{"segment":[{"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'>M\u1ed9t \u0111\u1ed3 ch\u01a1i c\u00f3 d\u1ea1ng g\u1ed3m m\u1ed9t h\u00ecnh n\u00f3n v\u00e0 m\u1ed9t n\u1eeda h\u00ecnh c\u1ea7u. H\u00ecnh n\u00f3n c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y b\u1eb1ng b\u00e1n k\u00ednh h\u00ecnh c\u1ea7u v\u00e0 b\u1eb1ng $R.$ Bi\u1ebft di\u1ec7n t\u00edch xung quanh c\u1ee7a ph\u1ea7n h\u00ecnh n\u00f3n b\u1eb1ng di\u1ec7n t\u00edch n\u1eeda m\u1eb7t c\u1ea7u. T\u00ednh chi\u1ec1u cao c\u1ee7a h\u00ecnh n\u00f3n theo $R$ ","select":["A. $h=\\dfrac{\\sqrt{3}}{2}R$ ","B. $h=\\dfrac{\\sqrt{3}}{3}R$ ","C. $h=\\sqrt{3}R$ ","D. $h=\\sqrt{2}R$"],"hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai24/lv3/img\/H943_K2.png' \/><\/center><br\/>Di\u1ec7n t\u00edch xung quanh h\u00ecnh n\u00f3n l\u00e0: ${{S}_{n}}=\\pi R.SA$ <br\/>Di\u1ec7n t\u00edch n\u1eeda m\u1eb7t c\u1ea7u l\u00e0: ${{S}_{c}}=\\dfrac{1}{2}.4\\pi {{R}^{2}}=2\\pi {{R}^{2}}$ <br\/>Theo gi\u1ea3 thi\u1ebft, ta c\u00f3:<br\/> ${{S}_{n}}={{S}_{c}}\\Leftrightarrow \\pi R.SA=2\\pi {{R}^{2}}\\Leftrightarrow SA=2R$ <br\/>Tam gi\u00e1c $SAO$ vu\u00f4ng t\u1ea1i $O.$ \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pytago, ta c\u00f3: <br\/>$S{{O}^{2}}+O{{A}^{2}}=S{{A}^{2}}\\Rightarrow SO=\\sqrt{S{{A}^{2}}-O{{A}^{2}}}=\\sqrt{4{{R}^{2}}-{{R}^{2}}}=\\sqrt{3}R$ <br\/>V\u1eady chi\u1ec1u cao c\u1ee7a h\u00ecnh n\u00f3n l\u00e0 $\\sqrt{3}R$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span> <\/span>","column":2}]}],"id_ques":1741},{"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":"<span class='basic_left'>Cho m\u1ed9t h\u00ecnh tr\u1ee5, m\u1ed9t h\u00ecnh n\u00f3n v\u00e0 m\u1ed9t h\u00ecnh c\u1ea7u c\u00f3 th\u1ec3 t\u00edch b\u1eb1ng nhau. B\u00e1n k\u00ednh \u0111\u00e1y c\u1ee7a h\u00ecnh tr\u1ee5, b\u00e1n k\u00ednh \u0111\u00e1y c\u1ee7a h\u00ecnh n\u00f3n v\u00e0 b\u00e1n k\u00ednh c\u1ee7a h\u00ecnh c\u1ea7u \u0111\u1ec1u b\u1eb1ng $R.$ T\u00ednh c\u00e1c chi\u1ec1u cao ${{h}_{1}}$ c\u1ee7a h\u00ecnh tr\u1ee5 v\u00e0 $h_2$ c\u1ee7a h\u00ecnh n\u00f3n theo $R$","select":["A. ${{h}_{1}}=\\dfrac{4}{3}R$ v\u00e0 ${{h}_{2}}=4R$","B. ${{h}_{1}}=4R$ v\u00e0 ${{h}_{2}}=\\dfrac{4}{3}R$","C. ${{h}_{1}}=2R$ v\u00e0 ${{h}_{2}}=\\dfrac{2}{3}R$","D. ${{h}_{1}}=\\dfrac{2}{3}R$ v\u00e0 ${{h}_{2}}=2R$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ednh th\u1ec3 t\u00edch c\u00e1c h\u00ecnh<br\/>B\u01b0\u1edbc 2: Cho th\u1ec3 t\u00edch ba h\u00ecnh b\u1eb1ng nhau, t\u1eeb \u0111\u00f3 gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh t\u00ecm $h_1$ v\u00e0 $h_2$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh tr\u1ee5 l\u00e0 $V_{\\text{tr\u1ee5}}=\\pi {{R}^{2}}{{h}_{1}}$ <br\/>Th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh n\u00f3n l\u00e0 $V_{\\text{n\u00f3n}}=\\dfrac{1}{3}\\pi {{R}^{2}}{{h}_{2}}$ <br\/>Th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh c\u1ea7u l\u00e0 $V_{\\text{c\u1ea7u}}=\\dfrac{4}{3}\\pi {{R}^{3}}$ <br\/>Ta c\u00f3 :<br\/>$\\begin{aligned} & \\,\\,\\,\\,\\,V_{\\text{tr\u1ee5}}=V_{\\text{n\u00f3n}}=V_{\\text{c\u1ea7u}} \\\\ & \\Leftrightarrow \\pi {{R}^{2}}{{h}_{1}}=\\dfrac{1}{3}\\pi {{R}^{2}}{{h}_{2}}=\\dfrac{4}{3}\\pi {{R}^{3}} \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & \\pi {{R}^{2}}{{h}_{1}}=\\dfrac{1}{3}\\pi {{R}^{2}}{{h}_{2}} \\\\ & \\dfrac{1}{3}\\pi {{R}^{2}}{{h}_{2}}=\\dfrac{4}{3}\\pi {{R}^{3}} \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & {{h}_{1}}=\\dfrac{{{h}_{2}}}{3} \\\\ & {{h}_{2}}=4R \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & {{h}_{1}}=\\dfrac{4}{3}R \\\\ & {{h}_{2}}=4R \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span> <\/span>","column":2}]}],"id_ques":1742},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["3"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>M\u1ed9t h\u00ecnh n\u00f3n c\u00f3 chi\u1ec1u cao $3$$cm,$ b\u00e1n k\u00ednh \u0111\u00e1y b\u1eb1ng b\u00e1n k\u00ednh c\u1ee7a m\u1ed9t h\u00ecnh c\u1ea7u. Bi\u1ebft di\u1ec7n t\u00edch xung quanh h\u00ecnh n\u00f3n b\u1eb1ng n\u1eeda di\u1ec7n t\u00edch m\u1eb7t c\u1ea7u. T\u00ednh b\u00e1n k\u00ednh \u0111\u00e1y c\u1ee7a h\u00ecnh n\u00f3n.<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $r=\\sqrt{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$ $(cm)$<\/span>","hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai24/lv3/img\/H943_K3.png' \/><\/center><br\/>G\u1ecdi $r$ l\u00e0 b\u00e1n k\u00ednh h\u00ecnh n\u00f3n, c\u0169ng l\u00e0 b\u00e1n k\u00ednh h\u00ecnh c\u1ea7u. \u0110\u01b0\u1eddng sinh c\u1ee7a h\u00ecnh n\u00f3n b\u1eb1ng $\\sqrt{{{r}^{2}}+9}$ $(cm)$<br\/>Di\u1ec7n t\u00edch xung quanh h\u00ecnh n\u00f3n b\u1eb1ng $\\pi r\\sqrt{{{r}^{2}}+9}$ $(cm^2)$<br\/>Di\u1ec7n t\u00edch m\u1eb7t c\u1ea7u l\u00e0 $4\\pi {{r}^{2}}$ $(cm^2)$<br\/>Theo \u0111\u1ec1 ra, ta c\u00f3:<br\/>$\\begin{align} & \\,\\,\\,\\,\\,\\,\\pi r\\sqrt{{{r}^{2}}+9}=\\dfrac{1}{2}.4\\pi {{r}^{2}} \\\\ & \\Leftrightarrow \\sqrt{{{r}^{2}}+9}=2r \\\\ & \\Leftrightarrow {{r}^{2}}+9=4{{r}^{2}} \\\\ & \\Leftrightarrow {{r}^{2}}=3 \\\\ & \\Leftrightarrow r=\\sqrt{3} \\\\ \\end{align}$<br\/> V\u1eady b\u00e1n k\u00ednh \u0111\u00e1y c\u1ee7a h\u00ecnh n\u00f3n l\u00e0 $\\sqrt{3}$ $cm$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0 $3.$ <\/span><\/span>"}]}],"id_ques":1743},{"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"Th\u00e0nh ph\u1ed1 Xt\u1ed1c \u2013 kh\u00f4m (Th\u1ee5y \u0110i\u1ec3n) \u1edf v\u0129 \u0111\u1ed9 $60^o$ B\u1eafc. \u0110\u1ed9 d\u00e0i v\u0129 tuy\u1ebfn \u0111i qua Xt\u1ed1c \u2013 kh\u00f4m b\u1eb1ng m\u1ea5y ph\u1ea7n \u0111\u1ed9 d\u00e0i X\u00edch \u0110\u1ea1o?","select":["A. $\\dfrac{1}{3}$","B. $\\dfrac{1}{2}$","C. $3$","D. $2$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>Ta coi th\u00e0nh ph\u1ed1 Xt\u1ed1c-kh\u00f4m l\u00e0 m\u1ed9t \u0111i\u1ec3m c\u00f3 v\u0129 \u0111\u1ed9 $60^o$ B\u1eafc tr\u00ean qu\u1ea3 \u0111\u1ecba c\u1ea7u. T\u1eeb \u0111\u00f3 ta t\u00ednh \u0111\u1ed9 d\u00e0i v\u0129 tuy\u1ebfn \u0111i qua $B$ v\u00e0 \u0111\u1ed9 d\u00e0i x\u00edch \u0111\u1ea1o<br\/><b>Ghi nh\u1edb:<\/b> V\u0129 \u0111\u1ed9 c\u1ee7a m\u1ed9t \u0111i\u1ec3m l\u00e0 g\u00f3c t\u1ea1o b\u1edfi \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua t\u00e2m \u0111\u1ecba c\u1ea7u v\u00e0 \u0111i\u1ec3m \u0111\u00f3 v\u1edbi m\u1eb7t ph\u1eb3ng t\u1ea1o b\u1edfi x\u00edch \u0111\u1ea1o. \u0110\u01b0\u1eddng t\u1ea1o b\u1edfi c\u00e1c \u0111i\u1ec3m c\u00f3 c\u00f9ng v\u0129 \u0111\u1ed9 g\u1ecdi l\u00e0 v\u0129 tuy\u1ebfn.<br\/><span class='basic_green'>Gi\u1ea3i<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai24/lv3/img\/H943_K1.png' \/><\/center><br\/>Tr\u00ean \u0111\u01b0\u1eddng tr\u00f2n l\u1edbn \u0111i qua Xt\u1ed1c \u2013 kh\u00f4m (\u0111i\u1ec3m $B$ tr\u00ean h\u00ecnh v\u1ebd), ta c\u00f3 $IB$ l\u00e0 b\u00e1n k\u00ednh c\u1ee7a v\u0129 tuy\u1ebfn $60^o,$ $\\widehat{AOB}={{60}^{o}}$ <br\/>\u0110\u1ed9 d\u00e0i v\u0129 tuy\u1ebfn \u0111i qua B b\u1eb1ng $2\\pi .IB$ <br\/>\u0110\u1ed9 d\u00e0i X\u00edch \u0110\u1ea1o b\u1eb1ng $2\\pi .OB$ <br\/>X\u00e9t tam gi\u00e1c vu\u00f4ng $IOB$ c\u00f3 $\\widehat{IOB}=\\widehat{AOI}-\\widehat{AOB}={{90}^{o}}-{{60}^{o}}={{30}^{o}}$ <br\/>T\u1ec9 s\u1ed1 ph\u1ea3i t\u00ecm b\u1eb1ng $\\dfrac{2\\pi .IB}{2\\pi .OB}=\\dfrac{IB}{OB}=\\sin \\widehat{IOB}=\\sin {{30}^{o}}=\\dfrac{1}{2}$ <br\/>V\u1eady \u0111\u1ed9 d\u00e0i v\u0129 tuy\u1ebfn \u0111i qua Xt\u1ed1c \u2013 kh\u00f4m b\u1eb1ng n\u1eeda \u0111\u1ed9 d\u00e0i X\u00edch \u0110\u1ea1o.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B.<\/span><\/span>","column":2}]}],"id_ques":1744},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["49"],["21,8"],["174,1","174"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>Cho tam gi\u00e1c $ABC$ \u0111\u1ec1u c\u1ea1nh $6cm,$ \u0111\u01b0\u1eddng cao $AH.$ Quay tam gi\u00e1c $ABC$ v\u00e0 c\u00e1c \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp, \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c \u0111\u00f3 v\u00f2ng quanh \u0111\u01b0\u1eddng cao $AH,$ ta \u0111\u01b0\u1ee3c m\u1ed9t h\u00ecnh n\u00f3n v\u00e0 hai h\u00ecnh c\u1ea7u. T\u00ednh th\u1ec3 t\u00edch m\u1ed7i h\u00ecnh (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t)<br\/><b>\u0110\u00e1p s\u1ed1:<\/b>$V_{n\u00f3n}\\approx$_input_ $(cm^3)$; $V_{c\u1ea7u\\, nh\u1ecf}\\approx$_input_ $(cm^3);$ $V_{c\u1ea7u \\,l\u1edbn}\\approx$_input_ $(cm^3)$<\/span>","hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai24/lv3/img\/H943_K4.png' \/><\/center><br\/>G\u1ecdi $r$ l\u00e0 b\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp, $R $ l\u00e0 b\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c $ABC. $<br\/>$\\Delta AHC$ vu\u00f4ng t\u1ea1i $H$ n\u00ean $AH=AC.\\sin {{60}^{o}}=6.\\dfrac{\\sqrt{3}}{2}=3\\sqrt{3}\\left( cm \\right)$ <br\/>$r=OH=\\dfrac{1}{3}AH=\\dfrac{1}{3}.3\\sqrt{3}=\\sqrt{3}\\left( cm \\right)$ <br\/>$R=OA=2OH=2r=2\\sqrt{3}$$(cm)$<br\/>$HC=\\dfrac{BC}{2}=\\dfrac{6}{2}=3\\left( cm \\right)$ <br\/>Th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh n\u00f3n l\u00e0 $\\dfrac{1}{3}\\pi .H{{C}^{2}}.AH=\\dfrac{1}{3}\\pi {{.3}^{2}}.3\\sqrt{3}=9\\sqrt{3}\\pi \\approx 49\\left( c{{m}^{3}} \\right)$ <br\/>Th\u1ec3 t\u00edch h\u00ecnh c\u1ea7u nh\u1ecf l\u00e0: $\\dfrac{4}{3}\\pi {{r}^{3}}=\\dfrac{4}{3}\\pi .{{\\left( \\sqrt{3} \\right)}^{3}}=4\\sqrt{3}\\pi \\approx 21,8\\left( c{{m}^{3}} \\right)$ <br\/>Th\u1ec3 t\u00edch h\u00ecnh c\u1ea7u l\u1edbn l\u00e0: $\\dfrac{4}{3}\\pi {{R}^{3}}=\\dfrac{4}{3}\\pi .{{\\left( 2\\sqrt{3} \\right)}^{3}}=32\\sqrt{3}\\pi \\approx 174,1\\left( c{{m}^{3}} \\right)$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $49;$ $21,8;$$174,1.$<\/span><\/span>"}]}],"id_ques":1745},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["3"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>M\u1ed9t h\u00ecnh n\u00f3n c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y b\u1eb1ng $1$$cm,$ chi\u1ec1u cao b\u1eb1ng \u0111\u01b0\u1eddng k\u00ednh m\u1ed9t h\u00ecnh c\u1ea7u. Di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n h\u00ecnh n\u00f3n b\u1eb1ng di\u1ec7n t\u00edch m\u1eb7t c\u1ea7u. T\u00ednh chi\u1ec1u cao h\u00ecnh n\u00f3n.<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $h=\\sqrt{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$ $(cm)$<\/span>","hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai24/lv3/img\/H943_K5.png' \/><\/center><br\/>G\u1ecdi $h$ l\u00e0 chi\u1ec1u cao c\u1ee7a h\u00ecnh n\u00f3n. \u0110\u01b0\u1eddng sinh c\u1ee7a h\u00ecnh n\u00f3n b\u1eb1ng $\\sqrt{{{h}^{2}}+1}$ <br\/>Di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n c\u1ee7a h\u00ecnh n\u00f3n l\u00e0: $\\pi .1.\\sqrt{{{h}^{2}}+1}+\\pi .{{1}^{2}}=\\pi \\left( \\sqrt{{{h}^{2}}+1}+1 \\right)\\left( c{{m}^{2}} \\right)$ <br\/>V\u00ec chi\u1ec1u cao n\u00f3n b\u1eb1ng \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a h\u00ecnh c\u1ea7u n\u00ean b\u00e1n k\u00ednh h\u00ecnh c\u1ea7u l\u00e0 $\\dfrac{h}{2}\\left( cm \\right)$ <br\/>Di\u1ec7n t\u00edch m\u1eb7t c\u1ea7u b\u1eb1ng $4\\pi {{\\left( \\dfrac{h}{2} \\right)}^{2}}=\\pi {{h}^{2}}\\left( c{{m}^{2}} \\right)$ <br\/>Theo \u0111\u1ec1 b\u00e0i, ta c\u00f3:<br\/>$\\begin{align} & \\,\\,\\,\\,\\,\\,\\pi \\left( \\sqrt{{{h}^{2}}+1}+1 \\right)=\\pi {{h}^{2}} \\\\ & \\Leftrightarrow \\sqrt{{{h}^{2}}+1}+1={{h}^{2}} \\\\ & \\Leftrightarrow {{h}^{2}}+1={{\\left( {{h}^{2}}-1 \\right)}^{2}}\\, (\\text{do}\\,h^2 \\ge 1) \\\\ & \\Leftrightarrow {{h}^{2}}+1={{h}^{4}}-2{{h}^{2}}+1 \\\\ & \\Leftrightarrow {{h}^{4}}=3{{h}^{2}} \\\\ & \\Leftrightarrow {{h}^{2}}\\left( {{h}^{2}}-3 \\right)=0 \\\\ & \\Leftrightarrow {{h}^{2}}-3=0\\,\\,\\,\\,\\,\\,\\,\\left( \\text{do}\\,{{h}^{2}}\\ne 0 \\right) \\\\ & \\Leftrightarrow h=\\sqrt{3} \\\\ \\end{align}$<br\/>V\u1eady chi\u1ec1u cao c\u1ee7a h\u00ecnh n\u00f3n l\u00e0 $\\sqrt{3}\\left( cm \\right)$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0 $3.$ <\/span><\/span>"}]}],"id_ques":1746},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["25"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>C\u00f3 $15$ qu\u1ea3 bi-a h\u00ecnh c\u1ea7u \u0111\u1eb7t n\u1eb1m tr\u00ean m\u1eb7t b\u00e0n sao cho ch\u00fang \u0111\u01b0\u1ee3c d\u1ed3n kh\u00edt trong m\u1ed9t khung h\u00ecnh tam gi\u00e1c \u0111\u1ec1u c\u00f3 chu vi b\u1eb1ng $858$ $mm$ nh\u01b0 h\u00ecnh v\u1ebd ($O$ l\u00e0 t\u00e2m qu\u1ea3 bi-a, $H$ v\u00e0 $K$ l\u00e0 ch\u00e2n \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c h\u1ea1 t\u1eeb t\u00e2m xu\u1ed1ng $BC$). T\u00ednh b\u00e1n k\u00ednh c\u1ee7a m\u1ed7i qu\u1ea3 bi-a<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai24/lv3/img\/H943_K6.png' \/><\/center><br\/><b>\u0110\u00e1p s\u1ed1:<\/b> _input_$(mm)$<\/span>","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/>B\u01b0\u1edbc 1: G\u1ecdi $r$ l\u00e0 b\u00e1n k\u00ednh m\u1ed7i qu\u1ea3 bi-a. T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh c\u1ee7a tam gi\u00e1c $ABC$<br\/>B\u01b0\u1edbc 2: T\u00ednh $BH,\\, CK$ theo $r$<br\/>B\u01b0\u1edbc 3: T\u00ednh $BC$ theo $r.$ K\u1ebft h\u1ee3p v\u1edbi $BC$ t\u00ednh \u0111\u01b0\u1ee3c \u1edf b\u01b0\u1edbc 1, ta t\u00ecm $r$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai24/lv3/img\/H943_K6.png' \/><\/center><br\/>G\u1ecdi $r$ l\u00e0 b\u00e1n k\u00ednh m\u1ed7i qu\u1ea3 bi-a.<br\/> C\u1ea1nh c\u1ee7a tam gi\u00e1c \u0111\u1ec1u $ABC$ b\u1eb1ng $858:3=286$$(mm)$<br\/>X\u00e9t $\\Delta OBH$ vu\u00f4ng t\u1ea1i $H,$ c\u00f3 $\\widehat{OBH}=\\dfrac{\\widehat{B}}{2}=\\dfrac{{{60}^{o}}}{2}={{30}^{o}}.$ Suy ra $\\widehat{BOH}={{60}^{o}}.$<br\/> Ta c\u00f3:<br\/>$BH=OH.tg{\\widehat{BOH}}=OH.tg{{60}^{o}}=r\\sqrt{3}$ $(mm)$<br\/>Suy ra: $KC=BH=r\\sqrt{3}\\left( mm \\right)$<br\/>Ta c\u00f3: $BC=BH+HK+KC=$$r\\sqrt{3}+8r+r\\sqrt{3}=2\\left( 4+\\sqrt{3} \\right)r$ $(mm)$<br\/>Suy ra: $r=\\dfrac{BC}{2\\left( 4+\\sqrt{3} \\right)}=\\dfrac{286}{2\\left( 4+\\sqrt{3} \\right)}=11\\left( 4-\\sqrt{3} \\right)\\approx 25\\left( mm \\right)$ <br\/>V\u1eady b\u00e1n k\u00ednh m\u1ed7i qu\u1ea3 bi-a b\u1eb1ng $25$$mm.$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0 $25.$ <\/span><\/span>"}]}],"id_ques":1747},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["24"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>M\u1ed9t qu\u1ea3 b\u00f3ng h\u00ecnh c\u1ea7u b\u00e1n k\u00ednh $13$$cm$ n\u1ed5i tr\u00ean m\u1eb7t h\u1ed3, \u0111\u1ec9nh c\u1ee7a qu\u1ea3 b\u00f3ng cao h\u01a1n m\u1eb7t h\u1ed3 $18$$cm. $ T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh b\u1edfi qu\u1ea3 b\u00f3ng v\u00e0 m\u1eb7t h\u1ed3.<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\pi$$(cm)$<\/span>","hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai24/lv3/img\/H943_K8.png' \/><\/center><br\/>G\u1ecdi $I $ v\u00e0 $r$ l\u00e0 t\u00e2m v\u00e0 b\u00e1n k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh b\u1edfi qu\u1ea3 b\u00f3ng v\u00e0 m\u1eb7t h\u1ed3.<br\/>Ta c\u00f3 kho\u1ea3ng c\u00e1ch t\u1eeb t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n \u0111\u1ebfn \u0111\u1ec9nh qu\u1ea3 b\u00f3ng l\u00e0 $18$$cm$ n\u00ean $OI=18-13=5$$(cm)$<br\/>$\\Delta IOA$ vu\u00f4ng t\u1ea1i $I$ n\u00ean $r=IA=\\sqrt{O{{A}^{2}}-O{{I}^{2}}}=\\sqrt{{{13}^{2}}-{{5}^{2}}}=12\\left( cm \\right)$ <br\/>Suy ra \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh b\u1edfi qu\u1ea3 b\u00f3ng v\u00e0 m\u1eb7t h\u1ed3 l\u00e0: $2.\\pi .12=24\\pi \\left( cm \\right)$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0 $24.$ <\/span><\/span>"}]}],"id_ques":1748},{"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":"M\u1ed9t qu\u1ea3 b\u00f3ng h\u00ecnh c\u1ea7u \u0111\u1eb7t tr\u00ean m\u1eb7t \u0111\u1ea5t c\u00f3 b\u00f3ng c\u1ee7a n\u00f3 nh\u01b0 h\u00ecnh v\u1ebd, trong \u0111\u00f3 \u0111\u1ed9 d\u00e0i l\u1edbn nh\u1ea5t c\u1ee7a b\u00f3ng l\u00e0 $BC=1m.$ Bi\u1ebft m\u1ed9t c\u1ed9t cao $1$$m$ l\u00fac \u0111\u00f3 c\u00f3 b\u00f3ng d\u00e0i $2$$m.$ T\u00ednh b\u00e1n k\u00ednh qu\u1ea3 b\u00f3ng.<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai24/lv3/img\/H943_K7.png' \/><\/center><\/span>","select":["A. $R=\\sqrt{5}+2$$(m)$","B. $R=\\sqrt{5}-2$$(m)$","C. $R=\\sqrt{5}+1$$(m)$","D. $R=\\sqrt{5}-1$$(m)$"],"hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai24/lv3/img\/H943_K9.png' \/><\/center><br\/>G\u1ecdi $R$ l\u00e0 b\u00e1n k\u00ednh qu\u1ea3 b\u00f3ng.<br\/> Ta c\u00f3 tia s\u00e1ng ti\u1ebfp x\u00fac v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n $(O)$ t\u1ea1i $E.$ G\u1ecdi $A$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $BO$ v\u00e0 $CE.$<br\/>V\u00ec khi c\u1ed9t cao $1$$m$ c\u00f3 b\u00f3ng d\u00e0i $2m$ n\u00ean $AB=BC:2=0,5$$(m)$<br\/>$\\Delta ABC$ vu\u00f4ng t\u1ea1i $A$ n\u00ean $AC=\\sqrt{A{{B}^{2}}+B{{C}^{2}}}=\\sqrt{0,{{5}^{2}}+{{1}^{2}}}=\\dfrac{\\sqrt{5}}{2}\\left( m \\right)$ <br\/>D\u1ec5 th\u1ea5y $\\Delta AEO$ \u0111\u1ed3ng d\u1ea1ng v\u1edbi $\\Delta ABC\\,\\,\\,\\left( g.g \\right)$ n\u00ean $\\dfrac{OA}{CA}=\\dfrac{OE}{BC}\\Leftrightarrow \\dfrac{AB-R}{AC}=\\dfrac{R}{BC}$ <br\/>Thay s\u1ed1 v\u00e0o \u0111\u1eb3ng th\u1ee9c, ta c\u00f3:<br\/>$\\dfrac{0,5-R}{\\dfrac{\\sqrt{5}}{2}}=\\dfrac{R}{1}\\Leftrightarrow 2\\left( 0,5-R \\right)=\\sqrt{5}R\\Leftrightarrow R=\\sqrt{5}-2\\left( m \\right)$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B.<\/span><\/span>","column":2}]}],"id_ques":1749},{"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 $A,$ \u0111\u01b0\u1eddng cao $AH.$ Cho ba n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp ba tam gi\u00e1c vu\u00f4ng $ABH,$ $ACH$ v\u00e0 $ABC$ l\u1ea7n l\u01b0\u1ee3t quay quanh c\u00e1c c\u1ea1nh $AB,$ $AC,$ $BC.$ G\u1ecdi $S_1;$$S_2;$$S_3$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 di\u1ec7n t\u00edch c\u00e1c m\u1eb7t c\u1ea7u ngo\u1ea1i ti\u1ebfp ba tam gi\u00e1c vu\u00f4ng $ABH,$ $ACH$ v\u00e0 $ABC.$ <br\/>Kh\u1eb3ng \u0111\u1ecbnh $S_1=S_2+S_3$, <b>\u0111\u00fang<\/b> hay <b>sai<\/b>?<\/span>","select":["A. \u0110\u00fang","B. Sai"],"hint":"T\u00ednh $S_1;$ $S_2$ v\u00e0 $S_3$ r\u1ed3i \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pytago \u0111\u1ec3 ki\u1ec3m tra kh\u1eb3ng \u0111\u1ecbnh c\u1ee7a b\u00e0i to\u00e1n. ","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai24/lv3/img\/H943_K10.png' \/><\/center><br\/>Ta c\u00f3 b\u00e1n k\u00ednh ba m\u1eb7t c\u1ea7u ngo\u1ea1i ti\u1ebfp ba tam gi\u00e1c vu\u00f4ng $ABH,$ $ACH$ v\u00e0 $ABC$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 $\\dfrac{AB}{2};\\dfrac{AC}{2};\\dfrac{BC}{2}$ <br\/> ${{S}_{1}}=4\\pi .{{\\left( \\dfrac{AB}{2} \\right)}^{2}}=\\pi .A{{B}^{2}}$ <br\/>${{S}_{2}}=4\\pi .{{\\left( \\dfrac{AC}{2} \\right)}^{2}}=\\pi .A{{C}^{2}}$<br\/>${{S}_{3}}=4\\pi .{{\\left( \\dfrac{BC}{2} \\right)}^{2}}=\\pi .B{{C}^{2}}$<br\/>Do tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$ n\u00ean $A{{B}^{2}}+A{{C}^{2}}=B{{C}^{2}}$ <br\/>Suy ra: ${{S}_{1}}+{{S}_{2}}=\\pi .A{{B}^{2}}+\\pi .A{{C}^{2}}=\\pi \\left( A{{B}^{2}}+A{{C}^{2}} \\right)=\\pi B{{C}^{2}}={{S}_{3}}$ <br\/>V\u1eady $S_1+S_2=S_3$<br\/>Suy ra kh\u1eb3ng \u0111\u1ecbnh b\u00e0i to\u00e1n l\u00e0 sai.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B.<\/span><\/span>","column":2}]}],"id_ques":1750}],"lesson":{"save":0,"level":3}}