C\u00e2u 1:<\/b>\u00a0M\u1ed9t h\u00ecnh ch\u00f3p c\u00f3 21 m\u1eb7t. H\u1ecfi h\u00ecnh ch\u00f3p \u0111\u00f3 c\u00f3 bao nhi\u00eau c\u1ea1nh?<\/p>\n

A. 41\u00a0\u00a0\u00a0B. 21\u00a0\u00a0\u00a0C. 42\u00a0\u00a0\u00a0D. 40<\/p>\n

C\u00e2u 2:<\/b>\u00a0Trong c\u00e1c m\u1ec7nh \u0111\u1ec1 sau \u0111\u00e2y m\u1ec7nh \u0111\u1ec1 n\u00e0o \u0111\u00fang?<\/p>\n

Kh\u1ed1i \u0111a di\u1ec7n c\u00f3 c\u00e1c m\u1eb7t l\u00e0 nh\u1eefng \u0111a gi\u00e1c c\u00f3 s\u1ed1 c\u1ea1nh l\u1ebb th\u00ec<\/p>\n

A. S\u1ed1 c\u00e1c m\u1eb7t l\u00e0 s\u1ed1 l\u1ebb \u00a0\u00a0\u00a0C. S\u1ed1 c\u00e1c c\u1ea1nh l\u00e0 s\u1ed1 l\u1ebb<\/p>\n

B. S\u1ed1 c\u00e1c m\u1eb7t l\u00e0 s\u1ed1 ch\u1eb5n\u00a0\u00a0\u00a0D. S\u1ed1 c\u00e1c \u0111\u1ec9nh l\u00e0 s\u1ed1 l\u1ebb<\/p>\n

C\u00e2u 3:<\/b>\u00a0Cho h\u00ecnh ch\u00f3p S.ABCD c\u00f3 SA \u22a5 (ABCD) , ABCD l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt v\u00e0 SA = AD = 2a . G\u00f3c gi\u1eefa m\u1eb7t ph\u1eb3ng (SBC) v\u00e0 m\u1eb7t \u0111\u00e1y l\u00e0 60^{o<\/sup>\u00a0. G\u1ecdi M l\u00e0 trung \u0111i\u1ec3m BC. Th\u1ec3 t\u00edch kh\u1ed1i ch\u00f3p S.ABMD l\u00e0<\/p>\n}

<\/p>\n

C\u00e2u 4:<\/b>\u00a0Cho l\u0103ng tr\u1ee5 ABC.A’B’C’ c\u00f3 \u0111\u1ed9 d\u00e0i c\u1ea1nh b\u00ean b\u1eb1ng 2a, \u0111\u00e1y ABC l\u00e0 tam gi\u00e1c \u0111\u1ec1u AB = a v\u00e0 h\u00ecnh chi\u1ebfu vu\u00f4ng g\u00f3c c\u1ee7a \u0111\u1ec9nh A\u2019 tr\u00ean m\u1eb7t ph\u1eb3ng (ABC) l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u1ea1nh BC. Th\u1ec3 t\u00edch kh\u1ed1i l\u0103ng tr\u1ee5 l\u00e0:<\/p>\n

<\/p>\n

C\u00e2u 5:<\/b>\u00a0Cho l\u0103ng tr\u1ee5 tam gi\u00e1c ABC.A’B’C’ . Bi\u1ebft th\u1ec3 t\u00edch c\u1ee7a kh\u1ed1i ch\u00f3p ABB\u2019C\u2019 l\u00e0 3a^{3<\/sup>\u00a0. Th\u1ec3 t\u00edch kh\u1ed1i l\u0103ng tr\u1ee5 l\u00e0:<\/p>\n}

A. a^{3<\/sup>\u00a0\u00a0\u00a0B. 3a3<\/sup>\u00a0\u00a0\u00a0C. 9a3<\/sup>\u00a0\u00a0\u00a0\u00a0D. 6a3<\/sup><\/p>\n}

C\u00e2u 6:<\/b>\u00a0Cho h\u00ecnh ch\u00f3p S.ABCD c\u00f3 \u0111\u00e1y l\u00e0 h\u00ecnh vu\u00f4ng c\u1ea1nh a. M\u1eb7t b\u00ean (SAD) l\u00e0 tam gi\u00e1c c\u00e2n \u0111\u1ec9nh S v\u00e0 n\u1eb1m trong m\u1eb7t ph\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi \u0111\u00e1y. G\u00f3c gi\u1eefa c\u1ea1nh b\u00ean SB v\u00e0 m\u1eb7t \u0111\u00e1y l\u00e0 60^{o<\/sup>\u00a0. T\u00ednh th\u1ec3 t\u00edch kh\u1ed1i ch\u00f3p S.ABCD<\/p>\n}

 <\/p>\n

<\/p>\n

H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n\n\n\n

1-D<\/td>\n

2-B<\/td>\n

3-D<\/td>\n

4-B<\/td>\n

5-C<\/td>\n

6-D<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 1:\u00a0M\u1ed9t h\u00ecnh ch\u00f3p c\u00f3 21 m\u1eb7t. H\u1ecfi h\u00ecnh ch\u00f3p \u0111\u00f3 c\u00f3 bao nhi\u00eau c\u1ea1nh? A. 41\u00a0\u00a0\u00a0B. 21\u00a0\u00a0\u00a0C. 42\u00a0\u00a0\u00a0D. 40 C\u00e2u 2:\u00a0Trong c\u00e1c m\u1ec7nh \u0111\u1ec1 sau \u0111\u00e2y m\u1ec7nh \u0111\u1ec1 n\u00e0o \u0111\u00fang? Kh\u1ed1i \u0111a di\u1ec7n c\u00f3 c\u00e1c m\u1eb7t l\u00e0 nh\u1eefng \u0111a gi\u00e1c c\u00f3 s\u1ed1 c\u1ea1nh l\u1ebb th\u00ec A. S\u1ed1 c\u00e1c m\u1eb7t l\u00e0 s\u1ed1 l\u1ebb \u00a0\u00a0\u00a0C. S\u1ed1 […]<\/p>\n","protected":false},"author":3,"featured_media":26774,"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