{"segment":[{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho tam gi\u00e1c \u0111\u1ec1u $ABC$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O; R)$, \u0111\u01b0\u1eddng k\u00ednh $AD$. G\u1ecdi $E$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u1ea1nh $AC$, tia $DE$ c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n \u1edf $F$. T\u00ednh $DE$ theo $R$. ","select":["A. $\\dfrac{3R}{2}$ ","B. $\\dfrac{R\\sqrt{7}}{2}$ ","C. $\\dfrac{R\\sqrt{7}}{14}$","D. $\\dfrac{R\\sqrt{21}}{7}$"],"hint":"N\u1ed1i $C$ v\u1edbi $D$, \u0111\u1eb7t $AB = x$ t\u00ednh \u0111\u1ed9 d\u00e0i c\u00e1c c\u1ea1nh $EC, DC$ theo $R$ r\u1ed3i \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pytago t\u00ednh c\u1ea1nh $DE$","explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai18/lv3/img\/h936_K1.png' \/><\/center> <br\/> Do tam gi\u00e1c $ABC$ \u0111\u1ec1u n\u00ean $O$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c <br\/>$\\Rightarrow BE=\\dfrac{3}{2}BO=\\dfrac{3R}{2}$ <br\/> Gi\u1ea3 s\u1eed $AB=x\\Rightarrow AE=\\dfrac{x}{2}$ <br\/> Ta c\u00f3: $BE\\bot AC$ (t\u00ednh ch\u1ea5t tam gi\u00e1c \u0111\u1ec1u) <br\/> X\u00e9t $\\Delta ABE$ vu\u00f4ng t\u1ea1i $E$ c\u00f3: <br\/>$A{{B}^{2}}=A{{E}^{2}}+B{{E}^{2}}$ (\u0111\u1ecbnh l\u00ed Pitago) <br\/>$\\Leftrightarrow {{x}^{2}}=\\dfrac{{{x}^{2}}}{4}+\\dfrac{9{{R}^{2}}}{4}\\\\ \\Leftrightarrow \\dfrac{3{{x}^{2}}}{4}=\\dfrac{9{{R}^{2}}}{4}\\\\ \\Leftrightarrow {{x}^{2}}=3{{R}^{2}}\\Rightarrow x=R\\sqrt{3}$ <br\/>$\\Rightarrow AC=AB=R\\sqrt{3};AE=EC=\\dfrac{R\\sqrt{3}}{2}$ <br\/> Ta c\u00f3: $\\widehat{ACD}={{90}^{o}}$ (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n) <br\/> X\u00e9t $\\Delta ACD$ vu\u00f4ng t\u1ea1i $C$ c\u00f3: <br\/>$DC=\\sqrt{A{{D}^{2}}-A{{C}^{2}}}=\\sqrt{4{{R}^{2}}-3{{R}^{2}}}=R$ <br\/> X\u00e9t $\\Delta DCE$ vu\u00f4ng t\u1ea1i $C$ c\u00f3: <br\/>$DE=\\sqrt{E{{C}^{2}}+D{{C}^{2}}}=\\sqrt{\\dfrac{3{{R}^{2}}}{4}+{{R}^{2}}}=\\dfrac{R\\sqrt{7}}{2}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B <\/span><\/span>","column":4}]}],"id_ques":1581},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"Cho \u0111a gi\u00e1c \u0111\u1ec1u $n$ c\u1ea1nh, \u0111\u1ed9 d\u00e0i c\u1ea1nh b\u1eb1ng $a$. B\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp \u0111a gi\u00e1c b\u1eb1ng: ","select":["A. $a\\sin \\dfrac{{{180}^{o}}}{n}$","B. $a tg \\dfrac{{{180}^{o}}}{n}$ ","C. $\\dfrac{a}{2tg \\dfrac{{{180}^{o}}}{n}}$","D. $\\dfrac{a}{2\\sin \\dfrac{{{180}^{o}}}{n}}$"],"explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai18/lv3/img\/h936_K2.png' \/><\/center> <br\/> G\u1ecdi $O$ l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp \u0111a gi\u00e1c \u0111\u1ec1u $n$ c\u1ea1nh. $AB$ l\u00e0 m\u1ed9t c\u1ea1nh c\u1ee7a \u0111a gi\u00e1c <br\/>$\\Rightarrow \\Delta AOB$ c\u00e2n t\u1ea1i $O$ v\u00e0 $\\widehat{AOB}=\\dfrac{{{360}^{o}}}{n}$ <br\/> K\u1ebb $OH\\bot AB\\Rightarrow H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB$ v\u00e0 $OH$ l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a$\\widehat{AOB}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/>$\\Rightarrow \\widehat{AOH}=\\dfrac{\\widehat{AOB}}{2}=\\dfrac{{{180}^{o}}}{n}$ (t\u00ednh ch\u1ea5t ph\u00e2n gi\u00e1c) <br\/> X\u00e9t $\\Delta AOH$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: <br\/>$AH=OA.\\sin \\widehat{AOH}=R.\\sin \\dfrac{{{180}^{o}}}{n}$ <br\/>$\\Rightarrow AB=2R.sin\\dfrac{{{180}^{o}}}{n}\\\\ \\Rightarrow R=\\dfrac{AB}{2.\\sin \\dfrac{{{180}^{o}}}{n}}=\\dfrac{a}{2.\\sin \\dfrac{{{180}^{o}}}{n}}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D <\/span><\/span>","column":2}]}],"id_ques":1582},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'> Cho b\u00e1t gi\u00e1c \u0111\u1ec1u $ABCDEFGH$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O;R)$. G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AC$ v\u00e0 $BF$. Ch\u1ee9ng minh r\u1eb1ng $BI=\\dfrac{R\\left( 2-\\sqrt{2} \\right)}{2}$.","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[4],[1],[6],[5],[3],[2]]],"list":[{"point":10,"left":["$a=2R.\\sin \\dfrac{{{180}^{o}}}{n}=2R.\\sin {{45}^{o}}=R\\sqrt{2}$"," D\u1ec5 th\u1ea5y $ACEG$ l\u00e0 h\u00ecnh vu\u00f4ng, g\u1ecdi $K$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $BF$ v\u00e0 $GE$ ","$\\Rightarrow BI=KF=\\dfrac{BF-IK}{2}=\\dfrac{2R-R\\sqrt{2}}{2}=\\dfrac{R\\left( 2-\\sqrt{2} \\right)}{2}$"," $\\Rightarrow IK=CE=AG=R\\sqrt{2}$"," $\\Rightarrow$ C\u1ea1nh h\u00ecnh vu\u00f4ng $ACEG$ l\u00e0: ","\u00c1p d\u1ee5ng k\u1ebft qu\u1ea3 c\u1ee7a b\u00e0i t\u1eadp tr\u00ean, ta c\u00f3: <br\/> B\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp \u0111a gi\u00e1c \u0111\u1ec1u $n$ c\u1ea1nh, \u0111\u1ed9 d\u00e0i m\u1ed7i c\u1ea1nh b\u1eb1ng $a$ l\u00e0 $\\dfrac{a}{2\\sin \\dfrac{{{180}^{o}}}{n}}$"],"top":100,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai18/lv3/img\/h936_K3.png' \/><\/center> <br\/> D\u1ec5 th\u1ea5y $ACEG$ l\u00e0 h\u00ecnh vu\u00f4ng, g\u1ecdi $K$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $BF$ v\u00e0 $GE$ <br\/> \u00c1p d\u1ee5ng k\u1ebft qu\u1ea3 c\u1ee7a b\u00e0i t\u1eadp tr\u00ean, ta c\u00f3: <br\/> B\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp \u0111a gi\u00e1c \u0111\u1ec1u $n$ c\u1ea1nh, \u0111\u1ed9 d\u00e0i m\u1ed7i c\u1ea1nh b\u1eb1ng $a$ l\u00e0 $\\dfrac{a}{2\\sin \\dfrac{{{180}^{o}}}{n}}$ <br\/> $\\Rightarrow$ C\u1ea1nh h\u00ecnh vu\u00f4ng $ACEG$ l\u00e0: <br\/> $a=2R.\\sin \\dfrac{{{180}^{o}}}{n}=2R.\\sin {{45}^{o}}=R\\sqrt{2}$ <br\/> $\\Rightarrow IK=CE=AG=R\\sqrt{2}$ <br\/>$\\Rightarrow BI=KF=\\dfrac{BF-IK}{2}=\\dfrac{2R-R\\sqrt{2}}{2}=\\dfrac{R\\left( 2-\\sqrt{2} \\right)}{2}$ <\/span>"}]}],"id_ques":1583},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{1}{4}$","B. $\\dfrac{1}{3}$","C. $\\dfrac{1}{2}$"],"ques":" Cho tam gi\u00e1c \u0111\u1ec1u $ABC$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O; R)$ v\u00e0 tam gi\u00e1c \u0111\u1ec1u $IJK$ ngo\u1ea1i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n \u0111\u00f3. Bi\u1ebft $A, B, C$ l\u1ea7n l\u01b0\u1ee3t n\u1eb1m tr\u00ean c\u00e1c c\u1ea1nh $KJ, JI, IK$. T\u1ec9 s\u1ed1 di\u1ec7n t\u00edch$\\dfrac{{{S}_{ABC}}}{{{S}_{IJK}}}$ l\u00e0 ?","explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai18/lv3/img\/h936_K4.png' \/><\/center> G\u1ecdi $D$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AI$ v\u00e0 $BC$ <br\/> Ta c\u00f3 $O$ l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c $ABC$ v\u00e0 $IJK$ <br\/>$\\Rightarrow AO=\\dfrac{2}{3}AD$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) <br\/>$\\Rightarrow AD=\\dfrac{3}{2}AO=\\dfrac{3}{2}R$ <br\/> $BC=2R.\\sin \\dfrac{{{180}^{o}}}{3}=2R.sin{{60}^{o}}=R\\sqrt{3}$ <br\/>$\\Rightarrow {{S}_{ABC}}=\\dfrac{1}{2}AD.BC=\\dfrac{1}{2}.\\dfrac{3R}{2}.R\\sqrt{3}=\\dfrac{3{{R}^{2}}\\sqrt{3}}{4}$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch) <br\/> Ta c\u00f3:$AO=\\dfrac{1}{3}AI$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) <br\/>$\\Rightarrow AI=3AO=3R$ <br\/> B\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c $IJK$ l\u00e0: <br\/>$IO=2AO=2R$ <br\/> $KJ=2.IO.\\sin \\dfrac{{{180}^{o}}}{3}=2.2R.sin{{60}^{o}}=2R\\sqrt{3}$ <br\/>$\\Rightarrow {{S}_{IJK}}=\\dfrac{1}{2}AI.KJ=\\dfrac{1}{2}.3R.2R\\sqrt{3}=3{{R}^{2}}\\sqrt{3}$ <br\/>$\\Rightarrow \\dfrac{{{S}_{ABC}}}{{{S}_{IJK}}}=\\dfrac{3{{R}^{2}}\\sqrt{3}}{4}:3{{R}^{2}}\\sqrt{3}=\\dfrac{1}{4}$ <\/span><\/span> "}]}],"id_ques":1584},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"M\u1ed9t \u0111a gi\u00e1c \u0111\u1ec1u n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O; 2cm)$. Bi\u1ebft \u0111\u1ed9 d\u00e0i m\u1ed7i c\u1ea1nh c\u1ee7a n\u00f3 l\u00e0 $2\\sqrt{3}cm.$ T\u00ednh di\u1ec7n t\u00edch c\u1ee7a \u0111a gi\u00e1c \u0111\u00f3. ","select":["A. $3\\sqrt{3}\\,\\left( c{{m}^{2}} \\right)$","B. $4\\sqrt{3}\\,\\left( c{{m}^{2}} \\right)$ ","C. $12\\,\\left( c{{m}^{2}} \\right)$","D. $27\\,\\left( c{{m}^{2}} \\right)$"],"explain":" <span class='basic_left'> Ta c\u00f3 $a$ l\u00e0 \u0111\u1ed9 d\u00e0i m\u1ed9t c\u1ea1nh c\u1ee7a \u0111a gi\u00e1c \u0111\u1ec1u, $n$ l\u00e0 s\u1ed1 c\u1ea1nh c\u1ee7a \u0111a gi\u00e1c \u0111\u1ec1u, $R$ l\u00e0 b\u00e1n k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp \u0111a gi\u00e1c <br\/> \u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c $R=\\dfrac{a}{2\\sin \\dfrac{{{180}^{o}}}{n}}\\Rightarrow \\sin \\dfrac{{{180}^{o}}}{n}=\\dfrac{a}{2R}=\\dfrac{2\\sqrt{3}}{2.2}=\\dfrac{\\sqrt{3}}{2}$ <br\/>$\\Rightarrow \\dfrac{{{180}^{o}}}{n}={{60}^{o}}\\Rightarrow n=\\dfrac{{{180}^{o}}}{{{60}^{o}}}=3$ <br\/> Suy ra \u0111a gi\u00e1c \u0111\u00e3 cho l\u00e0 tam gi\u00e1c \u0111\u1ec1u <br\/> \u0110\u01b0\u1eddng cao trong tam gi\u00e1c \u0111\u1ec1u c\u1ea1nh $a$ l\u00e0 $\\dfrac{a\\sqrt{3}}{2}$ <br\/>$\\Rightarrow S=\\dfrac{1}{2}ah=\\dfrac{1}{2}.2\\sqrt{3}.2\\sqrt{3}.\\dfrac{\\sqrt{3}}{2}=3\\sqrt{3}\\,\\left( c{{m}^{2}} \\right)$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A <\/span><\/span>","column":4}]}],"id_ques":1585},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["0,28"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" <span class='basic_left'> \u0110\u01b0\u1eddng k\u00ednh b\u00e1nh xe c\u1ee7a m\u1ed9t xe \u0111\u1ea1p l\u00e0 $73cm$. H\u1ecfi xe \u0111\u00f3 \u0111i \u0111\u01b0\u1ee3c bao nhi\u00eau $km$ n\u1ebfu b\u00e1nh xe quay \u0111\u01b0\u1ee3c $1200$ v\u00f2ng? (l\u00e0m tr\u00f2n \u0111\u1ebfn s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 hai) <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> Qu\u00e3ng \u0111\u01b0\u1eddng xe \u0111i \u0111\u01b0\u1ee3c l\u00e0: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\,(km)$","explain":" <span class='basic_left'> B\u00e1nh xe l\u0103n \u0111\u01b0\u1ee3c m\u1ed9t v\u00f2ng t\u1ee9c l\u00e0 n\u00f3 \u0111\u00e3 \u0111i m\u1ed9t \u0111\u1ed9 d\u00e0i b\u1eb1ng chu vi c\u1ee7a b\u00e1nh xe <br\/> Chu vi c\u1ee7a b\u00e1nh xe \u0111\u1ea1p l\u00e0: $C=\\pi d=73.\\pi \\,(cm)$ <br\/> Qu\u00e3ng \u0111\u01b0\u1eddng xe \u0111\u00f3 \u0111i \u0111\u01b0\u1ee3c khi b\u00e1nh xe quay $1200$ v\u00f2ng l\u00e0: $73\\pi .1200=87600\\pi \\approx 275203,5\\,\\left( cm \\right)$ <br\/> \u0110\u1ed5i $275203,5\\,cm \\approx 0,28\\,\\left( km \\right)$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $0,28$ <\/span><\/span> "}]}],"id_ques":1586},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["6366"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" <span class='basic_left'> M\u1ed7i kinh tuy\u1ebfn c\u1ee7a tr\u00e1i \u0111\u1ea5t l\u00e0 m\u1ed9t n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n l\u1edbn c\u1ee7a tr\u00e1i \u0111\u1ea5t m\u00e0 hai m\u00fat \u0111\u01b0\u1eddng k\u00ednh l\u00e0 B\u1eafc c\u1ef1c v\u00e0 Nam c\u1ef1c. Bi\u1ebft r\u1eb1ng, $1km$ b\u1eb1ng $\\dfrac{1}{20000}$ \u0111\u1ed9 d\u00e0i c\u1ee7a m\u1ed9t kinh tuy\u1ebfn. <br\/> B\u00e1n k\u00ednh tr\u00e1i \u0111\u1ea5t l\u00e0: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\,(km)$ <br\/> (K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb)","explain":" <span class='basic_left'> G\u1ecdi b\u00e1n k\u00ednh tr\u00e1i \u0111\u1ea5t l\u00e0 $R$ th\u00ec \u0111\u1ed9 d\u00e0i kinh tuy\u1ebfn Tr\u00e1i \u0110\u1ea5t l\u00e0: <br\/>$ \\dfrac{1}{20000}\\pi R=1\\\\ \\Leftrightarrow \\pi R=20000\\\\ \\Leftrightarrow R=\\dfrac{20000}{\\pi }\\approx 6366\\,\\left( km \\right)$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $6366$ <\/span><\/span> "}]}],"id_ques":1587},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Ch\u1ecdn m\u00e1y k\u00e9o c\u00f4ng nghi\u1ec7p c\u00f3 hai b\u00e1nh xe, b\u00e1nh sau l\u1edbn h\u01a1n b\u00e1nh tr\u01b0\u1edbc. Khi b\u01a1m c\u0103ng, b\u00e1nh sau c\u00f3 b\u00e1n k\u00ednh $0,75m$, b\u00e1nh tr\u01b0\u1edbc c\u00f3 b\u00e1n k\u00ednh $0,5m$. H\u1ecfi n\u1ebfu m\u00e1y k\u00e9o \u0111i \u0111\u01b0\u1ee3c $471m$ th\u00ec b\u00e1nh sau v\u00e0 b\u00e1nh tr\u01b0\u1edbc l\u1ea7n l\u01b0\u1ee3t l\u0103n \u0111\u01b0\u1ee3c bao nhi\u00eau v\u00f2ng? (l\u1ea5y $\\pi\\approx 3,14$)","select":["A. $100$ v\u00f2ng v\u00e0 $150$ v\u00f2ng","B. $120$ v\u00f2ng v\u00e0 $140$ v\u00f2ng ","C. $100$ v\u00f2ng v\u00e0 $120$ v\u00f2ng","D. $120$ v\u00f2ng v\u00e0 $150$ v\u00f2ng"],"explain":" <span class='basic_left'>\u0110\u1ed9 d\u00e0i b\u00e1nh sau c\u1ee7a m\u00e1y k\u00e9o l\u00e0:${{C}_{1}}=2\\pi R\\approx 2.3,14.0,75\\approx 4,71\\,(m)$ <br\/> \u0110\u1ed9 d\u00e0i b\u00e1nh tr\u01b0\u1edbc c\u1ee7a m\u00e1y k\u00e9o l\u00e0:${{C}_{2}}=2\\pi R\\approx 2.3,14.0,5\\approx 3,14\\,(m)$ <br\/> Khi m\u00e1y k\u00e9o \u0111i \u0111\u01b0\u1ee3c $471m$ th\u00ec b\u00e1nh sau quay \u0111\u01b0\u1ee3c s\u1ed1 v\u00f2ng l\u00e0:$471:4,71=100$ (v\u00f2ng) <br\/> Khi m\u00e1y k\u00e9o \u0111i \u0111\u01b0\u1ee3c $471m$ th\u00ec b\u00e1nh tr\u01b0\u1edbc quay \u0111\u01b0\u1ee3c s\u1ed1 v\u00f2ng l\u00e0:$471:3,14=150$ (v\u00f2ng)<br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A <\/span><\/span>","column":2}]}],"id_ques":1588},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["2.580.000"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<span class='basic_left'> Tr\u00e1i \u0110\u1ea5t quay quanh M\u1eb7t Tr\u1eddi theo m\u1ed9t qu\u1ef9 \u0111\u1ea1o g\u1ea7n tr\u00f2n. Gi\u1ea3 thi\u1ebft qu\u1ef9 \u0111\u1ea1o n\u00e0y tr\u00f2n v\u00e0 c\u00f3 b\u00e1n k\u00ednh kho\u1ea3ng $150$ tri\u1ec7u kil\u00f4m\u00e9t. C\u1ee9 h\u1ebft m\u1ed9t n\u0103m th\u00ec Tr\u00e1i \u0110\u1ea5t quay \u0111\u01b0\u1ee3c m\u1ed9t v\u00f2ng quanh M\u1eb7t Tr\u1eddi. Bi\u1ebft m\u1ed9t n\u0103m c\u00f3 $365$ ng\u00e0y, h\u00e3y t\u00ednh qu\u00e3ng \u0111\u01b0\u1eddng \u0111i \u0111\u01b0\u1ee3c c\u1ee7a Tr\u00e1i \u0110\u1ea5t sau m\u1ed9t ng\u00e0y (l\u00e0m tr\u00f2n \u0111\u1ebfn \u0111\u01a1n v\u1ecb $10000 km$) <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> M\u1ed9t ng\u00e0y Tr\u00e1i \u0110\u1ea5t \u0111i \u0111\u01b0\u1ee3c qu\u00e3ng \u0111\u01b0\u1eddng l\u00e0: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\,(km)$","explain":" <span class='basic_left'> \u0110\u1ed9 d\u00e0i c\u1ee7a qu\u1ef9 \u0111\u1ea1o l\u00e0: <br\/> $C=2\\pi R\\approx 2.3,14.150.000.000=942.000.000\\,\\left( km \\right)$ <br\/> M\u1ed7i ng\u00e0y Tr\u00e1i \u0110\u1ea5t \u0111i \u0111\u01b0\u1ee3c qu\u00e3ng \u0111\u01b0\u1eddng l\u00e0: <br\/> $\\dfrac{942.000.000}{365}\\approx 2.580.000\\,\\left( km \\right)$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $2.580.000$ <\/span><\/span> "}]}],"id_ques":1589},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O;6cm)$ v\u00e0 \u0111\u01b0\u1eddng tr\u00f2n $(O\u2019)$ c\u1eaft nhau \u1edf $M$ v\u00e0 $N$ $(O$ v\u00e0 $O\u2019$ thu\u1ed9c hai n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd $NM)$. Bi\u1ebft $OM\\bot O'M;ON\\bot O'N$ v\u00e0 $OO'=10cm.$ T\u00ednh t\u1ed5ng \u0111\u1ed9 d\u00e0i c\u00e1c cung nh\u1ecf $MN$ c\u1ee7a c\u00e1c \u0111\u01b0\u1eddng tr\u00f2n $(O)$ v\u00e0 $(O\u2019)$. ","select":["A. $13,3\\,\\left( cm \\right)$","B. $21,4\\,\\left( cm \\right)$ ","C. $26,6\\,\\left( cm \\right)$","D. $39,9\\,\\left( cm \\right)$"],"explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai18/lv3/img\/h936_K10.png' \/><\/center> <br\/> X\u00e9t $\\Delta OMN$ vu\u00f4ng t\u1ea1i $M$ c\u00f3: <br\/>$O'M=\\sqrt{OO{{'}^{2}}-O{{M}^{2}}}=\\sqrt{{{10}^{2}}-{{6}^{2}}}=8\\,\\left( cm \\right)$ <br\/>$\\sin \\widehat{MO'O}=\\dfrac{OM}{OO'}=\\dfrac{6}{10}=0,6\\\\ \\Rightarrow \\widehat{MO'O}\\approx {{37}^{o}}$ <br\/>$\\Rightarrow \\widehat{MO'N}=2\\widehat{MO'O}={{2.37}^{o}}={{74}^{o}}$ <br\/> T\u1ee9 gi\u00e1c $MONO\u2019$ c\u00f3 $M$ v\u00e0 $N$ nh\u00ecn $OO\u2019$ d\u01b0\u1edbi g\u00f3c ${{90}^{o}}$ n\u00ean n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n <br\/>$\\Rightarrow \\widehat{MON}={{180}^{o}}-\\widehat{MO'N}={{180}^{o}}-{{74}^{o}}={{106}^{o}}$ <br\/> T\u1ed5ng \u0111\u1ed9 d\u00e0i c\u00e1c cung nh\u1ecf $MN$ c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n $(O)$ v\u00e0 $(O\u2019)$ l\u00e0: <br\/>$\\Rightarrow l=\\dfrac{\\pi .6.106}{180}+\\dfrac{\\pi .8.74}{180}\\approx \\dfrac{3,14.53}{15}+\\dfrac{3,14.148}{45}=21,4\\,\\left( cm \\right)$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B <\/span><\/span>","column":4}]}],"id_ques":1590}],"lesson":{"save":0,"level":3}}