C\u00e2u 62:<\/b>\u00a0Cho h\u00ecnh l\u0103ng tr\u1ee5 ABC.A’B’C’ . G\u1ecdi E, F l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AA\u2019, BB\u2019. I l\u00e0 \u0111i\u1ec3m t\u00f9y \u00fd tr\u00ean CC\u2019. T\u00ednh th\u1ec3 t\u00edch V c\u1ee7a h\u00ecnh ch\u00f3p I.ABFE, bi\u1ebft th\u1ec3 t\u00edch c\u1ee7a l\u0103ng tr\u1ee5 \u0111\u00e3 cho b\u1eb1ng 9.<\/p>\n
A. V= 1\/3 \u00a0\u00a0\u00a0B. V= 3\/2 \u00a0\u00a0\u00a0C. V=3 \u00a0\u00a0\u00a0D.V=6<\/p>\n
C\u00e2u 63:<\/b>\u00a0T\u00ednh th\u1ec3 t\u00edch V c\u1ee7a kh\u1ed1i l\u0103ng tr\u1ee5 ABC.A’B’C’ , c\u00f3 c\u1ea1nh b\u00ean b\u1eb1ng a, AB=AC=2a. (ABB’A’) \u22a5 (ABC), g\u00f3c BA’A = g\u00f3c BAC = 90o<\/sup><\/p>\n A. V = \u221a3a3<\/sup>\/2\u00a0\u00a0\u00a0B. V = \u221a3a3<\/sup>\u00a0\u00a0\u00a0C. V = 2a3<\/sup>\u00a0\u00a0\u00a0D. V = 3a3<\/sup><\/p>\n C\u00e2u 64:<\/b>\u00a0Cho h\u00ecnh l\u0103ng tr\u1ee5 ABC.A’B’C’ . G\u1ecdi E, F l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AA\u2019, BB\u2019. I l\u00e0 \u0111i\u1ec3m t\u00f9y \u00fd tr\u00ean CC\u2019. G\u1ecdi V\u2019 l\u00e0 th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh ch\u00f3p I.ABFE, V l\u00e0 th\u1ec3 t\u00edch c\u1ee7a l\u0103ng tr\u1ee5 ABC.A’B’C’ . T\u00ednh t\u1ec9 s\u1ed1 k = V’\/V<\/p>\n A. k = 1\/2\u00a0\u00a0\u00a0B. k = 1\/3 \u00a0\u00a0\u00a0C. k = 1\/4\u00a0\u00a0\u00a0D. k = 1\/6<\/p>\n C\u00e2u 65:<\/b>\u00a0Cho h\u00ecnh h\u1ed9p BACD.A’B’C’D’ c\u00f3 \u0111\u00e1y l\u00e0 h\u00ecnh thoi c\u1ea1nh b\u1eb1ng a, g\u00f3c BAD = 60o<\/sup>\u00a0, h\u00ecnh chi\u1ebfu c\u1ee7a A\u2019 l\u00ean m\u1eb7t \u0111\u00e1y tr\u00f9ng v\u1edbi trung \u0111i\u1ec3m c\u1ee7a AC, c\u1ea1nh A\u2019A t\u1ea1o v\u1edbi \u0111\u00e1y m\u1ed9t g\u00f3c 60o<\/sup>\u00a0. T\u00ednh th\u1ec3 t\u00edch V c\u1ee7a h\u00ecnh h\u1ed9p \u0111\u00e3 cho<\/p>\n <\/p>\n C\u00e2u 66:<\/b>\u00a0T\u00ednh t\u1ed5ng c\u00e1c kho\u1ea3ng c\u00e1ch h t\u1eeb m\u1ed9t \u0111i\u1ec3m trong c\u1ee7a m\u1ed9t t\u1ee9 di\u1ec7n \u0111\u1ec1u c\u1ea1nh a \u0111\u1ebfn c\u00e1c m\u1eb7t c\u1ee7a n\u00f3.<\/p>\n <\/p>\n C\u00e2u 67:<\/b>\u00a0Cho h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng ABC.A’B’C’ , c\u00f3 AB = AC = 2, g\u00f3c BAC = 120o<\/sup>\u00a0. Bi\u1ebft th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u00e3 cho b\u1eb1ng \u221a3a3<\/sup>\u00a0. H\u00e3y t\u00ednh g\u00f3c \u03b1 gi\u1eefa hai m\u1eb7t ph\u1eb3ng (ABC) v\u00e0 (A\u2019BC)<\/p>\n A. \u03b1 = 15o<\/sup>\u00a0\u00a0\u00a0\u00a0B. \u03b1 = 30o<\/sup>\u00a0\u00a0\u00a0C. \u03b1 = 45o<\/sup>\u00a0\u00a0\u00a0\u00a0D. \u03b1 = 60o<\/sup><\/p>\n C\u00e2u 68:<\/b>\u00a0Cho h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng ABC.A’B’C’ ,c\u00f3 ABC = AC = 2a, g\u00f3c BAC = 120o<\/sup>\u00a0. Bi\u1ebft th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u00e3 cho b\u1eb1ng \u221a3a3<\/sup>\u00a0. H\u00e3y t\u00ednh kho\u1ea3ng c\u00e1ch h t\u1eeb B\u2019 \u0111\u1ebfn m\u1eb7t (A\u2019BC)<\/p>\n <\/p>\n C\u00e2u 69:<\/b>\u00a0Cho h\u00ecnh l\u0103ng tr\u1ee5 tam gi\u00e1c \u0111\u1ec1u ABC.A’B’C’ , c\u00f3 AB = 2, AA\u2019 = 3. L\u1ea5y \u0111i\u1ec3m E tr\u00ean c\u1ea1nh BB\u2019 sao cho EB\u2019=2EB. M\u1eb7t ph\u1eb3ng qua A\u2019E, song song v\u1edbi BC c\u1eaft c\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng CC\u2019, AB, AC l\u1ea7n l\u01b0\u1ee3t t\u1ea1i F, M, N. T\u00ednh t\u1ec9 s\u1ed1 k gi\u1eefa th\u1ec3 t\u00edch h\u00ecnh l\u0103ng tr\u1ee5 ABC.A’B’C’ v\u00e0 th\u1ec3 t\u00edch h\u00ecnh ch\u00f3p A’.AMN<\/p>\n A. k = 1\/2\u00a0\u00a0\u00a0B. k = 2\/3 \u00a0\u00a0\u00a0C. k = 3\/4 \u00a0\u00a0\u00a0D. k = 4\/3<\/p>\n C\u00e2u 70:<\/b>\u00a0Cho h\u00ecnh l\u1eadp ph\u01b0\u01a1ng ABCD.A’B’C’D’ c\u00f3 c\u1ea1nh b\u1eb1ng a. G\u1ecdi M, N l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u00e1c c\u1ea1nh A’B’.C’D’ . T\u00ednh t\u1ec9 s\u1ed1 k gi\u1eefa th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh ch\u00f3p D’.CNMP v\u00e0 th\u1ec3 t\u00edch h\u00ecnh l\u1eadp ph\u01b0\u01a1ng ABCD.A’B’C’D’<\/p>\n A. k = 1\/8 \u00a0\u00a0\u00a0B. k = 1\/6 \u00a0\u00a0\u00a0C. k = 1\/4 \u00a0\u00a0\u00a0D. k = 1\/3<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 62:<\/b><\/p>\n G\u1ecdi M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a CC\u2019. Ta c\u00f3<\/p>\n <\/p>\n C\u00e2u 64:<\/b><\/p>\n VI.ABFE<\/sub>\u00a0= VC.ABFE<\/sub>\u00a0= (2\/3)VABC.EFM<\/sub>\u00a0= (1\/3)VABC.A’B’C’<\/sub><\/p>\n C\u00e2u 68:<\/b><\/p>\n h = d(A, (A’BC))<\/p>\n C\u00e2u 69:<\/b><\/p>\n \u0110\u1ec3 \u00fd r\u1eb1ng:<\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 62:\u00a0Cho h\u00ecnh l\u0103ng tr\u1ee5 ABC.A’B’C’ . G\u1ecdi E, F l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AA\u2019, BB\u2019. I l\u00e0 \u0111i\u1ec3m t\u00f9y \u00fd tr\u00ean CC\u2019. T\u00ednh th\u1ec3 t\u00edch V c\u1ee7a h\u00ecnh ch\u00f3p I.ABFE, bi\u1ebft th\u1ec3 t\u00edch c\u1ee7a l\u0103ng tr\u1ee5 \u0111\u00e3 cho b\u1eb1ng 9. A. V= 1\/3 \u00a0\u00a0\u00a0B. V= 3\/2 \u00a0\u00a0\u00a0C. V=3 \u00a0\u00a0\u00a0D.V=6 C\u00e2u 63:\u00a0T\u00ednh th\u1ec3 […]<\/p>\n","protected":false},"author":3,"featured_media":28836,"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":"\n\n\n
\n 62-C<\/td>\n 63-B<\/td>\n 64-B<\/td>\n 65-B<\/td>\n 66-B<\/td>\n 67-C<\/td>\n 68-D<\/td>\n 69-D<\/td>\n 70-A<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n