 BH vu\u00f4ng g\u00f3c v\u1edbi SD t\u1ea1i H. Khi \u0111\u00f3 ta c\u00f3 BH \u22a5 (SCD).<\/p>\n

Suy ra: d(SC; AB) = d(AB, (SCD)) = d(B; (SCD))<\/p>\n

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

C\u00e2u 11:\u00a0Cho h\u00ecnh ch\u00f3p S.ABC c\u00f3 SA = SB = SC = a v\u00e0 tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A. G\u00f3c gi\u1eefa SB v\u1edbi \u0111\u00e1y l\u00e0 60o\u00a0. Th\u1ec3 t\u00edch kh\u1ed1i c\u1ea7u ngo\u1ea1i ti\u1ebfp h\u00ecnh ch\u00f3p l\u00e0 : C\u00e2u 12:\u00a0Cho l\u0103ng tr\u1ee5 \u0111\u1ee9ng ABC.A’B’C’ c\u00f3 \u0111\u00e1y ABC l\u00e0 tam gi\u00e1c vu\u00f4ng c\u00e2n t\u1ea1i A, BC = […]<\/p>\n","protected":false},"author":3,"featured_media":28635,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[157],"tags":[1450,1448],"yoast_head":"\nB\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m H\u00ecnh h\u1ecdc 12 \u2013 Ch\u01b0\u01a1ng 2 - B\u00e0i 2: M\u1eb7t c\u1ea7u (Ph\u1ea7n 5)<\/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\/bai-tap-trac-nghiem-hinh-hoc-12-chuong-2-bai-2-mat-cau-phan-5\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m H\u00ecnh h\u1ecdc 12 \u2013 Ch\u01b0\u01a1ng 2 - B\u00e0i 2: M\u1eb7t c\u1ea7u (Ph\u1ea7n 5)\" \/>\n<meta property=\"og:description\" content=\"C\u00e2u 11:\u00a0Cho h\u00ecnh ch\u00f3p S.ABC c\u00f3 SA = SB = SC = a v\u00e0 tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A. 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