{"segment":[{"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":[[["a"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"Cho h\u00ecnh ch\u00f3p t\u1ee9 gi\u00e1c \u0111\u1ec1u $S.ABCD$ c\u00f3 c\u1ea1nh \u0111\u00e1y b\u1eb1ng $a\\,cm$. Cho bi\u1ebft $SA\\bot SC$. T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh b\u00ean.<br\/><b>\u0110\u00e1p s\u1ed1: <\/b> _input_ ($cm$)","hint":"Ch\u1ee9ng minh $\\Delta SAC=\\Delta BAC$","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai22/lv3/img\/H8_B22_K1.1.png' \/><\/center>V\u00ec $S.ABCD$ l\u00e0 h\u00ecnh t\u1ee9 gi\u00e1c \u0111\u1ec1u n\u00ean:<br\/>- $ABCD$ l\u00e0 h\u00ecnh vu\u00f4ng. <br\/>- C\u00e1c m\u1eb7t b\u00ean $SAB;SBC;SCD$ v\u00e0 $SDA$ l\u00e0 c\u00e1c tam gi\u00e1c c\u00e2n t\u1ea1i $S$ v\u00e0 b\u1eb1ng nhau.<br\/>X\u00e9t tam gi\u00e1c $SAC$ c\u00f3 $SA=SC$ v\u00e0 $SA\\bot SC$ (gi\u1ea3 thi\u1ebft). <br\/>Do \u0111\u00f3, $SAC$ l\u00e0 tam gi\u00e1c vu\u00f4ng c\u00e2n t\u1ea1i $S$ suy ra $\\widehat{SAC}=\\widehat{SCA}=45^o$<br\/>Ta l\u1ea1i c\u00f3: $ABCD$ l\u00e0 h\u00ecnh vu\u00f4ng n\u00ean $\\Delta ABC$ l\u00e0 tam gi\u00e1c vu\u00f4ng c\u00e2n t\u1ea1i $B$, suy ra $\\widehat{BAC}=\\widehat{BCA}=45^o$<br\/>X\u00e9t hai tam gi\u00e1c vu\u00f4ng ${BAC}$ v\u00e0 $SAC$ c\u00f3:<br\/> $\\bullet \\,AC$ chung.<br\/>$\\bullet\\,\\widehat{BAC}=\\widehat{SAC}=45^o$ <br\/>V\u1eady $\\Delta {BAC}=\\Delta{SAC}$ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn)<br\/>Suy ra, $SA=BA\\\\ SC=BC$ (c\u1eb7p c\u1ea1nh t\u01b0\u1eddng \u1ee9ng)<br\/>M\u1eb7t kh\u00e1c ta c\u0169ng c\u00f3: $SA=SC$, suy ra $SA=SC=AB=a\\,(cm)$<br\/>V\u1eady \u0111\u1ed9 d\u00e0i c\u1ea1nh b\u00ean l\u00e0 $a\\,(cm)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $a$<\/span><br\/><b> Nh\u1eadn x\u00e9t:<\/b> Ngo\u00e0i c\u00e1ch gi\u1ea3i tr\u00ean ta c\u00f3 th\u1ec3 s\u1eed d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago \u0111\u1ec3 t\u00ednh tr\u1ef1c ti\u1ebfp \u0111\u1ed9 d\u00e0i c\u1ea1nh b\u00ean. Nh\u01b0 sau:<br\/>S\u1eed d\u1ee5ng c\u00e1c k\u1ebft qu\u1ea3 \u0111\u00e3 ch\u1ee9ng m\u00ecnh: $SAC$ vu\u00f4ng c\u00e2n t\u1ea1i S v\u00e0 $ABC$ vu\u00f4ng c\u00e2n t\u1ea1i $B$<br\/>X\u00e9t $\\Delta BAC$ vu\u00f4ng c\u00e2n t\u1ea1i $B$ c\u00f3 $BA=BC=a\\,cm$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: $AC=\\sqrt{BA^2+BC^2}=\\sqrt{2a^2}=a\\sqrt{2}\\,(cm)$<br\/>X\u00e9t tam gi\u00e1c $SAC$ vu\u00f4ng c\u00e2n t\u1ea1i $S$. \u0110\u1eb7t $SA=SC=x\\,(cm)$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: <br\/>$AC^2=SA^2+SC^2\\Rightarrow 2a^2=2x^2\\Leftrightarrow x=a$<br\/>V\u1eady \u0111\u1ed9 d\u00e0i c\u1ea1nh b\u00ean l\u00e0 $a\\,(cm)$<\/span>"}]}],"id_ques":1890},{"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/bai22/lv3/img\/2.jpg' \/><\/center>Cho h\u00ecnh ch\u00f3p t\u1ee9 gi\u00e1c \u0111\u1ec1u $S.ABCD$ c\u00f3 \u0111\u1ed9 d\u00e0i c\u1ea1nh b\u00ean b\u1eb1ng \u0111\u1ed9 d\u00e0i hai \u0111\u01b0\u1eddng ch\u00e9o \u1edf \u0111\u00e1y v\u00e0 b\u1eb1ng $a\\,cm$. G\u1ecdi $M,\\,N$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $SA,\\,SB$. T\u1ee9 gi\u00e1c $MNCD$ l\u00e0 h\u00ecnh g\u00ec?","select":["A. H\u00ecnh thang","B. H\u00ecnh thang c\u00e2n","C. H\u00ecnh ch\u1eef nh\u1eadt","D. H\u00ecnh b\u00ecnh h\u00e0nh."],"hint":"$MN$ l\u00e0 \u0111\u01b0\u1eddng g\u00ec trong tam gi\u00e1c $SAB$.","explain":" <span class='basic_left'><br\/><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> Ch\u1ee9ng minh $MN\/\/CD$<br\/><b>B\u01b0\u1edbc 2:<\/b> T\u00ednh $CM$ v\u00e0 $DN$<br\/><b>B\u01b0\u1edbc 3:<\/b> K\u1ebft lu\u1eadn<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai22/lv3/img\/H8_B22_K1.3.png' \/><\/center><br\/>X\u00e9t tam gi\u00e1c $SAB$ c\u00f3 $M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $SA$ v\u00e0 $N$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $SB$ n\u00ean $MN\/\/AB$<br\/>Ta l\u1ea1i c\u00f3 $ABCD$ l\u00e0 h\u00ecnh vu\u00f4ng n\u00ean $AB\/\/CD$.<br\/>Do v\u1eady $MN\/\/CD$<br\/>V\u1eady $MNCD$ l\u00e0 h\u00ecnh thang. (*)<br\/>M\u1eb7t kh\u00e1c.<br\/>X\u00e9t tam gi\u00e1c $SAC$ c\u00f3 $SA=SC=AC=a\\,(cm)$ n\u00ean $SAC$ l\u00e0 tam gi\u00e1c \u0111\u1ec1u.<br\/>Ta c\u00f3: $CM$ l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn \u1ee9ng v\u1edbi c\u1ea1nh $SA$ c\u1ee7a tam gi\u00e1c $SAC$ suy ra $CM\\bot SA$.<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago v\u00e0o tam gi\u00e1c vu\u00f4ng $SCM$ ta c\u00f3 $CM=\\sqrt {SC^2-SM^2}=\\sqrt{a^2-\\dfrac{a^2}{4}}=\\dfrac{a\\sqrt {3}}{2}\\,(cm)$ (1)<br\/>T\u01b0\u01a1ng t\u1ef1 ta c\u0169ng c\u00f3: $\\Delta SBD$ l\u00e0 tam gi\u00e1c \u0111\u1ec1u c\u00f3 $DN$ l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn \u0111\u1ed3ng th\u1eddi l\u00e0 \u0111\u01b0\u1eddng cao.<br\/>Ta d\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c $DN=\\dfrac{a\\sqrt 3}{2}$ (2)<br\/>T\u1eeb (1) v\u00e0 (2), suy ra $CM=DN$ (**)<br\/>T\u1eeb (*) v\u00e0 (**), ta c\u00f3 $MNCD$ l\u00e0 h\u00ecnh thang c\u00e2n (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft)<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":1891},{"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 tam gi\u00e1c \u0111\u1ec1u c\u00f3 chi\u1ec1u cao l\u00e0 $20 cm$ v\u00e0 \u0111\u1ed9 d\u00e0i c\u1ea1nh \u0111\u00e1y l\u00e0 $18 cm$.<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai22/lv3/img\/H8_B22_D1.9.png' \/><\/center><br\/>T\u00ednh di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh ch\u00f3p tam gi\u00e1c \u0111\u1ec1u.<br\/>(<i> K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 hai <\/i>)","select":["A. $579,47\\,cm^2$","B. $579,69\\,cm^2$","C. $557,82\\,cm^2$","D. $1115,64\\,cm^2$"],"hint":"T\u00ednh $CH$. S\u1eed d\u1ee5ng t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m tam gi\u00e1c.","explain":"<span class='basic_left'><br\/>X\u00e9t $S.ABC$ l\u00e0 h\u00ecnh ch\u00f3p tam gi\u00e1c \u0111\u1ec1u.<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\u1ee7a 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 $SO \\bot mp(ABC) \\,\\Rightarrow SO\\bot HO$<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai22/lv3/img\/H8_B22_D1.9.png' \/><\/center>V\u00ec $SH$ l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c c\u00e2n $SAB$ n\u00ean $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB$. Suy ra $AH=9\\,cm$<br\/>X\u00e9t tam gi\u00e1c $ACH$ vu\u00f4ng t\u1ea1i $H$ c\u00f3 $AH=9\\,cm$ v\u00e0 $AC=18\\,cm$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3:<br\/>$AC^2=AH^2+CH^2\\\\ \\Rightarrow CH^2=18^2-9^2\\\\ \\Leftrightarrow CH^2=243\\\\ \\Leftrightarrow CH=9\\sqrt{3}\\,(cm)$<br\/>Ta c\u00f3 $ABC$ l\u00e0 tam gi\u00e1c \u0111\u1ec1u n\u00ean $O$ l\u00e0 giao \u0111i\u1ec3m ba \u0111\u01b0\u1eddng trung tr\u1ef1c \u0111\u1ed3ng th\u1eddi l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c $ABC$. <br\/>Suy ra $HO=\\dfrac{1}{3}CH=3\\sqrt{3}\\,(cm)$ <br\/> X\u00e9t tam gi\u00e1c $SHO$ vu\u00f4ng t\u1ea1i $O$. <br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3:<br\/>$SO^2+OH^2=SH^2\\\\ \\Rightarrow SH^2=20^2+(3\\sqrt{3})^2\\\\ \\Leftrightarrow SH^2=427\\\\ \\Leftrightarrow SH \\approx 20,66 \\,(cm)$<br\/>Di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh ch\u00f3p \u0111\u1ec1u $S.ABC$ l\u00e0 $\\dfrac{3.AB}{2}.SH=\\dfrac{3.18}{2}.20,66\\approx557,82\\,(cm^2)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}]}],"id_ques":1892},{"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 ch\u00f3p tam gi\u00e1c \u0111\u1ec1u c\u00f3 \u0111\u1ed9 d\u00e0i c\u1ea1nh \u0111\u00e1y l\u00e0 $18\\sqrt 2 cm$ c\u00e1c m\u1eb7t b\u00ean l\u00e0 c\u00e1c tam gi\u00e1c vu\u00f4ng c\u00e2n t\u1ea1i $S$. T\u00ednh chi\u1ec1u cao c\u1ee7a h\u00ecnh ch\u00f3p tam gi\u00e1c \u0111\u1ec1u \u0111\u00f3.<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai22/lv3/img\/H8_B22_K1.9.png' \/><\/center><br\/>","select":["A. $108\\,cm$","B. $6\\sqrt{3}\\,cm$","C. $6\\sqrt{2}\\,cm$","D. $3\\sqrt{3}\\,cm$"],"hint":"T\u00ednh \u0111\u1ed9 d\u00e0i $SC$ v\u00e0 $CH$. \u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m tam gi\u00e1c ","explain":"<span class='basic_left'><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 b\u00ean<br\/><b>B\u01b0\u1edbc 2:<\/b> T\u00ednh $CH$, suy ra \u0111\u1ed9 d\u00e0i $CO$<br\/><b>B\u01b0\u1edbc 3:<\/b> T\u00ednh \u0111\u01b0\u1eddng cao.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai22/lv3/img\/H8_B22_K1.9.png' \/><\/center>X\u00e9t $S.ABC$ l\u00e0 h\u00ecnh ch\u00f3p tam gi\u00e1c \u0111\u1ec1u.<br\/>+) $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC$<br\/>+) $SO$ l\u00e0 \u0111\u01b0\u1eddng cao<br\/>- V\u00ec $SO$ l\u00e0 \u0111\u01b0\u1eddng cao n\u00ean $SO \\bot mp(ABC) \\,\\Rightarrow SO\\bot HO$<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai22/lv3/img\/H8_B22_K1.9a.png' \/><\/center>X\u00e9t m\u1eb7t b\u00ean $SBC$ vu\u00f4ng c\u00e2n t\u1ea1i $S$.<br\/>\u0110\u1eb7t $SC=SB=x\\,cm$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3:<br\/>$SC^2+SB^2=AB^2\\\\ \\Leftrightarrow 2x^2=648\\\\ \\Leftrightarrow x^2=324\\\\ \\Leftrightarrow x=18\\,(cm)$<br\/>X\u00e9t tam gi\u00e1c $CAH$ vu\u00f4ng t\u1ea1i $H$ c\u00f3 $AC=18\\sqrt {2}\\,cm$; $AH=9\\sqrt{2}\\,cm$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: $AC^2=CH^2+HA^2\\\\ \\Leftrightarrow CH^2=AC^2-HA^2\\\\ \\Leftrightarrow CH^2=486\\\\ \\Leftrightarrow CH=9\\sqrt{6}\\,(cm)$<br\/>V\u00ec O l\u00e0 t\u00e2m c\u1ee7a tam gi\u00e1c \u0111\u1ec1u $ABC$ n\u00ean $O $ l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c $ABC$. Suy ra $CO=\\dfrac{2}{3}CH=6\\sqrt{6}\\,(cm)$<br\/>X\u00e9t tam gi\u00e1c $SOC$ vu\u00f4ng t\u1ea1i $O$ c\u00f3 $SC=18\\,cm$ v\u00e0 $OC=6\\sqrt{6}\\,cm$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago, ta c\u00f3:<br\/>$SC^2=SO^2+OC^2\\\\ \\Leftrightarrow SO^2=SC^2-CO^2\\\\ \\Leftrightarrow SO^2=18^2- (6\\sqrt{6})^2\\\\ \\Leftrightarrow SO^2=108\\\\ \\Leftrightarrow SO=6\\sqrt{3}\\,(cm)$ <br\/>V\u1eady chi\u1ec1u cao c\u1ee7a h\u00ecnh ch\u00f3p l\u00e0 $6\\sqrt{3}\\,cm$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":1893},{"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 ch\u00f3p tam gi\u00e1c \u0111\u1ec1u c\u1ea1nh b\u00ean b\u1eb1ng $37 cm$ v\u00e0 \u0111\u1ed9 d\u00e0i c\u1ea1nh \u0111\u00e1y l\u00e0 $12\\sqrt {3} cm$. T\u00ednh th\u1ec3 t\u00edch h\u00ecnh ch\u00f3p tam gi\u00e1c \u0111\u1ec1u.<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai22/lv3/img\/H8_B22_D1.12.png' \/><\/center>","select":["A. $630\\sqrt{3}\\,cm^3$","B. $1260\\sqrt{3}\\,cm^3$","C. $444\\sqrt{3}\\,cm^3$","D. $1260\\,cm^3$"],"hint":"T\u00ednh $CH$. S\u1eed d\u1ee5ng t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m tam gi\u00e1c.","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 $CH$<br\/><b>B\u01b0\u1edbc 2:<\/b> \u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m tam gi\u00e1c t\u00ednh $CO$<br\/><b>B\u01b0\u1edbc 3:<\/b> T\u00ednh di\u1ec7n t\u00edch $ABC$<br\/><b>B\u01b0\u1edbc 4:<\/b> T\u00ednh th\u1ec3 t\u00edch h\u00ecnh ch\u00f3p<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai22/lv3/img\/H8_B22_D1.12.png' \/><\/center><br\/>X\u00e9t $S.ABC$ l\u00e0 h\u00ecnh ch\u00f3p tam gi\u00e1c \u0111\u1ec1u.<br\/>+) $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB$<br\/>+) $SO$ l\u00e0 \u0111\u01b0\u1eddng cao<br\/>- V\u00ec $SO$ l\u00e0 \u0111\u01b0\u1eddng cao n\u00ean $SO \\bot mp(ABC) \\,\\Rightarrow SO\\bot HO$<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai22/lv3/img\/H8_B22_D1.12a.png' \/><\/center>X\u00e9t tam gi\u00e1c $ABC$ \u0111\u1ec1u $CH\\bot AB\\Rightarrow HA=HB=6\\sqrt{3}$<br\/>X\u00e9t tam gi\u00e1c $ACH$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 $AH=6\\sqrt{3}\\,cm$ v\u00e0 $AC=12\\sqrt{3}\\,cm$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3:<br\/>$AC^2=AH^2+CH^2\\\\ \\Rightarrow CH^2=324\\\\ \\Leftrightarrow CH=18\\,(cm)$<br\/>Ta c\u00f3 $ABC$ l\u00e0 tam gi\u00e1c \u0111\u1ec1u n\u00ean $O$ l\u00e0 giao \u0111i\u1ec3m ba \u0111\u01b0\u1eddng trung tr\u1ef1c \u0111\u1ed3ng th\u1eddi l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c $ABC$. Suy ra $CO=\\dfrac{2}{3}CH=12\\,(cm)$ <br\/> X\u00e9t tam gi\u00e1c $SCO$ vu\u00f4ng t\u1ea1i $O$. \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3:<br\/>$SO^2+OC^2=SC^2\\\\ \\Rightarrow SO^2=37^2-12^2=1225\\\\ \\Leftrightarrow SO=35\\,(cm)$<br\/>Di\u1ec7n t\u00edch $ABC$ l\u00e0 $\\dfrac{1}{2}.12\\sqrt{3}.18=108\\sqrt{3}\\,(cm^2)$<br\/>Th\u1ec3 t\u00edch h\u00ecnh ch\u00f3p l\u00e0 $\\dfrac{1}{3}.108\\sqrt{3}.35=1260\\sqrt{3}\\,(cm^3)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":1894},{"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/bai22/lv3/img\/2.jpg' \/><\/center>T\u00ednh di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh ch\u00f3p l\u1ee5c gi\u00e1c \u0111\u1ec1u c\u00f3 chi\u1ec1u cao $13\\,cm$ v\u00e0 \u0111\u1ed9 d\u00e0i c\u1ea1nh \u0111\u00e1y l\u00e0 $6\\,cm$","select":["A. $126\\,cm^2$","B. $252\\,cm^2$","C. $504\\,cm^2$","D. \u0110\u00e1p \u00e1n kh\u00e1c"],"hint":"T\u00ednh $OH$ r\u1ed3i t\u00ednh $SH$","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> Ch\u1ee9ng minh t\u00e2m $O$ chia l\u1ee5c gi\u00e1c $ABCDEF$ th\u00e0nh $6$ l\u1ee5c gi\u00e1c \u0111\u1ec1u.<br\/><b>B\u01b0\u1edbc 2:<\/b> T\u00ednh s\u1ed1 \u0111o $OH$<br\/><b>B\u01b0\u1edbc 3:<\/b> T\u00ednh $SH$<br\/><b>B\u01b0\u1edbc 4:<\/b> T\u00ednh di\u1ec7n t\u00edch xung quanh<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai22/lv3/img\/H8_B22_K1.4.png' \/><\/center><br\/>Ta c\u00f3: $S.ABCDEF$ l\u00e0 h\u00ecnh ch\u00f3p l\u1ee5c gi\u00e1c \u0111\u1ec1u n\u00ean $ABCDEF$ l\u00e0 l\u1ee5c gi\u00e1c \u0111\u1ec1u.<br\/>G\u1ecdi $O$ l\u00e0 t\u00e2m c\u1ee7a l\u1ee5c gi\u00e1c \u0111\u1ec1u $ABCDEF$, suy ra $O$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a c\u00e1c \u0111\u01b0\u1eddng ch\u00e9o.<br\/>D\u1ec5 d\u00e0ng ch\u1ec9 ra \u0111\u01b0\u1ee3c $OAB;OBC;OCD;ODE;OEF$ v\u00e0 $OFA$ l\u00e0 c\u00e1c tam gi\u00e1c \u0111\u1ec1u c\u1ea1nh $6 cm$<br\/>X\u00e9t tam gi\u00e1c $OHA$ vu\u00f4ng t\u1ea1i $H$ c\u00f3 $OA=6\\,cm,\\,AH=3\\,cm$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: $OA^2=OH^2+AH^2\\\\ \\Leftrightarrow OH^2=OA^2-AH^2\\\\ \\Rightarrow OH^2=36-9=27\\\\ \\Leftrightarrow OH=3\\sqrt3\\,(cm)$<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai22/lv3/img\/H8_B22_K1.4a.png' \/><\/center>X\u00e9t tam gi\u00e1c $SOH$ vu\u00f4ng t\u1ea1i $O$ c\u00f3 $SO=12\\,cm$ v\u00e0 $OH=3\\sqrt3 \\,cm$.<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: <br\/>$SH^2=SO^2+OH^2=169+27=196\\Leftrightarrow SH=14\\,cm$<br\/>Di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh ch\u00f3p $S.ABCDEF$ l\u00e0 $\\dfrac{6.AB}{2}.SH=\\dfrac{6.6}{2}.14=252\\,(cm^2)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":1895},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $12\\sqrt {7}$","B. $5\\sqrt {7}$","C. $8\\sqrt {7}$"],"ques":"Cho h\u00ecnh ch\u00f3p t\u1ee9 gi\u00e1c \u0111\u1ec1u $S.ABCD$ c\u00f3 c\u1ea1nh b\u00ean l\u00e0 $5\\,cm$. Di\u1ec7n t\u00edch xung quanh b\u1eb1ng $48\\, cm^2$. T\u00ednh th\u1ec3 t\u00edch h\u00ecnh ch\u00f3p.<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> ?($cm^3$)","hint":"G\u1ecdi \u0111\u1ed9 d\u00e0i c\u1ea1nh $BC$ v\u00e0 trung \u0111o\u1ea1n l\u00e0 \u1ea9n. T\u1eeb d\u1eef ki\u1ec7n \u0111\u1ec1 b\u00e0i l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh theo \u1ea9n.","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai22/lv3/img\/H8_B22_K1.5.png' \/><\/center>\u0110\u1eb7t $AB=2a\\,cm$ v\u00e0 trung \u0111o\u1ea1n $SH=d\\,cm$ ($a < d$)<br\/>Khi \u0111\u00f3 $S_{xq}=\\dfrac{2a.4.d}{2}=4ad=48\\Leftrightarrow 2ad=24\\,(cm^2)$ (1)<br\/>X\u00e9t tam gi\u00e1c $SAH$ vu\u00f4ng t\u1ea1i $H$ c\u00f3 $SA=5\\,cm;\\,AH=a\\,cm$ v\u00e0 $SH=d\\,cm$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: <br\/>$SA^2=SH^2+AH^2\\\\ \\Rightarrow 25=d^2+a^2\\\\ \\Leftrightarrow a^2+2ad+d^2=25+24\\,\\,(\\text{C\u1ed9ng hai v\u1ebf v\u1edbi}\\,2ad=24 )\\\\ \\Leftrightarrow (a+d)^2=49\\\\ \\Leftrightarrow a+d=7\\,\\,(2)$<br\/>T\u1eeb (1) v\u00e0 (2) ta c\u00f3: <br\/>$\\left\\{ \\begin{align} & ad=12 \\\\ & a+d=7 \\\\ \\end{align} \\right.$<br\/>Thay $d=7-a$ v\u00e0 $ad=12$ ta c\u00f3:<br\/>$a(7-a)=12\\\\ \\Leftrightarrow a^2-7a+12=0\\\\ \\Leftrightarrow a^2-3a-4a+12=0\\\\ \\Leftrightarrow (a-3)(a-4)=0\\\\ \\Leftrightarrow \\left\\{ \\begin{align} & a=3 \\\\ & a=4 \\\\ \\end{align} \\right. $ <br\/><b> Tr\u01b0\u1eddng h\u1ee3p 1: <\/b>V\u1edbi $a=3\\Leftrightarrow d=4$ (th\u1ecfa m\u00e3n)<br\/>V\u1eady $AB=6\\,cm$ v\u00e0 $SH=4\\,cm$<br\/>X\u00e9t tam gi\u00e1c $SOH$ vu\u00f4ng t\u1ea1i $O$.<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3:<br\/>$SO=\\sqrt{SH^2-OH^2}=\\sqrt{4^2-3^2}=\\sqrt{7}\\,(cm)$<br\/>V\u1eady th\u1ec3 t\u00edch h\u00ecnh ch\u00f3p $S.ABCD$ l\u00e0 $\\dfrac{1}{3}.\\sqrt{7}.6^2=12\\sqrt{7}\\,(cm^3)$<br\/><b> Tr\u01b0\u1eddng h\u1ee3p 2: <\/b>V\u1edbi $a=4\\Leftrightarrow d=3$ (lo\u1ea1i )<br\/>V\u1eady th\u1ec3 t\u00edch h\u00ecnh ch\u00f3p l\u00e0 $12\\sqrt 7 \\,cm^3$<\/span>"}]}],"id_ques":1896},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai22/lv3/img\/2.jpg' \/><\/center>M\u1ed9t h\u00ecnh ch\u00f3p t\u1ee9 gi\u00e1c \u0111\u1ec1u c\u00f3 c\u1ea1nh $9\\,cm$, s\u1ed1 \u0111o th\u1ec3 t\u00edch (\u0111\u01a1n v\u1ecb $cm^3$) b\u1eb1ng $\\dfrac{3}{4}$ s\u1ed1 \u0111o di\u1ec7n t\u00edch xung quanh (\u0111\u01a1n v\u1ecb $cm^2$). T\u00ednh chi\u1ec1u cao v\u00e0 trung \u0111o\u1ea1n c\u1ee7a h\u00ecnh ch\u00f3p.<br\/>\u0110\u1ed9 d\u00e0i chi\u1ec1u cao v\u00e0 trung \u0111o\u1ea1n l\u1ea7n l\u01b0\u1ee3t l\u00e0:","select":["A. $\\dfrac{3\\sqrt{3}}{2}\\,cm$ v\u00e0 $3\\sqrt{3}\\,cm$ ","B. $\\dfrac{3\\sqrt{3}}{2}\\,cm$ v\u00e0 $\\sqrt{3}\\,cm$","C. $\\dfrac{\\sqrt{3}}{2}\\,cm$ v\u00e0 $3\\sqrt{3}\\,cm$","D. $\\dfrac{\\sqrt{3}}{2}\\,cm$ v\u00e0 $\\sqrt{3}\\,cm$"],"hint":"G\u1ecdi \u0111\u1ed9 d\u00e0i trung \u0111o\u1ea1n v\u00e0 chi\u1ec1u cao l\u00e0 \u1ea9n r\u1ed3i bi\u1ec3u di\u1ec5n c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh theo \u1ea9n","explain":" <span class='basic_left'>G\u1ecdi h\u00ecnh ch\u00f3p t\u1ee9 gi\u00e1c \u0111\u1ec1u c\u00f3 di\u1ec7n t\u00edch \u0111\u00e1y l\u00e0 $S$ ($cm^2$), n\u1eeda chu vi l\u00e0 $p\\,(cm)$, chi\u1ec1u cao $h$ ($cm$) v\u00e0 trung \u0111o\u1ea1n $d$ ($cm$).<br\/>Ta c\u00f3:<br\/>Th\u1ec3 t\u00edch h\u00ecnh ch\u00f3p \u0111\u1ec1u l\u00e0 $V=\\dfrac{1}{3}.S.h=\\dfrac{1}{3}.81.h=27h\\,(cm^3)$<br\/>Di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh h\u00ecnh ch\u00f3p l\u00e0 $ S_{xq}=p.d= \\dfrac{9.4}{2}.d=18d \\,(cm^2) $<br\/> Theo \u0111\u1ec1 b\u00e0i ta c\u00f3 s\u1ed1 \u0111o th\u1ec3 t\u00edch (\u0111\u01a1n v\u1ecb $ cm^3 $) b\u1eb1ng $ \\dfrac{3}{4} $ s\u1ed1 \u0111o di\u1ec7n t\u00edch xung quanh (\u0111\u01a1n v\u1ecb $ cm^2 $) n\u00ean:<br\/> $V=\\dfrac{3}{4}.S_{xq}\\Rightarrow 27h=\\dfrac{3}{4}.18d\\Leftrightarrow 108h=54d\\Leftrightarrow \\dfrac{h}{1}=\\dfrac{d}{2}\\Leftrightarrow \\dfrac{h^2}{1}=\\dfrac{d^2}{4}$<br\/>Ta l\u1ea1i c\u00f3 $d^2-h^2=\\dfrac{81}{4}$<br\/>$\\dfrac{d^2}{4}=\\dfrac{h^2}{1}=\\dfrac{d^2-h^2}{3}=\\dfrac{27}{4}$ (t\u00ednh ch\u1ea5t d\u1eaby t\u1ec9 s\u1ed1 b\u1eb1ng nhau)<br\/>V\u1eady $d^2=27\\Leftrightarrow d=3\\sqrt{3}\\,(cm)\\\\ h^2=\\dfrac{27}{4}\\Leftrightarrow h=\\dfrac{3\\sqrt{3}}{2}\\,(cm)$.<br\/>V\u1eady h\u00ecnh ch\u00f3p \u0111\u1ec1u c\u00f3 chi\u1ec1u cao l\u00e0 $\\dfrac{3\\sqrt{3}}{2}\\,(cm)$ v\u00e0 trung \u0111o\u1ea1n $3\\sqrt{3}\\,(cm)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":2}]}],"id_ques":1897},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"select":["A. $\\dfrac{\\sqrt{3}}{8}$","B. $\\dfrac{2\\sqrt{3}}{8}$","C. $\\dfrac{3\\sqrt{3}}{8}$"],"ques":"M\u1ed9t h\u00ecnh ch\u00f3p tam gi\u00e1c \u0111\u1ec1u c\u00f3 th\u1ec3 t\u00edch b\u1eb1ng $24\\,cm^3$, c\u1ea1nh \u0111\u00e1y b\u1eb1ng $16\\,cm$. T\u00ednh chi\u1ec1u cao h\u00ecnh ch\u00f3p.<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> ?($cm$)<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai22/lv3/img\/H8_B22_K1.8.png' \/><\/center>","hint":"T\u00ednh di\u1ec7n t\u00edch \u0111\u00e1y.","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai22/lv3/img\/H8_B22_K1.8.png' \/><\/center>- X\u00e9t $S.ABC$ l\u00e0 h\u00ecnh ch\u00f3p tam gi\u00e1c \u0111\u1ec1u <br\/>+) $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB$<br\/>+) $SO$ l\u00e0 \u0111\u01b0\u1eddng cao<br\/>- V\u00ec $SO$ l\u00e0 \u0111\u01b0\u1eddng cao n\u00ean $SO \\bot mp(ABC) \\,\\Rightarrow SO\\bot HO$<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai22/lv3/img\/H8_B22_K1.8a.png' \/><\/center>X\u00e9t tam gi\u00e1c $CBH$ vu\u00f4ng t\u1ea1i $H$ c\u00f3 $CB=16\\,cm$ v\u00e0 $BH=8\\,cm$<br\/>T\u1eeb \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: $CH=\\sqrt{CB^2-BH^2}=\\sqrt{16^2-8^2}=8\\sqrt{3}\\,(cm)$<br\/>Di\u1ec7n t\u00edch tam gi\u00e1c $ABC$ l\u00e0 $\\dfrac{1}{2}.8\\sqrt {3}.16=64\\sqrt{3}\\,cm^2$<br\/>V\u1eady chi\u1ec1u cao h\u00ecnh ch\u00f3p l\u00e0 $\\dfrac{3.24}{64\\sqrt {3}}=\\dfrac{3\\sqrt{3}}{8}\\,(cm)$<\/span>"}]}],"id_ques":1898},{"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":[[["125,09"]]],"list":[{"point":10,"width":100,"ques":"Cho h\u00ecnh ch\u00f3p tam gi\u00e1c \u0111\u1ec1u $S.ABC$ c\u00f3 c\u1ea1nh b\u00ean l\u00e0 $13\\,cm$ v\u00e0 chi\u1ec1u cao b\u1eb1ng $12\\,cm$. L\u1ea5y \u0111i\u1ec3m $A'$ thu\u1ed9c $SA$ sao cho $SA'=\\dfrac{1}{3}SA$. Qua $A'$ d\u1ef1ng m\u1eb7t ph\u1eb3ng $(P)$ song song v\u1edbi \u0111\u00e1y m\u1eb7t ph\u1eb3ng \u0111\u00e1y $(ABC)$ c\u1eaft $SB;\\,SC$ l\u1ea7n l\u01b0\u1ee3t t\u1ea1i $B';\\,C'$. $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB$, $SH$ c\u1eaft $A'B'$ t\u1ea1i $H'$. T\u00ednh th\u1ec3 t\u00edch h\u00ecnh ch\u00f3p c\u1ee5t $A'B'C'.ABC$<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai22/lv3/img\/H8_B22_K1.7.png' \/><\/center><b> \u0110\u00e1p s\u1ed1: <\/b> _input_ ($cm^3$)<br\/>(<i> K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 hai<\/i>)","hint":"T\u00ednh th\u1ec3 t\u00edch hai h\u00ecnh ch\u00f3p $S.ABC$ v\u00e0 $S.A'B'C'$. \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Ta-let.","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 \u0111\u00e1y<br\/><b>B\u01b0\u1edbc 2:<\/b> T\u00ednh th\u1ec3 t\u00edch h\u00ecnh ch\u00f3p $S.ABC$<br\/><b>B\u01b0\u1edbc 3:<\/b> T\u00ednh c\u00e1ch k\u00edch th\u01b0\u1edbc h\u00ecnh ch\u00f3p $S.A'B'C'$<br\/><b>B\u01b0\u1edbc 4:<\/b> T\u00ednh th\u1ec3 t\u00edch h\u00ecnh ch\u00f3p c\u1ee5t $A'B'C'.ABC$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>- V\u00ec $S.ABC$ l\u00e0 h\u00ecnh 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\/>+) $SO$ l\u00e0 \u0111\u01b0\u1eddng cao<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/bai22/lv3/img\/H8_B22_K1.7.png' \/><\/center>X\u00e9t tam gi\u00e1c $SOC$ vu\u00f4ng t\u1ea1i $O$ c\u00f3 $SO=12\\,cm$ v\u00e0 $SC=13\\,cm$<br\/>T\u1eeb \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: <br\/>$OC=\\sqrt{SC^2-SO^2}=\\sqrt{169-144}=5\\,(cm)$<br\/>Ta c\u00f3: $ABC$ l\u00e0 tam gi\u00e1c \u0111\u1ec1u c\u00f3 t\u00e2m $O$ n\u00ean $O$ l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c. <br\/>\u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m tam gi\u00e1c ta c\u00f3 $CO=\\dfrac{2}{3}CH\\Leftrightarrow CH=\\dfrac{3}{2}OC=\\dfrac{15}{2}\\,(cm)$<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai22/lv3/img\/H8_B22_K1.7a.png' \/><\/center>X\u00e9t tam gi\u00e1c $CHB$ vu\u00f4ng t\u1ea1i $H$. <br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: <br\/>$CB^2=CH^2+HB^2\\Leftrightarrow x^2=\\dfrac{15^2}{4}+\\dfrac{x^2}{4}\\Leftrightarrow x=5\\sqrt{3}\\,(cm)$<br\/>Di\u1ec7n t\u00edch tam gi\u00e1c $ABC$ l\u00e0 $\\dfrac{1}{2}.CH.AB=\\dfrac{1}{2}.\\dfrac{15}{2}.5\\sqrt{3}=\\dfrac{75\\sqrt{3}}{4}\\,(cm^2)$<br\/>Th\u1ec3 t\u00edch h\u00ecnh ch\u00f3p $S.ABC$ l\u00e0 $\\dfrac{1}{3}.12.\\dfrac{75\\sqrt{3}}{4}=75\\sqrt{3}\\approx 129,9\\,(cm^3)$<br\/> Theo gi\u1ea3 thi\u1ebft v\u00ec $(P)\/\/mp(ABC)$ n\u00ean $A'B'\/\/AB; \\,B'C'\/\/BC;\\,A'C'\/\/AC$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Ta-let, d\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c $A'B'C'$ \u0111\u1ec1u c\u1ea1nh b\u1eb1ng $\\dfrac{1}{3}AB=\\dfrac{5\\sqrt{3}}{3}\\,(cm)$ v\u00e0 \u0111\u01b0\u1eddng cao $C'H'=\\dfrac{1}{3}CH=\\dfrac{5}{2}\\,(cm)$<br\/>Do \u0111\u00f3 $S_{A'B'C'}=\\dfrac{1}{2}.C'H'.A'B'=\\dfrac{1}{2}.\\dfrac{5}{2}.\\dfrac{5\\sqrt{3}}{3}=\\dfrac{25\\sqrt{3}}{12}\\,(cm^2)$<br\/>Ta c\u0169ng c\u00f3 $SO'=\\dfrac{1}{3}SO=4\\,(cm)$<br\/>V\u1eady th\u1ec3 t\u00edch h\u00ecnh ch\u00f3p $S.A'B'C'=\\dfrac{1}{3}.SO'.S_{A'B'C'}=\\dfrac{1}{3}.4.\\dfrac{25\\sqrt{3}}{12}\\approx 4,81\\,(cm^3) $<br\/>V\u1eady th\u1ec3 t\u00edch h\u00ecnh ch\u00f3p c\u1ee5t $A'B'C'.ABC$ l\u00e0 $125,09\\,(cm^3)$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $125,09 $<\/span><\/span>"}]}],"id_ques":1899}],"lesson":{"save":0,"level":3}}