{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"C\u00f3 bao nhi\u00eau d\u1ea5u ch\u1ea5m tr\u00f2n c\u1ea7n \u0111i\u1ec1n v\u00e0o c\u00e1c m\u1eb7t $?1; ?2; ?3$ trong h\u00ecnh khai tri\u1ec3n c\u1ee7a vi\u00ean x\u00fac x\u1eafc h\u00ecnh l\u1eadp ph\u01b0\u01a1ng, bi\u1ebft r\u1eb1ng t\u1ed5ng hai s\u1ed1 \u1edf hai m\u1eb7t \u0111\u1ed1i di\u1ec7n c\u1ee7a vi\u00ean x\u00fac x\u1eafc b\u1eb1ng $7$.<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai23/lv3/img\/H8_B23_K1.9.png' \/><\/center><br\/>S\u1ed1 d\u1ea5u ch\u1ea5m tr\u00f2n c\u1ea7n \u0111i\u1ec1n theo th\u1ee9 t\u1ef1 l\u00e0:","select":["A. $2;3;1$","B. $1;2;3$","C. $3;2;1$","D. $2;1;3$"],"hint":"C\u00f3 th\u1ec3 c\u1eaft h\u00ecnh khai tri\u1ec3n b\u1eb1ng gi\u1ea5y r\u1ed3i gh\u00e9p l\u1ea1i.","explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai23/lv3/img\/H8_B23_K1.9a.png' \/><\/center><br\/>V\u00ec t\u1ed5ng hai m\u1eb7t \u0111\u1ed1i di\u1ec7n l\u00e0 $7$ ch\u1ea5m tr\u00f2n n\u00ean ta c\u00f3 th\u1ec3 t\u00ecm ra c\u00e1c m\u1eb7t \u0111\u1ed1i di\u1ec7n v\u1edbi c\u00e1c m\u1eb7t ghi s\u1ed1 nh\u01b0 sau:<br\/>- M\u1eb7t $?1$ k\u1ec1 v\u1edbi m\u1eb7t ghi $4$ ch\u1ea5m v\u00e0 m\u1eb7t $6$ ch\u1ea5m n\u00ean \u0111\u1ed1i di\u1ec7n v\u1edbi m\u1eb7t ghi $5$ ch\u1ea5m. Suy ra m\u1eb7t $?1$ l\u00e0 m\u1eb7t ghi $2$ ch\u1ea5m.<br\/>- M\u1eb7t $?2$ k\u1ec1 v\u1edbi m\u1eb7t ghi $6$ ch\u1ea5m v\u00e0 m\u1eb7t ghi $5$ ch\u1ea5m n\u00ean \u0111\u1ed1i di\u1ec7n m\u1eb7t ghi $4$ ch\u1ea5m. Suy ra m\u1eb7t $?2$ ghi $3$ ch\u1ea5m.<br\/>- M\u1eb7t $?3$ ghi $1$ ch\u1ea5m.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":2}]}],"id_ques":1920},{"time":24,"part":[{"time":3,"title":"Cho h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt $ABCD.A\u2019B\u2019C\u2019D\u2019$. Tr\u00ean c\u00e1c c\u1ea1nh $AA\u2019,\\, DD\u2019,\\, BB\u2019$ v\u00e0 $CC\u2019$ l\u1ea7n l\u01b0\u1ee3t l\u1ea5y c\u00e1c \u0111i\u1ec3m $F, \\,E,\\, G, \\,H$ sao cho $AF=DE=\\dfrac{2}{3}DD\u2019$; $BG=CH=\\dfrac{1}{3} CC'$. Ch\u1ee9ng minh $mp(ADHG)\/\/mp(FEC\u2019B\u2019)$.","title_trans":"","temp":"sequence","correct":[[[3],[1],[4],[2]]],"list":[{"point":10,"image":"https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai23/lv3/img\/H8_B23_K1.1.png","left":["X\u00e9t m\u1eb7t ph\u1eb3ng $(ADHG)$ c\u00f3 $HG$ v\u00e0 $DH$ c\u1eaft nhau t\u1ea1i $H$.<br\/>X\u00e9t m\u1eb7t ph\u1eb3ng $(FEC'B')$ c\u00f3 $B'C'$ v\u00e0 $EC'$ c\u1eaft nhau t\u1ea1i $C'$","T\u1ee9 gi\u00e1c $BCHG$ c\u00f3 $BG=CH$; $BG\/\/CH$ n\u00ean $BCHG$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh, suy ra $HG\/\/BC$.<br\/>M\u1eb7t kh\u00e1c, $BC\/\/B'C'$ n\u00ean $HG\/\/B'C'$.","Do v\u1eady, $mp(ADHG)\/\/mp(FEC'B)$","T\u1ee9 gi\u00e1c $DHC'E$ c\u00f3 $DE\/\/HC'$ v\u00e0 $DE=HC'$ n\u00ean $DHC'E$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh, suy ra $DH=EC'$."],"top":100,"hint":"Ch\u1ee9ng minh hai \u0111\u01b0\u1eddng th\u1eb3ng c\u1eaft nhau trong m\u1eb7t ph\u1eb3ng n\u00e0y l\u1ea7n l\u01b0\u1ee3t song song v\u1edbi hai \u0111\u01b0\u1eddng th\u1eb3ng c\u1eaft nhau trong m\u1eb7t ph\u1eb3ng kia.","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai23/lv3/img\/H8_B23_K1.1.png' \/><\/center>T\u1ee9 gi\u00e1c $BCHG$ c\u00f3 $BG=CH$; $BG\/\/CH$ n\u00ean $BCHG$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft)<br\/>Suy ra $HG\/\/BC$<br\/>M\u1eb7t kh\u00e1c, $BC\/\/B'C'$ n\u00ean $HG\/\/B'C'$<br\/>T\u1ee9 gi\u00e1c $DHC'E$ c\u00f3 $DE\/\/HC'$ v\u00e0 $DE=HC'$ n\u00ean $DHC'E$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh.<br\/>Suy ra $DH=EC'$. <br\/>X\u00e9t m\u1eb7t ph\u1eb3ng $(ADHG)$ c\u00f3 $HG$ v\u00e0 $DH$ c\u1eaft nhau t\u1ea1i $H$.<br\/>X\u00e9t m\u1eb7t ph\u1eb3ng $(FEC'B')$ c\u00f3 $B'C'$ v\u00e0 $EC'$ c\u1eaft nhau t\u1ea1i $C'$<br\/>Do v\u1eady, $mp(ADHG)\/\/mp(FEC'B)$<\/span>"}]}],"id_ques":1921},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"Cho h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt $ABCD.A\u2019B\u2019C\u2019D\u2019$. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt $ADC\u2019B\u2019$ bi\u1ebft $AB= 28\\,cm,\\, B'D^2=3709, DD\u2019=45\\,cm$.","select":["A. $1192,5\\,cm^2$","B. $1590\\,cm^2$","C. $2385\\,cm^2$","D. $3129\\,cm^2$"],"hint":"","explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai23/lv3/img\/H8_B23_K1.2.png' \/><\/center>X\u00e9t tam gi\u00e1c $AA'B'$ vu\u00f4ng t\u1ea1i $A'$ c\u00f3: $AA'=DD'=45\\,cm$ v\u00e0 $A'B'=AB=28\\,cm$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: <br\/>$AA'^2+A'B'^2=AB'^2\\\\ \\Leftrightarrow AB'=\\sqrt{AA'^2+A'B'^2}\\\\ \\Rightarrow AB'=\\sqrt{45^2+28^2}\\\\ \\Leftrightarrow AB'=\\sqrt{2809}=53\\,(cm)$<br\/>Ta c\u00f3: $AD\\bot AA';\\,AD\\bot AB$, suy ra $AD \\bot mp(AA'B'B) \\Rightarrow AD\\bot AB'$<br\/>X\u00e9t tam gi\u00e1c $ADB'$ vu\u00f4ng t\u1ea1i $A$ c\u00f3: $AB'=53\\,cm$ v\u00e0 $DB'^2=3709$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: <br\/>$AD^2+AB'^2=DB'^2\\\\ \\Leftrightarrow AD=\\sqrt{DB'^2-AB'^2}\\\\ \\Rightarrow AD=\\sqrt{3709-53^2}\\\\ \\Leftrightarrow AD=\\sqrt{900}=30\\,(cm)$<br\/>V\u1eady di\u1ec7n t\u00edch $ADC\u2019B\u2019$ b\u1eb1ng $AD.AB'=30.53=1590\\,(cm)^2$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":1922},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["3240"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"H\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt $ABCD.A\u2019B\u2019C\u2019D\u2019$ c\u00f3 \u0111\u00e1y l\u00e0 h\u00ecnh vu\u00f4ng. Di\u1ec7n t\u00edch m\u1eb7t ch\u00e9o $BDD\u2019B\u2019$ b\u1eb1ng $360 cm^2$, $M;\\, N$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AA\u2019$ v\u00e0 $CC\u2019$. T\u00ednh th\u1ec3 t\u00edch h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt bi\u1ebft $MN=18\\,cm$.<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai23/lv3/img\/H8_B23_K1.3.png' \/><\/center><b> \u0110\u00e1p s\u1ed1: <\/b> $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\,(cm^3)$","hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai23/lv3/img\/H8_B23_K1.3a.png' \/><\/center>X\u00e9t t\u1ee9 gi\u00e1c $AMNC$ c\u00f3 $AM\/\/CN$ v\u00e0 $AM=CN=\\dfrac{1}{2}AA'$<br\/>N\u00ean $AMNC$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh, suy ra $AC=MN=18\\,cm$<br\/>Ta c\u00f3, $ABCD$ l\u00e0 h\u00ecnh vu\u00f4ng n\u00ean $AC=BD=18\\,cm$ (t\u00ednh ch\u1ea5t h\u00ecnh vu\u00f4ng)<br\/>Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng $ABCD$ l\u00e0 $\\dfrac{1}{2}.18^2=162\\,(cm^2)$<br\/>V\u1eady th\u1ec3 t\u00edch h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt b\u1eb1ng $BB'. S_{ABCD}=20.162=3240\\,(cm^3)$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $3240$<\/span><br\/><b> Nh\u1eadn x\u00e9t:<\/b> Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng c\u00f3 th\u1ec3 t\u00ednh theo c\u00f4ng th\u1ee9c t\u00ednh di\u1ec7n t\u00edch h\u00ecnh thoi b\u1eb1ng $\\dfrac{1}{2}$ t\u00edch hai \u0111\u01b0\u1eddng ch\u00e9o<\/span>"}]}],"id_ques":1923},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"H\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng $ABCD.A\u2019B\u2019C\u2019D'$ c\u00f3 \u0111\u00e1y l\u00e0 h\u00ecnh thoi c\u1ea1nh $2a$. G\u00f3c $\\widehat {AC\u2019A\u2019}=45^o$. Di\u1ec7n t\u00edch m\u1eb7t b\u00ean $ABB\u2019A\u2019=6a^2$. T\u00ednh th\u1ec3 t\u00edch h\u00ecnh l\u0103ng tr\u1ee5.","select":["A. $\\dfrac{3a^3\\sqrt{7}}{2}$ (\u0111\u01a1n v\u1ecb th\u1ec3 t\u00edch)","B. $\\dfrac{3a^2\\sqrt{7}}{2}$ (\u0111\u01a1n v\u1ecb th\u1ec3 t\u00edch)","C. $\\dfrac{9a^2\\sqrt{7}}{2}$ (\u0111\u01a1n v\u1ecb th\u1ec3 t\u00edch)","D. $\\dfrac{18a^2\\sqrt{7}}{2}$ (\u0111\u01a1n v\u1ecb th\u1ec3 t\u00edch)"],"hint":"","explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai23/lv3/img\/H8_B23_K1.4.png' \/><\/center>X\u00e9t m\u1eb7t b\u00ean $ABB\u2019A\u2019$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 $A\u2019B\u2019=2a$ (gi\u1ea3 thi\u1ebft). <br\/>Suy ra $AA\u2019=S_{ABB\u2019A\u2019}:A\u2019B\u2019=6a^2:2a=3a$<br\/>X\u00e9t tam gi\u00e1c $AA\u2019C\u2019$ vu\u00f4ng t\u1ea1i $A\u2019 $ c\u00f3:<br\/>+ $\\widehat {AC\u2019A\u2019}=45^o$<br\/>Suy ra tam gi\u00e1c $AA\u2019C$ vu\u00f4ng c\u00e2n t\u1ea1i $A\u2019$.<br\/> N\u00ean ta c\u00f3, $A\u2019C\u2019=AA\u2019=3a$.<br\/>G\u1ecdi $O$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a hai \u0111\u01b0\u1eddng ch\u00e9o $A\u2019C\u2019$ v\u00e0 $B\u2019D\u2019$. <br\/>V\u00ec $A\u2019B\u2019C\u2019D\u2019$ l\u00e0 h\u00ecnh thoi n\u00ean $OA\u2019\\bot OD'$. <br\/>X\u00e9t tam gi\u00e1c $A\u2019OD\u2019$ vu\u00f4ng t\u1ea1i $O$ c\u00f3 $OA\u2019=\\dfrac{1}{2}A\u2019C\u2019=\\dfrac{3a}{2}$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3:<br\/> $OD\u2019^2=A\u2019D\u2019^2-OA\u2019^2\\\\ \\Rightarrow OD\u2019^2=4a^2-\\dfrac{9a^2}{4}\\\\ \\Leftrightarrow OD'^2=\\dfrac{7a^2}{4}\\\\ \\Leftrightarrow OD'=\\dfrac{a\\sqrt{7}}{2}$<br\/>Ta c\u00f3: $B\u2019D\u2019=2OD\u2019=a\\sqrt{7}$.<br\/>Di\u1ec7n t\u00edch h\u00ecnh thoi $A\u2019B\u2019C\u2019D\u2019$ l\u00e0<br\/> $\\dfrac{1}{2}.A\u2019C\u2019.B\u2019D\u2019=\\dfrac{1}{2}.3a.a\\sqrt{7}=\\dfrac{3a^2\\sqrt{7}}{2}$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/>V\u1eady th\u1ec3 t\u00edch h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng $ABCDF.A\u2019B\u2019C\u2019D\u2019$ l\u00e0 $3a.\\dfrac{3a^2\\sqrt{7}}{2}=\\dfrac{9a^2\\sqrt{7}}{2}$ (\u0111\u01a1n v\u1ecb th\u1ec3 t\u00edch)<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span>","column":2}]}],"id_ques":1924},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $72\\sqrt{3}$","B. $71\\sqrt{3}$","C. $70\\sqrt{3}$"],"ques":"T\u00ednh th\u1ec3 t\u00edch l\u0103ng tr\u1ee5 \u0111\u1ee9ng tam gi\u00e1c \u0111\u1ec1u $ABC.A'B'C'$ c\u00f3 c\u1ea1nh \u1edf \u0111\u00e1y l\u00e0 $6 cm$ v\u00e0 $AB\u2019=10 cm$.<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> ?($cm^3$)","hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai23/lv3/img\/H8_B23_K1.5.png' \/><\/center>X\u00e9t tam gi\u00e1c $AA\u2019B\u2019 $ vu\u00f4ng t\u1ea1i $A\u2019 $ c\u00f3 $A\u2019B\u2019=6\\,cm$ v\u00e0 $AB\u2019=10 cm$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3:<br\/> $AB\u2019^2=AA\u2019^2+A\u2019B\u2019^2\\\\ \\Rightarrow AA\u2019=\\sqrt{10^2-6^2}=\\sqrt{64}=8\\,(cm)$<br\/>X\u00e9t tam gi\u00e1c $ABC$ \u0111\u1ec1u.<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai23/lv3/img\/H8_B23_K1.5a.png' \/><\/center> <br\/>G\u1ecdi $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC$. Ta c\u00f3 $AH\\bot BC$ (t\u00ednh ch\u1ea5t tam gi\u00e1c \u0111\u1ec1u)<br\/>X\u00e9t tam gi\u00e1c $ABH$ vu\u00f4ng t\u1ea1i $H$ c\u00f3 $AB=6\\,cm$ v\u00e0 $BH=\\dfrac{1}{2}.BC=3\\,cm$<br\/> Theo \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3 $AH=\\sqrt{AB^2-BH^2}=\\sqrt{6^2-3^2}=\\sqrt{27}=3\\sqrt{3}\\,(cm)$<br\/>Di\u1ec7n t\u00edch tam gi\u00e1c $ABC$ l\u00e0 $\\dfrac{1}{2}.AH.BC=\\dfrac{1}{2}.3\\sqrt{3}.6=9\\sqrt{3}\\,(cm^2)$<br\/>Th\u1ec3 t\u00edch h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng l\u00e0 $AA\u2019.S_{ABC}=8.9\\sqrt{3}=72\\sqrt{3}\\,(cm^3)$<\/span>"}]}],"id_ques":1925},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng $ABCD.A\u2019B\u2019C\u2019D\u2019$ c\u00f3 \u0111\u00e1y l\u00e0 h\u00ecnh thang c\u00e2n, \u0111\u00e1y nh\u1ecf $DC=a$ \u0111\u00e1y l\u1edbn $AB=3a$ v\u00e0 $AC\\bot BC$. T\u00ednh th\u1ec3 t\u00edch l\u0103ng tr\u1ee5, bi\u1ebft $AC'=a\\sqrt {31}$<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai23/lv3/img\/H8_B23_K1.6.png' \/><\/center>","select":["A. $10a^3\\sqrt{3}$ (\u0111\u01a1n v\u1ecb th\u1ec3 t\u00edch)","B. $20a^3\\sqrt{3}$ (\u0111\u01a1n v\u1ecb th\u1ec3 t\u00edch)","C. $10a^3\\sqrt{2}$ (\u0111\u01a1n v\u1ecb th\u1ec3 t\u00edch)","D. $15a^2\\sqrt{3}$ (\u0111\u01a1n v\u1ecb th\u1ec3 t\u00edch)"],"hint":"","explain":"<span class='basic_left'><br\/>X\u00e9t h\u00ecnh thang c\u00e2n $ABCD$, h\u1ea1 $CH \\bot AB$.<br\/>Ta c\u00f3: $BH=\\dfrac{AB-DC}{2}=\\dfrac{3a-a}{2}=a$<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai23/lv3/img\/H8_B23_K1.6a.png' \/><\/center>X\u00e9t tam gi\u00e1c $ABC$ v\u00e0 tam gi\u00e1c $CBH$ c\u00f3:<br\/> $\\widehat {H}=\\widehat {C}=90^o$ (gi\u1ea3 thi\u1ebft);<br\/> G\u00f3c $B$ chung.<br\/>$\\Rightarrow \\Delta ABC \\backsim CBH$ (g.g)<br\/>Ta c\u00f3: $\\dfrac{AB}{CB}=\\dfrac{BC}{BH}$ (C\u00e1c c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7)<br\/>Suy ra ta c\u00f3: $\\dfrac{3a}{CB}=\\dfrac{BC}{a}\\\\ \\Leftrightarrow BC^2=3a^2\\\\ \\Leftrightarrow BC=a\\sqrt{3}$<br\/>X\u00e9t tam gi\u00e1c $BHC$ vu\u00f4ng t\u1ea1i $H$ c\u00f3 $BH=a, BC=a\\sqrt{3}$.<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: <br\/>$BC^2=BH^2+CH^2\\\\ \\Rightarrow CH^2=3a^2-a^2=2a^2\\\\ \\Leftrightarrow CH=a\\sqrt{2}$<br\/>Di\u1ec7n t\u00edch h\u00ecnh thang $ABCD$ l\u00e0 $\\dfrac{1}{2}.(AB+CD).CH=\\dfrac{1}{2}.(a+3a).a\\sqrt{2}=2a^2\\sqrt{2}$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/>X\u00e9t tam gi\u00e1c $ACH$ vu\u00f4ng t\u1ea1i $H$ c\u00f3 $AH=2a$, $CH=a\\sqrt{2}$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: <br\/>$AC^2=AH^2+HC^2\\\\ \\Rightarrow AC^2=4a^2+2a^2\\\\ \\Leftrightarrow AC=a\\sqrt{6}$<br\/>X\u00e9t tam gi\u00e1c $ACC'$ vu\u00f4ng t\u1ea1i $C$ c\u00f3: $AC'=a\\sqrt{31}; AC=a\\sqrt{6}$.<br\/>Theo \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3:<br\/>$CC'=\\sqrt{AC'^2-AC^2}\\\\ \\Leftrightarrow CC'=\\sqrt{31a^2-6a^2}=25a^2\\\\ \\Leftrightarrow CC'=5a$<br\/>V\u1eady th\u1ec3 t\u00edch h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng l\u00e0 $CC'.S_{ABCD}=5a.2a^2\\sqrt{2}=10a^3\\sqrt{2}$ (\u0111\u01a1n v\u1ecb th\u1ec3 t\u00edch)<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C <\/span><\/span>","column":2}]}],"id_ques":1926},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"width":50,"type_input":"","input_hint":["sqrt","frac"],"ques":"Cho h\u00ecnh ch\u00f3p tam gi\u00e1c \u0111\u1ec1u $S.ABC$. G\u1ecdi $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB$. T\u00ednh th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh ch\u00f3p bi\u1ebft $CH=\\sqrt 3 cm$ v\u00e0 c\u1ea1nh b\u00ean l\u00e0 $2\\sqrt 2\\,cm$.","select":["A. $\\dfrac{2\\sqrt{15}}{3}\\,cm^3$","B. $\\dfrac{2\\sqrt{5}}{3}\\,cm^3$","C. $\\dfrac{4\\sqrt{15}}{3}\\,cm^3$","D. $\\dfrac{2\\sqrt{5}}{3}\\,cm^3$"],"hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai23/lv3/img\/H8_B23_K1.7.png' \/><\/center>- V\u00ec $S.ABC$ l\u00e0 ch\u00f3p tam gi\u00e1c \u0111\u1ec1u n\u00ean $ABC$ l\u00e0 tam gi\u00e1c \u0111\u1ec1u v\u00e0 c\u00e1c m\u1eb7t b\u00ean l\u00e0 c\u00e1c tam gi\u00e1c c\u00e2n.<br\/>+) $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB$<br\/>+) $O$ l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n \u0111i qua c\u00e1c \u0111\u1ec9nh c\u1ea3 tam gi\u00e1c $ABC$ (giao \u0111i\u1ec3m ba \u0111\u01b0\u1eddng trung tr\u1ef1c)<br\/>- V\u00ec $SO$ l\u00e0 \u0111\u01b0\u1eddng cao n\u00ean $SO \\bot mp(ABC) \\,\\Rightarrow SO\\bot HO$<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai23/lv3/img\/H8_B23_K1.7a.png' \/><\/center>Gi\u1ea3 s\u1eed tam gi\u00e1c \u0111\u1ec1u $ABC$ c\u1ea1nh $x$.<br\/> Ta c\u00f3 tam gi\u00e1c $ABC$ \u0111\u1ec1u, $H$ l\u00e0 trung \u0111i\u1ec3m $AB$ n\u00ean $CH$ vu\u00f4ng g\u00f3c v\u1edbi $AB$.<br\/>X\u00e9t tam gi\u00e1c $ACH$ vu\u00f4ng t\u1ea1i $H$. \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3:<br\/> $x^2-\\dfrac{x^2}{4}=CH^2\\\\ \\Leftrightarrow \\dfrac{3}{4}.x^2=3\\\\ \\Leftrightarrow x^2=4\\\\ \\Leftrightarrow x=2\\,(cm)$.<br\/>Di\u1ec7n t\u00edch tam gi\u00e1c $ABC$ l\u00e0 $\\dfrac{1}{2}.CH.AB=\\dfrac{1}{2}.\\sqrt {3}.2=\\sqrt{3}\\,(cm^2)$<br\/>V\u00ec $ABC$ l\u00e0 tam gi\u00e1c \u0111\u1ec1u, $O$ l\u00e0 giao \u0111i\u1ec3m ba \u0111\u01b0\u1eddng trung tr\u1ef1c n\u00ean $O$ l\u00e0 giao \u0111i\u1ec3m ba \u0111\u01b0\u1eddng trung tuy\u1ebfn.<br\/> Suy ra $CO=\\dfrac{2}{3}CH=\\dfrac{2\\sqrt{3}}{3}\\,(cm)$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c)<br\/>X\u00e9t tam gi\u00e1c $SCO$ vu\u00f4ng t\u1ea1i $O$ c\u00f3: <br\/>$SO^2=SC^2-CO^2\\\\ \\Rightarrow SO^2=8-\\left(\\dfrac{2\\sqrt{3}}{3}\\right)^2 \\\\ \\Leftrightarrow SO=\\dfrac{2\\sqrt{15}}{3}\\,(cm)$<br\/>V\u1eady th\u1ec3 t\u00edch h\u00ecnh ch\u00f3p l\u00e0 $\\dfrac{1}{3}.SO.S_{ABC}=\\dfrac{1}{3}.\\dfrac{2\\sqrt{15}}{3}.\\sqrt{3}=\\dfrac{2\\sqrt{5}}{3}\\,(cm^3)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":1927},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0110\u00fang ho\u1eb7c Sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"H\u00ecnh ch\u00f3p c\u1ee5t $A\u2019B\u2019C\u2019.ABC$ c\u00f3 th\u1ec3 t\u00edch l\u00e0 $\\dfrac{520}{3}\\,cm^3$. Hai \u0111\u00e1y l\u00e0 hai tam gi\u00e1c \u0111\u1ec1u \u0111\u1ed3ng d\u1ea1ng v\u1edbi t\u1ec9 s\u1ed1 l\u00e0 $\\dfrac{1}{3}$. C\u00f3 di\u1ec7n t\u00edch \u0111\u00e1y l\u1edbn $ABC$ l\u00e0 $90 cm^2$. Chi\u1ec1u cao h\u00ecnh ch\u00f3p c\u1ee5t l\u00e0 $6\\,cm$.<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai23/lv3/img\/H8_B23_K1.8.png' \/><\/center>","select":["\u0110\u00fang","Sai"],"hint":"T\u00ednh t\u1ec9 l\u1ec7 $\\dfrac{SO'}{SO}$.","explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai23/lv3/img\/H8_B23_K1.8a.png' \/><\/center>Ta c\u00f3: $\\Delta A\u2019B\u2019C\u2019$ \u0111\u1ed3ng d\u1ea1ng v\u1edbi $\\Delta ABC$ theo t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng l\u00e0 $\\dfrac{1}{3}$. Suy ra, $\\dfrac{S_{A\u2019B\u2019C\u2019}}{S_{ABC}}=\\dfrac{1}{9}$.<br\/>V\u1eady $S_{A\u2019B\u2019C\u2019}=10\\,cm^2$.<br\/>G\u1ecdi $O\u2019$ l\u00e0 t\u00e2m tam gi\u00e1c \u0111\u1ec1u $A\u2019B\u2019C\u2019$ v\u00e0 $O$ l\u00e0 t\u00e2m tam gi\u00e1c \u0111\u1ec1u $ABC$.<br\/>C\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng $AA\u2019; BB\u2019; CC\u2019$ v\u00e0 $OO\u2019 $ c\u1eaft nhau t\u1ea1i $S$.<br\/> X\u00e9t m\u1eb7t b\u00ean $SAB$ c\u00f3 $A\u2019B\u2019\/\/AB$ v\u00e0 $A\u2019B\u2019=\\dfrac{1}{3} AB$.<br\/>Theo h\u1ec7 qu\u1ea3 \u0111\u1ecbnh l\u00fd Ta-let ta c\u00f3: $\\dfrac{SA\u2019}{SA}=\\dfrac{SB\u2019}{SB}=\\dfrac{1}{3}$<br\/>Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u00f3: $\\dfrac{SA\u2019}{SA}=\\dfrac{SB\u2019}{SB}=\\dfrac{SC\u2019}{SC}=\\dfrac{1}{3}$<br\/>G\u1ecdi $H\u2019$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $A\u2019B\u2019$ v\u00e0 $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB$.<br\/> Ta c\u00f3 $S;\\,H\u2019;\\,H$ th\u1eb3ng h\u1eb1ng. <br\/>Ta c\u0169ng d\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c $\\dfrac{SH\u2019}{SH}=\\dfrac{1}{3}$.<br\/>Ta c\u00f3: $O'$ l\u00e0 t\u00e2m c\u1ee7a tam gi\u00e1c $A\u2019B\u2019C\u2019$ n\u00ean $C';\\,O\u2019;\\,H\u2019$ th\u1eb3ng h\u00e0ng. <br\/>T\u01b0\u01a1ng t\u1ef1 c\u00f3 $C;\\, H;\\, O$ th\u1eb3ng h\u00e0ng.<br\/> X\u00e9t tam gi\u00e1c $SHC$ c\u00f3 $\\dfrac{SC\u2019}{SC}=\\dfrac{SH\u2019}{SH}=\\dfrac{1}{3}$, suy ra $C\u2019H\u2019\/\/CH$.<br\/> X\u00e9t tam gi\u00e1c $SOC$ c\u00f3 $O\u2019C\u2019\/\/OC$; $\\dfrac{SC\u2019}{SC}=\\dfrac{1}{3}$. <br\/>Theo h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00fd Ta-let ta c\u00f3: $\\dfrac{SO\u2019}{SO}=\\dfrac{1}{3}$.<br\/>\u0110\u1eb7t $SO\u2019=x\\Leftrightarrow SO=3x$.<br\/>Ta c\u00f3: $V_{A'B'C'.ABC}=V_{S.ABC}-V_{S.A'B'C'}\\\\ \\Leftrightarrow \\dfrac{520}{3}=\\dfrac{1}{3}.SO.S_{ABC}-\\dfrac{1}{3}.SO'.S_{A'B'C'}\\\\ \\Leftrightarrow \\dfrac{520}{3}=\\dfrac{1}{3}.3x.90-\\dfrac{1}{3}.x.10\\\\ \\Leftrightarrow 520=270x-10x\\\\ \\Leftrightarrow x=2\\,(cm)$<br\/>V\u1eady $SO'=2\\,cm\\Rightarrow SO=6\\,cm\\Rightarrow OO'=4\\,cm$<br\/>V\u1eady h\u00ecnh ch\u00f3p c\u1ee5t c\u00f3 chi\u1ec1u cao l\u00e0 $4\\,cm$<br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 Sai<\/span><\/span>","column":2}]}],"id_ques":1928},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho h\u00ecnh ch\u00f3p t\u1ee9 gi\u00e1c \u0111\u1ec1u c\u00f3 c\u1ea1nh \u0111\u00e1y l\u00e0 $a\\,\\sqrt 2 \\,cm $ v\u00e0 c\u1ea1nh b\u00ean b\u1eb1ng $a\\sqrt 5 cm$. T\u00ednh th\u1ec3 t\u00edch h\u00ecnh ch\u00f3p.<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai23/lv3/img\/H8_B23_K1.10.png' \/><\/center>","select":["A. $4a^3\\,cm^3$","B. $8a^3\\,cm^3$","C. $\\dfrac{4a^3}{3}\\,cm^3$","D. $\\dfrac{8a^3}{3}\\,cm^3$"],"hint":"","explain":" <span class='basic_left'>- V\u00ec $S.ABCD$ l\u00e0 ch\u00f3p t\u1ee9 gi\u00e1c \u0111\u1ec1u n\u00ean $ABCD$ l\u00e0 h\u00ecnh vu\u00f4ng v\u00e0 c\u00e1c m\u1eb7t b\u00ean l\u00e0 c\u00e1c tam gi\u00e1c c\u00e2n.<br\/>+) $O$ l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n \u0111i qua c\u00e1c \u0111\u1ec9nh c\u1ee7a h\u00ecnh vu\u00f4ng $ABCD$ (giao \u0111i\u1ec3m hai \u0111\u01b0\u1eddng ch\u00e9o)<br\/>- V\u00ec $SO$ l\u00e0 \u0111\u01b0\u1eddng cao n\u00ean $SO \\bot mp(ABCD) \\,\\Rightarrow SO\\bot AO$<br\/>Ta c\u00f3: Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng $ABCD$ l\u00e0 $AB^2=(a\\sqrt{2})^2=2a^2$ ($cm^2$)<br\/>X\u00e9t tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $B$ c\u00f3 $AB=BC=a\\sqrt{2}\\,cm$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: $AC^2=AB^2+BC^2\\\\ \\Rightarrow AC^2=2a^2+2a^2 \\\\ \\Leftrightarrow AC=2a$ <br\/>X\u00e9t tam gi\u00e1c $SOA$ vu\u00f4ng t\u1ea1i $O$ c\u00f3 $SA=a\\sqrt{5}\\,(cm);\\,AO=\\dfrac{1}{2}AC=a\\,(cm)$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3:<br\/>$SA^2=SO^2+OA^2\\\\ \\Leftrightarrow SO=\\sqrt{SA^2-AO^2}\\\\ \\Leftrightarrow SO=\\sqrt{5a^2-a^2}\\\\ \\Leftrightarrow SO=2a\\,(cm)$<br\/>V\u1eady th\u1ec3 t\u00edch h\u00ecnh ch\u00f3p l\u00e0 $\\dfrac{1}{3}.SO.S_{ABCD}=\\dfrac{1}{3}.2a.2a^2=\\dfrac{4a^3}{3}\\,(cm^3)$<br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}]}],"id_ques":1929}],"lesson":{"save":0,"level":3}}