{"segment":[{"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["t","t","f","t","t"]],"list":[{"point":10,"image":"https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai24/lv3/img\/H8_B24_K1.1.png","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["$AC \\bot CC'$","$AD \\bot mp(DCC'D')$","$mp(ABB'A) \/\/ mp(ADD'A')$","$A'C=B'D$","$AC \/\/ mp(A'B'C'D')$"],"hint":"","explain":["\u0110\u00fang, v\u00ec $ACC'A'$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt.","<br\/> \u0110\u00fang v\u00ec $AD\\bot DC$; $AD\\bot DD'$.","<br\/> Sai v\u00ec $AB$ giao $AD$","<br\/> \u0110\u00fang v\u00ec $A'BCD'$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt. Suy ra $A'C=B'D$","<br\/>\u0110\u00fang v\u00ec $\\left\\{\\begin{aligned}&AC\/\/A'C'\\\\ &AC\\notin\\,mp(A'B'C'D')\\\\ \\end{aligned}\\right. \\Rightarrow AC \/\/ mp(A'B'C'D')$"]}],"id_ques":1930},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"S\u1ed1 m\u1eb7t, s\u1ed1 c\u1ea1nh v\u00e0 s\u1ed1 \u0111\u1ec9nh c\u1ee7a h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng tam gi\u00e1c l\u00e0:","select":["A. $3$ m\u1eb7t, $6$ c\u1ea1nh v\u00e0 $6$ \u0111\u1ec9nh.","B. $5$ m\u1eb7t, $6$ c\u1ea1nh v\u00e0 $6$ \u0111\u1ec9nh.","C. $5$ m\u1eb7t, $9$ c\u1ea1nh v\u00e0 $6$ \u0111\u1ec9nh.","D. $3$ m\u1eb7t, $9$ c\u1ea1nh v\u00e0 $6$ \u0111\u1ec9nh."],"hint":"","explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai24/lv3/img\/H8_B24_K1.2.png' \/><\/center>H\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng tam gi\u00e1c c\u00f3 3 c\u1ea1nh \u1edf \u0111\u00e1y n\u00ean c\u00f3 $5$ m\u1eb7t ($2$ m\u1eb7t \u0111\u00e1y v\u00e0 $3$ m\u1eb7t b\u00ean); $9$ c\u1ea1nh ($6$ c\u1ea1nh \u1edf hai \u0111\u00e1y v\u00e0 $3$ c\u1ea1nh b\u00ean) v\u00e0 $6$ \u0111\u1ec9nh.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}],"id_ques":1931},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["5"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"H\u00ecnh ch\u00f3p c\u00f3 $10$ c\u1ea1nh th\u00ec \u0111a gi\u00e1c \u1edf \u0111\u00e1y c\u00f3 bao nhi\u00eau \u0111\u1ec9nh?<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> _input_(\u0111\u1ec9nh)","hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai24/lv3/img\/H8_B24_K1.3.png' \/><\/center>Trong h\u00ecnh ch\u00f3p s\u1ed1 c\u1ea1nh b\u1eb1ng $2$ l\u1ea7n s\u1ed1 \u0111\u1ec9nh c\u1ee7a \u0111a gi\u00e1c \u1edf \u0111\u00e1y n\u00ean n\u1ebfu h\u00ecnh ch\u00f3p c\u00f3 $10$ c\u1ea1nh th\u00ec \u0111a gi\u00e1c \u1edf \u0111\u00e1y c\u00f3 $5$ c\u1ea1nh.<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $5$<\/span><\/span>"}],"id_ques":1932},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["60"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"Cho h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt $ABCD.A\u2019B\u2019C\u2019D\u2019$ c\u00f3 $A'C=5\\sqrt{2}\\,cm$; $AD=3\\,cm$. T\u00ednh th\u1ec3 t\u00edch h\u00ecnh h\u1ed9p bi\u1ebft $\\Delta AA'C'$ c\u00e2n t\u1ea1i $A$. <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai24/lv3/img\/H8_B24_K1.4.png' \/><\/center><br\/><b> \u0110\u00e1p s\u1ed1: <\/b> _input_($cm^3$)","hint":"","explain":"<span class='basic_left'>Ta c\u00f3 $ABCD.A'B'C'D'$ l\u00e0 h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt n\u00ean ta c\u00f3: $AA' \\bot mp(ABCD) \\Leftrightarrow \\Delta AA'C$ vu\u00f4ng t\u1ea1i $A$.<br\/>Theo \u0111\u1ec1 b\u00e0i, ta l\u1ea1i c\u00f3 $\\Delta AA'C$ c\u00e2n, suy ra $AC=AA'$<br\/>X\u00e9t tam gi\u00e1c $AA'C$ vu\u00f4ng t\u1ea1i $A$ ta c\u00f3: <br\/>$AA'^2+AC^2=A'C^2 \\Rightarrow 2AA'^2=(5\\sqrt{2})^2\\\\ \\Leftrightarrow AA'^2=25 \\Leftrightarrow AA'=5\\,(cm)$<br\/>X\u00e9t tam gi\u00e1c $ADC$ vu\u00f4ng t\u1ea1i $D$ c\u00f3 $AC=5\\,cm$ v\u00e0 $AD=3\\,cm$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago, ta c\u00f3:<br\/>$AC^2=AD^2+DC^2\\\\ \\Rightarrow 5^2=3^2+DC^2 \\Leftrightarrow DC^2=16 \\Leftrightarrow DC=4\\,(cm)$<br\/>V\u1eady di\u1ec7n t\u00edch \u0111\u00e1y $ABCD$ b\u1eb1ng $AD.DC=3.4=12\\,(cm^2)$<br\/>Th\u1ec3 t\u00edch h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt l\u00e0 $AA'.S_{ABCD}=5.12=60\\,(cm^3)$<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $60$<\/span><\/span>"}],"id_ques":1933},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"M\u1ed9t h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt c\u00f3 \u0111\u00e1y l\u00e0 h\u00ecnh vu\u00f4ng v\u00e0 th\u1ec3 t\u00edch l\u00e0 $128 cm^3$. T\u00ednh chi\u1ec1u cao c\u1ee7a h\u00ecnh h\u1ed9p, bi\u1ebft $AC=4\\sqrt{2}\\,cm$.","select":["A. $4\\,cm$","B. $4\\sqrt{2}\\,cm$","C. $8\\,cm$","D. $8\\sqrt{2}\\,cm$"],"hint":"T\u00ednh c\u1ea1nh \u0111\u00e1y r\u1ed3i 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/bai24/lv3/img\/H8_B24_K1.5.png' \/><\/center>X\u00e9t h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt $ABCD.A'B'C'D'$ c\u00f3 hai \u0111\u00e1y l\u00e0 h\u00ecnh vu\u00f4ng.<br\/>X\u00e9t tam gi\u00e1c $ABC$ vu\u00f4ng c\u00e2n t\u1ea1i $B$ c\u00f3 $AC=4\\sqrt{2}\\,cm$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago, ta c\u00f3: $AC^2=AB^2+BC^2 \\Leftrightarrow AC^2=2AB^2\\\\ \\Rightarrow 2AB^2=32 \\Leftrightarrow AB^2=16\\\\ \\Leftrightarrow AB=4\\,(cm)$<br\/>Di\u1ec7n t\u00edch \u0111\u00e1y $ABCD$ l\u00e0 $AB^2=16\\,(cm^2)$<br\/>V\u1eady chi\u1ec1u cao l\u0103ng tr\u1ee5 l\u00e0 $V_{ABCD.A'B'C'D'}:S_{ABCD}=128:16=8\\,(cm)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}],"id_ques":1934},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Cho h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng $ABC.A'B'C'$ c\u00f3 \u0111\u00e1y l\u00e0 tam gi\u00e1c \u0111\u1ec1u. G\u1ecdi $M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC$. Tam gi\u00e1c $A'AM$ l\u00e0 tam gi\u00e1c g\u00ec n\u1ebfu $4AA'^2=3AB^2$.","select":["A. Tam gi\u00e1c vu\u00f4ng","B. Tam gi\u00e1c c\u00e2n","C. Tam gi\u00e1c \u0111\u1ec1u ","D. Tam gi\u00e1c vu\u00f4ng c\u00e2n"],"hint":"","explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai24/lv3/img\/H8_B24_K1.10.png' \/><\/center>X\u00e9t tam gi\u00e1c $ABC$ \u0111\u1ec1u, c\u00f3 $M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC$ n\u00ean $AM\\bot BC$.<br\/>X\u00e9t tam gi\u00e1c $ABM$ vu\u00f4ng t\u1ea1i $M$.<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3:<br\/>$AB^2=AM^2+BM^2\\\\ \\Leftrightarrow AM^2=AB^2-\\dfrac{AB^2}{4}\\\\ \\Leftrightarrow AM^2=\\dfrac{3AB^2}{4}$<br\/>Ta l\u1ea1i c\u00f3: $4AA'^2=3AB^2 \\Leftrightarrow AA'^2=\\dfrac{3AB^2}{4} \\Leftrightarrow AA'=AM $<br\/>V\u1eady tam gi\u00e1c $ AA'M$ c\u00e2n t\u1ea1i $A$<br\/>M\u1eb7t kh\u00e1c, ta l\u1ea1i c\u00f3: $AA'\\bot AC;\\,AA'\\bot AB$. Suy ra $AA'\\bot mp(ABC)\\Leftrightarrow AA' \\bot AM$<br\/>V\u1eady tam gi\u00e1c $AA'M$ vu\u00f4ng c\u00e2n t\u1ea1i $A$.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><\/span>","column":2}],"id_ques":1935},{"title":"L\u1ef1a ch\u1ecdn \u0110\u00fang ho\u1eb7c Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"H\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng \u0111\u00e1y l\u00e0 tam gi\u00e1c vu\u00f4ng c\u00e2n c\u00f3 c\u1ea1nh huy\u1ec1n $8 cm$ v\u00e0 \u0111\u01b0\u1eddng ch\u00e9o m\u1eb7t b\u00ean l\u00e0 $12\\,cm$ th\u00ec c\u00f3 th\u1ec3 t\u00edch l\u00e0 $64\\sqrt{7}\\,cm^3$","select":["\u0110\u00fang","Sai"],"hint":"D\u1ef1a v\u00e0o \u0111\u1ecbnh l\u00fd Pytago, t\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh g\u00f3c vu\u00f4ng c\u1ee7a tam gi\u00e1c \u1edf \u0111\u00e1y.","explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai24/lv3/img\/H8_B24_K1.6.png' \/><\/center>X\u00e9t h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng $ABC.A'B'C'$ c\u00f3 hai \u0111\u00e1y l\u00e0 hai tam gi\u00e1c vu\u00f4ng.<br\/> X\u00e9t tam gi\u00e1c $ABC$ vu\u00f4ng c\u00e2n t\u1ea1i $B$, c\u00f3 $BA=BC$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: <br\/>$AC^2=BA^2+BC^2 \\Leftrightarrow AC^2=2AB^2 \\Leftrightarrow 8^2=2AB^2\\\\ \\Leftrightarrow AB^2=32 \\Leftrightarrow AB=4\\sqrt{2}\\,(cm)$<br\/>X\u00e9t tam gi\u00e1c $AA'B' $ vu\u00f4ng t\u1ea1i $A'$ c\u00f3 $A'B'=4\\sqrt{2}\\,cm$ v\u00e0 $AB'=12\\,cm$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago, ta c\u00f3: $AB'^2=AA'^2+A'B'^2 \\Leftrightarrow 12^2=AA'^2+(4\\sqrt{2})^2 \\Leftrightarrow AA'^2=144-32\\\\ \\Leftrightarrow AA'^2=112 \\Leftrightarrow AA'=4\\sqrt{7}\\,(cm)$<br\/>Ta c\u00f3 di\u1ec7n t\u00edch \u0111\u00e1y $ABC$ l\u00e0 $\\dfrac{1}{2}.BC.BA=\\dfrac{1}{2}.(4\\sqrt{2})^2=16\\,(cm^2)$<br\/>V\u1eady th\u1ec3 t\u00edch h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng l\u00e0 $AA'.S_{ABC}=4\\sqrt{7}.16=64\\sqrt{7}\\,(cm^3)$<br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0110\u00fang.<\/span><\/span>","column":2}],"id_ques":1936},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho h\u00ecnh ch\u00f3p c\u1ee5t tam gi\u00e1c \u0111\u1ec1u $ABC.A'B'C'$ c\u00f3 c\u1ea1nh b\u00ean b\u1eb1ng $20\\,cm$, c\u1ea1nh \u0111\u00e1y l\u1edbn b\u1eb1ng $42\\,cm$; c\u1ea1nh \u0111\u00e1y nh\u1ecf b\u1eb1ng $22\\,cm$; $A'H\\bot AB$. T\u00ednh di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh ch\u00f3p c\u1ee5t.<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai24/lv3/img\/H8_B24_K1.8.png' \/><\/center>","select":["A. $960\\sqrt{3}\\,cm^2$","B. $960\\sqrt{2}\\,cm^2$","C. $1920\\sqrt{3}\\,cm^2$","D. \u0110\u00e1p \u00e1n kh\u00e1c"],"hint":"T\u00ednh chi\u1ec1u cao c\u1ea1nh b\u00ean","explain":"<span class='basic_left'>X\u00e9t m\u1eb7t b\u00ean $ABB'A'$, ta c\u00f3: <br\/>$AH=\\dfrac{AB-A'B'}{2}=\\dfrac{42-22}{2}=10\\,(cm)$<br\/>X\u00e9t $\\Delta A'AH$ vu\u00f4ng t\u1ea1i $H$ c\u00f3 $A'A=20\\,cm$, $AH=10\\,cm$.<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3:<br\/>$A'H^2=AA'^2-AH^2=20^2-10^2=300\\Rightarrow A'H=10\\sqrt{3}\\,(cm)$<br\/>Di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh ch\u00f3p l\u00e0 <br\/>$S_{xq}=\\dfrac{(22+42).10\\sqrt{3}}{2}.3=960\\sqrt{3}\\,(cm^2)$<br\/>V\u1eady di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh ch\u00f3p c\u1ee5t l\u00e0 $960\\sqrt{3}\\,cm^2$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":2}],"id_ques":1937},{"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 t\u1ee9 gi\u00e1c \u0111\u1ec1u $S.ABCD$ c\u00f3 $SA=AB=4\\,cm$. S\u1ed1 \u0111o g\u00f3c $SAC$ v\u00e0 di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh ch\u00f3p \u0111\u1ec1u l\u00e0:<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai24/lv3/img\/H8_B24_K1.7a.png' \/><\/center>","select":["A. $\\widehat{SAC}=45^o$ v\u00e0 $S_{xq}=8\\sqrt{3}\\,cm^2$","B. $\\widehat{SAC}=45^o$ v\u00e0 $S_{xq}=16\\sqrt{3}\\,cm^2$","C. $\\widehat{SAC}=60^o$ v\u00e0 $S_{xq}=8\\sqrt{3}\\,cm^2$","D. $\\widehat{SAC}=60^o$ v\u00e0 $S_{xq}=16\\sqrt{3}\\,cm^2$"],"hint":"So s\u00e1nh \u0111\u1ed9 d\u00e0i $SO$ v\u00e0 $AC$","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 $SO \\bot mp(ABCD) \\,\\Rightarrow SO\\bot AC$<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai24/lv3/img\/H8_B24_K1.7.png' \/><\/center> <br\/>X\u00e9t tam gi\u00e1c $ABC$ vu\u00f4ng c\u00e2n t\u1ea1i $B$.<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: $AC^2=AB^2+BC^2 \\Leftrightarrow AC^2=4^2+4^2\\\\ \\Leftrightarrow AC^2=32 \\Leftrightarrow AC=4\\sqrt{2}\\,(cm)$<br\/>X\u00e9t $\\Delta SOA$ vu\u00f4ng t\u1ea1i $O$ c\u00f3: $SA=4\\,cm$ v\u00e0 $OA=\\dfrac{1}{2}AC=2\\sqrt{2}\\,cm$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago, ta c\u00f3: $SA^2=SO^2+OA^2 \\Leftrightarrow 16=SO^2+8\\\\ \\Leftrightarrow SO^2=8 \\Leftrightarrow SO=2\\sqrt{2}\\,(cm)$<br\/>X\u00e9t tam gi\u00e1c $SAC$ c\u00e2n t\u1ea1i $S$, $O$ l\u00e0 trung \u0111i\u1ec3m $AC$ c\u00f3: $SO=OA=OC=2\\sqrt{2}\\,cm$<br\/>V\u1eady tam gi\u00e1c $SAC$ vu\u00f4ng c\u00e2n t\u1ea1i $S$, suy ra $\\widehat {SAC}=45^o$.<br\/>G\u1ecdi $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB$. <br\/>V\u00ec $SAB$ c\u00e2n t\u1ea1i $S$ n\u00ean $SH\\bot AB$<br\/>X\u00e9t tam gi\u00e1c $SAH$ vu\u00f4ng t\u1ea1i $H$ c\u00f3 $SA=4\\,cm; \\,AH=2\\,cm$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago, ta c\u00f3: $SH^2=SA^2-AH^2 \\Leftrightarrow SH^2=16-4\\\\ \\Leftrightarrow SH^2=12 \\Leftrightarrow SH=2\\sqrt{3}\\,(cm)$<br\/>V\u1eady di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh ch\u00f3p $S.ABCD$ l\u00e0 $p.h=\\dfrac{4.AB}{2}. SH=2.4.2\\sqrt{3}=16\\sqrt{3}\\,(cm^2)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}],"id_ques":1938},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"Cho h\u00ecnh ch\u00f3p tam gi\u00e1c \u0111\u1ec1u $S.ABC$; $H$ l\u00e0 trung \u0111i\u1ec3m $AB$ v\u00e0 $SO$ l\u00e0 \u0111\u01b0\u1eddng cao. T\u00ednh th\u1ec3 t\u00edch h\u00ecnh ch\u00f3p bi\u1ebft h\u00ecnh ch\u00f3p c\u00f3 t\u1ea5t c\u1ea3 c\u00e1c c\u1ea1nh \u0111\u1ec1u b\u1eb1ng $2 cm$.","select":["A. $\\dfrac{\\sqrt{24}}{\\sqrt 3}\\,cm^3$","B. $\\dfrac{\\sqrt{8}}{\\sqrt 3}\\,cm^3$","C. $\\dfrac{\\sqrt{2}}{3}\\,cm^3$","D. $\\dfrac{\\sqrt{8}}{3}\\,cm^3$"],"hint":"\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago \u0111\u1ec3 t\u00ednh $CH$ v\u00e0 $SO$","explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai24/lv3/img\/H8_B24_K1.9.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\/>+) $O$ t\u00e2m c\u1ee7a tam gi\u00e1c $ABC$ (giao \u0111i\u1ec3m c\u1ee7a ba \u0111\u01b0\u1eddng trung tr\u1ef1c)<br\/>- V\u00ec $SO$ l\u00e0 \u0111\u01b0\u1eddng cao n\u00ean $SO \\bot mp(ABC) \\,\\Rightarrow SO\\bot CH$<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai24/lv3/img\/H8_B24_K1.9a.png' \/><\/center>X\u00e9t tam gi\u00e1c $AHC$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: $AC=2\\,cm$ v\u00e0 $AH=1\\,cm$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3:<br\/>$HC=\\sqrt{AC^2-AH^2} \\Rightarrow HC=\\sqrt{4-1} \\Leftrightarrow HC=\\sqrt{3}\\,(cm)$<br\/>Di\u1ec7n t\u00edch tam gi\u00e1c $ABC$ l\u00e0 $\\dfrac{1}{2}.CH.AB=\\sqrt 3\\,(cm^2)$<br\/>V\u00ec $O$ l\u00e0 t\u00e2m c\u1ee7a tam gi\u00e1c \u0111\u1ec1u n\u00ean $O$ \u0111\u1ed3ng th\u1eddi l\u00e0 giao \u0111i\u1ec3m c\u1ee7a ba \u0111\u01b0\u1eddng trung tuy\u1ebfn.<br\/>Theo t\u00ednh ch\u1ea5t v\u1ec1 tr\u1ecdng t\u00e2m tam gi\u00e1c, ta c\u00f3 $CO=\\dfrac{2}{3}CH=\\dfrac{2\\sqrt 3}{3}\\,(cm)$<br\/>X\u00e9t tam gi\u00e1c $SOC$ vu\u00f4ng t\u1ea1i $O$ c\u00f3: $SC=2 cm$ v\u00e0 $CO=\\dfrac{2\\sqrt 3}{3}\\,cm$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3: <br\/>$SO=\\sqrt{SC^2-CO^2}\\\\ \\Rightarrow SO=\\sqrt{4-\\dfrac{4}{3}}=\\sqrt{\\dfrac{8}{3}}\\,(cm)$<br\/>V\u1eady th\u1ec3 t\u00edch $S.ACB$ l\u00e0 $\\dfrac{1}{3}.\\sqrt{3}.\\sqrt{\\dfrac{8}{3}}=\\dfrac{\\sqrt{8}}{3}\\,(cm^3)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span>","column":2}],"id_ques":1939}],"id_ques":0}],"lesson":{"save":1,"level":3,"time":44}}