B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m H\u00ecnh h\u1ecdc 12 \u2013 Ch\u01b0\u01a1ng 2 \u2013 B\u00e0i 2: M\u1eb7t c\u1ea7u (Ph\u1ea7n 3)<\/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-4-2\/\" \/>\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 \u2013 B\u00e0i 2: M\u1eb7t c\u1ea7u (Ph\u1ea7n 3)\" \/>\n<meta property=\"og:description\" content=\"C\u00e2u 1:\u00a0Cho m\u1eb7t c\u1ea7u (S) t\u00e2m O b\u00e1n k\u00ednh R v\u00e0 m\u1ed9t m\u1eb7t ph\u1eb3ng (P). K\u00ed hi\u1ec7u h l\u00e0 kho\u1ea3ng c\u00e1ch t\u1eeb O \u0111\u1ebfn m\u1eb7t ph\u1eb3ng (P). 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{"id":28674,"date":"2018-04-27T15:19:32","date_gmt":"2018-04-27T08:19:32","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=28674"},"modified":"2018-04-27T15:22:13","modified_gmt":"2018-04-27T08:22:13","slug":"bai-tap-trac-nghiem-hinh-hoc-12-chuong-2-bai-2-mat-cau-phan-4-2","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/bai-tap-trac-nghiem-hinh-hoc-12-chuong-2-bai-2-mat-cau-phan-4-2\/","title":{"rendered":"B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m H\u00ecnh h\u1ecdc 12 \u2013 Ch\u01b0\u01a1ng 2 \u2013 B\u00e0i 2: M\u1eb7t c\u1ea7u (Ph\u1ea7n 3)"},"content":{"rendered":"

C\u00e2u 1:<\/b>\u00a0Cho m\u1eb7t c\u1ea7u (S) t\u00e2m O b\u00e1n k\u00ednh R v\u00e0 m\u1ed9t m\u1eb7t ph\u1eb3ng (P). K\u00ed hi\u1ec7u h l\u00e0 kho\u1ea3ng c\u00e1ch t\u1eeb O \u0111\u1ebfn m\u1eb7t ph\u1eb3ng (P). M\u1eb7t ph\u1eb3ng (P) v\u00e0 m\u1eb7t c\u1ea7u (S) c\u00f3 \u0111i\u1ec3m chung n\u1ebfu v\u00e0 ch\u1ec9 n\u1ebfu :<\/p>\n

A. h < R \u00a0\u00a0\u00a0B. h = R\u00a0\u00a0\u00a0C. h \u2264 R \u00a0\u00a0\u00a0D. h \u2265 R<\/p>\n

C\u00e2u 2:<\/b>\u00a0Trong kh\u00f4ng gian cho \u0111\u01b0\u1eddng th\u1eb3ng \u0394 v\u00e0 \u0111i\u1ec3m O c\u00e1ch \u0394 m\u1ed9t kho\u1ea3ng b\u1eb1ng 20cm. M\u1eb7t c\u1ea7u (S) t\u00e2m O c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng \u0394 theo m\u1ed9t d\u00e2y c\u00f3 \u0111\u1ed9 d\u00e0i 30cm c\u00f3 b\u00e1n k\u00ednh r b\u1eb1ng :<\/p>\n

A. r = 45cm\u00a0\u00a0\u00a0B. r = 30cm \u00a0\u00a0\u00a0C. r = 25cm \u00a0\u00a0\u00a0D. r = 20cm<\/p>\n

C\u00e2u 3:<\/b>\u00a0Cho h\u00ecnh ch\u00f3p tam gi\u00e1c \u0111\u1ec1u S.ABC c\u00f3 SA t\u1ea1o v\u1edbi \u0111\u00e1y m\u1ed9t g\u00f3c b\u1eb1ng 30^{o<\/sup>\u00a0v\u00e0 SA=2a. Trong c\u00e1c \u0111i\u1ec3m S, B, C \u0111i\u1ec3m n\u00e0o n\u1eb1m trong m\u1eb7t c\u1ea7u t\u00e2m A b\u00e1n k\u00ednh 3a.<\/p>\n}

A. Kh\u00f4ng \u0111i\u1ec3m n\u00e0o\u00a0\u00a0\u00a0C. Ch\u1ec9 hai \u0111i\u1ec3m B v\u00e0 C<\/p>\n

B. Ch\u1ec9 \u0111i\u1ec3m S\u00a0\u00a0\u00a0D. C\u1ea3 ba \u0111i\u1ec3m<\/p>\n

C\u00e2u 4:<\/b>\u00a0Cho h\u00ecnh ch\u00f3p S.ABCD c\u00f3 \u0111\u00e1y ABCD l\u00e0 h\u00ecnh vu\u00f4ng c\u00f3 c\u1ea1nh b\u1eb1ng 2a, \uf044SAB l\u00e0 tam gi\u00e1c \u0111\u1ec1u. B\u00e1n k\u00ednh m\u1eb7t c\u1ea7u t\u00e2m A c\u1eaft SB theo m\u1ed9t d\u00e2y c\u00f3 \u0111\u1ed9 d\u00e0i a l\u00e0:<\/p>\n

A. a\u221a13\/2\u00a0\u00a0\u00a0B. 2a\u00a0\u00a0\u00a0C. 2a\u221a2\u00a0\u00a0\u00a0D. a\u221a3<\/p>\n

C\u00e2u 5:<\/b>\u00a0Cho h\u00ecnh ch\u00f3p S.ABCD c\u00f3 \u0111\u00e1y ABCD l\u00e0 h\u00ecnh vu\u00f4ng, SA vu\u00f4ng g\u00f3c v\u1edbi \u0111\u00e1y v\u00e0 AB=a. G\u00f3c gi\u1eefa c\u1ea1nh b\u00ean SD v\u00e0 \u0111\u00e1y b\u1eb1ng 45^{o<\/sup>\u00a0. B\u00e1n k\u00ednh m\u1eb7t c\u1ea7u t\u00e2m S ti\u1ebfp x\u00fac v\u1edbi BD theo a l\u00e0:<\/p>\n}

A. a\u221a6\/2\u00a0\u00a0\u00a0B. a\u221a6\/4\u00a0\u00a0\u00a0C. a\u221a3\u00a0\u00a0\u00a0D. a<\/p>\n

H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n\n\n\n
1-C<\/td>\n 2-C<\/td>\n 3-B<\/td>\n 4-A<\/td>\n 5-A<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
C\u00e2u 1:<\/b><\/p>\n
T\u1eeb v\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a m\u1ed9t m\u1eb7t ph\u1eb3ng v\u00e0 m\u1eb7t c\u1ea7u ta c\u00f3 m\u1eb7t ph\u1eb3ng (P) c\u00f3 \u0111i\u1ec3m chung v\u1edbi m\u1eb7t c\u1ea7u (S) khi v\u00e0 ch\u1ec9 khi m\u1eb7t ph\u1eb3ng (P) ti\u1ebfp x\u00fac ho\u1eb7c c\u1eaft m\u1eb7t c\u1ea7u (S)<\/p>\n
C\u00e2u 2:<\/b><\/p>\n
$\"\"$ <\/p>\n
C\u00e2u 3:<\/b><\/p>\n
G\u1ecdi O l\u00e0 t\u00e2m c\u1ee7a tam gi\u00e1c \u0111\u1ec1u ABC. Ta c\u00f3:<\/p>\n
g\u00f3c SAO = 30^{o<\/sup>\u00a0=> AO = a\u221a3 => AB = AC = 3a<\/p>\n}
C\u00e2u 4:<\/b><\/p>\n
G\u1ecdi S(A;r) l\u00e0 m\u1eb7t c\u1ea7u t\u00e2m A c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng SB theo m\u1ed9t d\u00e2y c\u00f3 \u0111\u1ed9 d\u00e0i a, khi \u0111\u00f3 ta c\u00f3:<\/p>\n
$\"\"$ <\/p>\n
G\u1ecdi H l\u00e0 trung \u0111i\u1ec3m c\u1ee7a SB. Do tam gi\u00e1c SAB \u0111\u1ec1u n\u00ean AH \u22a5 SB hay AH l\u00e0 kho\u1ea3ng c\u00e1ch t\u1eeb A \u0111\u1ebfn SB. X\u00e9t tam gi\u00e1c \u0111\u1ec1u SAB ta c\u00f3 :<\/p>\n
<\/p>\n
$\"\"$ <\/p>\n
C\u00e2u 5:<\/b><\/p>\n
G\u1ecdi S(S;r) l\u00e0 m\u1eb7t c\u1ea7u t\u00e2m S ti\u1ebfp x\u00fac v\u1edbi \u0111\u01b0\u1eddng th\u1eb3ng BD, khi \u0111\u00f3 ta c\u00f3 r = d(S; BD).<\/p>\n
T\u1eeb gi\u1ea3 thi\u1ebft ta c\u00f3 :<\/p>\n
$\"\"$ <\/p>\n
=> SA = AD = AB = a.<\/p>\n
G\u1ecdi O l\u00e0 t\u00e2m c\u1ee7a h\u00ecnh vu\u00f4ng ABCD. Khi \u0111\u00f3 ta c\u00f3 SO \u22a5 BD hay SO l\u00e0 kho\u1ea3ng c\u00e1ch t\u1eeb S \u0111\u1ebfn BD. X\u00e9t tam gi\u00e1c vu\u00f4ng SAO ta c\u00f3 :<\/p>\n
$\"\"$ <\/p>\n
$\"\"$ <\/p>\n","protected":false},"excerpt":{"rendered":"
C\u00e2u 1:\u00a0Cho m\u1eb7t c\u1ea7u (S) t\u00e2m O b\u00e1n k\u00ednh R v\u00e0 m\u1ed9t m\u1eb7t ph\u1eb3ng (P). K\u00ed hi\u1ec7u h l\u00e0 kho\u1ea3ng c\u00e1ch t\u1eeb O \u0111\u1ebfn m\u1eb7t ph\u1eb3ng (P). 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