{"segment":[{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<span class='basic_left'> M\u1ed9t \u0111\u01b0\u1eddng h\u1ea7m c\u00f3 m\u1eb7t c\u1eaft vu\u00f4ng g\u00f3c l\u00e0 m\u1ed9t ph\u1ea7n \u0111\u01b0\u1eddng tr\u00f2n nh\u01b0 h\u00ecnh b\u00ean d\u01b0\u1edbi. M\u1eb7t c\u1eaft n\u00e0y c\u00f3 chi\u1ec1u cao l\u1edbn nh\u1ea5t l\u00e0 $5m$, chi\u1ec1u r\u1ed9ng $10\\sqrt{3}m.$ T\u00ednh di\u1ec7n t\u00edch m\u1eb7t c\u1eaft vu\u00f4ng g\u00f3c c\u1ee7a \u0111\u01b0\u1eddng h\u1ea7m \u0111\u00f3. <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai19/lv3/img\/h937_K1.png' \/><\/center>","select":["A. $\\dfrac{50\\pi }{3}\\,\\left( {{m}^{2}} \\right)$ ","B. $\\dfrac{100\\pi }{2}-15\\sqrt{3}\\,\\left( {{m}^{2}} \\right)$ ","C. $\\dfrac{100\\pi }{3}-25\\sqrt{3}\\,\\left( {{m}^{2}} \\right)$","D. $\\dfrac{50\\pi }{3}-25\\sqrt{3}\\,\\left( {{m}^{2}} \\right)$"],"hint":"D\u1ef1ng ph\u1ea7n c\u00f2n l\u1ea1i c\u1ee7a h\u00ecnh tr\u00f2n. Di\u1ec7n t\u00edch c\u1ea7n t\u00ecm b\u1eb1ng hi\u1ec7u c\u1ee7a di\u1ec7n t\u00edch qu\u1ea1t tr\u00f2n v\u00e0 di\u1ec7n t\u00edch tam gi\u00e1c.","explain":" <span class='basic_left'><center> D\u1ef1ng ph\u1ea7n c\u00f2n l\u1ea1i c\u1ee7a h\u00ecnh tr\u00f2n v\u00e0 l\u1ea5y c\u00e1c \u0111i\u1ec3m nh\u01b0 h\u00ecnh v\u1ebd sau: <br\/> <img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai19/lv3/img\/h937_K1.1.png' \/><\/center> <br\/> G\u1ecdi $S$ l\u00e0 di\u1ec7n t\u00edch c\u1ea7n t\u00ecm th\u00ec: $S={{S}_{\\text{qu\u1ea1t OAD}}}-{{S}_{\\Delta AOD}}$ <br\/> Ta c\u00f3: $\\left\\{ \\begin{align} & BH=5cm \\\\ & \\widehat{BAC}={{90}^{o}}\\,\\left( \\text{gi\u1ea3 thi\u1ebft} \\right) \\\\ \\end{align} \\right.$ <br\/> $AH=\\dfrac{AD}{2}=\\dfrac{10\\sqrt{3}}{2}=5\\sqrt{3}\\,\\left( m \\right)$ (\u0111\u1ecbnh l\u00ed \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) <br\/> X\u00e9t $\\Delta AHB$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: <br\/> $A{{B}^{2}}=A{{H}^{2}}+H{{B}^{2}}$ (\u0111\u1ecbnh l\u00ed Pitago) <br\/> $\\Rightarrow A{{B}^{2}}={{\\left( 5\\sqrt{3} \\right)}^{2}}+{{5}^{2}}=100$ <br\/> $\\Rightarrow AB=10\\,\\left( m \\right)$ <br\/> X\u00e9t $\\Delta ABC$ vu\u00f4ng t\u1ea1i $A$, \u0111\u01b0\u1eddng cao $AH$ c\u00f3: <br\/> $A{{B}^{2}}=BH.BC$ (h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\Rightarrow BC=\\dfrac{A{{B}^{2}}}{BH}=\\dfrac{100}{5}=20\\,\\left( m \\right)$ <br\/> $\\Rightarrow $ B\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n l\u00e0 $10m$ <br\/> X\u00e9t $\\Delta OAH$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: <br\/> $\\sin \\widehat{AOH}=\\dfrac{AH}{AO}=\\dfrac{5\\sqrt{3}}{10}=\\dfrac{\\sqrt{3}}{2}$ <br\/> $\\Rightarrow \\widehat{AOH}={{60}^{o}}\\Rightarrow \\widehat{AOD}=2.\\widehat{AOH}={{2.60}^{o}}={{120}^{o}}$ <br\/> Di\u1ec7n t\u00edch qu\u1ea1t tr\u00f2n b\u00e1n k\u00ednh $10m$ cung ${{120}^{o}}$ l\u00e0: <br\/> ${{S}_{\\text{qu\u1ea1t AOD}}}=\\dfrac{\\pi {{R}^{2}}n}{360}=\\dfrac{\\pi {{.10}^{2}}.120}{360}=\\dfrac{100\\pi }{3}\\,\\left( {{m}^{2}} \\right)$ <br\/> Di\u1ec7n t\u00edch tam gi\u00e1c $AOD$ l\u00e0: ${{S}_{\\Delta AOD}}=\\dfrac{1}{2}OH.AD=\\dfrac{1}{2}.5.10\\sqrt{3}=25\\sqrt{3}\\,\\left( {{m}^{2}} \\right)$ <br\/> Di\u1ec7n t\u00edch m\u1eb7t c\u1eaft vu\u00f4ng g\u00f3c c\u1ee7a \u0111\u01b0\u1eddng h\u1ea7m l\u00e0: $S={{S}_{\\text{qu\u1ea1t AOD}}}-{{S}_{\\Delta AOD}}=\\dfrac{100\\pi }{3}-25\\sqrt{3}\\,\\left( {{m}^{2}} \\right)$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C <\/span><\/span>","column":2}]}],"id_ques":1611},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["625"],["8"]]],"list":[{"point":10,"width":50,"ques":"<span class='basic_left'> M\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n v\u00e0 m\u1ed9t n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n ti\u1ebfp x\u00fac nhau v\u00e0 ti\u1ebfp x\u00fac v\u1edbi hai \u0111\u01b0\u1eddng th\u1eb3ng song song c\u00e1ch nhau $10cm$, nh\u01b0 h\u00ecnh v\u1ebd d\u01b0\u1edbi \u0111\u00e2y (\u0111\u01b0\u1eddng n\u1ed1i t\u00e2m vu\u00f4ng g\u00f3c v\u1edbi hai \u0111\u01b0\u1eddng th\u1eb3ng song song). H\u00e3y t\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a t\u00edch c\u00e1c di\u1ec7n t\u00edch h\u00ecnh tr\u00f2n v\u00e0 n\u1eeda h\u00ecnh tr\u00f2n. <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai19/lv3/img\/h937_K2.png' \/><\/center> <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> Gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a t\u00edch c\u00e1c di\u1ec7n t\u00edch l\u00e0: $\\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\pi^2}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$ $(cm^2)$","hint":"S\u1eed d\u1ee5ng B\u1ea5t \u0111\u1eb3ng th\u1ee9c C\u00f4-si \u0111\u1ec3 t\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a t\u00edch c\u00e1c di\u1ec7n t\u00edch","explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai19/lv3/img\/h937_K2.png' \/><\/center> <br\/> G\u1ecdi b\u00e1n k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n l\u00e0 $x$ th\u00ec b\u00e1n k\u00ednh c\u1ee7a n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n l\u00e0 $10 \u2013 2x$ ($0 < x < 5$) <br\/> Di\u1ec7n t\u00edch h\u00ecnh tr\u00f2n l\u00e0: $\\pi {{x}^{2}}\\,\\left( c{{m}^{2}} \\right)$ <br\/> Di\u1ec7n t\u00edch n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n l\u00e0: $\\dfrac{\\pi {{\\left( 10-2x \\right)}^{2}}}{2}\\,\\left( c{{m}^{2}} \\right)$ <br\/> T\u00edch c\u00e1c di\u1ec7n t\u00edch l\u00e0: $\\dfrac{{{\\left[ \\pi .x.\\left( 10-2x \\right) \\right]}^{2}}}{2}=2{{\\pi }^{2}}.{{\\left[ x.\\left( 5-x \\right) \\right]}^{2}}$ <br\/> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c C\u00f4 \u2013 si ta c\u00f3: $\\sqrt{x.\\left( 5-x \\right)}\\le \\dfrac{x+5-x}{2}=\\dfrac{5}{2}$ <br\/> $\\Rightarrow x\\left( 5-x \\right)\\le \\dfrac{25}{4}$ <br\/> $\\Rightarrow 2{{\\pi }^{2}}.{{\\left[ x.\\left( 5-x \\right) \\right]}^{2}}\\le 2{{\\pi }^{2}}.{{\\left( \\dfrac{25}{4} \\right)}^{2}}=\\dfrac{625{{\\pi }^{2}}}{8}$ <br\/> Suy ra gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a t\u00edch c\u00e1c di\u1ec7n t\u00edch b\u1eb1ng $\\dfrac{625{{\\pi }^{2}}}{8}$ <br\/> D\u1ea5u \u201c=\u201d x\u1ea3y ra khi $x=5-x\\Leftrightarrow x=\\dfrac{5}{2}\\,\\left( cm \\right)$ <br\/> <span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $625$ v\u00e0 $8$ <\/span><\/span><\/span> "}]}],"id_ques":1612},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"Cho hai \u0111\u01b0\u1eddng tr\u00f2n $\\left( O;\\,4cm \\right)$ v\u00e0 $\\left( O';\\,4cm \\right)$ c\u1eaft nhau \u1edf $A$ v\u00e0 $B$. T\u00ednh di\u1ec7n t\u00edch ph\u1ea7n chung c\u1ee7a hai h\u00ecnh tr\u00f2n $(O)$ v\u00e0 $(O\u2019)$, bi\u1ebft $OO'=4cm$. ","select":["A. $\\dfrac{32\\pi }{3}\\,\\left( c{{m}^{2}} \\right)$","B. $\\dfrac{32\\pi -24\\sqrt{3}}{3}\\,\\left( c{{m}^{2}} \\right)$ ","C. $\\dfrac{16\\pi }{3}-8\\sqrt{3}\\,\\left( c{{m}^{2}} \\right)$","D. $\\dfrac{16\\pi -24\\sqrt{3}}{3}\\,\\left( c{{m}^{2}} \\right)$"],"explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai19/lv3/img\/h937_K3.png' \/><\/center> <br\/> G\u1ecdi $S$ l\u00e0 di\u1ec7n t\u00edch c\u1ea7n t\u00ecm th\u00ec: $S = S_{\\text{qu\u1ea1t AOB}}+S_{\\text{qu\u1ea1t AO'B}}-{S}_{OAO'B}$ <br\/> D\u1ec5 th\u1ea5y c\u00e1c tam gi\u00e1c $OAO\u2019; OBO\u2019$ \u0111\u1ec1u <br\/> $\\Rightarrow \\widehat{AOO'}=\\widehat{AO'O}=\\widehat{BOO'}=\\widehat{BO'O}={{60}^{o}}$ <br\/> $\\Rightarrow \\widehat{AOB}=\\widehat{AO'B}={{120}^{o}}$ <br\/> ${{S}_{\\text{qu\u1ea1t} AOB}}=\\dfrac{\\pi .O{{A}^{2}}.n}{360}=\\dfrac{\\pi {{.4}^{2}}.120}{360}=\\dfrac{16\\pi }{3}\\,\\left( c{{m}^{2}} \\right)$ <br\/> ${{S}_{\\text{qu\u1ea1t} AO'B}}=\\dfrac{\\pi .O'{{A}^{2}}.n}{360}=\\dfrac{\\pi {{.4}^{2}}.120}{360}=\\dfrac{16\\pi }{3}\\,\\left( c{{m}^{2}} \\right)$ <br\/> Ta c\u00f3: $OA=O'A=OB=O'B=4cm$ <br\/> $\\Rightarrow OAO'B$ l\u00e0 h\u00ecnh thoi <br\/> $\\Rightarrow \\left\\{ \\begin{align} & OO'\\bot AB \\\\ & OH=\\dfrac{1}{2}OO'=2\\left( cm \\right) \\\\ \\end{align} \\right.$ (t\u00ednh ch\u1ea5t h\u00ecnh thoi) <br\/> X\u00e9t $\\Delta OHA$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: <br\/> $AH=\\sqrt{O{{A}^{2}}-O{{H}^{2}}}=\\sqrt{{{4}^{2}}-{{2}^{2}}}=2\\sqrt{3}\\,\\left( cm \\right)$ <br\/> ${{S}_{OAO'B}}=\\dfrac{1}{2}AB.OO'=OO'.AH=4.2\\sqrt{3}=8\\sqrt{3}\\,\\left( c{{m}^{2}} \\right)$ <br\/> $\\Rightarrow S=\\dfrac{16\\pi }{3}+\\dfrac{16\\pi }{3}-8\\sqrt{3}=\\dfrac{32\\pi -24\\sqrt{3}}{3}\\,\\left( c{{m}^{2}} \\right)$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B <\/span><\/span>","column":2}]}],"id_ques":1613},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$. V\u1ebd n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp $\\Delta ABC$ r\u1ed3i v\u1ebd c\u00e1c n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh $AB$, $AC$ n\u1eb1m ngo\u00e0i tam gi\u00e1c \u0111\u00f3. <br\/> Ch\u1ee9ng minh r\u1eb1ng di\u1ec7n t\u00edch h\u00ecnh gi\u1edbi h\u1ea1n b\u1edfi ba n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n n\u00f3i tr\u00ean b\u1eb1ng di\u1ec7n t\u00edch tam gi\u00e1c $ABC$.","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[5],[6],[4],[2],[1],[3]]],"list":[{"point":10,"left":["$=\\dfrac{\\pi \\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \\right)}{8}+{{S}_{\\Delta ABC}}$","$={{S}_{\\Delta ABC}}\\,\\left( do\\,{{b}^{2}}+{{c}^{2}}={{a}^{2}}\\Rightarrow {{b}^{2}}+{{c}^{2}}-{{a}^{2}}=0 \\right)$ "," $=\\dfrac{\\pi {{c}^{2}}}{8}+\\dfrac{\\pi {{b}^{2}}}{8}-\\dfrac{\\pi {{a}^{2}}}{8}+{{S}_{\\Delta ABC}}$ ","Khi \u0111\u00f3: $S={{S}_{2}}+{{S}_{3}}-{{S}_{1}}+{{S}_{\\Delta ABC}}$"," G\u1ecdi $S$ l\u00e0 di\u1ec7n t\u00edch \u0111\u01b0\u1ee3c gi\u1edbi h\u1ea1n b\u1edfi ba \u0111\u01b0\u1eddng tr\u00f2n ${{S}_{1}};{{S}_{2}};{{S}_{3}}$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 di\u1ec7n t\u00edch c\u1ee7a c\u00e1c n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh $BC=a;AB=c;AC=b$"," $=\\dfrac{\\pi {{\\left( \\dfrac{c}{2} \\right)}^{2}}}{2}+\\dfrac{\\pi {{\\left( \\dfrac{b}{2} \\right)}^{2}}}{2}-\\dfrac{\\pi {{\\left( \\dfrac{a}{2} \\right)}^{2}}}{2}+{{S}_{\\Delta ABC}}$"],"top":100,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai19/lv3/img\/h937_K4.png' \/><\/center> <br\/> G\u1ecdi $S$ l\u00e0 di\u1ec7n t\u00edch \u0111\u01b0\u1ee3c gi\u1edbi h\u1ea1n b\u1edfi ba \u0111\u01b0\u1eddng tr\u00f2n <br\/> ${{S}_{1}};{{S}_{2}};{{S}_{3}}$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 di\u1ec7n t\u00edch c\u1ee7a c\u00e1c n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh $BC=a;AB=c;AC=b$ <br\/> Khi \u0111\u00f3: $S={{S}_{2}}+{{S}_{3}}-{{S}_{1}}+{{S}_{\\Delta ABC}}$ <br\/> $=\\dfrac{\\pi {{\\left( \\dfrac{c}{2} \\right)}^{2}}}{2}+\\dfrac{\\pi {{\\left( \\dfrac{b}{2} \\right)}^{2}}}{2}-\\dfrac{\\pi {{\\left( \\dfrac{a}{2} \\right)}^{2}}}{2}+{{S}_{\\Delta ABC}}$ <br\/> $=\\dfrac{\\pi {{c}^{2}}}{8}+\\dfrac{\\pi {{b}^{2}}}{8}-\\dfrac{\\pi {{a}^{2}}}{8}+{{S}_{\\Delta ABC}}$ <br\/> $=\\dfrac{\\pi \\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \\right)}{8}+{{S}_{\\Delta ABC}}$ <br\/> $={{S}_{\\Delta ABC}}\\,\\left( do\\,{{b}^{2}}+{{c}^{2}}={{a}^{2}}\\Rightarrow {{b}^{2}}+{{c}^{2}}-{{a}^{2}}=0 \\right)$ <\/span>"}]}],"id_ques":1614},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["9"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" <span class='basic_left'> Cho n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n $(O)$ \u0111\u01b0\u1eddng k\u00ednh $AB$. G\u1ecdi $M$ l\u00e0 m\u1ed9t \u0111i\u1ec3m tr\u00ean n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n, k\u1ebb $MH\\bot AB$sao cho $MH = 6cm; BH= 4cm$. \u1ede ph\u00eda trong c\u1ee7a n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n $(O)$ v\u1ebd c\u00e1c n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m $I$ \u0111\u01b0\u1eddng k\u00ednh $AH$, n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m $K$ \u0111\u01b0\u1eddng k\u00ednh $BH$. Di\u1ec7n t\u00edch ph\u1ea7n gi\u1edbi h\u1ea1n b\u1edfi ba n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n l\u00e0: _input_ $\\pi (cm^2)$ ","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai19/lv3/img\/h937_K5.png' \/><\/center> <br\/> G\u1ecdi $S$ l\u00e0 di\u1ec7n t\u00edch c\u1ea7n t\u00ecm. Khi \u0111\u00f3: $S=\\dfrac{1}{2}\\left( {{S}_{\\left( O \\right)}}-{{S}_{\\left( I \\right)}}-{{S}_{\\left( K \\right)}} \\right)$ <br\/> Ta c\u00f3: $\\widehat{AMB}={{90}^{o}}$ (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n) <br\/> X\u00e9t $\\Delta AMB$ vu\u00f4ng t\u1ea1i $M$ c\u00f3 $MH\\bot AB$ n\u00ean: <br\/> $M{{H}^{2}}=BH.AH$ (h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\Rightarrow AH=\\dfrac{M{{H}^{2}}}{BH}=\\dfrac{{{6}^{2}}}{4}=9\\,\\left( cm \\right)$ <br\/> Do \u0111\u00f3 $AB=AH+HB=9+4=13\\left( cm \\right)$ <br\/> ${{S}_{\\left( O \\right)}}=\\pi .{{\\left( \\dfrac{AB}{2} \\right)}^{2}}=\\pi .{{\\left( \\dfrac{13}{2} \\right)}^{2}}=\\dfrac{169\\pi }{4}\\,\\left( c{{m}^{2}} \\right)$ <br\/> ${{S}_{\\left( I \\right)}}=\\pi .{{\\left( \\dfrac{AH}{2} \\right)}^{2}}=\\pi .{{\\left( \\dfrac{9}{2} \\right)}^{2}}=\\dfrac{81\\pi }{4}\\,\\left( c{{m}^{2}} \\right)$ <br\/> ${{S}_{\\left( K \\right)}}=\\pi .{{\\left( \\dfrac{BH}{2} \\right)}^{2}}=\\pi .{{\\left( \\dfrac{4}{2} \\right)}^{2}}=4\\pi \\,\\left( c{{m}^{2}} \\right)$ <br\/> $\\Rightarrow S=\\dfrac{1}{2}\\left( \\dfrac{169\\pi }{4}-\\dfrac{81\\pi }{4}-4\\pi \\right)=9\\pi \\,\\left( c{{m}^{2}} \\right)$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $9$ <\/span><\/span> "}]}],"id_ques":1615},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["12"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<span class='basic_left'> Cho tam gi\u00e1c c\u00e2n $ABC$, $\\widehat{A}={{120}^{o}},\\,AB=AC=4cm.$ Qua $C$ k\u1ebb $CH\\bot AB$ t\u1ea1i $H$. V\u1ebd \u0111\u01b0\u1eddng tr\u00f2n $\\left( A;AH \\right)$ v\u00e0 \u0111\u01b0\u1eddng tr\u00f2n $\\left( A;AB \\right)$. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh v\u00e0nh kh\u0103n n\u1eb1m gi\u1eefa hai \u0111\u01b0\u1eddng tr\u00f2n tr\u00ean. <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> Di\u1ec7n t\u00edch c\u1ea7n t\u00ecm l\u00e0 _input_$\\pi \\, (cm^2)$","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai19/lv3/img\/h937_K6.png' \/><\/center> <br\/> G\u1ecdi $S$ l\u00e0 di\u1ec7n t\u00edch c\u1ea7n t\u00ecm. Khi \u0111\u00f3 $S={{S}_{\\left( A;AB \\right)}}-{{S}_{\\left( A;AH \\right)}}$ <br\/> Ta c\u00f3: $\\widehat{HAC}+\\widehat{BAC}={{180}^{o}}$ (hai g\u00f3c k\u1ec1 b\u00f9) <br\/> $\\Rightarrow \\widehat{HAC}={{180}^{o}}-\\widehat{BAC}={{180}^{o}}-{{120}^{o}}={{60}^{o}}$ <br\/> X\u00e9t $\\Delta HAC$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: <br\/> $AH=AC.cos\\widehat{HAC}=4.cos{{60}^{o}}=2\\left( cm \\right)$ <br\/> ${{S}_{\\left( A;AB \\right)}}=\\pi .A{{B}^{2}}=\\pi {{.4}^{2}}=16\\pi \\,\\left( c{{m}^{2}} \\right)$ <br\/> ${{S}_{\\left( A;AH \\right)}}=\\pi .A{{H}^{2}}=\\pi {{.2}^{2}}=4\\pi \\,\\left( c{{m}^{2}} \\right)$ <br\/> $\\Rightarrow S=16\\pi -4\\pi =12\\pi \\,\\left( c{{m}^{2}} \\right)$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $12$ <\/span><\/span> "}]}],"id_ques":1616},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["24"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh $AD = 12 cm$. Tr\u00ean $AD$ l\u1ea5y hai \u0111i\u1ec3m $B$ v\u00e0 $C$ sao cho $AB=BC=CD$. V\u1ebd c\u00e1c n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh $AB, AC, BD, CD$ nh\u01b0 h\u00ecnh v\u1ebd sau. <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai19/lv3/img\/h937_K7.png' \/><\/center> <br\/> Di\u1ec7n t\u00edch ph\u1ea7n kh\u00f4ng in \u0111\u1eadm l\u00e0 _input_$\\pi \\, (cm^2)$","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai19/lv3/img\/h937_K7.png' \/><\/center> <br\/> Do $AD=12cm\\Rightarrow AB=BC=CD=4cm$ <br\/> Di\u1ec7n t\u00edch n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh $CD$ l\u00e0: <br\/> ${{S}_{1}}=\\dfrac{1}{2}\\pi {{\\left( \\dfrac{CD}{2} \\right)}^{2}}=\\dfrac{1}{2}\\pi {{.2}^{2}}=2\\pi \\,\\left( c{{m}^{2}} \\right)$ <br\/> Di\u1ec7n t\u00edch n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh $BD$ l\u00e0: <br\/> ${{S}_{2}}=\\dfrac{1}{2}\\pi {{\\left( \\dfrac{BD}{2} \\right)}^{2}}=\\dfrac{1}{2}\\pi {{.4}^{2}}=8\\pi \\,\\left( c{{m}^{2}} \\right)$ <br\/> Di\u1ec7n t\u00edch b\u1ecb gi\u1edbi h\u1ea1n b\u1edfi 2 n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh $CD$ v\u00e0 \u0111\u01b0\u1eddng k\u00ednh $BD$ l\u00e0: <br\/> ${{S}_{3}}={{S}_{2}}-{{S}_{1}}=8\\pi -2\\pi =6\\pi \\,\\left( c{{m}^{2}} \\right)$ <br\/> Di\u1ec7n t\u00edch \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh $AD$ l\u00e0: <br\/> ${{S}_{4}}=\\pi {{\\left( \\dfrac{AD}{2} \\right)}^{2}}=\\pi {{.6}^{2}}=36\\pi \\,\\left( c{{m}^{2}} \\right)$ <br\/> Di\u1ec7n t\u00edch c\u1ea7n t\u00ecm l\u00e0: $S={{S}_{4}}-2{{S}_{3}}=36\\pi -2.6\\pi =24\\pi \\,\\left( c{{m}^{2}} \\right)$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $24$ <\/span><\/span> "}]}],"id_ques":1617},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O; 5cm)$ v\u00e0 \u0111i\u1ec3m $M$ n\u1eb1m b\u00ean ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n, bi\u1ebft $OM=10cm$. Qua $M$ v\u1ebd hai ti\u1ebfp tuy\u1ebfn $MA$ v\u00e0 $MB$ ($A,B$ l\u00e0 c\u00e1c ti\u1ebfp \u0111i\u1ec3m). T\u00ednh di\u1ec7n t\u00edch ph\u1ea7n t\u1ee9 gi\u00e1c $AMBO$ n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n. ","select":["A. $25-\\dfrac{25\\pi }{3}\\,\\left( c{{m}^{2}} \\right)$","B. $25\\sqrt{2}-\\dfrac{25\\pi }{3}\\,\\left( c{{m}^{2}} \\right)$ ","C. $25\\sqrt{3}-\\dfrac{25}{3}\\,\\left( c{{m}^{2}} \\right)$","D. $25\\sqrt{3}-\\dfrac{25\\pi }{3}\\,\\left( c{{m}^{2}} \\right)$"],"explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai19/lv3/img\/h937_K8.png' \/><\/center> <br\/> G\u1ecdi $H$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AB$ v\u00e0 $OM$ <br\/> Ta c\u00f3: $OA=OB=5cm$ (gi\u1ea3 thi\u1ebft) <br\/> $MA=MB$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> $\\Rightarrow OM$ l\u00e0 trung tr\u1ef1c c\u1ee7a $AB$ <br\/> $\\Rightarrow OM\\bot AB$ <br\/> X\u00e9t $\\Delta OAM$ vu\u00f4ng t\u1ea1i $A$ c\u00f3: <br\/> $\\cos \\widehat{AOM}=\\dfrac{OA}{OM}=\\dfrac{5}{10}=\\dfrac{1}{2}$ <br\/> $\\Rightarrow \\widehat{AOM}={{60}^{o}}$ <br\/> $\\Rightarrow \\widehat{AOB}=2\\widehat{AOM}={{2.60}^{o}}={{120}^{o}}$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> X\u00e9t $\\Delta AHO$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: <br\/> $AH=AO.\\sin \\widehat{AOH}=5.sin{{60}^{o}}=\\dfrac{5\\sqrt{3}}{2}\\,\\left( cm \\right)$ <br\/> G\u1ecdi $S$ l\u00e0 di\u1ec7n t\u00edch c\u1ea7n t\u00ecm, khi \u0111\u00f3: <br\/> $S={{S}_{AMBO}}-{{S}_{\\text{qu\u1ea1t} AOB}}\\\\ = \\dfrac{1}{2}AB.OM-\\dfrac{\\pi .O{{A}^{2}}.n}{360}$ <br\/> $=AH.OM-\\dfrac{\\pi {{.5}^{2}}.120}{360}=\\dfrac{5\\sqrt{3}}{2}.10-\\dfrac{25\\pi }{3}\\\\=25\\sqrt{3}-\\dfrac{25\\pi }{3}\\,\\left( c{{m}^{2}} \\right)$<br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D<\/span><\/span>","column":2}]}],"id_ques":1618},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho h\u00ecnh thang $ABCD,\\widehat{A}=\\widehat{B}={{90}^{o}}.$ M\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n $(O)$ n\u1ed9i ti\u1ebfp h\u00ecnh thang, ti\u1ebfp x\u00fac v\u1edbi \u0111\u00e1y nh\u1ecf $BC$ t\u1ea1i $M$. Ti\u1ebfp x\u00fac v\u1edbi \u0111\u00e1y l\u1edbn $AD$ t\u1ea1i $N$. Bi\u1ebft $BC=\\dfrac{3}{2}r$ ($r$ l\u00e0 b\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n $(O)$). T\u00ednh di\u1ec7n t\u00edch ph\u1ea7n h\u00ecnh thang $ABCD$ n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n $(O)$.","select":["A. $\\left( \\dfrac{9}{2}+\\pi \\right){{r}^{2}}$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)","B. $\\left( 9-\\pi \\right){{r}^{2}}$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch) ","C. $\\left( \\dfrac{9}{2}-\\pi \\right){{r}^{2}}$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)","D. $\\left( \\dfrac{3}{2}-\\pi \\right){{r}^{2}}$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)"],"explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai19/lv3/img\/h937_K9.png' \/><\/center><br\/> G\u1ecdi $E$ l\u00e0 ti\u1ebfp \u0111i\u1ec3m c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n v\u00e0 c\u1ea1nh $AB$ <br\/> D\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c $EBMO; EANO$ l\u00e0 h\u00ecnh vu\u00f4ng <br\/> $\\Rightarrow BM=AN=OE=EB=EA=r$ <br\/> $\\Rightarrow AB=2r;MC=BC-BM=\\dfrac{r}{2}$ <br\/> Ta c\u00f3: $\\widehat{BCD}+\\widehat{CDA}={{180}^{o}}$ (t\u1ed5ng hai g\u00f3c trong c\u00f9ng ph\u00eda) <br\/> $\\widehat{OCD}+\\widehat{ODC}=\\dfrac{1}{2}\\left( \\widehat{BCD}+\\widehat{CDA} \\right)$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> $=\\dfrac{1}{2}{{.180}^{o}}={{90}^{o}}$ <br\/> $\\Rightarrow \\widehat{COD}={{90}^{o}}$ <br\/> X\u00e9t $\\Delta OMC$ v\u00e0 $\\Delta DNO$ c\u00f3: <br\/> $\\widehat{OMC}=\\widehat{DNO}={{90}^{o}}$ <br\/> $\\widehat{ODN}=\\widehat{COM}\\,\\left( \\text{c\u00f9ng ph\u1ee5 v\u1edbi}\\,\\widehat{NOD} \\right)$ <br\/> $\\Rightarrow \\Delta OMC\\backsim \\Delta DNO\\,\\left( g.g \\right)$ <br\/> $\\Rightarrow \\dfrac{OM}{ND}=\\dfrac{MC}{NO}$ (c\u1eb7p t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng) <br\/> $\\Rightarrow ND=\\dfrac{OM.ON}{MC}=\\dfrac{r.r}{\\dfrac{r}{2}}=2r$ <br\/> $\\Rightarrow AD=AN+ND=r+2r=3r$ <br\/> G\u1ecdi $S$ l\u00e0 di\u1ec7n t\u00edch c\u1ea7n t\u00ecm, khi \u0111\u00f3: <br\/> $S={{S}_{ABCD}}-{{S}_{\\left( O \\right)}}\\\\=\\dfrac{1}{2}.AB.\\left( BC+AD \\right)-\\pi {{r}^{2}}\\\\=\\dfrac{1}{2}.2r.\\left( \\dfrac{3}{2}r+3r \\right)-\\pi {{r}^{2}}\\\\=\\left( \\dfrac{9}{2}-\\pi \\right){{r}^{2}}\\,\\text{(\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C <\/span><\/span>","column":2}]}],"id_ques":1619},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho h\u00ecnh v\u1ebd, <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai19/lv3/img\/h937_K10.png' \/><\/center> <br\/> Bi\u1ebft di\u1ec7n t\u00edch mi\u1ec1n t\u00f4 \u0111en l\u00e0 $86c{{m}^{2}}.$ Di\u1ec7n t\u00edch h\u00ecnh tr\u00f2n l\u00e0: ","select":["A. $100\\pi \\,\\left( c{{m}^{2}} \\right)$","B. $12\\pi \\,\\left( c{{m}^{2}} \\right)$ ","C. $50\\pi \\,\\left( c{{m}^{2}} \\right)$","D. $75\\pi \\,\\left( c{{m}^{2}} \\right)$"],"explain":" <span class='basic_left'> G\u1ecdi \u0111\u1ed9 d\u00e0i c\u1ea1nh h\u00ecnh vu\u00f4ng l\u00e0 $2a\\,(cm)$ <br\/> $\\Rightarrow $ B\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n l\u00e0 $a\\,\\left( cm \\right)$ <br\/> Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng l\u00e0: <br\/> ${{S}_{hv}}={{\\left( 2a \\right)}^{2}}=4{{a}^{2}}\\,\\left( c{{m}^{2}} \\right)$ <br\/> Di\u1ec7n t\u00edch h\u00ecnh tr\u00f2n l\u00e0: <br\/> ${{S}_{ht}}=\\pi {{a}^{2}}\\,\\left( c{{m}^{2}} \\right)$ <br\/> Khi \u0111\u00f3: <br\/> ${{S}_{hv}}-{{S}_{ht}}= 86\\\\ \\Leftrightarrow 4{{a}^{2}}-\\pi {{a}^{2}}= 86 \\\\ \\Rightarrow {{a}^{2}}=\\dfrac{86}{4-\\pi }\\approx 100$ <br\/> $\\Rightarrow {{S}_{ht}}=100\\pi \\,\\left( c{{m}^{2}} \\right)$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A <\/span><\/span>","column":4}]}],"id_ques":1620}],"lesson":{"save":0,"level":3}}