{"segment":[{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Cho \u0111\u01b0\u1eddng tr\u00f2n $(O; 6cm)$ v\u00e0 m\u1ed9t \u0111i\u1ec3m $A$ c\u00e1ch $O$ m\u1ed9t kho\u1ea3ng b\u1eb1ng $10 cm$ th\u00ec $A$","select":["A. N\u1eb1m tr\u00ean \u0111\u01b0\u1eddng tr\u00f2n","B. N\u1eb1m trong \u0111\u01b0\u1eddng tr\u00f2n","C. N\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n"],"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv1/img\/h926_D1.1.png' \/><\/center> Ta c\u00f3: $OA = 10 cm$ <br\/> $ \\Rightarrow OA > R$ <br\/> V\u1eady \u0111i\u1ec3m $A$ n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n<br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 C<\/span><\/span>","column":3}]}],"id_ques":1241},{"time":9,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2"],["1"],["0"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<table><tr><th>S\u1ed1 th\u1ee9 t\u1ef1<br><\/th><th>V\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a $(O;R)$ v\u00e0 $(O';r)$ bi\u1ebft $R > r$<br><\/th><th>S\u1ed1 \u0111i\u1ec3m chung<br><\/th><\/tr><tr><td>1<\/td><td>C\u1eaft nhau<\/td><td>_input_<\/td><\/tr><tr><td>2<\/td><td>Ti\u1ebfp x\u00fac ngo\u00e0i<\/td><td>_input_<\/td><\/tr><tr><td>3<\/td><td>\u0110\u1ed3ng t\u00e2m<\/td><td>_input_<\/td><\/tr><\/table>","explain":"<span class='basic_left'>Hai \u0111\u01b0\u1eddng tr\u00f2n c\u1eaft nhau c\u00f3 hai \u0111i\u1ec3m chung <br\/> Hai \u0111\u01b0\u1eddng tr\u00f2n ti\u1ebfp x\u00fac ngo\u00e0i nhau c\u00f3 m\u1ed9t \u0111i\u1ec3m chung <br\/> Hai \u0111\u01b0\u1eddng tr\u00f2n \u0111\u1ed3ng t\u00e2m m\u00e0 $R > r$ th\u00ec kh\u00f4ng c\u00f3 \u0111i\u1ec3m chung <br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $2$; $1$; $0$<\/span> <\/span>"}]}],"id_ques":1242},{"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'> Cho \u0111o\u1ea1n th\u1eb3ng $OO' = 5cm$. Hai \u0111\u01b0\u1eddng tr\u00f2n $(O; 2 cm)$ v\u00e0 $(O', r)$ \u1edf ngo\u00e0i nhau n\u1ebfu:<\/span>","select":["A. $3 cm < r < 7 cm$","B. $r < 3 cm$","C. $r = 7 cm$","D. $r > 3 cm$"],"explain":"<span class='basic_left'> Hai \u0111\u01b0\u1eddng tr\u00f2n \u1edf ngo\u00e0i nhau n\u1ebfu $R + r < OO'$ <br\/> $\\Rightarrow r < OO' - R $ <br\/> $\\Rightarrow r < 5 - 2 $ <br\/> $\\Rightarrow r < 3 cm$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B. <\/span><\/span>","column":2}]}],"id_ques":1243},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"\u0110\u01b0\u1eddng tr\u00f2n t\u00e2m $A$ c\u00f3 b\u00e1n k\u00ednh b\u1eb1ng $3cm$ l\u00e0 t\u1eadp h\u1ee3p c\u00e1c \u0111i\u1ec3m","select":["A. C\u00e1ch \u0111\u1ec1u $A$","B. C\u00e1ch $A$ m\u1ed9t kho\u1ea3ng b\u1eb1ng $3cm$","C. N\u1eb1m tr\u00ean m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n ","d. N\u1eb1m tr\u00ean m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng c\u00e1ch $A$ m\u1ed9t kho\u1ea3ng b\u1eb1ng $3cm$ "],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv1/img\/h926_D1.png' \/><\/center><br\/> \u0110\u01b0\u1eddng tr\u00f2n t\u00e2m $A$ c\u00f3 b\u00e1n k\u00ednh b\u1eb1ng $3cm$ l\u00e0 t\u1eadp h\u1ee3p c\u00e1c \u0111i\u1ec3m c\u00e1ch $A$ m\u1ed9t kho\u1ea3ng b\u1eb1ng $3cm$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B. <\/span>","column":1}]}],"id_ques":1244},{"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":[[["20"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<span class='basic_left'>Cho \u0111\u01b0\u1eddng tr\u00f2n $(O;5cm)$, d\u00e2y cung $BC$ b\u1eb1ng $8cm$. L\u1ea5y \u0111i\u1ec3m $A$ thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n sao cho $OA$ vu\u00f4ng g\u00f3c v\u1edbi $BC$. T\u00ednh \u0111\u1ed9 d\u00e0i d\u00e2y cung $AB$. <br\/><br\/> <b> \u0110\u00e1p s\u1ed1:<\/b> $AB = \\sqrt{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}} (cm)$<\/span>","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv1/img\/h926_D2.png' \/><\/center><span class='basic_left'> G\u1ecdi $H$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $OA$ v\u00e0 $BC$ <br\/> Ta c\u00f3: $OA\\bot BC \\Rightarrow BH= HC$ (\u0111\u1ecbnh l\u00ed \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) <br\/> $\\Rightarrow BH=\\dfrac{BC}{2}=\\dfrac{8}{2}$$=4\\,(cm)$ <br\/>X\u00e9t $\\Delta OBH$ vu\u00f4ng t\u1ea1i $H$ <br\/> $OB^2 = OH^2 + HB^2$ (\u0111\u1ecbnh l\u00ed Pitago) <br\/> $OH=\\sqrt{O{{B}^{2}}-H{{B}^{2}}}$ <br\/> $=\\sqrt{{{5}^{2}}-{{4}^{2}}}$ <br\/> $=3(cm)$ <br\/> $OA=5\\,(cm) \\Rightarrow HA = OA - OH =2\\,(cm)$ <br\/> $\\Delta BHA$ vu\u00f4ng t\u1ea1i $H$<br\/> $AB^2 = BH^2 + AH^2$ (\u0111\u1ecbnh l\u00ed Pitago) <br\/> $AB=\\sqrt{B{{H}^{2}}+A{{H}^{2}}}$ <br\/> $=\\sqrt{{{4}^{2}}+{{2}^{2}}} $ <br\/> $=\\sqrt{20}\\, (cm)$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $20.$ <\/span><\/span>"}]}],"id_ques":1245},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O; R)$ v\u00e0 d\u00e2y $MN = R\\sqrt{3}$. $Mx$ l\u00e0 tia ti\u1ebfp tuy\u1ebfn t\u1ea1i $M$ c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n $(O)$. S\u1ed1 \u0111o c\u1ee7a $\\widehat{xMN}$ l\u00e0: <\/span>","select":["A. $90^o$","B. $120^o$","C. $60^o$","D. C\u1ea3 B v\u00e0 C \u0111\u1ec1u \u0111\u00fang"],"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv1/img\/h926_D5.png' \/><\/center> G\u1ecdi $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $MN$ <br\/> $\\Rightarrow OH\\bot MN$ (\u0111\u1ecbnh l\u00ed \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) <br\/> $MH= \\dfrac{MN}{2}=\\dfrac{R\\sqrt{3}}{2}$ (\u0111\u1ecbnh l\u00ed \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) <br\/> Ta c\u00f3 tia $Mx$ l\u00e0 ti\u1ebfp tuy\u1ebfn c\u1ee7a $(O)$<br\/>X\u00e9t $\\Delta MHO$ vu\u00f4ng t\u1ea1i $H$ <br\/> $\\cos \\widehat{OMH}=\\dfrac{MH}{MO}=\\dfrac{R\\sqrt{3}}{2R}$$=\\dfrac{\\sqrt{3}}{2}$ <br\/> $\\Rightarrow \\widehat{OMH}={{30}^{o}}$ <br\/> Khi \u0111\u00f3 c\u00f3 hai tr\u01b0\u1eddng h\u1ee3p x\u1ea3y ra: <br\/> Tr\u01b0\u1eddng h\u1ee3p 1: $Mx$ n\u1eb1m tr\u00ean n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd $MO$ kh\u00f4ng ch\u1ee9a tia $MN$ <br\/> $\\Rightarrow\\widehat{xMN} = 90^o + 30^o $$=120 ^o$ <br\/> Tr\u01b0\u1eddng h\u1ee3p 2: $Mx$ n\u1eb1m tr\u00ean n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd $MO$ ch\u1ee9a tia $MN$ <br\/> $\\Rightarrow\\widehat{xMN} = 90^o - 30^o $$=60 ^o$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D. <\/span><\/span>","column":2}]}],"id_ques":1246},{"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'> Cho hai \u0111\u01b0\u1eddng tr\u00f2n $(O; 2cm)$. T\u1eeb \u0111i\u1ec3m $A$ sao cho $OA = 4cm$ v\u1ebd hai ti\u1ebfp tuy\u1ebfn $AB,\\, AC$ \u0111\u1ebfn \u0111\u01b0\u1eddng tr\u00f2n $(O)$ ($B, C$ l\u00e0 c\u00e1c ti\u1ebfp \u0111i\u1ec3m). Chu vi tam gi\u00e1c $ABC$ l\u00e0: <\/span>","select":["A. $6\\sqrt{3}\\,cm$","B. $5\\sqrt{3}\\,cm$","C. $4\\sqrt{3}\\,cm$","D. $2\\sqrt{3}\\,cm$"],"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv1/img\/h926_D6.png' \/><\/center> G\u1ecdi $H$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AO$ v\u00e0 $BC$ <br\/> $OB=OC$$\\Rightarrow C$ thu\u1ed9c trung tr\u1ef1c c\u1ee7a $BC$ <br\/> $AB=AC$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) (1)<br\/> $\\Rightarrow A$ thu\u1ed9c trung tr\u1ef1c c\u1ee7a $BC$ <br\/> $\\Rightarrow AO$ l\u00e0 trung tr\u1ef1c c\u1ee7a $BC$ <br\/> $\\Rightarrow AO\\bot BH;$$BH=HC$ <br\/>X\u00e9t $\\Delta AOB$ vu\u00f4ng t\u1ea1i $B$;$BH\\bot AO$ <br\/> $\\bullet OB^2 = OH^2 + HB^2$ (\u0111\u1ecbnh l\u00ed Pitago) <br\/> $\\Rightarrow AB=\\sqrt{O{{A}^{2}}-O{{B}^{2}}}$ <br\/> $=\\sqrt{16-4}$ <br\/> $=\\sqrt{12} =2\\sqrt{3}\\,(cm)$ (2) <br\/> $\\bullet BH.AO=AB.BO$ (h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\Rightarrow BH=\\dfrac{OB.AB}{AO}=\\dfrac{2.2\\sqrt{3}}{4}=\\sqrt{3}\\, (cm)$ <br\/> $\\Rightarrow BC=2BH $$=2\\sqrt{3}\\, (cm)$ (3) <br\/>T\u1eeb (1), (2), (3) $\\Rightarrow$ Chu vi tam gi\u00e1c $ABC$ l\u00e0: <br\/> $AB + BC + AC = 3.AB$$=6\\sqrt{3}\\,(cm)$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A. <\/span><\/span>","column":4}]}],"id_ques":1247},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Cho hai \u0111\u01b0\u1eddng tr\u00f2n $(O; 3cm)$ v\u00e0 $(O'; 4cm)$, bi\u1ebft $OO' = 7cm$. V\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a $(O)$ v\u00e0 $(O')$ l\u00e0: ","select":["A. Ti\u1ebfp x\u00fac trong","B. Ti\u1ebfp x\u00fac ngo\u00e0i","C. C\u1eaft nhau","D. \u1ede ngo\u00e0i nhau"],"explain":"Ta c\u00f3 $7 = 3 + 4$ <br\/> T\u1ee9c l\u00e0 $OO' = R + r$ <br\/> V\u1eady hai \u0111\u01b0\u1eddng tr\u00f2n ti\u1ebfp x\u00fac ngo\u00e0i <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B. <\/span><\/span>","column":2}]}],"id_ques":1248},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"<span class='basic_left'> Tam gi\u00e1c \u0111\u1ec1u c\u00f3 c\u1ea1nh b\u1eb1ng $8cm$ th\u00ec b\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp tam gi\u00e1c $r$ b\u1eb1ng:<\/span>","select":["A. $2\\sqrt{3}\\,cm$","B. $4\\sqrt{3}\\,cm$","C. $\\dfrac{2\\sqrt{3}}{3}\\,cm$","D. $\\dfrac{4\\sqrt{3}}{3}\\,cm$"],"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv1/img\/h926_D7.1.png' \/><\/center> X\u00e9t $\\Delta BAC$ \u0111\u1ec1u c\u00f3 \u0111\u01b0\u1eddng cao $h$<br\/> Ta c\u00f3 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp tr\u00f9ng v\u1edbi tr\u1ecdng t\u00e2m tam gi\u00e1c <br\/> \u0110\u01b0\u1eddng cao trong tam gi\u00e1c \u0111\u1ec1u c\u1ea1nh $a$ c\u00f3 \u0111\u1ed9 d\u00e0i b\u1eb1ng $\\dfrac{a\\sqrt{3}}{2}$ <br\/> $\\Rightarrow h = \\dfrac{AC\\sqrt{3}}{2}$$=4\\sqrt{3} (cm)$ <br\/> Ta c\u00f3 $r=\\dfrac{1}{3}.h $ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m tam gi\u00e1c) <br\/> $=\\dfrac{4\\sqrt{3}}{3}\\, (cm)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D. <\/span><\/span>","column":2}]}],"id_ques":1249},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho b\u1ed1n c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<span class='basic_left'> Cho \u0111i\u1ec3m $I$ tr\u00ean \u0111\u01b0\u1eddng tr\u00f2n $(O; R)$, \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $OI$ c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n $(O)$ t\u1ea1i $A$ v\u00e0 $B$. <br\/> <b> C\u00e2u 1: <\/b> T\u00ednh $AB$ theo $R$.<\/span>","select":["A. $R\\sqrt{3}$","B. $3R\\sqrt{3}$","C. $R\\sqrt{2}$","D. $2R\\sqrt{2}$"],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv1/img\/h926_D14.png' \/><\/center><span class='basic_left'> G\u1ecdi $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $OI$ <br\/> $\\Rightarrow OH=\\dfrac{R}{2}$ <br\/> $\\Delta OHA$ vu\u00f4ng t\u1ea1i $H$ <br\/> $OA^2 = AH^2 + OH^2$ (\u0111\u1ecbnh l\u00ed Pitago) <br\/> $\\Rightarrow AH=\\sqrt{O{{A}^{2}}-O{{H}^{2}}}$ <br\/> $=\\sqrt{{{R}^{2}}-\\dfrac{{{R}^{2}}}{4}}$ <br\/> $=\\dfrac{R\\sqrt{3}}{2}$ <br\/> M\u00e0 $OI \\bot AB$ (t\u00ednh ch\u1ea5t trung tr\u1ef1c) <br\/> $\\Rightarrow HA = HB$ (\u0111\u1ecbnh l\u00ed \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) <br\/> $\\Rightarrow AB=2AH$$=R\\sqrt{3}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A. <\/span> <\/span>","column":4}]}],"id_ques":1250},{"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 \u0111i\u1ec3m $I$ tr\u00ean \u0111\u01b0\u1eddng tr\u00f2n $(O; R)$, \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $OI$ c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n $(O)$ t\u1ea1i $A$ v\u00e0 $B$. <br\/> <b> C\u00e2u 2: <\/b> T\u1ee9 gi\u00e1c $OAIB$ l\u00e0 h\u00ecnh g\u00ec?<\/span>","select":["A. H\u00ecnh b\u00ecnh h\u00e0nh","B. H\u00ecnh ch\u1eef nh\u1eadt","C. H\u00ecnh thoi","D. H\u00ecnh vu\u00f4ng"],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv1/img\/h926_D14.1.png' \/><\/center><span class='basic_left'>Ta c\u00f3 $AB$ l\u00e0 trung tr\u1ef1c c\u1ee7a $OI$ (gi\u1ea3 thi\u1ebft) <br\/> Suy ra $HO = HI$ <br\/> $HA=HB$ (ch\u1ee9ng minh tr\u00ean) <br\/> $\\Rightarrow OAIB$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <br\/> M\u00e0 $OI\\bot AB$ (gi\u1ea3 thi\u1ebft). <br\/> $\\Rightarrow OAIB$ l\u00e0 h\u00ecnh thoi (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C. <\/span> <\/span>","column":4}]}],"id_ques":1251},{"time":24,"part":[{"time":3,"title":"<span class='basic_left'> Cho \u0111i\u1ec3m $I$ tr\u00ean \u0111\u01b0\u1eddng tr\u00f2n $(O; R)$, \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $OI$ c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n $(O)$ t\u1ea1i $A$ v\u00e0 $B$. <br\/> <b> C\u00e2u 3: <\/b> Hai ti\u1ebfp tuy\u1ebfn k\u1ebb t\u1eeb $A$ v\u00e0 $B$ c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n $(O)$ c\u1eaft nhau t\u1ea1i $C$. Ch\u1ee9ng minh ba \u0111i\u1ec3m $O, I, C$ th\u1eb3ng h\u00e0ng. <\/span>","title_trans":"S\u1eafp x\u1ebfp th\u1ee9 t\u1ef1 c\u00e1c b\u01b0\u1edbc ch\u1ee9ng minh","temp":"sequence","correct":[[[1],[3],[5],[4],[2]]],"list":[{"point":5,"image":"https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv1/img\/h926_D14.2.png","left":["Ta c\u00f3: $CA=CB$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) $\\Rightarrow C$ thu\u1ed9c trung tr\u1ef1c c\u1ee7a $AB$","$\\Rightarrow OC$ l\u00e0 trung tr\u1ef1c c\u1ee7a $AB$$\\Rightarrow OC\\bot AB$ (1)","T\u1eeb (1), (2) $\\Rightarrow C,I,O$ th\u1eb3ng h\u00e0ng "," $ OI\\bot AB$ (gi\u1ea3 thi\u1ebft) (2)","$OA=OB=R$ $\\Rightarrow O$ thu\u1ed9c trung tr\u1ef1c c\u1ee7a $AB$"],"top":55,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv1/img\/h926_D14.2.png' \/><\/center>Ta c\u00f3: $CA=CB$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> $\\Rightarrow C$ thu\u1ed9c trung tr\u1ef1c c\u1ee7a $AB$ <br\/> $OA=OB=R$ <br\/> $\\Rightarrow O$ thu\u1ed9c trung tr\u1ef1c c\u1ee7a $AB$ <br\/> $\\Rightarrow OC$ l\u00e0 trung tr\u1ef1c c\u1ee7a $AB$ <br\/> $\\Rightarrow OC\\bot AB$ (1) <br\/> $ OI\\bot AB$ (gi\u1ea3 thi\u1ebft) (2) <br\/> T\u1eeb (1), (2) $\\Rightarrow C,I,O$ th\u1eb3ng h\u00e0ng <\/span>"}]}],"id_ques":1252},{"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 \u0111i\u1ec3m $I$ tr\u00ean \u0111\u01b0\u1eddng tr\u00f2n $(O; R)$, \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $OI$ c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n $(O)$ t\u1ea1i $A$ v\u00e0 $B$. <br\/> <b> C\u00e2u 4: <\/b> T\u00ednh chu vi tam gi\u00e1c $ABC$.<\/span>","select":["A. $R\\sqrt{3}+2R$","B. $3R\\sqrt{3}$","C. $R\\sqrt{3}+2R\\sqrt{2}$","D. $2R\\sqrt{3}$"],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv1/img\/h926_D14.2.png' \/><\/center><span class='basic_left'> Theo c\u00e2u 1: $OH = \\dfrac{R}{2}; AH = \\dfrac{R\\sqrt{3}}{2}; AB = R\\sqrt{3} $ <br\/> Theo c\u00e2u 2: $AC = BC$ <br\/> X\u00e9t $\\Delta AHO$ v\u00e0 $\\Delta AOC$ <br\/> $\\widehat{H}=\\widehat{A}={{90}^{o}}$ <br\/> $\\widehat{O}$ chung <br\/> $\\Rightarrow \\Delta AHO\\sim \\Delta CAO$ (g.g) <br\/> $\\Rightarrow \\dfrac{AH}{AC}=\\dfrac{HO}{AO}$ <br\/> $\\Rightarrow AC=\\dfrac{AH.AO}{HO}$ <br\/> $=\\dfrac{R.\\dfrac{R\\sqrt{3}}{2}}{\\dfrac{R}{2}}$ <br\/> $=R\\sqrt{3}$ <br\/> $\\Rightarrow $ Chu vi tam gi\u00e1c $ABC$ l\u00e0: <br\/> $ AB + BC + CA = 3AC=3R\\sqrt{3}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B. <\/span> <\/span>","column":2}]}],"id_ques":1253},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<span class='basic_left'>Cho \u0111\u01b0\u1eddng tr\u00f2n $(O; 25 cm)$ v\u00e0 hai d\u00e2y $MN \/\/ PQ$. <br\/> <b> C\u00e2u 1: <\/b> D\u00e2y l\u1edbn nh\u1ea5t c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n c\u00f3 s\u1ed1 \u0111o b\u1eb1ng:<\/span>","select":["A. $ 50\\, cm$","B. $25\\, cm$","C. $20 \\,cm$","D. $625\\, cm$"],"explain":"D\u00e2y cung l\u1edbn nh\u1ea5t l\u00e0 \u0111\u01b0\u1eddng k\u00ednh <br\/> V\u1eady d\u00e2y cung l\u1edbn nh\u1ea5t b\u1eb1ng $50\\, cm$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A. <\/span> <\/span>","column":4}]}],"id_ques":1254},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O; 25 cm)$ v\u00e0 hai d\u00e2y $MN \/\/ PQ$. <br\/> <b> C\u00e2u 2: <\/b> Bi\u1ebft d\u00e2y $MN $ d\u00e0i $ 40 cm$ v\u00e0 $PQ $ d\u00e0i $48 cm$ ($MN$ v\u00e0 $PQ$ kh\u00e1c ph\u00eda so v\u1edbi $O$). Khi \u0111\u00f3 kho\u1ea3ng c\u00e1ch gi\u1eefa d\u00e2y $MN$ v\u00e0 $PQ$ l\u00e0: <\/span>","select":["A. $ 22\\, cm$","B. $8\\,cm$","C. $15\\, cm$","D. $7\\,cm$"],"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv1/img\/h926_D10.png' \/> <\/center> Qua t\u00e2m $O$, k\u1ebb $HK \\bot MN$ $\\Rightarrow HK\\bot PQ$ <br\/> $\\Rightarrow \\left\\{ \\begin{align} & MH=\\dfrac{MN}{2}=20\\,(cm) \\\\ & PK=\\dfrac{PQ}{2}=24 \\,(cm) \\\\ \\end{align} \\right.$ (\u0111\u1ecbnh l\u00ed \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) <br\/> X\u00e9t $ \\Delta OHM$ vu\u00f4ng t\u1ea1i $H$ <br\/> $OM^2 = OH^2 + HM^2$ (\u0111\u1ecbnh l\u00ed Pitago) <br\/> $OH=\\sqrt{O{{M}^{2}}-H{{M}^{2}}}$ <br\/> $=\\sqrt{{{25}^{2}}-{{20}^{2}}}$ <br\/> $=15\\,(cm) $ <br\/> X\u00e9t $\\Delta OPK$ vu\u00f4ng t\u1ea1i $K$ <br\/> $OP^2 = OK^2 + KP^2$ (\u0111\u1ecbnh l\u00ed Pitago) <br\/> $OK=\\sqrt{O{{P}^{2}}-P{{K}^{2}}}$ <br\/> $=\\sqrt{{{25}^{2}}-{{24}^{2}}}$ <br\/> $=7\\,(cm) $ <br\/> Ta c\u00f3 $HK$ l\u00e0 kho\u1ea3ng c\u00e1ch gi\u1eefa hai d\u00e2y $MN$ v\u00e0 $PQ$ <br\/> $\\Rightarrow HK = OH+OK = 22\\,(cm)$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A. <\/span> <\/span>","column":4}]}],"id_ques":1255},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv1/img\/h926_D12.png'<\/center> <br\/>H\u00ecnh \u1ea3nh l\u00edp xe \u0111\u1ea1p trong th\u1ef1c t\u1ebf li\u00ean quan \u0111\u1ebfn v\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a hai \u0111\u01b0\u1eddng tr\u00f2n? ","select":["A. Hai \u0111\u01b0\u1eddng tr\u00f2n \u0111\u1ed3ng t\u00e2m","B. Hai \u0111\u01b0\u1eddng tr\u00f2n c\u1eaft nhau","C. Hai \u0111\u01b0\u1eddng tr\u00f2n ti\u1ebfp x\u00fac nhau","D. Hai \u0111\u01b0\u1eddng tr\u00f2n \u1edf ngo\u00e0i nhau"],"explain":" H\u00ecnh \u1ea3nh l\u00edp xe \u0111\u1ea1p trong th\u1ef1c t\u1ebf g\u1ed3m hai \u0111\u01b0\u1eddng tr\u00f2n r\u1eddi nhau th\u1ec3 hi\u1ec7n v\u1ecb tr\u00ed t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a hai \u0111\u01b0\u1eddng tr\u00f2n \u1edf ngo\u00e0i nhau <br\/> <span class='basic_pink'> \u0110\u00e1p \u00e1n l\u00e0 D <\/span>","column":2}]}],"id_ques":1256},{"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 g\u00f3c $xOy$ b\u1eb1ng $120^o$, \u0111\u01b0\u1eddng tr\u00f2n $(O)$ c\u1eaft tia $Ox, Oy$ l\u1ea7n l\u01b0\u1ee3t t\u1ea1i $B$ v\u00e0 $C$. Ti\u1ebfp tuy\u1ebfn c\u1ee7a $(O)$ t\u1ea1i $B$ v\u00e0 $C$ c\u1eaft nhau t\u1ea1i $A$. G\u1ecdi $H$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $OA$ v\u00e0 $BC$. <br\/> <b> C\u00e2u 1: <\/b> Tam gi\u00e1c $ABC$ l\u00e0 tam gi\u00e1c g\u00ec? <\/span>","select":["A. Tam gi\u00e1c c\u00e2n","B. Tam gi\u00e1c vu\u00f4ng","C. Tam gi\u00e1c vu\u00f4ng c\u00e2n","D. Tam gi\u00e1c \u0111\u1ec1u"],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv1/img\/h926_D13.png' \/><\/center><span class='basic_left'> Ta c\u00f3 $AB=AC$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> Suy ra $\\Delta ABC$ c\u00e2n t\u1ea1i $A$<br\/> $OA$ l\u00e0 ph\u00e2n gi\u00e1c $\\widehat{BOC}$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> $\\widehat{BOC}={{120}^{o}}$ $\\Rightarrow \\widehat{AOB} = 60 ^o$ <br\/> $AB$ l\u00e0 ti\u1ebfp tuy\u1ebfn $(O)$ $\\Rightarrow \\widehat{OBA}=90^o$ <br\/> $\\Delta OBA$ c\u00f3: $\\widehat{OBA}=90^o; \\widehat{AOB} = 60 ^o$ <br\/> $\\Rightarrow \\widehat{OAB} = 30 ^o$ (\u0111\u1ecbnh l\u00ed t\u1ed5ng ba g\u00f3c c\u1ee7a tam gi\u00e1c)<br\/> $\\Rightarrow \\widehat{CAB} = 2\\widehat{OAB} = 60 ^o$ <br\/> $ \\Rightarrow \\Delta ABC$ \u0111\u1ec1u <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D. <\/span> <\/span>","column":2}]}],"id_ques":1257},{"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":[[["AH","HA"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<span class='basic_left'> Cho g\u00f3c $xOy$ b\u1eb1ng $120^o$, \u0111\u01b0\u1eddng tr\u00f2n $(O)$ c\u1eaft tia $Ox, Oy$ l\u1ea7n l\u01b0\u1ee3t t\u1ea1i $B$ v\u00e0 $C$. Ti\u1ebfp tuy\u1ebfn c\u1ee7a $(O)$ t\u1ea1i $B$ v\u00e0 $C$ c\u1eaft nhau t\u1ea1i $A$. G\u1ecdi $H$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $OA$ v\u00e0 $BC$. <br\/> <b> C\u00e2u 2: <\/b> Ta ch\u1ee9ng minh \u0111\u01b0\u1ee3c $BH^2 = OH.$ _input_<\/span>","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv1/img\/h926_D13.png' \/><\/center><span class='basic_left'> D\u1ec5 d\u00e0ng ch\u1ee9ng minh $AO$ l\u00e0 trung tr\u1ef1c c\u1ee7a $BC$ <br\/> Suy ra $AO \\bot BC$ <br\/> X\u00e9t $\\Delta ABO$ vu\u00f4ng t\u1ea1i $B;$$BH\\bot AO$ <br\/> $\\Rightarrow B{{H}^{2}}=OH.AH$ (h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng)<br\/> <span class='basic_pink'>V\u1eady t\u1eeb c\u1ea7n \u0111i\u1ec1n l\u00e0 $AH$ <\/span><\/span>"}]}],"id_ques":1258},{"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 g\u00f3c $xOy$ b\u1eb1ng $120^o$, \u0111\u01b0\u1eddng tr\u00f2n $(O)$ c\u1eaft tia $Ox, Oy$ l\u1ea7n l\u01b0\u1ee3t t\u1ea1i $B$ v\u00e0 $C$. Ti\u1ebfp tuy\u1ebfn c\u1ee7a $(O)$ t\u1ea1i $B$ v\u00e0 $C$ c\u1eaft nhau t\u1ea1i $A$. G\u1ecdi $H$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $OA$ v\u00e0 $BC$. <br\/> <b> C\u00e2u 3: <\/b> T\u00ednh \u0111\u1ed9 d\u00e0i $AB$ theo $R$.<\/span>","select":["A. $R$","B. $2R$","C. $R\\sqrt{3}$","D. $2R\\sqrt{3}$"],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv1/img\/h926_D13.png' \/><\/center><span class='basic_left'>Ta c\u00f3: $OA$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BOC}$ (t\u00ednh ch\u1ea5t hai ti\u1ebfp tuy\u1ebfn c\u1eaft nhau) <br\/> $\\Rightarrow \\widehat{BOA} = \\widehat{COA}= \\dfrac{\\widehat{BOC}}{2}=\\dfrac{120^o}{2}=60^o$ <br\/> X\u00e9t $\\Delta ABO$ vu\u00f4ng t\u1ea1i $B$ ta c\u00f3:<br\/> $AB = OB. \\tan\\widehat{BOA} $ (h\u1ec7 th\u1ee9c v\u1ec1 c\u1ea1nh v\u00e0 g\u00f3c trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\Rightarrow AB= R.\\tan\\widehat{60^o}=R\\sqrt{3}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C. <\/span> <\/span>","column":4}]}],"id_ques":1259},{"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 g\u00f3c $xOy$ b\u1eb1ng $120^o$, \u0111\u01b0\u1eddng tr\u00f2n $(O)$ c\u1eaft tia $Ox, Oy$ l\u1ea7n l\u01b0\u1ee3t t\u1ea1i $B$ v\u00e0 $C$. Ti\u1ebfp tuy\u1ebfn c\u1ee7a $(O)$ t\u1ea1i $B$ v\u00e0 $C$ c\u1eaft nhau t\u1ea1i $A$. G\u1ecdi $H$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $OA$ v\u00e0 $BC$. <br\/> <b> C\u00e2u 4: <\/b> V\u1ebd \u0111\u01b0\u1eddng k\u00ednh $CD$ c\u1ee7a $(O)$. T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $BCD$ theo b\u00e1n k\u00ednh $R$ c\u1ee7a $(O)$.<\/span>","select":["A. $\\dfrac{R^2}{2}$ \u0111vdt","B. $\\dfrac{R^2}{4}$ \u0111vdt","C. $\\dfrac{R^2\\sqrt{3}}{2}$ \u0111vdt","D. $\\dfrac{R^2\\sqrt{3}}{4}$ \u0111vdt"],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai11/lv1/img\/h926_D13.1.png' \/><\/center><span class='basic_left'>X\u00e9t $\\Delta BCD$ c\u00f3: <br\/> $OB = OC = OD$$\\Rightarrow\\widehat{CBD}={{90}^{o}}$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) <br\/> $ CD=2R $; $ BC=R\\sqrt{3}$ (theo c\u00e2u 3)<br\/> $\\Delta CBD$ vu\u00f4ng t\u1ea1i $B$ <br\/> $CD^2 = BD^2 + BC^2$ (\u0111\u1ecbnh l\u00ed Pitago) <br\/> $BD=\\sqrt{C{{D}^{2}}-B{{C}^{2}}}$ <br\/> $=\\sqrt{4{{R}^{2}}-3{{R}^{2}}}$ <br\/> $=R$ <br\/> ${{S}_{BCD}}=\\dfrac{1}{2}BC.BD$ <br\/> $=\\dfrac{1}{2}R.R\\sqrt{3}$ <br\/> $=\\dfrac{{{R}^{2}}\\sqrt{3}}{2}$ (\u0111vdt) <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C. <\/span> <\/span>","column":4}]}],"id_ques":1260}],"lesson":{"save":0,"level":1}}