{"segment":[{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<span class='basic_left'> T\u00e2m $O$ c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp tam gi\u00e1c $ABC$ n\u1eb1m tr\u00ean \u0111\u01b0\u1eddng n\u00e0o? <\/span>","select":["A. \u0110\u01b0\u1eddng ph\u00e2n gi\u00e1c g\u00f3c $A$","B. \u0110\u01b0\u1eddng ph\u00e2n gi\u00e1c g\u00f3c ngo\u00e0i t\u1ea1i $B$","C. \u0110\u01b0\u1eddng ph\u00e2n gi\u00e1c g\u00f3c ngo\u00e0i t\u1ea1i $C$"],"explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv1/img\/h924_D2.2.png' \/><\/center>T\u00e2m $O$ c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp tam gi\u00e1c l\u00e0 giao \u0111i\u1ec3m c\u1ee7a ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong<br\/> V\u1eady t\u00e2m $O$ thu\u1ed9c \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c g\u00f3c $A$<br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 A<\/span><\/span>","column":1}]}],"id_ques":1181},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<span class='basic_left'>\u0110\u01b0\u1eddng tr\u00f2n ti\u1ebfp x\u00fac v\u1edbi ba c\u1ea1nh c\u1ee7a m\u1ed9t tam gi\u00e1c \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0:<\/span>","select":["A. \u0110\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c","B. \u0110\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp tam gi\u00e1c","C. \u0110\u01b0\u1eddng tr\u00f2n ti\u1ebfp x\u00fac v\u1edbi tam gi\u00e1c "],"explain":"\u0110\u01b0\u1eddng tr\u00f2n ti\u1ebfp x\u00fac v\u1edbi ba c\u1ea1nh c\u1ee7a m\u1ed9t tam gi\u00e1c \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp tam gi\u00e1c (\u0111\u1ecbnh ngh\u0129a) <br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 B<\/span>","column":1}]}],"id_ques":1182},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<span class='basic_left'>T\u00e2m $O$ c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n b\u00e0ng ti\u1ebfp trong g\u00f3c $A$ c\u1ee7a tam gi\u00e1c $ABC$ n\u1eb1m tr\u00ean \u0111\u01b0\u1eddng<\/span>","select":["A. Ph\u00e2n gi\u00e1c g\u00f3c $A$","B. Ph\u00e2n gi\u00e1c g\u00f3c $B$","C. Ph\u00e2n gi\u00e1c g\u00f3c ngo\u00e0i t\u1ea1i $A$ ","D. Ph\u00e2n gi\u00e1c g\u00f3c $C$"],"hint":"\u0110\u01b0\u1eddng tr\u00f2n ti\u1ebfp x\u00fac v\u1edbi m\u1ed9t c\u1ea1nh c\u1ee7a tam gi\u00e1c v\u00e0 ti\u1ebfp x\u00fac v\u1edbi ph\u1ea7n k\u00e9o d\u00e0i c\u1ee7a hai c\u1ea1nh kia g\u1ecdi l\u00e0 \u0111\u01b0\u1eddng tr\u00f2n b\u00e0ng ti\u1ebfp","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv1/img\/h924_D1.png' \/><\/center><span class='basic_left'>T\u00e2m $O$ c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n b\u00e0ng ti\u1ebfp trong g\u00f3c $A$ c\u1ee7a tam gi\u00e1c $ABC$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c g\u00f3c $A$ v\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c g\u00f3c ngo\u00e0i t\u1ea1i $B$ (ho\u1eb7c $C)$. V\u1eady $O$ thu\u1ed9c \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c g\u00f3c $A$<br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 A<\/span><\/span>","column":2}]}],"id_ques":1183},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["t","f","t"]],"list":[{"point":5,"ques":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv1/img\/h924_D12.png' \/><\/center> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O)$ n\u1ed9i ti\u1ebfp tam gi\u00e1c $ABC$ v\u00e0 ti\u1ebfp x\u00fac v\u1edbi c\u00e1c c\u1ea1nh $AB, BC, AC$ l\u1ea7n l\u01b0\u1ee3t t\u1ea1i $D, E, F$. C\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00fang hay sai? <\/span>","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["$AB, AC, BC$ l\u00e0 c\u00e1c ti\u1ebfp tuy\u1ebfn c\u1ee7a $(O)$ ","$AB = AF + EC$","$\\Delta {ADO} = \\Delta{AFO}$"],"explain":["<span class='basic_left'> \u0110\u00fang, v\u00ec $AB, AC, BC$ ti\u1ebfp x\u00fac v\u1edbi $(O)$ v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi b\u00e1n k\u00ednh n\u00ean l\u00e0 c\u00e1c ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n<br\/><\/span>","<br\/> <span class='basic_left'> Sai, v\u00ec $AD = AF; BE = BD$ (t\u00ednh ch\u1ea5t) <br\/> $\\Rightarrow AB = AD + DB$ $ = AF+ BD $ <br\/><\/span> "," <br\/> <span class='basic_left'> \u0110\u00fang, v\u00ec ta c\u00f3: <br\/> $AD = AF$ <br\/> $\\widehat{AFO}=\\widehat{ADO}=90^o$ <br\/> $AO$ chung <br\/>$\\Rightarrow \\Delta{ADO}=\\Delta{AFO}$ (c.g.c)<\/span>"]}]}],"id_ques":1184},{"time":24,"part":[{"time":3,"title":"N\u1ed1i t\u1eeb ho\u1eb7c c\u1ee5m t\u1eeb \u1edf c\u1ed9t b\u00ean tr\u00e1i v\u1edbi c\u1ed9t b\u00ean ph\u1ea3i \u0111\u1ec3 \u0111\u01b0\u1ee3c c\u00e2u ho\u00e0n ch\u1ec9nh","title_trans":"","audio":"","temp":"matching","correct":[["2","3","1"]],"list":[{"point":5,"image":"","left":["T\u00e2m c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp tam gi\u00e1c","T\u00e2m c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c","T\u00e2m c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n b\u00e0ng ti\u1ebfp trong g\u00f3c $B$ c\u1ee7a $\\Delta ABC$"],"right":["L\u00e0 giao \u0111i\u1ec3m c\u1ee7a \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c g\u00f3c $B$ v\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c g\u00f3c ngo\u00e0i t\u1ea1i $C$ ","L\u00e0 giao \u0111i\u1ec3m c\u1ee7a ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong tam gi\u00e1c","L\u00e0 giao \u0111i\u1ec3m c\u1ee7a ba \u0111\u01b0\u1eddng trung tr\u1ef1c trong tam gi\u00e1c"],"top":100,"explain":"<span class='basic_left'>T\u00e2m c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp tam gi\u00e1c l\u00e0 giao \u0111i\u1ec3m c\u1ee7a ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong tam gi\u00e1c <br\/>T\u00e2m c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c l\u00e0 giao \u0111i\u1ec3m c\u1ee7a ba \u0111\u01b0\u1eddng trung tr\u1ef1c trong tam gi\u00e1c<br\/>T\u00e2m c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n b\u00e0ng ti\u1ebfp trong g\u00f3c $B$ c\u1ee7a $\\Delta ABC$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c g\u00f3c $B$ v\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c g\u00f3c ngo\u00e0i t\u1ea1i $C$<\/span>"}]}],"id_ques":1185},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<span class='basic_left'>Cho tam gi\u00e1c $ABC$ ngo\u1ea1i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O)$. C\u00e1c ti\u1ebfp \u0111i\u1ec3m tr\u00ean c\u1ea1nh $AB, BC, CA$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 $D, E, F$. <br\/> <b> C\u00e2u 1: <\/b> Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang?<\/span>","select":["A. $AB + AC - BC =2AD$","B. $AB + AC - BC =2DE$","C. $AB + AC - BC =2EF$ ","D. $AB + AC - BC =2FD$"],"explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv1/img\/h924_D12.png' \/><\/center><span class='basic_left'>Ta c\u00f3 $AB; BC; AC$ l\u00e0 c\u00e1c ti\u1ebfp tuy\u1ebfn c\u1ee7a $(O)$ <br\/> $\\Rightarrow\\left\\{ \\begin{align} & AE=AF \\\\ & BD=BF \\\\ & CE=CD \\\\ \\end{align} \\right.$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> $AB+AC-BC<br\/>$ <br\/> $=\\left( AD+DB \\right)$$+\\left( AF+FC \\right)$$-\\left( BE+EC \\right)$ <br\/> $=AD+DB+AD$$+FC-BD-FC$ <br\/> $=2AD$ <br\/> $=2AF$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A. <\/span><\/span>","column":2}]}],"id_ques":1186},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv1/img\/h924_D12.png' \/><\/center> <br\/> <b> C\u00e2u 1: <\/b> Cho tam gi\u00e1c $ABC$ ngo\u1ea1i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O)$. C\u00e1c ti\u1ebfp \u0111i\u1ec3m tr\u00ean c\u1ea1nh $AB, BC, CA$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 $D, E, F$. <br\/> <b> C\u00e2u 2: <\/b> Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang? <\/span>","select":["A. $2BE + CA - BA = BD$","B. $2BE + CA - BA = AB$ ","C. $2BE + CA - BA = AC$ ","D. $2BE + CA - BA = BC$ "],"explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv1/img\/h924_D12.png' \/><\/center><span class='basic_left'> Ta c\u00f3: $\\left\\{ \\begin{align} & AD=AF \\\\ & BD=BE \\\\ & CE=CF \\\\ \\end{align} \\right.$ (theo c\u00e2u 1) <br\/> X\u00e9t $2BE + CA - BA $ <br\/> $=BE+BD+CF$ $+FA-BD-DA$ <br\/> $=BE+BD+EC$ $+FA-BD-AF$ <br\/> $=BE+EC$ <br\/> $=BC$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D. <\/span><\/span>","column":2}]}],"id_ques":1187},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<span class='basic_left'>Cho tam gi\u00e1c $ABC$ ngo\u1ea1i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O)$. C\u00e1c ti\u1ebfp \u0111i\u1ec3m tr\u00ean c\u1ea1nh $AB, BC, CA$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 $D, E, F$. <br\/> <b> C\u00e2u 3: <\/b> Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang? <\/span>","select":["A. $AC+CB - 2CF = AF$ ","B. $AC+CB - 2CF = AB$ ","C. $AC+CB - 2CF = AE$ ","D. $AC+CB - 2CF = AC$ "],"explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv1/img\/h924_D12.png' \/><\/center><span class='basic_left'>Ta c\u00f3:$AD=AF;$ $BD=BE;$ $CE=CF$ (theo c\u00e2u 1) <br\/> X\u00e9t $AC+CB - 2CF$ <br\/> $=AF+FC+CE$$+EB - CF - CE$ <br\/> $=AF+EB$ <br\/> $=AD+DB$ <br\/> $=AB$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span>","column":2}]}],"id_ques":1188},{"time":24,"part":[{"title":"N\u1ebfu hai ti\u1ebfp tuy\u1ebfn c\u1ee7a m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n c\u1eaft nhau t\u1ea1i m\u1ed9t \u0111i\u1ec3m th\u00ec:","title_trans":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","temp":"true_false","correct":[["t","t","f"]],"list":[{"point":5,"image":"","ques":"","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["<span class='basic_left'>\u0110i\u1ec3m \u0111\u00f3 c\u00e1ch \u0111\u1ec1u hai ti\u1ebfp \u0111i\u1ec3m <\/span>","<span class='basic_left'>\u0110\u01b0\u1eddng th\u1eb3ng n\u1ed1i \u0111i\u1ec3m \u0111\u00f3 v\u1edbi t\u00e2m l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c t\u1ea1o b\u1edfi hai ti\u1ebfp tuy\u1ebfn<\/span>","<span class='basic_left'>\u0110i\u1ec3m \u0111\u00f3 c\u00f9ng v\u1edbi hai ti\u1ebfp \u0111i\u1ec3m lu\u00f4n t\u1ea1o th\u00e0nh m\u1ed9t tam gi\u00e1c \u0111\u1ec1u<\/span>"],"explain":["<span class='basic_left'> \u0110\u00fang, v\u00ec n\u1ebfu hai ti\u1ebfp tuy\u1ebfn c\u1ee7a m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n c\u1eaft nhau t\u1ea1i m\u1ed9t \u0111i\u1ec3m th\u00ec \u0111i\u1ec3m \u0111\u00f3 c\u00e1ch \u0111\u1ec1u hai ti\u1ebfp \u0111i\u1ec3m (t\u00ednh ch\u1ea5t)<\/span>"," <span class='basic_left'> \u0110\u00fang, v\u00ec n\u1ebfu hai ti\u1ebfp tuy\u1ebfn c\u1ee7a m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n c\u1eaft nhau t\u1ea1i m\u1ed9t \u0111i\u1ec3m th\u00ec \u0111\u01b0\u1eddng th\u1eb3ng n\u1ed1i \u0111i\u1ec3m \u0111\u00f3 v\u1edbi t\u00e2m l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c t\u1ea1o b\u1edfi hai ti\u1ebfp tuy\u1ebfn (t\u00ednh ch\u1ea5t)<\/span>","<span class='basic_left'> Sai, v\u00ec giao \u0111i\u1ec3m c\u1ee7a hai ti\u1ebfp tuy\u1ebfn v\u00e0 hai ti\u1ebfp \u0111i\u1ec3m ch\u1ec9 lu\u00f4n t\u1ea1o th\u00e0nh tam gi\u00e1c c\u00e2n<\/span>"]}]}],"id_ques":1189},{"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":[[["3"],["3"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<span class='basic_left'>Cho $\\Delta ABC$ \u0111\u1ec1u, ngo\u1ea1i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n b\u00e1n k\u00ednh $1 cm$. Di\u1ec7n t\u00edch tam gi\u00e1c $ABC$ b\u1eb1ng bao nhi\u00eau $cm^2$?<br\/><br\/> <b>\u0110\u00e1p \u00e1n: $S_{ABC}=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\sqrt{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}(cm^2)$ <\/b><\/span>","hint":"T\u00e2m \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp tam gi\u00e1c \u0111\u1ec1u tr\u00f9ng v\u1edbi tr\u1ecdng t\u00e2m tam gi\u00e1c","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv1/img\/h924_D3.png' \/><\/center><span class='basic_left'>G\u1ecdi $D$ l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp tam gi\u00e1c $ABC$ <br\/> $F, H, G$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 ch\u00e2n \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb c\u00e1c \u0111\u1ec9nh $A, B, C$ xu\u1ed1ng c\u00e1c c\u1ea1nh $BC;AC;AB$ <br\/> $\\Rightarrow D$ l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c $ABC$ (t\u00ednh ch\u1ea5t tam gi\u00e1c \u0111\u1ec1u) <br\/> $\\Rightarrow CG = 3DG = 3 \\,(cm)$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) <br\/> X\u00e9t $\\Delta CGA$ vu\u00f4ng t\u1ea1i $G$ c\u00f3:<br\/> $\\widehat{CAG} = 60^o$ (do $\\Delta ABC$ \u0111\u1ec1u) <br\/> $CG = AC.\\sin\\widehat{CAG}$ (h\u1ec7 th\u1ee9c v\u1ec1 c\u1ea1nh v\u00e0 g\u00f3c trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\Rightarrow AC = \\dfrac{CG}{\\sin\\widehat{CAG}} = \\dfrac{3}{\\dfrac{\\sqrt{3}}{2}} = 2\\sqrt{3} \\,(cm)$ <br\/> $\\Rightarrow AC = AB = 2\\sqrt{3} \\, (cm)$ ($\\Delta ABC$ \u0111\u1ec1u) <br\/> $S_{ABC}=\\dfrac{1}{2}CG.AB$ <br\/> $=\\dfrac{1}{2}. 3.2\\sqrt{3}$ <br\/> $=3\\sqrt{3}\\, (cm^2)$ <br\/> <span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $3\\sqrt{3}$<\/span><\/span>"}]}],"id_ques":1190},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"select":["A. $2\\sqrt{3}$","B. $5\\sqrt{3}$","C. $\\sqrt{3}$"],"ques":"<span class='basic_left'>Cho $\\Delta DEF$ \u0111\u1ec1u, c\u1ea1nh 6 cm. \u0110\u01b0\u1eddng tr\u00f2n $(I)$ n\u1ed9i ti\u1ebfp tam gi\u00e1c $DEF$. T\u00ednh b\u00e1n k\u00ednh $r$ c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n \u0111\u00f3. <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> $r = $ ?$(cm)$<\/span>","hint":"T\u00e2m \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp tam gi\u00e1c \u0111\u1ec1u tr\u00f9ng v\u1edbi tr\u1ecdng t\u00e2m tam gi\u00e1c","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv1/img\/h924_D4.png' \/><\/center><span class='basic_left'> <br\/> G\u1ecdi $I$ l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp tam gi\u00e1c $DEF$ <br\/> $H, M, N$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 ch\u00e2n \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb c\u00e1c \u0111\u1ec9nh $D, E, F$ xu\u1ed1ng c\u00e1c c\u1ea1nh $EF, DF, DE$ <br\/> $\\Rightarrow I$ l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c $DEF$ (t\u00ednh ch\u1ea5t tam gi\u00e1c \u0111\u1ec1u) <br\/> X\u00e9t $\\Delta DHE$ vu\u00f4ng t\u1ea1i $H$ c\u00f3:<br\/> $\\widehat{DEH} = 60^o$ (do $\\Delta DEF$ \u0111\u1ec1u) <br\/> $DH = DE.\\sin\\widehat{DEH}$ (h\u1ec7 th\u1ee9c v\u1ec1 c\u1ea1nh v\u00e0 g\u00f3c trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\Rightarrow DH = 6.\\dfrac{\\sqrt{3}}{2} = 3.\\sqrt{3} \\,(cm)$ <br\/> $\\Rightarrow r = IH =\\dfrac{DH}{3}$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m tam gi\u00e1c) <br\/> $=\\dfrac{3\\sqrt{3}}{3}=\\sqrt{3}\\, (cm)$<\/span>"}]}],"id_ques":1191},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["90"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'> Cho n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m $(O)$, \u0111\u01b0\u1eddng k\u00ednh $AB$. Tr\u00ean n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd $AB$, k\u1ebb $Ax, By$ vu\u00f4ng g\u00f3c v\u1edbi $AB$. Qua \u0111i\u1ec3m $M$ n\u1eb1m tr\u00ean n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n $(M$ kh\u00e1c $A $ v\u00e0 $B)$, k\u1ebb ti\u1ebfp tuy\u1ebfn v\u1edbi n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n c\u1eaft $Ax, By$ theo th\u1ee9 t\u1ef1 \u1edf $C$ v\u00e0 $D$.<br\/><b>C\u00e2u 1: <\/b> T\u00ednh g\u00f3c $COD$. <br\/><br\/><b>\u0110\u00e1p \u00e1n:<\/b> $\\widehat{COD}$=_input_$^o$<\/span>","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv1/img\/h924_D6.png' \/><\/center><span class='basic_left'> Ta c\u00f3: $CD$ l\u00e0 ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n ( gi\u1ea3 thi\u1ebft)<br\/> $\\Rightarrow OC$ l\u00e0 ph\u00e2n gi\u00e1c $\\widehat{AOM}$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> $\\Rightarrow \\widehat{COM}=\\dfrac{1}{2}\\widehat{AOM}$ <br\/> $OD$ l\u00e0 ph\u00e2n gi\u00e1c $\\widehat{BOM}$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> $\\Rightarrow \\widehat{DOM}=\\dfrac{1}{2}\\widehat{BOM}$ <br\/> $\\Rightarrow \\widehat{COD}=\\widehat{COM} + \\widehat{DOM} $ <br\/> $=\\dfrac{1}{2}\\widehat{AOM} + \\dfrac{1}{2}\\widehat{BOM} $ <br\/> $= \\dfrac{1}{2}\\widehat{AOB}= 90^o$<br\/> <span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $90.$<\/span><\/span>"}]}],"id_ques":1192},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv1/img\/h924_D6.png' \/><\/center> Cho n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m $(O; R)$, \u0111\u01b0\u1eddng k\u00ednh $AB$. Tr\u00ean n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd $AB$, k\u1ebb $Ax,\\, By$ vu\u00f4ng g\u00f3c v\u1edbi $AB$. Qua \u0111i\u1ec3m $M$ n\u1eb1m tr\u00ean n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n $(M\\ne A; B)$, k\u1ebb ti\u1ebfp tuy\u1ebfn v\u1edbi n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n c\u1eaft $Ax, \\,By$ theo th\u1ee9 t\u1ef1 \u1edf $C$ v\u00e0 $D$.<br\/> <b>C\u00e2u 2: <\/b> Ch\u1ee9ng minh $OC \\bot AM$. <\/span>","title_trans":"S\u1eafp x\u1ebfp th\u1ee9 t\u1ef1 c\u00e1c b\u01b0\u1edbc ch\u1ee9ng minh","temp":"sequence","correct":[[[4],[1],[3],[2]]],"list":[{"point":5,"left":["Suy ra $AM \\bot OC$","Ta c\u00f3: $CA =CM$ (t\u00ednh ch\u1ea5t) $\\Rightarrow$ C thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AM$","$\\Rightarrow OC$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AM$ ","M\u00e0 $OA = OM = R \\Rightarrow O$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AM$"],"top":55,"explain":"<span class='basic_left'>Ta c\u00f3: $CA =CM$ (t\u00ednh ch\u1ea5t) $\\Rightarrow C$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AM$ <br\/> M\u00e0 $OA = OM = R \\Rightarrow O$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AM$ <br\/> $\\Rightarrow OC$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AM$<br\/>Suy ra $AM \\bot OC$ <\/span>"}]}],"id_ques":1193},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<span class='basic_left'>Cho n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m $(O; R)$, \u0111\u01b0\u1eddng k\u00ednh $AB$. Tr\u00ean n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd $AB$, k\u1ebb $Ax,\\, By$ vu\u00f4ng g\u00f3c v\u1edbi $AB$. Qua \u0111i\u1ec3m $M$ n\u1eb1m tr\u00ean n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n $(M\\ne A; B)$, k\u1ebb ti\u1ebfp tuy\u1ebfn v\u1edbi n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n c\u1eaft $Ax, \\,By$ theo th\u1ee9 t\u1ef1 \u1edf $C$ v\u00e0 $D$.<br\/><b>C\u00e2u 3: <\/b> Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? <\/span>","select":["A. $ CD = AC+BD$","B. $ CD = AM + MB$","C. $ CD = AC.BD$","D. $ CD = AB+OM$"],"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv1/img\/h924_D6.png' \/><\/center>Ta c\u00f3: $CA = CM; DM = DB$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> $\\Rightarrow CD = CM + MD $$= AC + BD$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span>","column":2}]}],"id_ques":1194},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"<span class='basic_left'>Cho n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m $(O; R)$, \u0111\u01b0\u1eddng k\u00ednh $AB$. Tr\u00ean n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd $AB$, k\u1ebb $Ax,\\, By$ vu\u00f4ng g\u00f3c v\u1edbi $AB$. Qua \u0111i\u1ec3m $M$ n\u1eb1m tr\u00ean n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n $(M\\ne A; B)$, k\u1ebb ti\u1ebfp tuy\u1ebfn v\u1edbi n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n c\u1eaft $Ax, \\,By$ theo th\u1ee9 t\u1ef1 \u1edf $C$ v\u00e0 $D$.<br\/><b>C\u00e2u 4: <\/b> Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? <\/span>","select":["A. $ OM^2=DB.DM $","B. $ OM^2= CA.CM$","C. $ OM^2=AC.BD$"],"explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv1/img\/h924_D6.png' \/><\/center><span class='basic_left'>Ta c\u00f3: $CA = CM; DM = DB$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> $OA = OM = R$ <br\/> X\u00e9t $\\Delta OCD$ vu\u00f4ng t\u1ea1i $O$ (theo c\u00e2u 1) <br\/> $OM \\bot CD$ (t\u00ednh ch\u1ea5t ti\u1ebfp tuy\u1ebfn) <br\/> $\\Rightarrow OM^2 = CM.MD = AC.BD$ (h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng) <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span>","column":3}]}],"id_ques":1195},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"<span class='basic_left'>Tr\u00ean h\u00ecnh b\u00ean, tam gi\u00e1c $IKM$ ngo\u1ea1i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n $(O)$. H\u1ec7 th\u1ee9c n\u00e0o sau \u0111\u00e2y \u0111\u00fang?<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv1/img\/h924_D9.png' \/><\/center><\/span>","select":["A. $KI + KM - IM = 2IE$","B. $KI + MI - KM = 2 KD$","C. $MI + MK - IK = 2 M F$","D. $KI + MI - KM = 2 ME$"],"explain":"<span class='basic_left'>Ta c\u00f3: $IK, KM, MI$ l\u00e0 c\u00e1c ti\u1ebfp tuy\u1ebfn c\u1ee7a $(O)$ <br\/> $\\Rightarrow ID =IE;\\, $$KD = KF;\\,$$ MF = ME$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> $\\bullet KI + KM - IM $ <br\/> $= KD + ID + KF $$+ FM - IE - EM$ <br\/> $= KD + ID + KD $$+ FM - ID - FM$ <br\/> $ = 2KD=2KF$ <br\/> $\\bullet KI + MI - KM$ <br\/> $ = KD + DI + IE $$+ ME - KF - MF$ <br\/> $=ID + KD + ID$$ + ME - KD - ME$<br\/> $= 2ID =2IE$ <br\/> V\u1eady \u0111\u00e1p \u00e1n A, B, D sai <br\/> $\\bullet MI + MK - IK $ <br\/> $= ME + EI + MF $$+ FK - ID - DK$ <br\/> $=ME + IE + ME$$+KF - IE - KF$ <br\/> $= 2ME = 2MF$ <br\/> V\u1eady \u0111\u00e1p \u00e1n C \u0111\u00fang<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span>","column":2}]}],"id_ques":1196},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O; R)$ v\u00e0 m\u1ed9t \u0111i\u1ec3m $A$ n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n sao cho $OA = 2R.$ K\u1ebb ti\u1ebfp tuy\u1ebfn $AB, AC\\, (B, C$ l\u00e0 c\u00e1c ti\u1ebfp \u0111i\u1ec3m$)$. \u0110o\u1ea1n th\u1eb3ng $OA$ giao v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $I$. \u0110\u01b0\u1eddng th\u1eb3ng qua $O$ v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi $OB$ c\u1eaft $AC$ t\u1ea1i $K$. <br\/><br\/> <b> C\u00e2u 1:<\/b> B\u1ed1n \u0111i\u1ec3m $A, B, O, C$ c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n. Kh\u1eb3ng \u0111\u1ecbnh tr\u00ean \u0111\u00fang hay sai?<\/span>","select":["A. \u0110\u00fang","B. Sai"],"explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv1/img\/h924_D10.png' \/><\/center><span class='basic_left'>Ta c\u00f3: $OA=2R;OI=R \\Rightarrow IA=R.$ <br\/> Suy ra $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AO$ <br\/> $\\widehat{ABO}=\\widehat{ACO}={{90}^{o}}$ (t\u00ednh ch\u1ea5t ti\u1ebfp tuy\u1ebfn) <br\/> $\\Delta ABO$ vu\u00f4ng t\u1ea1i $B$ <br\/> $\\Rightarrow BI=IO=IA$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\Delta ACO$ vu\u00f4ng t\u1ea1i $C$ <br\/> $\\Rightarrow OI=CI=IA$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\Rightarrow IA=IB=IC=IO$ <br\/> V\u1eady b\u1ed1n \u0111i\u1ec3m $A, B, C, O$ c\u00f9ng thu\u1ed9c m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span>","column":2}]}],"id_ques":1197},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O; R)$ v\u00e0 m\u1ed9t \u0111i\u1ec3m $A$ n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n sao cho $OA = 2R.$ K\u1ebb ti\u1ebfp tuy\u1ebfn $AB, AC\\, (B, C$ l\u00e0 c\u00e1c ti\u1ebfp \u0111i\u1ec3m$)$. \u0110o\u1ea1n th\u1eb3ng $OA$ giao v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $I$. \u0110\u01b0\u1eddng th\u1eb3ng qua $O$ v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi $OB$ c\u1eaft $AC$ t\u1ea1i $K$. <br\/><br\/> <b> C\u00e2u 2:<\/b> Tam gi\u00e1c $AOK$ l\u00e0 tam gi\u00e1c g\u00ec?<\/span>","select":["A. Vu\u00f4ng","B. C\u00e2n","C. \u0110\u1ec1u"],"explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv1/img\/h924_D10.1.png' \/><\/center><span class='basic_left'>Ta c\u00f3: $AO$ l\u00e0 ph\u00e2n gi\u00e1c $\\widehat{BAC}$(t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> $\\Rightarrow \\widehat{{{A}_{1}}}=\\widehat{{{A}_{2}}}$ <br\/>M\u00e0 $\\widehat{{{O}_{1}}}=\\widehat{{{A}_{2}}}$ (c\u00f9ng ph\u1ee5 v\u1edbi $\\widehat{AOB}$) <br\/> $\\Rightarrow \\widehat{{{O}_{1}}}=\\widehat{{{A}_{1}}}$ <br\/> Suy ra $\\Delta AOK$ c\u00e2n (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n)<br\/> <span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 B<\/span><\/span>","column":3}]}],"id_ques":1198},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"fill_the_blank","correct":[[["90"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>Cho \u0111\u01b0\u1eddng tr\u00f2n $(O; R)$ v\u00e0 m\u1ed9t \u0111i\u1ec3m $A$ n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n sao cho $OA = 2R.$ K\u1ebb ti\u1ebfp tuy\u1ebfn $AB, AC\\, (B, C$ l\u00e0 c\u00e1c ti\u1ebfp \u0111i\u1ec3m$)$. \u0110o\u1ea1n th\u1eb3ng $OA$ giao v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $I$. \u0110\u01b0\u1eddng th\u1eb3ng qua $O$ v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi $OB$ c\u1eaft $AC$ t\u1ea1i $K$. <br\/> <b> C\u00e2u 3:<\/b> K\u00e9o d\u00e0i $KI$ c\u1eaft $AB$ t\u1ea1i $M$. Khi \u0111\u00f3 s\u1ed1 \u0111o $\\widehat{KIO}$ l\u00e0 _input_$^o$<\/span>","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv1/img\/h924_D10.2.png' \/><\/center><span class='basic_left'> Theo c\u00e2u 2: $\\Delta AKO$ c\u00e2n t\u1ea1i $K$ <br\/> $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $OA$ <br\/> $\\Rightarrow KI \\bot OA$ (trong tam gi\u00e1c c\u00e2n, \u0111\u01b0\u1eddng trung tuy\u1ebfn c\u0169ng l\u00e0 \u0111\u01b0\u1eddng cao) <br\/> Suy ra $\\widehat{OIK}=90^o$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $90.$<\/span><\/span>"}]}],"id_ques":1199},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O; R)$ v\u00e0 m\u1ed9t \u0111i\u1ec3m $A$ n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n sao cho $OA = 2R.$ K\u1ebb ti\u1ebfp tuy\u1ebfn $AB, AC \\,(B, C$ l\u00e0 c\u00e1c ti\u1ebfp \u0111i\u1ec3m$)$. \u0110o\u1ea1n th\u1eb3ng $OA$ giao v\u1edbi \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $I$. \u0110\u01b0\u1eddng th\u1eb3ng qua $O$ v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi $OB$ c\u1eaft $AC$ t\u1ea1i $K$. <br\/> <b> C\u00e2u 4:<\/b> Chu vi tam gi\u00e1c $AKM$ theo $R$ l\u00e0:<\/span>","select":["A. $R\\sqrt{5}$","B. $2R\\sqrt{5}$","C. $R\\sqrt{3}$","D. $2R\\sqrt{3}$"],"explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai9/lv1/img\/h924_D10.2.png' \/><\/center><span class='basic_left'> Ta c\u00f3: $OA = 2R, OB = R$ <br\/> X\u00e9t $\\Delta OAB$ vu\u00f4ng t\u1ea1i $B$ <br\/> $OA^2=OB^2+AB^2$ (\u0111\u1ecbnh l\u00ed Pitago) <br\/> $\\Rightarrow AB = \\sqrt{OA^2 -OB^2}$ <br\/> $\\Rightarrow AB=\\sqrt{4R^2-R^2}=R\\sqrt{3}$ <br\/> M\u00e0 $KM\\bot OI,$ $I\\in (O)$ (theo c\u00e2u 3) <br\/> $\\Rightarrow KM$ l\u00e0 ti\u1ebfp tuy\u1ebfn c\u1ee7a $(O)$ <br\/> $\\Rightarrow MI =MB; IK=KC$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> Chu vi c\u1ee7a tam gi\u00e1c AKM l\u00e0:<br\/> $ AM+MK+KA$ <br\/> $= AM + MI + IK + KA$ <br\/> $=AM+MB+AK+KC$ <br\/> $=AB+AC$ <br\/> $=2AB$ <br\/> $=2R\\sqrt{3}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span>","column":4}]}],"id_ques":1200}],"lesson":{"save":0,"level":1}}