{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["15"]]],"list":[{"point":10,"width":60,"type_input":"","ques":"<span class='basic_left'>M\u1ed9t h\u00ecnh n\u00f3n c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y b\u1eb1ng $20$ $cm,$ th\u1ec3 t\u00edch b\u1eb1ng $4$ l\u1ea7n di\u1ec7n t\u00edch xung quanh. T\u00ednh chi\u1ec1u cao c\u1ee7a h\u00ecnh n\u00f3n. <br\/><b>\u0110\u00e1p s\u1ed1: <\/b> _input_$(cm)$ <\/span>","hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai23/lv3/img\/H942_K1.png' \/><\/center>Th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh n\u00f3n l\u00e0: $V=\\dfrac{1}{3}\\pi {{r}^{2}}h=\\dfrac{1}{3}\\pi .{{20}^{2}}.h=\\dfrac{400}{3}\\pi h$ $(cm^3)$<br\/>\u0110\u01b0\u1eddng sinh c\u1ee7a h\u00ecnh n\u00f3n l\u00e0: $l=SA=\\sqrt{{{h}^{2}}+{{r}^{2}}}=\\sqrt{{{h}^{2}}+{{20}^{2}}}$ $(cm)$<br\/>Di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh n\u00f3n l\u00e0:<br\/>${{S}_{xq}}=\\pi rl=\\pi .20.\\sqrt{{{h}^{2}}+{{20}^{2}}}$ <br\/>V\u00ec th\u1ec3 t\u00edch b\u1eb1ng $4$ l\u1ea7n di\u1ec7n t\u00edch xung quanh n\u00ean ta c\u00f3:<br\/>$\\begin{align} & \\,\\,\\,\\,\\,V=4{{S}_{xq}} \\\\ & \\Leftrightarrow \\dfrac{400}{3}\\pi h=4.\\pi .20.\\sqrt{{{h}^{2}}+{{20}^{2}}} \\\\ & \\Leftrightarrow 5h=3\\sqrt{{{h}^{2}}+400} \\\\ & \\Leftrightarrow 25{{h}^{2}}=9\\left( {{h}^{2}}+400 \\right) \\\\ & \\Leftrightarrow 16{{h}^{2}}=3600 \\\\ & \\Leftrightarrow {{h}^{2}}=225 \\\\ & \\Leftrightarrow h=15 \\\\ \\end{align}$<br\/> V\u1eady chi\u1ec1u cao c\u1ee7a h\u00ecnh n\u00f3n l\u00e0 $15$$cm$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $15.$<\/span><\/span>"}]}],"id_ques":1711},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u (>,<,=) th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["<"],["<"]]],"list":[{"point":10,"width":60,"type_input":"","ques":"<span class='basic_left'>Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A,$ $AC=b,$ $AB=c,$ $ b < c. $ Quay $\\Delta ABC$ m\u1ed9t v\u00f2ng quanh c\u1ea1nh $AB,$ ta \u0111\u01b0\u1ee3c m\u1ed9t h\u00ecnh n\u00f3n c\u00f3 di\u1ec7n t\u00edch xung quanh $S_1$ v\u00e0 di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n $S_{1}^{'}$ . Quay $\\Delta ABC$ m\u1ed9t v\u00f2ng quanh c\u1ea1nh $AC,$ ta \u0111\u01b0\u1ee3c m\u1ed9t h\u00ecnh n\u00f3n c\u00f3 di\u1ec7n t\u00edch xung quanh $S_2$ v\u00e0 di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n $S_{2}^{'}.$ So s\u00e1nh $S_1$ v\u00e0 $S_2;$ $S_{1}^{'}$ v\u00e0 $S_{2}^{'}$ <br\/><b>\u0110\u00e1p s\u1ed1: <\/b> $S_1$ _input_ $S_2;$ $S_{1}^{'}$ _input_ $S_{2}^{'}$ <\/span>","hint":"","explain":"<span class='basic_left'><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai23/lv3/img\/H942_K2.png' \/>$\\hspace{2cm}$ <img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai23/lv3/img\/H942_K2a.png' \/><br\/><br\/>X\u00e9t h\u00ecnh n\u00f3n th\u1ee9 nh\u1ea5t v\u00e0 h\u00ecnh n\u00f3n th\u1ee9 hai, ta c\u00f3:<br\/>+ ${{S}_{1}}=\\pi b.BC$; ${{S}_{2}}=\\pi c.BC$<br\/>Suy ra: $\\dfrac{{{S}_{1}}}{{{S}_{2}}}=\\dfrac{b}{c} < 1\\,\\,\\,\\,\\,\\left( \\text{do}\\,b < c \\right)$<br\/>Do \u0111\u00f3 $S_1 < S_2$<br\/>+ $S_{1}^{'}={{S}_{1}}+\\pi {{b}^{2}};$ $S_{2}^{'}={{S}_{2}}+\\pi {{c}^{2}}$ <br\/>Do $ b < c $ n\u00ean $ b ^2 < c^2. $<br\/>L\u1ea1i c\u00f3 $ S_1 < S_2 $<br\/>Suy ra $ S_{1}^{'} < S_{2}^{'} $ <br\/><span class='basic_pink'>V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $<$ v\u00e0 $<$<\/span><\/span>"}]}],"id_ques":1712},{"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":"<span class='basic_left'>Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A,$ $BC =10cm,$ \u0111\u01b0\u1eddng cao $AH=4cm. $ Quay tam gi\u00e1c $ABC$ m\u1ed9t v\u00f2ng quanh c\u1ea1nh $BC.$ T\u00ednh th\u1ec3 t\u00edch h\u00ecnh t\u1ea1o th\u00e0nh.","select":["A. $V= \\dfrac{400\\pi }{3}\\left( c{{m}^{3}} \\right) $","B. $V= \\dfrac{160\\pi }{3}\\left( c{{m}^{3}} \\right) $","C. $V= \\dfrac{200\\pi }{3}\\left( c{{m}^{3}} \\right) $","D. $V= \\dfrac{80\\pi }{3}\\left( c{{m}^{3}} \\right) $"],"hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai23/lv3/img\/H942_K5.png' \/><\/center><br\/>Khi quay tam gi\u00e1c $ABC$ m\u1ed9t v\u00f2ng quanh c\u1ea1nh $BC,$ h\u00ecnh t\u1ea1o th\u00e0nh g\u1ed3m hai h\u00ecnh n\u00f3n c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 $AH$ v\u00e0 \u0111\u01b0\u1eddng cao theo th\u1ee9 t\u1ef1 l\u00e0 $HB$ v\u00e0 $HC.$<br\/>Th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh t\u1ea1o th\u00e0nh l\u00e0<br\/>$\\begin{align} & \\,\\,\\,\\,\\dfrac{1}{3}.\\pi .A{{H}^{2}}.BH+\\dfrac{1}{3}.\\pi .A{{H}^{2}}.CH \\\\ & =\\dfrac{1}{3}.\\pi .A{{H}^{2}}.\\left( BH+CH \\right) \\\\ & =\\dfrac{1}{3}.\\pi .A{{H}^{2}}.BC \\\\ & =\\dfrac{1}{3}.\\pi {{.4}^{2}}.10=\\dfrac{160\\pi }{3}\\left( c{{m}^{3}} \\right) \\\\ \\end{align}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span> <\/span>","column":2}]}],"id_ques":1713},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\sqrt{2}$","B. $2\\sqrt{2}$","C. $3\\sqrt{2}$"],"ques":"<span class='basic_left'>M\u1ed9t h\u00ecnh n\u00f3n c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y b\u1eb1ng $2$ $cm,$ \u0111\u01b0\u1eddng sinh b\u1eb1ng $6$$cm.$ M\u1ed9t m\u1eb7t ph\u1eb3ng song song v\u1edbi \u0111\u00e1y chia h\u00ecnh n\u00f3n th\u00e0nh m\u1ed9t h\u00ecnh n\u00f3n nh\u1ecf v\u00e0 m\u1ed9t h\u00ecnh n\u00f3n c\u1ee5t c\u00f3 di\u1ec7n t\u00edch xung quanh b\u1eb1ng nhau. T\u00ednh b\u00e1n k\u00ednh \u0111\u00e1y c\u1ee7a h\u00ecnh n\u00f3n nh\u1ecf.$(cm)$<\/span>","hint":"Do h\u00ecnh n\u00f3n nh\u1ecf v\u00e0 m\u1ed9t h\u00ecnh n\u00f3n c\u1ee5t c\u00f3 di\u1ec7n t\u00edch xung quanh b\u1eb1ng nhau n\u00ean di\u1ec7n t\u00edch xung quanh h\u00ecnh n\u00f3n nh\u1ecf b\u1eb1ng n\u1eeda di\u1ec7n t\u00edch xung quanh h\u00ecnh n\u00f3n l\u1edbn","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/>B\u01b0\u1edbc 1 :\u0110\u1eb7t $IB=x.$ T\u00ednh $SB$ theo $x$<br\/>B\u01b0\u1edbc 2: T\u00ednh di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh n\u00f3n nh\u1ecf v\u00e0 h\u00ecnh n\u00f3n l\u1edbn<br\/>B\u01b0\u1edbc 3: T\u1eeb di\u1ec7n t\u00edch xung quanh h\u00ecnh n\u00f3n nh\u1ecf b\u1eb1ng n\u1eeda di\u1ec7n t\u00edch xung quanh h\u00ecnh n\u00f3n l\u1edbn, ta t\u00ecm $x$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai23/lv3/img\/H942_K4.png' \/><\/center><br\/>\u0110\u1eb7t $IB=x$<br\/>Do $IB\/\/OA$ n\u00ean ta c\u00f3:<br\/>$\\begin{align} & \\dfrac{IB}{OA}=\\dfrac{SB}{SA} \\\\ & \\Rightarrow \\dfrac{x}{2}=\\dfrac{SB}{6}\\Leftrightarrow SB=3x\\,\\,\\left( 1 \\right) \\\\ \\end{align}$ <br\/>Di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh n\u00f3n nh\u1ecf l\u00e0 ${{S}_{1}}=\\pi .IB.SB=\\pi .x.SB$ <br\/>Di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh n\u00f3n l\u1edbn l\u00e0 ${{S}_{2}}=\\pi .OA.SA=\\pi .2.6=12\\pi $ <br\/>Do h\u00ecnh n\u00f3n nh\u1ecf v\u00e0 m\u1ed9t h\u00ecnh n\u00f3n c\u1ee5t c\u00f3 di\u1ec7n t\u00edch xung quanh b\u1eb1ng nhau n\u00ean di\u1ec7n t\u00edch xung quanh h\u00ecnh n\u00f3n nh\u1ecf b\u1eb1ng n\u1eeda di\u1ec7n t\u00edch xung quanh h\u00ecnh n\u00f3n l\u1edbn, t\u1ee9c<br\/>${{S}_{1}}=\\dfrac{{{S}_{2}}}{2}\\Leftrightarrow \\pi x.SB=\\dfrac{12\\pi }{2}\\Leftrightarrow SB=\\dfrac{6}{x}$ (2)<br\/>T\u1eeb (1) v\u00e0 (2), ta c\u00f3 $3x=\\dfrac{6}{x}\\Leftrightarrow {{x}^{2}}=2\\Leftrightarrow x=\\sqrt{2}$ <br\/>V\u1eady b\u00e1n k\u00ednh \u0111\u00e1y c\u1ee7a h\u00ecnh n\u00f3n nh\u1ecf l\u00e0 $\\sqrt{2}$ $(cm)$<\/span>"}]}],"id_ques":1714},{"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":"<span class='basic_left'>Cho h\u00ecnh thang vu\u00f4ng $ABCD$ c\u00f3 $\\widehat{A}=\\widehat{D}={{90}^{o}},AB=BC=a,\\widehat{C}={{60}^{o}}.$ T\u00ednh di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n c\u1ee7a h\u00ecnh t\u1ea1o th\u00e0nh khi quay h\u00ecnh thang m\u1ed9t v\u00f2ng xung quanh c\u1ea1nh $DC$","select":["A. $S_{tp}= \\dfrac{8\\sqrt{3}+3}{4}.\\pi {{a}^{2}}$(\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)","B. $S_{tp}= \\dfrac{3\\sqrt{3}}{2}.\\pi {{a}^{2}}$(\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)","C. $S_{tp}= \\dfrac{6\\sqrt{3}+3}{4}.\\pi {{a}^{2}}$(\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)","D. $S_{tp}= \\dfrac{6\\sqrt{3}-3}{4}.\\pi {{a}^{2}}$(\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>Ph\u00e2n t\u00edch<\/span><br\/>H\u00ecnh t\u1ea1o th\u00e0nh g\u1ed3m m\u1ed9t h\u00ecnh tr\u1ee5 v\u00e0 m\u1ed9t h\u00ecnh n\u00f3n<br\/>Di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n c\u1ee7a h\u00ecnh t\u1ea1o th\u00e0nh b\u1eb1ng t\u1ed5ng di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh tr\u1ee5, di\u1ec7n t\u00edch xung quanh h\u00ecnh n\u00f3n v\u00e0 di\u1ec7n t\u00edch \u0111\u00e1y h\u00ecnh tr\u1ee5<br\/><span class='basic_green'>L\u1eddi gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai23/lv3/img\/H942_K3.png' \/><\/center><br\/>X\u00e9t h\u00ecnh thang $ABCD,$ h\u1ea1 $BH\\bot CD$ <br\/>$\\Delta BHC$ vu\u00f4ng t\u1ea1i $H$ c\u00f3 $BH=BC.\\sin C=a.\\sin {{60}^{o}}=\\dfrac{a\\sqrt{3}}{2}$ <br\/>H\u00ecnh t\u1ea1o th\u00e0nh g\u1ed3m m\u1ed9t h\u00ecnh tr\u1ee5 v\u00e0 m\u1ed9t h\u00ecnh n\u00f3n<br\/>+ H\u00ecnh tr\u1ee5 c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 $BH,$ \u0111\u01b0\u1eddng cao $AB$<br\/>+ H\u00ecnh n\u00f3n c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 $BH,$ \u0111\u01b0\u1eddng sinh l\u00e0 $BC$<br\/>Di\u1ec7n t\u00edch xung quanh h\u00ecnh tr\u1ee5 l\u00e0 $2\\pi .BH.AB=2\\pi .\\dfrac{a\\sqrt{3}}{2}.a=\\pi {{a}^{2}}\\sqrt{3}$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/>Di\u1ec7n t\u00edch xung quanh h\u00ecnh n\u00f3n l\u00e0 $\\pi .BH.BC=\\pi \\dfrac{a\\sqrt{3}}{2}.a=\\dfrac{{{a}^{2}}\\sqrt{3}}{2}\\pi $(\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch) <br\/>Di\u1ec7n t\u00edch \u0111\u00e1y l\u00e0 $\\pi {{\\left( \\dfrac{a\\sqrt{3}}{2} \\right)}^{2}}=\\dfrac{3{{a}^{2}}}{4}\\pi $(\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch) <br\/>Di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n c\u1ee7a h\u00ecnh t\u1ea1o th\u00e0nh l\u00e0:<br\/> $\\pi {{a}^{2}}\\sqrt{3}+\\dfrac{{{a}^{2}}\\sqrt{3}}{2}\\pi +\\dfrac{3{{a}^{2}}}{4}\\pi =\\pi {{a}^{2}}\\left( \\sqrt{3}+\\dfrac{3}{4}+\\dfrac{\\sqrt{3}}{2} \\right)=\\dfrac{6\\sqrt{3}+3}{4}.\\pi {{a}^{2}}$(\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span> <\/span>","column":2}]}],"id_ques":1715},{"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":[[["24"],["10"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>Cho tam gi\u00e1c $OBC$ vu\u00f4ng t\u1ea1i $O$. N\u1ebfu quay tam gi\u00e1c \u0111\u00f3 quanh c\u1ea1nh $OB$ c\u1ed1 \u0111\u1ecbnh th\u00ec \u0111\u01b0\u1ee3c m\u1ed9t h\u00ecnh n\u00f3n c\u00f3 th\u1ec3 t\u00edch $800\\pi\\,cm^3 $ c\u00f2n n\u1ebfu quay tam gi\u00e1c \u0111\u00f3 quanh c\u1ea1nh $OC$ th\u00ec \u0111\u01b0\u1ee3c m\u1ed9t h\u00ecnh n\u00f3n c\u00f3 th\u1ec3 t\u00edch $1920\\pi\\,cm^3. $ T\u00ednh \u0111\u1ed9 d\u00e0i $OB$ v\u00e0 $OC$<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $OB=$ _input_$(cm);$ $OC=$_input_$(cm)$<\/span>","hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai23/lv3/img\/H942_K6.png' \/><\/center><br\/>+ H\u00ecnh n\u00f3n t\u1ea1o th\u00e0nh khi quay tam gi\u00e1c quanh $OB$ c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 $OC$ v\u00e0 \u0111\u01b0\u1eddng cao l\u00e0 $OB.$<br\/> Th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh n\u00f3n l\u00e0 $\\dfrac{1}{3}\\pi O{{C}^{2}}.OB=800\\pi $ (1)<br\/>+ H\u00ecnh n\u00f3n t\u1ea1o th\u00e0nh khi quay tam gi\u00e1c quanh $OC$ c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 $OB$ v\u00e0 \u0111\u01b0\u1eddng cao l\u00e0 $OC.$<br\/> Th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh n\u00f3n l\u00e0 $\\dfrac{1}{3}\\pi O{{B}^{2}}.OC=1920\\pi $ (2)<br\/>L\u1ea5y (1) chia cho (2) \u0111\u01b0\u1ee3c $\\dfrac{OC}{OB}=\\dfrac{5}{12}\\Rightarrow OB=\\dfrac{12}{5}OC$ <br\/>Thay $OB=\\dfrac{12}{5}OC$ v\u00e0o (1) \u0111\u01b0\u1ee3c:<br\/>$\\begin{align} & \\,\\,\\,\\,\\,\\,\\dfrac{1}{3}\\pi .O{{C}^{2}}.\\dfrac{12}{5}OC=800\\pi \\\\ & \\Leftrightarrow O{{C}^{3}}=1000 \\\\ & \\Leftrightarrow OC=10\\,(cm) \\\\ & \\Rightarrow OB=\\dfrac{12}{5}.10=24 \\,(cm)\\\\ \\end{align}$. <br\/> <span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $24$ v\u00e0 $10.$<\/span><\/span>"}]}],"id_ques":1716},{"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":"<span class='basic_left'>M\u1ed9t h\u00ecnh n\u00f3n \u0111\u1ec9nh $S$ c\u00f3 \u0111\u00e1y l\u00e0 \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m $O,$ \u0111\u01b0\u1eddng k\u00ednh $AB,$ $\\widehat{ASB}={{60}^{o}}.$ Qua trung \u0111i\u1ec3m $I$ c\u1ee7a $SO,$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi $AB,$ c\u1eaft m\u1eb7t xung quanh c\u1ee7a h\u00ecnh n\u00f3n \u1edf $C$ v\u00e0 $D.$ T\u00ednh th\u1ec3 t\u00edch h\u00ecnh n\u00f3n bi\u1ebft $CD=6cm$","select":["A. $V=36\\pi $$ (cm^3)$","B. $V=144\\pi $$ (cm^3)$","C. $V=216\\sqrt{3}\\pi $$ (cm^3)$","D. $V=72\\sqrt{3}\\pi $$ (cm^3)$"],"hint":"","explain":"<span class='basic_left'><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai23/lv3/img\/H942_K7.png' \/><br\/><br\/>D\u1ec5 th\u1ea5y $CD$ l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c $SAB$ n\u00ean $AB=2CD=12cm.$ Suy ra $AO=6(cm)$<br\/>X\u00e9t tam gi\u00e1c $SAB$ c\u00e2n t\u1ea1i $S,$ c\u00f3 $SO$ vu\u00f4ng g\u00f3c v\u1edbi $AB$ n\u00ean $SO$ c\u0169ng l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $ASB$<br\/>Suy ra $\\widehat{ASO}={{30}^{o}}$ <br\/>Tam gi\u00e1c $SAO$ vu\u00f4ng t\u1ea1i $O$ n\u00ean $SO=\\dfrac{AO}{tg {\\widehat{ASO}}}=\\dfrac{6}{tg{30^o}}=6\\sqrt{3}$ $(cm)$<br\/>Th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh n\u00f3n l\u00e0 $V=\\dfrac{1}{3}\\pi .{{OA}^{2}}.SO=\\dfrac{1}{3}\\pi {{.6}^{2}}.6\\sqrt{3}=72\\sqrt{3}\\pi $$ (cm^3)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span> <\/span>","column":2}]}],"id_ques":1717},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o c\u00e1c ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["201,1","201"],["377,3","377,1"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>T\u1eeb m\u1ed9t h\u00ecnh n\u00f3n b\u00e1n k\u00ednh \u0111\u00e1y $6$$cm$ v\u00e0 \u0111\u01b0\u1eddng sinh l\u00e0 $12$$cm,$ ng\u01b0\u1eddi ta c\u1eaft b\u1eb1ng m\u1ed9t m\u1eb7t ph\u1eb3ng song song v\u1edbi \u0111\u00e1y \u0111\u01b0\u1ee3c m\u1ed9t h\u00ecnh n\u00f3n c\u1ee5t c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y tr\u00ean l\u00e0 $2$$cm.$ T\u00ednh di\u1ec7n t\u00edch xung quanh v\u00e0 th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh n\u00f3n c\u1ee5t t\u1ea1o th\u00e0nh (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t)<br\/><b>\u0110\u00e1p s\u1ed1:<\/b>$S_{xq}\\approx$_input_$(cm^2)$; $V\\approx$_input_$(cm^3)$<\/span>","hint":"\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng \u0111\u1ec3 t\u00ecm \u0111\u01b0\u1eddng cao, \u0111\u01b0\u1eddng sinh c\u1ee7a h\u00ecnh n\u00f3n c\u1ee5t","explain":"<span class='basic_left'>X\u00e9t h\u00ecnh n\u00f3n nh\u01b0 h\u00ecnh v\u1ebd sau:<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai23/lv3/img\/H942_K9.png' \/><\/center><br\/>Tam gi\u00e1c $SAO$ vu\u00f4ng t\u1ea1i $O$ c\u00f3 $\\sin \\widehat{ASO}=\\dfrac{OA}{SA}=\\dfrac{6}{12}=0,5\\Rightarrow \\widehat{ASO}={{30}^{o}}$ <br\/>$SO=AO.cotg\\widehat{ASO}=6.cotg{{30}^{o}}=6\\sqrt{3}\\left( cm \\right)$ <br\/>Tam gi\u00e1c $SA\u2019O'$ vu\u00f4ng t\u1ea1i $O'$ c\u00f3 $SA'=\\dfrac{A'O'}{\\sin \\widehat{A'SO'}}=\\dfrac{2}{\\sin {{30}^{o}}}=4\\left( cm \\right)$ <br\/>$SO'=SA'.\\cos \\widehat{A'SO'}=4.\\cos {{30}^{o}}=2\\sqrt{3}$ $(cm) $<br\/>Suy ra $AA'=SA-SA'=12-4=8\\left( cm \\right)$ <br\/>$OO'=SO-SO'=6\\sqrt{3}-2\\sqrt{3}=4\\sqrt{3}$$(cm)$<br\/>Di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh n\u00f3n c\u1ee5t l\u00e0:<br\/> ${{S}_{xq}}=\\pi \\left( OA+O'A' \\right).AA'=\\pi \\left( 2+6 \\right).8=64\\pi \\approx 201,1$ $(cm^2)$<br\/>Th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh n\u00f3n c\u1ee5t l\u00e0:<br\/> $V=\\dfrac{1}{3}\\pi .OO'.\\left( O{{A}^{2}}+O'A{{'}^{2}}+OA.O'A' \\right)=\\dfrac{1}{3}\\pi .4\\sqrt{3}\\left( {{2}^{2}}+{{6}^{2}}+2.6 \\right)=\\dfrac{208\\pi \\sqrt{3}}{3}\\approx 377,3\\left( c{{m}^{3}} \\right)$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $201,1;$$377,3.$ <\/span><\/span>"}]}],"id_ques":1718},{"time":24,"part":[{"time":3,"title":"Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 $BC=a,\\,\\,AC=b,\\,AB=c$ v\u00e0 \u0111\u01b0\u1eddng cao $ AH=h.$ G\u1ecdi $V_1;$$ V_2;$ $V_3$ th\u1ee9 t\u1ef1 l\u00e0 th\u1ec3 t\u00edch c\u1ee7a nh\u1eefng h\u00ecnh sinh ra khi quay tam gi\u00e1c $ABC$ m\u1ed9t v\u00f2ng xung quanh c\u00e1c c\u1ea1nh $BC,$ $AB$ v\u00e0 $AC. $ Ch\u1ee9ng minh $\\dfrac{1}{V_{1}^{2}}=\\dfrac{1}{V_{2}^{2}}+\\dfrac{1}{V_{3}^{2}}$ ","title_trans":"H\u00e3y s\u1eafp x\u1ebfp c\u00e1c c\u00e2u sau \u0111\u1ec3 \u0111\u01b0\u1ee3c l\u1eddi gi\u1ea3i \u0111\u00fang","temp":"sequence","correct":[[[2],[4],[7],[5],[1],[3],[6]]],"list":[{"point":10,"image":"","left":["Quay tam gi\u00e1c quanh c\u1ea1nh AB ta \u0111\u01b0\u1ee3c h\u00ecnh n\u00f3n c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 $AC=b,$ \u0111\u01b0\u1eddng cao $AB=c$ n\u00ean ${{V}_{2}}=\\dfrac{1}{3}\\pi {{b}^{2}}c\\Rightarrow \\dfrac{1}{V_{2}^{2}}=\\dfrac{9}{{{\\pi }^{2}}{{b}^{4}}{{c}^{2}}}$ (2)","Quay tam gi\u00e1c quanh c\u1ea1nh $BC$ ta \u0111\u01b0\u1ee3c hai h\u00ecnh n\u00f3n c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 $h,$ \u0111\u01b0\u1eddng cao l\u1ea7n l\u01b0\u1ee3t l\u00e0 $HC $ v\u00e0 $HB$ n\u00ean<br\/> ${{V}_{1}}=\\dfrac{1}{3}\\pi {{h}^{2}}.HC+\\dfrac{1}{3}\\pi {{h}^{2}}.HB=\\dfrac{1}{3}\\pi {{h}^{2}}.a=\\dfrac{\\pi {{b}^{2}}{{c}^{2}}}{3a}$ ","K\u1ebft h\u1ee3p v\u1edbi (1), ta c\u00f3 $\\dfrac{1}{V_{1}^{2}}=\\dfrac{1}{V_{2}^{2}}+\\dfrac{1}{V_{3}^{2}}$","T\u01b0\u01a1ng t\u1ef1 $\\dfrac{1}{V_{3}^{2}}=\\dfrac{9}{{\\pi}^2 {{b}^{2}}{{c}^{4}}}$ (3)","\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng $ABC,$ ta c\u00f3: $bc=ah \\Rightarrow h=\\dfrac{bc}{a}$","Suy ra $\\dfrac{1}{V_{1}^{2}}=\\dfrac{9{{a}^{2}}}{{{\\pi }^{2}}{{b}^{4}}{{c}^{4}}}$ (1)","T\u1eeb (2) v\u00e0 (3), ta c\u00f3: $\\dfrac{1}{V_{2}^{2}}+\\dfrac{1}{V_{3}^{2}}=\\dfrac{9}{{{\\pi }^{2}}{{b}^{4}}{{c}^{2}}}+\\dfrac{9}{{{\\pi }^{2}}{{b}^{2}}{{c}^{4}}}=\\dfrac{9\\left( {{b}^{2}}+{{c}^{2}} \\right)}{{{\\pi }^{2}}{{b}^{4}}{{c}^{4}}}=\\dfrac{9{{a}^{2}}}{{{\\pi }^{2}}{{b}^{4}}{{c}^{4}}}$"],"top":120,"hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai23/lv3/img\/H942_K10.png' \/><\/center><br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng $ABC,$ ta c\u00f3: $bc=ah.$<br\/>Suy ra $h=\\dfrac{bc}{a}$ <br\/>+ Quay tam gi\u00e1c quanh c\u1ea1nh $BC$ ta \u0111\u01b0\u1ee3c hai h\u00ecnh n\u00f3n c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 $h,$ \u0111\u01b0\u1eddng cao l\u1ea7n l\u01b0\u1ee3t l\u00e0 $HC $ v\u00e0 $HB$ n\u00ean<br\/> ${{V}_{1}}=\\dfrac{1}{3}\\pi {{h}^{2}}.HC+\\dfrac{1}{3}\\pi {{h}^{2}}.HB=\\dfrac{1}{3}\\pi {{h}^{2}}(HB+HC)=\\dfrac{1}{3}\\pi {{h}^{2}}BC=\\dfrac{1}{3}\\pi {{h}^{2}}.a=\\dfrac{\\pi {{b}^{2}}{{c}^{2}}}{3a}$ <br\/>Suy ra $\\dfrac{1}{V_{1}^{2}}=\\dfrac{9{{a}^{2}}}{{{\\pi }^{2}}{{b}^{4}}{{c}^{4}}}$ (1)<br\/>+ Quay tam gi\u00e1c quanh c\u1ea1nh $AB$ ta \u0111\u01b0\u1ee3c h\u00ecnh n\u00f3n c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 $AC=b,$ \u0111\u01b0\u1eddng cao $AB=c$ n\u00ean<br\/> ${{V}_{2}}=\\dfrac{1}{3}\\pi {{b}^{2}}c\\Rightarrow \\dfrac{1}{V_{2}^{2}}=\\dfrac{9}{{{\\pi }^{2}}{{b}^{4}}{{c}^{2}}}$ (2)<br\/>+ T\u01b0\u01a1ng t\u1ef1 $\\dfrac{1}{V_{3}^{2}}=\\dfrac{9}{{\\pi}^2 {{b}^{2}}{{c}^{4}}}$ (3)<br\/>T\u1eeb (2) v\u00e0 (3), ta c\u00f3: $\\dfrac{1}{V_{2}^{2}}+\\dfrac{1}{V_{3}^{2}}=\\dfrac{9}{{{\\pi }^{2}}{{b}^{4}}{{c}^{2}}}+\\dfrac{9}{{{\\pi }^{2}}{{b}^{2}}{{c}^{4}}}=\\dfrac{9\\left( {{b}^{2}}+{{c}^{2}} \\right)}{{{\\pi }^{2}}{{b}^{4}}{{c}^{4}}}=\\dfrac{9{{a}^{2}}}{{{\\pi }^{2}}{{b}^{4}}{{c}^{4}}}$<br\/>K\u1ebft h\u1ee3p v\u1edbi (1), ta c\u00f3 $\\dfrac{1}{V_{1}^{2}}=\\dfrac{1}{V_{2}^{2}}+\\dfrac{1}{V_{3}^{2}}$<\/span>"}]}],"id_ques":1719},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<span class='basic_left'>Ng\u01b0\u1eddi ta d\u00f9ng n\u1eeda h\u00ecnh tr\u00f2n b\u00e1n k\u00ednh $15$$cm$ \u0111\u1ec3 u\u1ed1n l\u1ea1i th\u00e0nh m\u1eb7t xung quanh c\u1ee7a m\u1ed9t h\u00ecnh n\u00f3n. T\u00ednh th\u1ec3 t\u00edch h\u00ecnh n\u00f3n.<\/span>","select":["A. $V=\\dfrac{1125\\sqrt{3}\\pi }{4}\\left( c{{m}^{3}} \\right)$ ","B. $V=\\dfrac{1235\\sqrt{3}\\pi }{4}\\left( c{{m}^{3}} \\right)$ ","C. $V=\\dfrac{1125\\sqrt{3}\\pi }{8}\\left( c{{m}^{3}} \\right)$ ","D. $V=\\dfrac{1235\\sqrt{3}\\pi }{8}\\left( c{{m}^{3}} \\right)$ "],"hint":"Khi ta u\u1ed1n n\u1eeda h\u00ecnh tr\u00f2n th\u00e0nh m\u1eb7t xung quanh c\u1ee7a h\u00ecnh n\u00f3n th\u00ec chu vi n\u1eeda h\u00ecnh tr\u00f2n ch\u00ednh l\u00e0 chu vi \u0111\u01b0\u1eddng tr\u00f2n \u0111\u00e1y c\u1ee7a h\u00ecnh n\u00f3n v\u00e0 b\u00e1n k\u00ednh h\u00ecnh tr\u00f2n ch\u00ednh l\u00e0 \u0111\u01b0\u1eddng sinh c\u1ee7a h\u00ecnh n\u00f3n","explain":"<span class='basic_left'>Gi\u1ea3 s\u1eed h\u00ecnh n\u00f3n t\u1ea1o th\u00e0nh nh\u01b0 h\u00ecnh v\u1ebd<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai23/lv3/img\/H942_K8.png' \/><\/center><br\/>\u0110\u1ed9 d\u00e0i cung qu\u1ea1t c\u1ee7a n\u1eeda h\u00ecnh tr\u00f2n l\u00e0: $l=15\\pi $ $(cm)$<br\/>Suy ra chu vi \u0111\u01b0\u1eddng tr\u00f2n \u0111\u00e1y c\u1ee7a h\u00ecnh n\u00f3n t\u1ea1o th\u00e0nh l\u00e0 $15\\pi $ $(cm)$<br\/>G\u1ecdi $r $ l\u00e0 b\u00e1n k\u00ednh \u0111\u00e1y c\u1ee7a h\u00ecnh n\u00f3n<br\/>Ta c\u00f3: $2\\pi r=15\\pi \\Leftrightarrow r=7,5\\left( cm \\right)$ <br\/>Tam gi\u00e1c $SAO$ vu\u00f4ng t\u1ea1i $O$ n\u00ean<br\/> $S{{O}^{2}}=S{{A}^{2}}-O{{A}^{2}}={{15}^{2}}-7,{{5}^{2}}=168,75\\Rightarrow SO=7,5\\sqrt{3}$ <br\/>Th\u1ec3 t\u00edch h\u00ecnh n\u00f3n l\u00e0 $V=\\dfrac{1}{3}\\pi {{r}^{2}}.SO=\\dfrac{1}{3}\\pi .7,{{5}^{2}}.7,5\\sqrt{3}=\\dfrac{1125\\sqrt{3}\\pi }{8}\\left( c{{m}^{3}} \\right)$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 C<\/span><\/span>","column":2}]}],"id_ques":1720}],"lesson":{"save":0,"level":3}}