{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai21/lv3/img\/16.jpg' \/><\/center>M\u1ed9t h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng c\u00f3 t\u1ed5ng s\u1ed1 m\u1eb7t, s\u1ed1 c\u1ea1nh v\u00e0 s\u1ed1 \u0111\u1ec9nh l\u00e0 $602$. S\u1ed1 c\u1ea1nh c\u1ee7a \u0111a gi\u00e1c \u1edf \u0111\u00e1y l\u00e0 bao nhi\u00eau?","select":["A. $90$","B. $95$ ","C. $99$ ","D. $100$"],"hint":"Nh\u1edb l\u1ea1i m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa s\u1ed1 \u0111\u1ec9nh, c\u1ea1nh v\u00e0 m\u1eb7t c\u1ee7a l\u0103ng tr\u1ee5 \u0111\u1ee9ng v\u1edbi s\u1ed1 c\u1ea1nh c\u1ee7a \u0111a gi\u00e1c \u1edf \u0111\u00e1y.","explain":" <span class='basic_left'>G\u1ecdi s\u1ed1 c\u1ea1nh c\u1ee7a \u0111a gi\u00e1c \u1edf \u0111\u00e1y l\u00e0 $n$ ($n\\in \\mathbb N^*$) <br\/>H\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng c\u00f3 $n+2$ m\u1eb7t, $2n$ \u0111\u1ec9nh v\u00e0 $3n$ c\u1ea1nh.<br\/>V\u00ec t\u1ed5ng s\u1ed1 m\u1eb7t, s\u1ed1 c\u1ea1nh v\u00e0 s\u1ed1 \u0111\u1ec9nh l\u00e0 $602$ n\u00ean ta c\u00f3:<br\/>$n+2+2n+3n=602\\Leftrightarrow n=100$<br\/>V\u1eady \u0111a gi\u00e1c \u1edf \u0111\u00e1y c\u00f3 $100$ c\u1ea1nh.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><\/span>","column":2}]}],"id_ques":1860},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["4320"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"T\u00ednh th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng c\u00f3 k\u00edch th\u01b0\u1edbc cho b\u1edfi h\u00ecnh v\u1ebd sau:<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai21/lv3/img\/H8_B21_K1.12.png' \/><\/center><br\/><b> \u0110\u00e1p s\u1ed1: <\/b> _input_($cm^3$)","hint":"Chia h\u00ecnh l\u0103ng tr\u1ee5 th\u00e0nh hai h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt r\u1ed3i t\u00ednh.","explain":"<span class='basic_left'>K\u1ebb $PM'\\bot A'D'$ v\u00e0 $QN' \\bot B'C'$.<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai21/lv3/img\/H8_B21_K1.12a.png' \/><\/center><br\/>Khi \u0111\u00f3, th\u1ec3 t\u00edch h\u00ecnh l\u0103ng tr\u1ee5 ban \u0111\u1ea7u b\u1eb1ng t\u1ed5ng th\u1ec3 t\u00edch $ABNM.A'B'N'M'$ v\u00e0 $PQCD.M'N'C'D'$<br\/>X\u00e9t h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng $ABNM.A'B'N'M'$ l\u00e0 h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt c\u00f3 k\u00edch th\u01b0\u1edbc $12\\,cm;12\\,cm$ v\u00e0 $18\\,cm$<br\/>Do v\u1eady th\u1ec3 t\u00edch h\u00ecnh l\u0103ng tr\u1ee5 l\u00e0 $12.12.18=2592\\,(cm^3)$<br\/>D\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng $PQCD.M'N'C'D'$ l\u00e0 h\u00ecnh l\u1eadp ph\u01b0\u01a1ng c\u00f3 k\u00edch th\u01b0\u1edbc $12\\,cm$.<br\/>Do v\u1eady th\u1ec3 t\u00edch $PQCD.M'N'C'D'$ l\u00e0 $12^3=1728\\,(cm^3)$<br\/>V\u1eady th\u1ec3 t\u00edch h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng ban \u0111\u1ea7u l\u00e0 $2592+1728=4320\\,(cm^3)$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $4320$.<\/span><\/span>"}]}],"id_ques":1861},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai21/lv3/img\/2.jpg' \/><\/center>M\u1ed9t h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng $ABCD. A\u2019B\u2019C\u2019D\u2019$ c\u00f3 \u0111\u00e1y l\u00e0 h\u00ecnh thoi v\u1edbi c\u00e1c \u0111\u01b0\u1eddng ch\u00e9o c\u1ee7a \u0111\u00e1y $BD=16\\,cm$ v\u00e0 $AC=30\\, cm$. Di\u1ec7n t\u00edch xung quanh c\u1ee7a l\u0103ng tr\u1ee5 l\u00e0 $612\\,cm^2$. T\u00ednh th\u1ec3 t\u00edch l\u0103ng tr\u1ee5.","select":["A. $1080\\,cm^3$","B. $2160\\,cm^3$","C. $4320\\,cm^3$","D. $8640\\,cm^3$"],"hint":"\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago t\u00ednh \u0111\u1ed9 d\u00e0i m\u1ed7i c\u1ea1nh c\u1ee7a h\u00ecnh thoi.","explain":" <span class='basic_left'><br\/><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh h\u00ecnh thoi.<br\/><b>B\u01b0\u1edbc 2:<\/b> T\u00ednh chi\u1ec1u cao h\u00ecnh l\u0103ng tr\u1ee5<br\/><b>B\u01b0\u1edbc 3:<\/b> T\u00ednh di\u1ec7n t\u00edch \u0111\u00e1y v\u00e0 th\u1ec3 t\u00edch h\u00ecnh thoi.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai21/lv3/img\/H8_B21_K1.1.png' \/><\/center>G\u1ecdi $O $ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AC$ v\u00e0 $BD$. <br\/>V\u00ec $ABCD$ l\u00e0 h\u00ecnh thoi n\u00ean $AC\\bot BD$.<br\/>X\u00e9t tam gi\u00e1c $AOD$ vu\u00f4ng t\u1ea1i $O$ c\u00f3: $AO=\\dfrac{1}{2}AC=15\\,cm$ v\u00e0 $ OD=\\dfrac{1}{2}BD= 8\\,cm$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: <br\/>$AD^2=AO^2+OD^2\\\\ \\Rightarrow AD^2=15^2+8^2\\\\ \\Leftrightarrow AD^2=289\\\\ \\Leftrightarrow AD=17\\,(cm)$<br\/>Chu vi h\u00ecnh thoi $ABCD$ l\u00e0 $4.AD=68\\,(cm)$<br\/>Do di\u1ec7n t\u00edch xung quanh c\u1ee7a l\u0103ng tr\u1ee5 l\u00e0 $612\\,cm^2$ n\u00ean ta c\u00f3:<br\/> $AA\u2019. 68=612\\\\ \\Leftrightarrow AA\u2019=9\\,(cm)$.<br\/>Di\u1ec7n t\u00edch \u0111\u00e1y $ABCD$ l\u00e0 $\\dfrac{1}{2}.16.30=240\\,(cm^2)$<br\/>Th\u1ec3 t\u00edch c\u1ee7a l\u0103ng tr\u1ee5 l\u00e0 $AA\u2019.S_{ABCD}=9.240=2160\\,(cm^3)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B <\/span><\/span>","column":2}]}],"id_ques":1862},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["504"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"M\u1ed9t kh\u1ed1i g\u1ed7 h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt c\u00f3 k\u00edch th\u01b0\u1edbc nh\u01b0 h\u00ecnh v\u1ebd. Ng\u01b0\u1eddi ta c\u1eaft m\u1ed9t l\u00e1t c\u1eaft theo m\u1eb7t ph\u1eb3ng $BB\u2019M\u2019M$ \u0111\u01b0\u1ee3c h\u00ecnh l\u0103ng tr\u1ee5 $ABM.A\u2019B\u2019M\u2019$ c\u00f3 th\u1ec3 t\u00edch $216 cm^3$. V\u1edbi $M \\in AD$ sao cho $\\dfrac{AM}{MD}=\\dfrac{3}{2}$ v\u00e0 $MM'\/\/AA'$. T\u00ednh th\u1ec3 t\u00edch l\u0103ng tr\u1ee5 $BMDC.B\u2019M\u2019D\u2019C\u2019$.<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai21/lv3/img\/H8_B21_K1.6.png' \/><\/center><br\/><b> \u0110\u00e1p s\u1ed1: <\/b> _input_($cm^3$)","hint":"T\u00ednh th\u1ec3 t\u00edch h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng $ABCD.A\u2019B\u2019C\u2019B\u2019$","explain":"<span class='basic_left'><br\/><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago v\u00e0o tam gi\u00e1c $A'B'M'$ t\u00ednh $A'M'$.<br\/><b>B\u01b0\u1edbc 2:<\/b> T\u00ednh chi\u1ec1u cao $A'A$<br\/><b>B\u01b0\u1edbc 3:<\/b> T\u00ednh $AD$<br\/><b>B\u01b0\u1edbc 4:<\/b> T\u00ednh th\u1ec3 t\u00edch l\u0103ng tr\u1ee5 \u0111\u1ee9ng $ABCD.A'B'C'D'$<br\/><b>B\u01b0\u1edbc 5:<\/b> T\u00ednh th\u1ec3 t\u00edch l\u0103ng tr\u1ee5 $BMDC.B'M'C'D'$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai21/lv3/img\/H8_B21_K1.6.png' \/><\/center><br\/>V\u00ec $ABCD.A'B'C'D'$ l\u00e0 h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt n\u00ean $AD=A'D'$, $A'D'\\bot A'B'$<br\/>X\u00e9t tam gi\u00e1c $A\u2019B\u2019M\u2019$ vu\u00f4ng t\u1ea1i $A\u2019$.<br\/>\u00c1P d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3:<br\/>$A\u2019M\u2019^2=B\u2019M\u2019^2-A\u2019B\u2019^2\\\\ \\Rightarrow A\u2019M\u2019^2=10^2-6^2\\\\ \\Leftrightarrow A\u2019M\u2019^2=64\\\\ \\Leftrightarrow A\u2019M\u2019=8\\,(cm)$<br\/>Di\u1ec7n t\u00edch $A\u2019B\u2019M\u2019$ l\u00e0 $\\dfrac{1}{2}.6.8=24\\,(cm^2)$.<br\/>V\u00ec $ABM.A\u2019B\u2019M\u2019$ c\u00f3 th\u1ec3 t\u00edch l\u00e0 $216\\,cm^3$ n\u00ean chi\u1ec1u cao $A\u2019A$ l\u00e0 $216:24=9\\,(cm)$<br\/>Ta c\u00f3, $MM'\/\/AA'$ v\u00e0 $A'M'\/\/AM$ n\u00ean $AM=A'M'=8\\,(cm)$ (t\u00ednh ch\u1ea5t \u0111o\u1ea1n ch\u1eafn)<br\/>Ta c\u00f3, $\\dfrac{AM}{MD}=\\dfrac{3}{2}$ n\u00ean $\\dfrac{AM}{AD}=\\dfrac{3}{5}$. Suy ra, $AD=\\dfrac{5}{3}AM=\\dfrac{5}{3}.8=\\dfrac{40}{3}\\,(cm)$.<br\/>V\u1eady th\u1ec3 t\u00edch $ABCD.A\u2019B\u2019C\u2019D\u2019$ l\u00e0 $AD.AB.AA'=\\dfrac{40}{3}.6.9=720\\,(cm^3)$<br\/>V\u1eady th\u1ec3 t\u00edch $BMDC.B\u2019M\u2019D\u2019C\u2019$ l\u00e0 $720-216=504\\,(cm^3)$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $504$<\/span><\/span>"}]}],"id_ques":1863},{"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 thoi. Bi\u1ebft \u0111\u01b0\u1eddng cao $AA\u2019=15cm$ v\u00e0 c\u00e1c \u0111\u01b0\u1eddng ch\u00e9o $AC\u2019=17 cm$ v\u00e0 $DB\u2019= 25 cm$. T\u00ednh th\u1ec3 t\u00edch h\u00ecnh l\u0103ng tr\u1ee5.<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai21/lv3/img\/H8_B21_K1.3a.png' \/><\/center>","select":["A. $1000\\,cm^3$","B. $1100\\,cm^3$","C. $1200\\,cm^3$","D. $2400\\,cm^3$"],"hint":"Tam gi\u00e1c $ACC\u2019$ vu\u00f4ng t\u1ea1i $C$ v\u00e0 tam gi\u00e1c $DBB'$ vu\u00f4ng t\u1ea1i $B$","explain":" <span class='basic_left'><br\/><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago v\u00e0o hai tam gi\u00e1c vu\u00f4ng $ACC'$ v\u00e0 $DBB'$, t\u00ednh hai \u0111\u01b0\u1eddng ch\u00e9o c\u1ee7a h\u00ecnh thoi \u1edf \u0111\u00e1y.<br\/><b>B\u01b0\u1edbc 2:<\/b> T\u00ednh di\u1ec7n t\u00edch \u0111\u00e1y.<br\/><b>B\u01b0\u1edbc 3:<\/b> T\u00ednh th\u1ec3 t\u00edch l\u0103ng tr\u1ee5 \u0111\u1ee9ng<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai21/lv3/img\/H8_B21_K1.3a.png' \/><\/center>V\u00ec $ABCD.A'B'C'D'$ l\u00e0 l\u0103ng tr\u1ee5 \u0111\u1ee9ng n\u00ean $AA'=BB'=CC'=DD'=15\\,cm$<br\/>Ta c\u00f3, $CC'\\bot mp(ABCD)\\Leftrightarrow CC'\\bot AC\\,\\Leftrightarrow \\widehat {AC'C}=90^o$<br\/>X\u00e9t tam gi\u00e1c $ACC\u2019$ vu\u00f4ng t\u1ea1i $C$. <br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: $AC^2=AC\u2019^2-CC\u2019^2\\\\ \\Rightarrow AC^2=17^2-15^2\\\\ \\Leftrightarrow AC^2= 64\\\\ \\Leftrightarrow AC=8\\,(cm)$<br\/>Ta c\u00f3, $BB'\\bot mp(ABCD)\\Leftrightarrow BB'\\bot BD\\,\\Leftrightarrow \\widehat {DBB'}=90^o$<br\/>X\u00e9t tam gi\u00e1c $DBB'$ vu\u00f4ng t\u1ea1i $B$. <br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: $BD^2=DB\u2019^2-BB\u2019^2\\\\ \\Rightarrow BD^2 =25^2-15^2\\\\ \\Leftrightarrow BD^2 =400\\\\ \\Leftrightarrow BD=20\\,(cm)$<br\/>Di\u1ec7n t\u00edch \u0111\u00e1y $ABCD$ l\u00e0 $\\dfrac{1}{2}.AC.BD=\\dfrac{1}{2}.8.20=80\\,(cm^2)$<br\/>V\u1eady th\u1ec3 t\u00edch l\u0103ng tr\u1ee5 l\u00e0 $15.80=1200\\,(cm^3)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}]}],"id_ques":1864},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang ho\u1eb7c sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"H\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng tam gi\u00e1c $ABC.A\u2019B\u2019C\u2019$ c\u00f3 \u0111\u00e1y l\u00e0 tam gi\u00e1c \u0111\u1ec1u v\u1edbi $M$ l\u00e0 trung \u0111i\u1ec3m $BC$ v\u00e0 $AA\u2019=AM=a$ th\u00ec c\u00f3 di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n l\u00e0 $\\dfrac{7a^2}{\\sqrt 3}\\,cm^2$.<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai21/lv3/img\/H8_B21_K1.4.png' \/><\/center>","select":["\u0110\u00fang","Sai"],"hint":"T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh \u0111\u00e1y.<br\/>G\u1ecdi \u0111\u1ed9 d\u00e0i c\u1ea1nh tam gi\u00e1c \u0111\u1ec1u l\u00e0 $x$","explain":"<span class='basic_left'><br\/><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> G\u1ecdi \u0111\u1ed9 d\u00e0i c\u1ea1nh \u0111\u00e1y l\u00e0 $x$<br\/><b>B\u01b0\u1edbc 2:<\/b> L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh t\u00ecm $x$ theo $a$<br\/><b>B\u01b0\u1edbc 3:<\/b> T\u00ednh di\u1ec7n t\u00edch \u0111\u00e1y v\u00e0 di\u1ec7n t\u00edch xung quanh c\u1ee7a l\u0103ng tr\u1ee5<br\/><b>B\u01b0\u1edbc 4:<\/b> T\u00ednh di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n c\u1ee7a l\u0103ng tr\u1ee5.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai21/lv3/img\/H8_B21_K1.4a.png' \/><\/center>G\u1ecdi \u0111\u1ed9 d\u00e0i c\u1ea1nh tam gi\u00e1c \u0111\u1ec1u l\u00e0 $x$ ($cm,\\,x>0$)<br\/>Ta c\u00f3, tam gi\u00e1c $ABC$ \u0111\u1ec1u c\u00f3 $AB=x\\,(cm); \\,BM=\\dfrac{x}{2}\\,(cm)$<br\/>V\u00ec $AM$ l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn c\u1ee7a tam gi\u00e1c \u0111\u1ec1u $ABC$ n\u00ean \u0111\u1ed3ng th\u1eddi l\u00e0 \u0111\u01b0\u1eddng cao.<br\/>X\u00e9t tam gi\u00e1c $ABM$ vu\u00f4ng t\u1ea1i $M$. \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3:<br\/>$AB^2=AM^2+BM^2\\\\ \\Rightarrow x^2=a^2+\\dfrac{x^2}{4}\\\\ \\Leftrightarrow \\dfrac{3x^2}{4}=a^2\\\\ \\Leftrightarrow x=\\dfrac{2a}{\\sqrt{3}}\\,(cm) $<br\/>V\u1eady tam gi\u00e1c $ABC$ \u0111\u1ec1u c\u1ea1nh $\\dfrac{2a}{\\sqrt{3}}\\,(cm) $<br\/>Di\u1ec7n t\u00edch tam gi\u00e1c $ABC$ l\u00e0 $\\dfrac{1}{2}.AM.BC=\\dfrac{1}{2}.a.\\dfrac{2a}{\\sqrt{3}}=\\dfrac{a^2}{\\sqrt{3}}\\,(cm^2)$<br\/>Di\u1ec7n t\u00edch xung quanh c\u1ee7a l\u0103ng tr\u1ee5 l\u00e0 $3.\\dfrac{2a}{\\sqrt{3}}.a=\\dfrac{6a^2}{\\sqrt{3}}\\,(cm^2)$<br\/>V\u1eady di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n c\u1ee7a l\u0103ng tr\u1ee5 l\u00e0 $2.\\dfrac{a^2}{\\sqrt{3}}+\\dfrac{6a^2}{\\sqrt{3}}=\\dfrac{8a^2}{\\sqrt{3}}\\,(cm^2)$<br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 Sai<\/span><\/span>","column":2}]}],"id_ques":1865},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["10"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"M\u1ed9t h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ec1u c\u00f3 t\u1ed5ng s\u1ed1 m\u1eb7t, s\u1ed1 \u0111\u1ec9nh v\u00e0 s\u1ed1 c\u1ea1nh l\u00e0 $26$. Bi\u1ebft th\u1ec3 t\u00edch h\u00ecnh l\u0103ng tr\u1ee5 l\u00e0 $640\\,cm^3$. Di\u1ec7n t\u00edch xung quanh l\u00e0 $320 cm^2$. T\u00ednh chi\u1ec1u cao c\u1ee7a l\u0103ng tr\u1ee5.<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> Chi\u1ec1u cao c\u1ee7a l\u0103ng tr\u1ee5 l\u00e0 _input_ ($cm$)<br\/>(<i>L\u0103ng tr\u1ee5 \u0111\u1ec1u l\u00e0 l\u0103ng tr\u1ee5 c\u00f3 \u0111\u00e1y l\u00e0 \u0111a gi\u00e1c \u0111\u1ec1u<\/i>)","hint":"T\u00ecm s\u1ed1 c\u1ea1nh \u1edf \u0111\u00e1y. Vi\u1ebft bi\u1ec3u th\u1ee9c t\u00ednh di\u1ec7n t\u00edch xung quanh v\u00e0 th\u1ec3 t\u00edch theo c\u00e1c k\u00edch th\u01b0\u1edbc c\u1ee7a l\u0103ng tr\u1ee5 \u0111\u1ee9ng.","explain":"<span class='basic_left'><br\/><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> T\u00ecm s\u1ed1 c\u1ea1nh c\u1ee7a \u0111a gi\u00e1c \u1edf \u0111\u00e1y.<br\/><b>B\u01b0\u1edbc 2:<\/b> G\u1ecdi \u0111\u1ed9 d\u00e0i c\u1ea1nh \u0111\u00e1y v\u00e0 chi\u1ec1u cao l\u0103ng tr\u1ee5 l\u00e0 $a$ v\u00e0 $h$<br\/><b>B\u01b0\u1edbc 3:<\/b> Vi\u1ebft bi\u1ec3u th\u1ee9c t\u00ednh di\u1ec7n t\u00edch xung quanh v\u00e0 th\u1ec3 t\u00edch theo c\u00e1c k\u00edch th\u01b0\u1edbc c\u1ee7a l\u0103ng tr\u1ee5 \u0111\u1ee9ng.<br\/><b>B\u01b0\u1edbc 4:<\/b> Bi\u1ebfn \u0111\u1ed5i t\u00ecm c\u00e1c k\u00edch th\u01b0\u1edbc.<br\/><b>B\u01b0\u1edbc 5:<\/b> T\u00ecm chi\u1ec1u cao v\u00e0 k\u1ebft lu\u1eadn.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai21/lv3/img\/H8_B21_K1.7.png' \/><\/center><br\/>G\u1ecdi s\u1ed1 c\u1ea1nh c\u1ee7a \u0111a gi\u00e1c \u1edf \u0111\u00e1y c\u1ee7a h\u00ecnh l\u0103ng tr\u1ee5 l\u00e0 $n$ ($n\\in \\mathbb N^*$)<br\/>Khi \u0111\u00f3, h\u00ecnh l\u0103ng tr\u1ee5 c\u00f3 $2n$ \u0111\u1ec9nh, $3n$ c\u1ea1nh v\u00e0 $n+2$ m\u1eb7t. <br\/>V\u00ec h\u00ecnh l\u0103ng tr\u1ee5 c\u00f3 t\u1ed5ng s\u1ed1 m\u1eb7t, s\u1ed1 \u0111\u1ec9nh v\u00e0 s\u1ed1 c\u1ea1nh l\u00e0 $26$ n\u00ean ta c\u00f3:<br\/> $2n+3n+n+2=26\\\\ \\Leftrightarrow 6n=24\\\\ \\Leftrightarrow n=4$<br\/>V\u1eady l\u0103ng tr\u1ee5 l\u00e0 l\u0103ng tr\u1ee5 t\u1ee9 gi\u00e1c \u0111\u1ec1u hay l\u0103ng tr\u1ee5 l\u00e0 h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt c\u00f3 \u0111\u00e1y l\u00e0 h\u00ecnh vu\u00f4ng.<br\/>G\u1ecdi \u0111\u1ed9 d\u00e0i c\u1ea1nh h\u00ecnh vu\u00f4ng l\u00e0 $a, (cm,\\,a>0)$ v\u00e0 \u0111\u01b0\u1eddng cao l\u00e0 $h \\,(cm,\\,h > 0)$<br\/>Di\u1ec7n t\u00edch xung quanh c\u1ee7a l\u0103ng tr\u1ee5 l\u00e0 $4ah=320$ $(cm^2)$ (1)<br\/>Th\u1ec3 t\u00edch c\u1ee7a l\u0103ng tr\u1ee5 l\u00e0 $a^2.h=640\\,(cm^3)$ (2)<br\/>T\u1eeb (1), ta c\u00f3: $ah=\\dfrac{320}{4}=80$.<br\/>Thay gi\u00e1 tr\u1ecb c\u1ee7a $ah$ v\u00e0o (2), ta c\u00f3: $a.80=640\\Leftrightarrow a=8\\,(cm)$<br\/>V\u1eady \u0111\u1ed9 d\u00e0i c\u1ea1nh \u0111\u00e1y l\u00e0 $8\\, cm$.<br\/>Chi\u1ec1u cao c\u1ee7a l\u0103ng tr\u1ee5 l\u00e0 $80:8=10\\,(cm)$<br\/><span class='basic_pink'>V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $10$<\/span><\/span>"}]}],"id_ques":1866},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"M\u1ed9t h\u00ecnh l\u0103ng tr\u1ee5 l\u1ee5c gi\u00e1c \u0111\u1ec1u c\u00f3 \u0111\u1ed9 d\u00e0i m\u1ed7i c\u1ea1nh \u0111\u1ec1u b\u1eb1ng $2\\sqrt 3 cm$. T\u00ednh th\u1ec3 t\u00edch l\u0103ng tr\u1ee5 \u0111\u00f3.<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai21/lv3/img\/H8_B21_K1.5a.png' \/><\/center>","select":["A. $72\\sqrt{3}\\,cm^3$","B. $36\\sqrt{3}\\,cm^3$","C. $54\\,cm^3$","D. $108\\,cm^3$"],"hint":"Chia \u0111\u00e1y th\u00e0nh $6$ tam gi\u00e1c nh\u1ecf c\u00f3 c\u00f9ng di\u1ec7n t\u00edch, r\u1ed3i t\u00ednh di\u1ec7n t\u00edch \u0111\u00e1y.","explain":" <span class='basic_left'>L\u0103ng tr\u1ee5 l\u1ee5c gi\u00e1c \u0111\u1ec1u l\u00e0 l\u0103ng tr\u1ee5 c\u00f3 \u0111\u00e1y l\u00e0 h\u00ecnh l\u1ee5c gi\u00e1c \u0111\u1ec1u.<br\/>X\u00e9t h\u00ecnh l\u1ee5c gi\u00e1c \u0111\u1ec1u $ABCDEF$ c\u00f3 t\u00e2m l\u00e0 $O$.<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai21/lv3/img\/H8_B21_K1.5.png' \/><\/center><br\/>Ta c\u00f3 di\u1ec7n t\u00edch l\u1ee5c gi\u00e1c \u0111\u1ec1u $ABCDEF$ b\u1eb1ng $6$ l\u1ea7n di\u1ec7n t\u00edch tam gi\u00e1c \u0111\u1ec1u $ABO$.<br\/>K\u1ebb $OH \\bot AB$ ($H\\in AB$)<br\/>Ta c\u00f3, $OH$ l\u00e0 \u0111\u01b0\u1eddng cao \u0111\u1ed3ng th\u1eddi l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn c\u1ee7a tam gi\u00e1c $OAB$. Suy ra $AH=\\dfrac{AB}{2}=\\sqrt{3}\\,cm$<br\/>X\u00e9t tam gi\u00e1c $OAH$ vu\u00f4ng t\u1ea1i $H$. \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: <br\/>$OH^2=AO^2-AH^2\\\\ \\Rightarrow OH^2=(2\\sqrt{3})^2-(\\sqrt{3})^2\\\\ \\Leftrightarrow OH^2=9\\\\ \\Leftrightarrow OH=3\\,(cm)$<br\/>Di\u1ec7n t\u00edch tam gi\u00e1c $ABO$ l\u00e0 $\\dfrac{1}{2}OH.AB=\\dfrac{1}{2}.3.2\\sqrt{3}=3\\sqrt{3}\\,(cm^2)$<br\/>Di\u1ec7n t\u00edch l\u1ee5c gi\u00e1c \u0111\u1ec1u c\u1ea1nh $2\\sqrt{3}\\,cm$ l\u00e0 $6.3\\sqrt{3}=18\\sqrt{3}\\,(cm^2)$<br\/>V\u1eady th\u1ec3 t\u00edch h\u00ecnh l\u0103ng tr\u1ee5 l\u00e0 $18\\sqrt{3}.2\\sqrt {3}=108\\,(cm^3)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D <\/span><\/span>","column":2}]}],"id_ques":1867},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng $ABCD.A\u2019B\u2019C\u2019D\u2019$ c\u00f3 \u0111\u00e1y $ABCD$ l\u00e0 h\u00ecnh thang vu\u00f4ng t\u1ea1i $A$ v\u00e0 $D$. Bi\u1ebft $AB=AD=a\\,cm, \\widehat{BCD}=45^o$ v\u00e0 $AC\u2019=3a\\,cm$. T\u00ednh th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng.<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai21/lv3/img\/H8_B21_K1.8.png' \/><\/center>","select":["A. $3a^3\\,cm^3$","B. $a^3\\,cm^3$","C. $\\dfrac{a^3}{2}\\,cm^3$","D. $\\dfrac{a^3}{3}\\,cm^3$"],"hint":"K\u1ebb $BH\\bot DC$. T\u00ednh $DC$.","explain":"<span class='basic_left'><br\/><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> K\u1ebb $BH\\bot DC$. Ch\u1ee9ng minh tam gi\u00e1c $BHC$ vu\u00f4ng c\u00e2n.<br\/><b>B\u01b0\u1edbc 2:<\/b> T\u00ednh $DC$<br\/><b>B\u01b0\u1edbc 3:<\/b> T\u00ednh $AC$ v\u00e0 $CC'$<br\/><b>B\u01b0\u1edbc 4:<\/b> T\u00ednh di\u1ec7n t\u00edch \u0111\u00e1y r\u1ed3i t\u00ednh th\u1ec3 t\u00edch h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai21/lv3/img\/H8_B21_K1.8a.png' \/><\/center><br\/>K\u1ebb $BH\\bot DC$ ($H\\bot DC$)<br\/>Ta c\u00f3 $BH\\bot DC;\\,AD\\bot DC$ (gi\u1ea3 thi\u1ebft) suy ra $BH\/\/AD$. <br\/>L\u1ea1i c\u00f3 $AB\/\/DH$ (gi\u1ea3 thi\u1ebft), suy ra $ABHD$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh.<br\/>Suy ra $BH=AD=a$<br\/>X\u00e9t tam gi\u00e1c $BHC$ vu\u00f4ng t\u1ea1i $H$ c\u00f3 $\\widehat{C}=45^o$ n\u00ean $BHC$ l\u00e0 tam gi\u00e1c vu\u00f4ng c\u00e2n t\u1ea1i $H$.<br\/>Do \u0111\u00f3, $HC=BH=a$, suy ra $DC=2a$<br\/>X\u00e9t tam gi\u00e1c $ADC$ vu\u00f4ng t\u1ea1i $D$. <br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3:<br\/>$AC^2=AD^2+DC^2\\\\ \\Rightarrow AC^2=a^2+(2a)^2\\\\ \\Leftrightarrow AC^2=5a^2\\\\ \\Leftrightarrow AC=a\\sqrt{5}\\,(cm)$<br\/>X\u00e9t tam gi\u00e1c $ACC'$ vu\u00f4ng t\u1ea1i $C$.<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3:<br\/>$CC'^2=AC'^2-AC^2\\\\ \\Rightarrow CC'^2=(3a)^2-(a\\sqrt{5})^2\\\\ \\Leftrightarrow CC'^2=4a^2\\\\ \\Leftrightarrow CC'=2a\\,(cm)$<br\/>Di\u1ec7n t\u00edch h\u00ecnh thang $ABCD$ l\u00e0 $\\dfrac{1}{2}(AB+DC).AD=\\dfrac{1}{2}.(a+2a).a=\\dfrac{3a^2}{2}$ ($cm^2$)<br\/>V\u1eady th\u1ec3 t\u00edch h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng l\u00e0 $CC'.S_{ABCD}=2a.\\dfrac{3a^2}{2}=3a^3$ ($cm^3$)<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":2}]}],"id_ques":1868},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["432"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"Cho h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng $ABC.A'B'C'$ \u0111\u00e1y l\u00e0 c\u00e1c tam gi\u00e1c vu\u00f4ng t\u1ea1i $B$ v\u00e0 $B'$. M\u1eb7t b\u00ean $AA'C'C$ l\u00e0 h\u00ecnh vu\u00f4ng c\u1ea1nh $12\\,cm$ v\u00e0 $\\widehat{A'BC}=30^o$. T\u00ednh th\u1ec3 t\u00edch h\u00ecnh l\u0103ng tr\u1ee5. <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai21/lv3/img\/H8_B21_K1.14.png' \/><\/center><br\/><b> \u0110\u00e1p s\u1ed1: <\/b> _input_ ($cm^3$)","hint":"Trong tam gi\u00e1c vu\u00f4ng \u0111\u1ed9 d\u00e0i c\u1ea1nh g\u00f3c vu\u00f4ng \u0111\u1ed1i di\u1ec7n g\u00f3c $30^o$ c\u00f3 \u0111\u1ed9 d\u00e0i b\u1eb1ng n\u1eeda c\u1ea1nh huy\u1ec1n.","explain":"<span class='basic_left'><br\/><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> T\u00ednh $A'C$<br\/><b>B\u01b0\u1edbc 2:<\/b> Ch\u1ee9ng minh tam gi\u00e1c $A'BC$ vu\u00f4ng t\u1ea1i $B$<br\/><b>B\u01b0\u1edbc 3:<\/b> T\u00ednh $BC$ v\u00e0 $AB$<br\/><b>B\u01b0\u1edbc 4:<\/b> T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $ABC$ r\u1ed3i t\u00ednh th\u1ec3 t\u00edch l\u0103ng tr\u1ee5.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai21/lv3/img\/H8_B21_K1.14.png' \/><\/center><br\/>V\u00ec $AA'C'C$ l\u00e0 h\u00ecnh vu\u00f4ng n\u00ean tam gi\u00e1c $A'AC$ vu\u00f4ng c\u00e2n t\u1ea1i $A$, suy ra $A'C=12\\sqrt{2}\\,(cm)$<br\/>Ta c\u00f3, $ABC.A'B'C'$ l\u00e0 h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng tam gi\u00e1c c\u00f3 $CB\\bot AB$ v\u00e0 $CB\\bot BB'$ n\u00ean $CB\\bot mp(AA'B'B)$.<br\/>Do v\u1eady $CB\\bot A'B$ hay tam gi\u00e1c $A'BC$ vu\u00f4ng t\u1ea1i $B$.<br\/>Ta l\u1ea1i c\u00f3, trong tam gi\u00e1c vu\u00f4ng $A'BC$ c\u00f3 $\\widehat {BA'C}=30^o$, suy ra $BC=\\dfrac{A'C}{2}=6\\sqrt{2}\\,(cm)$<br\/>X\u00e9t tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $B$ <br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago, ta c\u00f3: <br\/>$BC^2+AB^2=AC^2\\\\ \\Rightarrow AB^2=12^2-(6\\sqrt{2})^2\\\\ \\Leftrightarrow AB^2=72\\\\ \\Leftrightarrow AB=6\\sqrt{2}\\,(cm)$<br\/>Di\u1ec7n t\u00edch tam gi\u00e1c $ABC$ l\u00e0 $\\dfrac{1}{2}AB.BC=\\dfrac{1}{2}.6\\sqrt{2}.6\\sqrt{2}=36\\,(cm^2)$<br\/>V\u1eady th\u1ec3 t\u00edch h\u00ecnh l\u0103ng tr\u1ee5 l\u00e0 $36.12=432\\,(cm^3)$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $432$<\/span><br\/><\/span>"}]}],"id_ques":1869}],"lesson":{"save":0,"level":3}}