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{"id":20539,"date":"2018-03-12T13:16:31","date_gmt":"2018-03-12T06:16:31","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=20539"},"modified":"2018-03-12T13:16:31","modified_gmt":"2018-03-12T06:16:31","slug":"hinh-hoc-chuong-2-mat-non-mat-tru-mat-cau-tap","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/hinh-hoc-chuong-2-mat-non-mat-tru-mat-cau-tap\/","title":{"rendered":"H\u00ecnh H\u1ecdc – Ch\u01b0\u01a1ng 2: M\u1eb7t n\u00f3n, m\u1eb7t tr\u1ee5, M\u1eb7t c\u1ea7u – \u00d4n t\u1eadp"},"content":{"rendered":"

B\u00e0i 1 (trang 50 SGK H\u00ecnh h\u1ecdc 12):\u00a0Cho ba \u0111i\u1ec3m A, B, C c\u00f9ng thu\u1ed9c m\u1ed9t m\u1eb7t c\u1ea7u sao cho (ACB)=90o<\/sup>.Trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau, kh\u1eb3ng \u0111\u1ecbnh n\u00e0o \u0111\u00fang?<\/strong><\/span><\/p>\n

a) \u0110\u01b0\u1eddng tr\u00f2n qua ba \u0111i\u1ec3m A, B, C n\u1eb1m tr\u00ean m\u1eb7t c\u1ea7u.<\/strong><\/span><\/p>\n

b) AB l\u00e0 m\u1ed9t \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a m\u1eb7t c\u1ea7u \u0111\u00e3 cho.<\/strong><\/span><\/p>\n

c) AB kh\u00f4ng ph\u1ea3i l\u00e0 \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a m\u1eb7t c\u1ea7u.<\/strong><\/span><\/p>\n

d) AB l\u00e0 \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n giao tuy\u1ebfn t\u1ea1o b\u1edfi m\u1eb7t c\u1ea7u v\u00e0 m\u1eb7t ph\u1eb3ng (ABC).<\/strong><\/span><\/p>\n

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

a) \u0110\u00fang<\/p>\n

b) Sai<\/p>\n

c) Sai<\/p>\n

d) \u0110\u00fang.<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

B\u00e0i 2 (trang 50 SGK H\u00ecnh h\u1ecdc 12):\u00a0Cho t\u1ee9 di\u1ec7n ABCD c\u00f3 c\u1ea1nh AD vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng (ABC) v\u00e0 c\u1ea1nh BD vu\u00f4ng g\u00f3c v\u1edbi c\u1ea1nh BC. Bi\u1ebft AB = AD = a. T\u00ednh di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh n\u00f3n v\u00e0 th\u1ec3 t\u00edch c\u1ee7a kh\u1ed1i n\u00f3n \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh khi quay \u0111\u01b0\u1eddng g\u1ea5p kh\u00fac BDA quanh c\u1ea1nh AB.<\/strong><\/span><\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

B\u00e0i 3 (trang 50 SGK H\u00ecnh h\u1ecdc 12):\u00a0M\u1ed9t h\u00ecnh ch\u00f3p c\u00f3 t\u1ea5t c\u1ea3 c\u00e1c c\u1ea1nh b\u00ean b\u1eb1ng nhau. Ch\u1ee9ng minh r\u1eb1ng h\u00ecnh ch\u00f3p \u0111\u00f3 n\u1ed9i ti\u1ebfp \u0111\u01b0\u1ee3c trong m\u1ed9t m\u1eb7t c\u1ea7u (c\u00e1c \u0111\u1ec9nh c\u1ee7a h\u00ecnh ch\u00f3p n\u1eb1m tr\u00ean m\u1eb7t c\u1ea7u).<\/strong><\/span><\/p>\n

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

Cho h\u00ecnh ch\u00f3p S.A1<\/sub>A2<\/sub>A3<\/sub>…An<\/sub>\u00a0c\u00f3 c\u00e1c c\u1ea1nh b\u00ean b\u1eb1ng nhau.<\/p>\n

Gi\u1ea3 s\u1eed I l\u00e0 h\u00ecnh chi\u1ebfu vu\u00f4ng g\u00f3c c\u1ee7a S tr\u00ean m\u1eb7t \u0111\u00e1y.<\/p>\n

Ta c\u00f3: SA1<\/sub>\u00a0= SA2<\/sub>\u00a0= SA3<\/sub>\u00a0= … = SAn<\/sub><\/p>\n

Suy ra \u0394SIA1<\/sub>= \u0394SIA2<\/sub>\u00a0= \u0394SIA3<\/sub>\u00a0= … = \u0394SIAn<\/sub><\/p>\n

Suy ra IA1<\/sub>\u00a0= IA2<\/sub>\u00a0= IA3<\/sub>\u00a0= … = IAn<\/sub><\/p>\n

\u0110a gi\u00e1c A1<\/sub>A2<\/sub>A3<\/sub>…An<\/sub>\u00a0l\u00e0 m\u1ed9t \u0111a gi\u00e1c n\u1ed9i ti\u1ebfp \u0111\u01b0\u1ee3c trong m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m I b\u00e1n k\u00ednh IA, tr\u1ee5c SI.<\/p>\n

Trong mp(SAI), \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a SA1<\/sub>\u00a0c\u1eaft SI t\u1ea1i O, ta c\u00f3:<\/p>\n

OS = OA1<\/sub>\u00a0(1)<\/p>\n

OA1<\/sub>\u00a0= OA2<\/sub>\u00a0= OA3<\/sub>\u00a0= … = OAn<\/sub> (2)<\/p>\n

T\u1eeb (1) v\u00e0 (2) suy ra OS = OA1<\/sub>\u00a0= OA2<\/sub>\u00a0= OA3<\/sub>\u00a0= … = OAn<\/sub><\/p>\n

V\u1eady h\u00ecnh ch\u00f3p S.A1<\/sub>A2<\/sub>A3<\/sub>…An<\/sub>\u00a0n\u1ed9i ti\u1ebfp \u0111\u01b0\u1ee3c trong m\u1ed9t m\u1eb7t c\u1ea7u.<\/p>\n

B\u00e0i 4 (trang 50 SGK H\u00ecnh h\u1ecdc 12):\u00a0H\u00ecnh ch\u00f3p S.ABC c\u00f3 m\u1ed9t m\u1eb7t c\u1ea7u ti\u1ebfp x\u00fac v\u1edbi c\u00e1c c\u1ea1nh b\u00ean SA, SB, SC. M\u1eb7t c\u1ea7u n\u00e0y c\u00f2n ti\u1ebfp x\u00fac v\u1edbi ba c\u1ea1nh AB, BC, CA t\u1ea1i trung \u0111i\u1ec3m c\u1ee7a m\u1ed7i c\u1ea1nh. Ch\u1ee9ng minh r\u1eb1ng h\u00ecnh ch\u00f3p \u0111\u00f3 l\u00e0 h\u00ecnh ch\u00f3p tam gi\u00e1c \u0111\u1ec1u.<\/strong><\/span><\/p>\n

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

\"\"<\/p>\n

G\u1ecdi M, N, P l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AB, BC, CA; I, J, K l\u00e0 ti\u1ebfp \u0111i\u1ec3m c\u1ee7a c\u00e1c c\u1ea1nh b\u00ean SA, SB, SC v\u1edbi m\u1eb7t c\u1ea7u.<\/p>\n

Ta c\u00f3: AM = AI v\u00e0 BM = BJ<\/p>\n

M\u00e0 AM = BM n\u00ean AI = BJ<\/p>\n

M\u1eb7t kh\u00e1c SI = SJ<\/p>\n

N\u00ean SI + AI = SJ + BJ<\/p>\n

V\u1eady SA = SB (1)<\/p>\n

T\u01b0\u01a1ng t\u1ef1, ta c\u00f3: SB = SC (2)<\/p>\n

T\u1eeb (1) v\u00e0 (2) => SA = SB = SC (3)<\/p>\n

M\u1eb7t kh\u00e1c BM = BN v\u00e0 CN = CP<\/p>\n

Suy ra AB = 2BM = BC = 2CN = 2CP = CA<\/p>\n

Khi \u0111\u00f3 ABC l\u00e0 tam gi\u00e1c \u0111\u1ec1u (4)<\/p>\n

T\u1eeb (3) v\u00e0 (4) suy ra S.ABC l\u00e0 h\u00ecnh ch\u00f3p tam gi\u00e1c \u0111\u1ec1u.<\/p>\n

B\u00e0i 5 (trang 50 SGK H\u00ecnh h\u1ecdc 12):\u00a0Cho t\u1ee9 di\u1ec7n \u0111\u1ec1u ABCD c\u1ea1nh a. G\u1ecdi H l\u00e0 h\u00ecnh chi\u1ebfu vu\u00f4ng g\u00f3c c\u1ee7a \u0111\u1ec9nh A xu\u1ed1ng m\u1eb7t ph\u1eb3ng (BCD).<\/strong><\/span><\/p>\n

\n

a) Ch\u1ee9ng minh H l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c BCD. T\u00ednh \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n AH.<\/strong><\/span><\/p>\n

b) T\u00ednh di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh tr\u1ee5 v\u00e0 th\u1ec3 t\u00edch c\u1ee7a kh\u1ed1i tr\u1ee5 c\u00f3 \u0111\u01b0\u1eddng tr\u00f2n \u0111\u00e1y ngo\u1ea1i ti\u1ebfp tam gi\u00e1c BCD v\u00e0 chi\u1ec1u cao AH.<\/strong><\/span><\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n<\/div>\n

B\u00e0i 6 (trang 50 SGK H\u00ecnh h\u1ecdc 12):\u00a0Cho h\u00ecnh vu\u00f4ng ABCD c\u1ea1nh a. T\u1eeb t\u00e2m O c\u1ee7a h\u00ecnh vu\u00f4ng d\u1ef1ng \u0111\u01b0\u1eddng th\u1eb3ng \u0394 vu\u00f4ng g\u00f3c v\u1edbi m\u1eb7t ph\u1eb3ng (ABCD). Tr\u00ean \u0394 l\u1ea5y \u0111i\u1ec3m S sao cho OS = a\/2 . 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.ABCD. T\u00ednh di\u1ec7n t\u00edch m\u1eb7t c\u1ea7u v\u00e0 th\u1ec3 t\u00edch kh\u1ed1i c\u1ea7u \u0111\u01b0\u1ee3c t\u1ea1o n\u00ean b\u1edfi m\u1eb7t c\u1ea7u \u0111\u00f3.<\/strong><\/span><\/p>\n

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

\"\" \"\" \"\"<\/p>\n

B\u00e0i 7 (trang 50 SGK H\u00ecnh h\u1ecdc 12):\u00a0Cho h\u00ecnh tr\u1ee5 c\u00f3 b\u00e1n k\u00ednh r, tr\u1ee5c OO’ = 2r v\u00e0 m\u1eb7t c\u1ea7u \u0111\u01b0\u1eddng k\u00ednh OO’.<\/strong><\/span><\/p>\n

a) H\u00e3y so s\u00e1nh di\u1ec7n t\u00edch m\u1eb7t c\u1ea7u v\u00e0 di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh tr\u1ee5.<\/strong><\/span><\/p>\n

b) H\u00e3y so s\u00e1nh th\u1ec3 t\u00edch kh\u1ed1i tr\u1ee5 v\u00e0 th\u1ec3 t\u00edch kh\u1ed1i c\u1ea7u \u0111\u01b0\u1ee3c t\u1ea1o n\u00ean b\u1edfi h\u00ecnh tr\u1ee5 v\u00e0 m\u1eb7t c\u1ea7u \u0111\u00e3 cho.<\/strong><\/span><\/p>\n

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

\"\"<\/p>\n

 <\/p>\n

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

B\u00e0i 1 (trang 50 SGK H\u00ecnh h\u1ecdc 12):\u00a0Cho ba \u0111i\u1ec3m A, B, C c\u00f9ng thu\u1ed9c m\u1ed9t m\u1eb7t c\u1ea7u sao cho (ACB)=90o.Trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau, kh\u1eb3ng \u0111\u1ecbnh n\u00e0o \u0111\u00fang? a) \u0110\u01b0\u1eddng tr\u00f2n qua ba \u0111i\u1ec3m A, B, C n\u1eb1m tr\u00ean m\u1eb7t c\u1ea7u. b) AB l\u00e0 m\u1ed9t \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a m\u1eb7t c\u1ea7u \u0111\u00e3 cho. c) AB […]<\/p>\n","protected":false},"author":3,"featured_media":20541,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1298],"tags":[1356,1355,1376],"yoast_head":"\nH\u00ecnh H\u1ecdc - Ch\u01b0\u01a1ng 2: M\u1eb7t n\u00f3n, m\u1eb7t tr\u1ee5, M\u1eb7t c\u1ea7u - \u00d4n 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-mat-non-mat-tru-mat-cau-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: M\u1eb7t n\u00f3n, m\u1eb7t tr\u1ee5, M\u1eb7t c\u1ea7u - \u00d4n t\u1eadp\" \/>\n<meta property=\"og:description\" content=\"B\u00e0i 1 (trang 50 SGK H\u00ecnh h\u1ecdc 12):\u00a0Cho ba \u0111i\u1ec3m A, B, C c\u00f9ng thu\u1ed9c m\u1ed9t m\u1eb7t c\u1ea7u sao cho (ACB)=90o.Trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau, kh\u1eb3ng \u0111\u1ecbnh n\u00e0o \u0111\u00fang? 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