Tính chất ba đường trung tuyến của tam giác-level3

Home Lớp 7 Toán lớp 7 - Sách kết nối tri thức Tính chất ba đường trung tuyến của tam giácBài tập nâng cao

{"common":{"save":0,"post_id":"1354","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"1911","post_id":"1354","mon_id":"0","chapter_id":"0","question":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","options":{"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":[[["90"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 $BC = 10cm$, c\u00e1c \u0111\u01b0\u1eddng trung tuy\u1ebfn $BD = 9cm$, trung tuy\u1ebfn $CE = 12cm$, $BD$ v\u00e0 $CE$ c\u1eaft nhau t\u1ea1i $G$. T\u00ednh g\u00f3c $\\widehat{BGC}$. <br\/> \u0110\u00e1p \u00e1n l\u00e0: <\/b>$\\widehat{BGC}$ = _input_ $^{o}$ ","hint":"","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> T\u00ednh $BG; CG$ <br\/> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $\\triangle{BGC}$ vu\u00f4ng t\u1ea1i $G$ <br\/> B\u01b0\u1edbc 3: <\/b> T\u00ednh $\\widehat{BGC}$ <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='img\/H7C3B20_K8.png' \/><\/center> <br\/> $\\blacktriangleright$ G\u1ecdi $G$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $BD$ v\u00e0 $CE$, suy ra $G$ l\u00e0 tr\u1ecdng t\u00e2m $\\triangle{ABC}$ <br\/> Theo t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m, ta c\u00f3: <br\/> $BG = \\dfrac{2}{3}BD = \\dfrac{2}{3}.9 = 6(cm)$ <br\/> $CG = \\dfrac{2}{3}CE = \\dfrac{2}{3}.12 = 8(cm)$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{BGC}$ c\u00f3: $BC = 10cm; BG = 6cm; CG = 8cm$ <br\/> Ta c\u00f3: $10^2 = 6^2 + 8^2$ hay $BC^2 = CG^2 + BG^2$ <br\/> $\\Rightarrow$ $\\triangle{BGC}$ vu\u00f4ng t\u1ea1i $G$ <br\/> $\\Rightarrow$ $\\widehat{BGC} = 90^{o}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $90$ <\/span> "}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:23:05"},{"id":"1912","post_id":"1354","mon_id":"0","chapter_id":"0","question":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","options":{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $ABC$ v\u1edbi trung tuy\u1ebfn $AD$. Qua $D$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi $AB$, qua $B$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi $AD$. Hai \u0111\u01b0\u1eddng th\u1eb3ng tr\u00ean c\u1eaft nhau t\u1ea1i $E$. Khi \u0111\u00f3 $D$ l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c $AEC$. \u0110\u00fang <\/b> hay sai <\/b>? ","select":["A. \u0110\u00daNG ","B. SAI "],"hint":"","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AE$ v\u00e0 $BC$ <br\/> Ch\u1ee9ng minh $DC = \\dfrac{2}{3}CI$ <br\/> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $CI$ l\u00e0 trung tuy\u1ebfn c\u1ee7a $\\triangle{ACE}$ <br\/> B\u01b0\u1edbc 3 <\/b> : T\u1eeb b\u01b0\u1edbc 1 v\u00e0 b\u01b0\u1edbc 1 suy ra \u0111i\u1ec1u c\u1ea7n ch\u1ee9ng minh <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='img\/H7C3B20_K9.png' \/><\/center> <br\/> $\\blacktriangleright$ G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AE$ v\u00e0 $BC$ <br\/> V\u00ec $AB \/\/ DE$ v\u00e0 $AD \/\/ BE$ (gt) <br\/> $\\Rightarrow$ $AD = BE; AB = DE$ (t\u00ednh ch\u1ea5t \u0111o\u1ea1n ch\u1eafn song song) <br\/> V\u00e0 $\\widehat{BAI} = \\widehat{DEI}$ (so le trong) <br\/> $\\widehat{ABI} = \\widehat{EDI}$ (so le trong) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABI}$ v\u00e0 $\\triangle{EDI}$ c\u00f3: <br\/> $\\begin{cases} \\widehat{ABI} = \\widehat{EDI} \\\\ AB = ED \\\\ \\widehat{BAI} = \\widehat{DEI} \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{ABI} = \\triangle{EDI}$ (g.c.g) <br\/> $\\Rightarrow$ $AI = EI$ v\u00e0 $IB = ID$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> V\u00ec $IB = ID(cmt)$ m\u00e0 $BD = DC$ $\\Rightarrow$ $DC = \\dfrac{2}{3}CI$ (1) <br\/> V\u00ec $AI = EI $ $\\Rightarrow$ $CI$ l\u00e0 trung tuy\u1ebfn c\u1ee7a $\\triangle{ACE}$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $D$ l\u00e0 tr\u1ecdng t\u00e2m tam gi\u00e1c $ACE$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0110\u00daNG <\/span>","column":2}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:23:05"},{"id":"1913","post_id":"1354","mon_id":"0","chapter_id":"0","question":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","options":{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $ABC$, \u0111\u01b0\u1eddng trung tuy\u1ebfn $AM$. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $MA$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $MD = MA$. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. $AB \/\/ CD; AB = CD$ ","B. $AB > CD; AC < BD$","C. $AC \/\/ BD; AC = BD$ ","D. A v\u00e0 C \u0111\u00fang"],"hint":"","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> X\u00e9t $\\triangle{AMB}$ v\u00e0 $\\triangle{CMD}$ <br\/> B\u01b0\u1edbc 2: <\/b> X\u00e9t $\\triangle{AMC}$ v\u00e0 $\\triangle{BMD}$ <br\/> B\u01b0\u1edbc 3: <\/b> D\u1ef1a v\u00e0o b\u01b0\u1edbc 1 v\u00e0 b\u01b0\u1edbc 2 \u0111\u1ec3 ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang. <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='img\/H7C3B20_K3.png' \/><\/center> <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{AMB}$ v\u00e0 $\\triangle{CMD}$ c\u00f3: <br\/> $\\begin{cases} AM = MD (gt) \\\\ \\widehat{M_{3}} = \\widehat{M_{4}} (\\text{\u0111\u1ed1i} \\hspace{0,2cm} \\text{\u0111\u1ec9nh}) \\\\ BM = MC (gt) \\end{cases}$<br\/> $\\Rightarrow$ $\\triangle{AMB} = \\triangle{DMC}$ (c.g.c) <br\/> $\\Rightarrow$ $AB = CD$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\widehat{B_{1}} = \\widehat{C_{1}}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) m\u00e0 hai g\u00f3c n\u00e0y \u1edf v\u1ecb tr\u00ed so le trong, b\u1eb1ng nhau n\u00ean $AB \/\/ CD$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{AMC}$ v\u00e0 $\\triangle{BMD}$ c\u00f3: <br\/> $\\begin{cases} AM = MD (gt) \\\\ MC = MB (gt) \\\\ \\widehat{M_{1}} = \\widehat{M_{2}} (\\text{\u0111\u1ed1i} \\hspace{0,2cm} \\text{\u0111\u1ec9nh}) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{AMC} = \\triangle{DMB}$ (c.g.c) <br\/> $\\Rightarrow$ $AC = BD$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\widehat{A_{1}} = \\widehat{D_{1}}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) m\u00e0 $\\widehat{A_{1}}$ v\u00e0 $\\widehat{D_{1}}$ \u1edf v\u1ecb tr\u00ed so le trong, b\u1eb1ng nhau <br\/> $\\Rightarrow$ $AC \/\/ BD$ <br\/> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D <\/span>","column":2}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:23:05"}]}