{"segment":[{"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":"H\u00e3y ch\u1ecdn m\u1ec7nh \u0111\u1ec1 \u0111\u00fang: Trung tuy\u1ebfn c\u1ee7a m\u1ed9t tam gi\u00e1c l\u00e0 m\u1ed9t \u0111o\u1ea1n th\u1eb3ng","select":["<span class='basic_left'> A. Chia di\u1ec7n t\u00edch tam gi\u00e1c th\u00e0nh hai ph\u1ea7n b\u1eb1ng nhau <\/span> ","<span class='basic_left'> B. Chia \u0111\u00f4i m\u1ed9t g\u00f3c c\u1ee7a tam gi\u00e1c <\/span> ","<span class='basic_left'>C. Vu\u00f4ng g\u00f3c v\u1edbi m\u1ed9t c\u1ea1nh v\u00e0 \u0111i qua trung \u0111i\u1ec3m c\u1ee7a c\u1ea1nh \u0111\u00f3 <\/span> ","<span class='basic_left'> D. Xu\u1ea5t ph\u00e1t t\u1eeb \u0111\u1ec9nh v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi c\u1ea1nh \u0111\u1ed1i di\u1ec7n <\/span> "],"hint":"","explain":"<span class='basic_left'> A - \u0110\u00fang v\u00ec: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D11.png' \/><\/center> <br\/> G\u1ecdi $AM$ l\u00e0 trung tuy\u1ebfn c\u1ee7a $\\triangle{ABC}$ <br\/> Ta c\u00f3: $S_{\\triangle{ABC}} = \\dfrac{1}{2} AH . BC$ (1) <br\/> $S_{\\triangle{ABM}} = \\dfrac{1}{2}AH . BM$ (2) <br\/> $S_{\\triangle{AMC}} = \\dfrac{1}{2}AH . MC$ (3) <br\/> M\u1eb7t kh\u00e1c $BM = MC = \\dfrac{1}{2}BC$ (v\u00ec $M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC$) (4) <br\/> T\u1eeb (1), (2), (3), (4) $\\Rightarrow$ $S_{\\triangle{ABM}} = S_{\\triangle{AMC}} = \\dfrac{1}{2}S_{\\triangle{ABC}}$ <br\/> B _ Sai v\u00ec \u0111\u01b0\u1eddng chia \u0111\u00f4i m\u1ed9t g\u00f3c c\u1ee7a tam gi\u00e1c l\u00e0 <b> \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c <\/b> <br\/> C - Sai v\u00ec \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c v\u1edbi m\u1ed9t c\u1ea1nh v\u00e0 \u0111i qua trung \u0111i\u1ec3m c\u1ee7a c\u1ea1nh \u0111\u00f3 l\u00e0 <b> \u0111\u01b0\u1eddng trung tr\u1ef1c <\/b> <br\/> D - Sai v\u00ec \u0111\u01b0\u1eddng xu\u1ea5t ph\u00e1t t\u1eeb \u0111\u1ec9nh v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi c\u1ea1nh \u0111\u1ed1i di\u1ec7n l\u00e0 <b> \u0111\u01b0\u1eddng cao <\/b> <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A <\/span> <br\/> <span class='basic_green'><i> Nh\u1eadn x\u00e9t: \u1ede b\u00e0i n\u00e0y ta \u0111\u00e3 s\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p so s\u00e1nh di\u1ec7n t\u00edch hai tam gi\u00e1c <br\/> +) N\u1ebfu hai tam gi\u00e1c c\u00f3 c\u00f9ng chi\u1ec1u cao, c\u1ea1nh \u0111\u00e1y c\u1ee7a tam gi\u00e1c n\u00e0y t\u1ec9 l\u1ec7 c\u1ea1nh \u0111\u00e1y c\u1ee7a tam gi\u00e1c kia theo t\u1ec9 s\u1ed1 k th\u00ec di\u1ec7n t\u00edch c\u1ee7a tam gi\u00e1c n\u00e0y c\u0169ng t\u1ec9 l\u1ec7 v\u1edbi di\u1ec7n t\u00edch c\u1ee7a tam gi\u00e1c kia theo t\u1ec9 s\u1ed1 k <br\/> +) N\u1ebfu hai tam gi\u00e1c c\u00f3 c\u00f9ng chung c\u1ea1nh \u0111\u00e1y, nh\u01b0ng chi\u1ec1u cao c\u1ee7a tam gi\u00e1c n\u00e0y t\u1ec9 l\u1ec7 v\u1edbi chi\u1ec1u cao c\u1ee7a tam gi\u00e1c kia theo t\u1ec9 s\u1ed1 k, th\u00ec di\u1ec7n t\u00edch c\u1ee7a tam gi\u00e1c n\u00e0y c\u0169ng t\u1ec9 l\u1ec7 v\u1edbi di\u1ec7n t\u00edch c\u1ee7a tam gi\u00e1c kia theo t\u1ec9 s\u1ed1 k <\/b> <\/span> ","column":1}]}],"id_ques":1891},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"N\u1ebfu $G$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c $PQR$ v\u00e0 $PX$ l\u00e0 m\u1ed9t trung tuy\u1ebfn c\u1ee7a tam gi\u00e1c th\u00ec t\u1ef7 s\u1ed1 $\\dfrac{PG}{GX}$ b\u1eb1ng:","select":["A. $1:1$ ","B. $3:1$ ","C. $2:1$","D. $3:2$"],"hint":"","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D12.png' \/><\/center> <br\/> V\u00ec $G$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c $PQR$ n\u00ean theo t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m ta c\u00f3: <br\/> $PG = \\dfrac{2}{3}PX$ <br\/> $\\Rightarrow$ $GX = \\dfrac{1}{3}PX$ <br\/> $\\Rightarrow$ $\\dfrac{PG}{GX} = \\dfrac{2}{3}PX : \\dfrac{1}{3}PX = 2$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C <\/span> ","column":4}]}],"id_ques":1892},{"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 $ABC$. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $BA$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $BD = BA$. Tr\u00ean c\u1ea1nh $BC$ l\u1ea5y \u0111i\u1ec3m $G$ sao cho $BG = \\dfrac{1}{3}BC$. Khi \u0111\u00f3 $G$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c $ACD$. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>? ","select":[" A. \u0110\u00daNG ","B. SAI "],"hint":"M\u1ed9t \u0111i\u1ec3m l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c khi v\u00e0 ch\u1ec9 khi n\u00f3 thu\u1ed9c \u0111\u01b0\u1eddng trung tuy\u1ebfn v\u00e0 \u0111\u1ed9 d\u00e0i t\u1eeb \u0111i\u1ec3m \u0111\u00f3 t\u1edbi \u0111\u1ec9nh b\u1eb1ng $\\dfrac{2}{3}$ \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng trung tuy\u1ebfn \u0111i qua \u0111\u1ec9nh \u1ea5y. ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $CB$ l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $M \\in BC$ v\u00e0 $CG = \\dfrac{2}{3}BC$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D22.png' \/><\/center> <br\/> Trong tam gi\u00e1c $ACD$ c\u00f3: <br\/> $AB = BD$ (gt) $\\Rightarrow$ $CB$ l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn (1) <br\/> M\u1eb7t kh\u00e1c ta c\u00f3: $BG = \\dfrac{1}{3}BC$ (gt) $\\Rightarrow$ $CG = \\dfrac{2}{3}BC$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $G$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c $ACD$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span> ","column":2}]}],"id_ques":1893},{"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 h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D13.png' \/><\/center> <br\/> Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["<span class='basic_left'> A. N\u1ebfu $PG = 6$ th\u00ec $GF = 4$ <\/span> ","<span class='basic_left'> B. N\u1ebfu $NG = 8$ th\u00ec $NE = 10 $<\/span> ","<span class='basic_left'> C. N\u1ebfu $MD = 9$ th\u00ec $GD = 6$ <\/span>","<span class='basic_left'> D. N\u1ebfu $MG = 4$ th\u00ec $MD = 6$; $GD = 2$ <\/span> "],"hint":"\u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t: Tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c c\u00e1ch m\u1ed7i \u0111\u1ec9nh m\u1ed9t kho\u1ea3ng b\u1eb1ng $\\dfrac{2}{3}$ \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng trung tuy\u1ebfn xu\u1ea5t ph\u00e1t t\u1eeb \u0111\u1ec9nh \u1ea5y ","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D13.png' \/><\/center> <br\/> V\u00ec $G$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c $MNP$ n\u00ean theo t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m ta c\u00f3: <br\/> +) $PG = \\dfrac{2}{3}PF$ $\\Rightarrow$ $\\dfrac{FG}{PF} = \\dfrac{1}{3}$ $\\Rightarrow$ $\\dfrac{FG}{PG} = \\dfrac{1}{2}$ <br\/> N\u1ebfu $PG = 6$ $\\Rightarrow$ $FG = \\dfrac{6}{2} = 3$ <br\/> <b> $\\Rightarrow$ \u0110\u00e1p \u00e1n A sai <\/b> <br\/> +) $NG = \\dfrac{2}{3}NE$ <br\/> N\u1ebfu $NG = 8$ $\\Rightarrow$ $NE = \\dfrac{8 . 3}{2} = 12$ <br\/> <b> $\\Rightarrow$ \u0110\u00e1p \u00e1n B sai <\/b> <br\/> +) $MG = \\dfrac{2}{3}MD$ $\\Rightarrow$ $GD = \\dfrac{1}{3}MD$ <br\/> N\u1ebfu $MD = 9$ $\\Rightarrow$ $GD = \\dfrac{9}{3} = 3$ <br\/> <b> $\\Rightarrow$ \u0110\u00e1p \u00e1n C sai <\/b> <br\/> +) $MG = \\dfrac{2}{3}MD$ $\\Rightarrow$ $MD = \\dfrac{3}{2}MG = \\dfrac{3}{2} . 4 = 6$ <br\/> $GD = \\dfrac{1}{3}MD = \\dfrac{1}{3} . 6 = 2$ <br\/> <b> $\\Rightarrow$ \u0110\u00e1p \u00e1n D \u0111\u00fang <\/b> <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D <\/span> ","column":1}]}],"id_ques":1894},{"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 $ABC$, $M$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC$. $G$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c $ABC$ n\u1ebfu: <br\/> ","select":["A. $G$ thu\u1ed9c \u0111\u01b0\u1eddng th\u1eb3ng $AM$ v\u00e0 $GM = \\dfrac{1}{2}GA$ ","B. $G$ thu\u1ed9c tia $MA$ v\u00e0 $AG = \\dfrac{2}{3}AM$ ","C. $G$ thu\u1ed9c \u0111o\u1ea1n th\u1eb3ng $AM$ v\u00e0 $MG = \\dfrac{1}{2}MA$ ","D. $G$ thu\u1ed9c tia $MA$ v\u00e0 $MG = \\dfrac{1}{2}AG$ "],"hint":"","explain":" <span class='basic_left'> A - Sai v\u00ec c\u00f3 th\u1ec3 s\u1ea3y ra tr\u01b0\u1eddng h\u1ee3p nh\u01b0 h\u00ecnh sau: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D02.png' \/><\/center> <br\/> B - Sai v\u00ec c\u00f3 th\u1ec3 x\u1ea3y ra tr\u01b0\u1eddng h\u1ee3p: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D03.png' \/><\/center> <br\/> C - Sai v\u00ec: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D04.png' \/><\/center> <br\/> D - \u0110\u00fang v\u00ec: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D01.png' \/><\/center> <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D <\/span> ","column":2}]}],"id_ques":1895},{"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 $G$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c $MNP$ v\u1edbi \u0111\u01b0\u1eddng trung tuy\u1ebfn $MI$. Trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau kh\u1eb3ng \u0111\u1ecbnh n\u00e0o \u0111\u00fang? ","select":["A. $\\dfrac{MG}{MI} = \\dfrac{1}{2} $ ","B. $\\dfrac{GI}{MG} = \\dfrac{2}{3} $ ","C. $\\dfrac{MG}{GI} = \\dfrac{2}{3} $ ","D. $\\dfrac{GI}{MI} = \\dfrac{1}{3} $ "],"hint":"D\u1ef1a v\u00e0o \u0111\u1ecbnh l\u00fd v\u1ec1 t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng trung tuy\u1ebfn trong tam gi\u00e1c","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D05.png' \/><\/center> <br\/> Theo \u0111\u1ecbnh l\u00fd v\u1ec1 t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng trung tuy\u1ebfn trong tam gi\u00e1c, ta c\u00f3: <br\/> +) $\\dfrac{MG}{MI} = \\dfrac{2}{3} $ <br\/> +) $\\dfrac{GI}{MG} = \\dfrac{1}{2}$ <br\/> +) $\\dfrac{MG}{GI} = 2$ <br\/> +) $\\dfrac{GI}{MI} = \\dfrac{1}{3}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D. $\\dfrac{GI}{MI} = \\dfrac{1}{3} $ <\/span> ","column":2}]}],"id_ques":1896},{"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":[[["2"],["3"],["1"],["3"],["1"],["2"]]],"list":[{"point":5,"width":50,"ques":" Cho h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D06.png' \/><\/center> <br\/> H\u00e3y \u0111i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o m\u1ed7i \u00f4 tr\u1ed1ng trong c\u00e1c \u0111\u1eb3ng th\u1ee9c sau: <br\/> $DG$ = <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>$DI$; $GI$ = <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>$DI$; $GI$ = <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>$DG$ ","hint":"D\u1ef1a v\u00e0o \u0111\u1ecbnh l\u00fd v\u1ec1 t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng trung tuy\u1ebfn trong tam gi\u00e1c","explain":"Theo \u0111\u1ecbnh l\u00fd v\u1ec1 t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng trung tuy\u1ebfn c\u1ee7a tam gi\u00e1c, ta c\u00f3: <br\/> $\\dfrac{DG}{DI} = \\dfrac{2}{3}$ hay $DG = \\dfrac{2}{3}DI$ <br\/> $\\Rightarrow$ $GI = \\dfrac{1}{3}DI$ <br\/> $GI = \\dfrac{1}{2}DG$"}]}],"id_ques":1897},{"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"],["2"],["3"],["2"]]],"list":[{"point":5,"width":50,"type_input":"","input_hint":["frac"],"ques":" Cho h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D06.png' \/><\/center> <br\/> H\u00e3y \u0111i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o m\u1ed7i \u00f4 tr\u1ed1ng trong c\u00e1c \u0111\u1eb3ng th\u1ee9c sau: <br\/> $EK$ = <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>$EG$; $EK$ = _input_ $GK$; $EG$ = _input_ $GK$ ","hint":"D\u1ef1a v\u00e0o \u0111\u1ecbnh l\u00fd v\u1ec1 t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng trung tuy\u1ebfn trong tam gi\u00e1c","explain":"Theo \u0111\u1ecbnh l\u00fd v\u1ec1 t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng trung tuy\u1ebfn c\u1ee7a tam gi\u00e1c, ta c\u00f3: <br\/> $\\dfrac{EG}{EK} = \\dfrac{2}{3}$ hay $EG = \\dfrac{2}{3}EK$ <br\/> $\\Rightarrow$ $EK = \\dfrac{3}{2}EG$ <br\/> $EK = 3GK$ <br\/> $EG = 2GK$"}]}],"id_ques":1898},{"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":[[["10"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd: \"Tr\u1ecdng t\u00e2m tam gi\u00e1c chia tam gi\u00e1c th\u00e0nh ba tam gi\u00e1c nh\u1ecf c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau\" \u0111\u1ec3 gi\u1ea3i b\u00e0i to\u00e1n sau: <br\/> Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 $AB = 5cm; BC = 13cm$. \u0110\u01b0\u1eddng trung tuy\u1ebfn $AM$ v\u00e0 $O$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c $ABC$. T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $BOC$. <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $S_{\\triangle{BOC}}$ = _input_ $cm^2$ ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $ABC$ r\u1ed3i t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $BOC$ d\u1ef1a v\u00e0o \u0111\u1ecbnh l\u00fd \u0111\u00e3 cho <br\/><br\/> <span class='basic_green'> B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D19.png' \/><\/center> <br\/> $\\blacktriangleright$ V\u00ec $O$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a $\\triangle{ABC}$, n\u00ean ta c\u00f3: <br\/> $S_{\\triangle{BOC}} = S_\\triangle{AOB} = \\triangle{AOC} = \\dfrac{1}{3}S_{\\triangle{ABC}}$ (tr\u1ecdng t\u00e2m tam gi\u00e1c chia tam gi\u00e1c th\u00e0nh ba tam gi\u00e1c nh\u1ecf c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau) <br\/> $\\blacktriangleright$ $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$, n\u00ean ta c\u00f3: <br\/> $\\begin{align} AB^2 + AC^2 &= BC^2 \\\\ \\Rightarrow AC^2 &= BC^2 - AB^2 \\\\ &= 13^2 - 5^2 \\\\ &= 169 - 25 \\\\ &= 144 \\end{align}$ <br\/> $\\Rightarrow$ $AC = 12(cm)$ <br\/> $S_{\\triangle{ABC}} = \\dfrac{1}{2}AB.AC = \\dfrac{1}{2}.5 . 12 = 30(cm^2)$ <br\/> $\\Rightarrow$ $S_\\triangle{BOC} = \\dfrac{1}{3}.30 = 10(cm^2)$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $10$ <\/span> "}]}],"id_ques":1899},{"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 $ABC$, \u0111\u01b0\u1eddng trung tuy\u1ebfn $AM$ v\u00e0 tr\u1ecdng t\u00e2m $G$. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $MA$ l\u1ea5y c\u00e1c \u0111i\u1ec3m $I$ v\u00e0 $K$ sao cho $MI = MG$; $IK = IG$. G\u1ecdi $N$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $CK$. <br\/> Khi \u0111\u00f3, ba \u0111i\u1ec3m $B; I; N$ th\u1eb3ng h\u00e0ng. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>? ","select":[" A. \u0110\u00daNG","B. SAI "],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $I$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c $BKC$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $B; I; N$ th\u1eb3ng h\u00e0ng <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D20.png' \/><\/center> <br\/> $\\triangle{KBC}$ c\u00f3 $KM$ l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn <br\/> V\u00e0 $IM = \\dfrac{1}{2}IG = \\dfrac{1}{2}IK$ (gt) <br\/> $\\Rightarrow$ $I$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c $BKC$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) <br\/> M\u00e0 $BN$ c\u0169ng l\u00e0 trung tuy\u1ebfn c\u1ee7a tam gi\u00e1c $BKC$ n\u00ean $I$ thu\u1ed9c $BN$ <br\/> Do \u0111\u00f3, ba \u0111i\u1ec3m $B; I; N$ th\u1eb3ng h\u00e0ng <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span> ","column":2}]}],"id_ques":1900},{"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"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 trung tuy\u1ebfn $BM = 3cm$, trung tuy\u1ebfn $CN = 4,5cm$. T\u00ednh \u0111\u1ed9 d\u00e0i $BC$, bi\u1ebft r\u1eb1ng \u0111\u1ed9 d\u00e0i \u0111\u00f3 (\u0111\u01a1n v\u1ecb: $cm$) l\u00e0 m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean l\u1ebb. <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $BC$ = _input_ $cm$ ","hint":"D\u1ef1a v\u00e0o b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> G\u1ecdi $G$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c $ABC$ <br\/> T\u00ednh $BG$ v\u00e0 $CG$ d\u1ef1a v\u00e0o t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m <br\/> <b> B\u01b0\u1edbc 2: <\/b> \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c v\u00e0 h\u1ec7 qu\u1ea3 cho tam gi\u00e1c $GBC$, t\u00ecm s\u1ed1 \u0111o c\u1ea1nh $BC$ <br\/><br\/> <span class='basic_green'> B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D21.png' \/><\/center> <br\/> $\\blacktriangleright$ G\u1ecdi $G$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $BM$ v\u00e0 $CN$ <br\/> Ta c\u00f3: $BG = \\dfrac{2}{3}BM = \\dfrac{2}{3}.3 = 2(cm)$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) <br\/> $CG = \\dfrac{2}{3}CN = \\dfrac{2}{3}.4,5 = 3(cm)$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) <br\/> $\\blacktriangleright$ \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c v\u00e0 h\u1ec7 qu\u1ea3 cho tam gi\u00e1c $GBC$, ta c\u00f3: <br\/> $CG - BG < BC < CG + BG$ <br\/> T\u1ee9c l\u00e0: $3 - 2 < BC < 3 + 2$ <br\/> $1 < BC < 5$ <br\/> V\u00ec $BC$ l\u00e0 s\u1ed1 l\u1ebb n\u00ean $BC = 3cm$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $3$ <\/span> "}]}],"id_ques":1901},{"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":[[["25"],["3"]]],"list":[{"point":5,"width":50,"type_input":"","input_hint":["frac"],"ques":" Bi\u1ebft r\u1eb1ng trong tam gi\u00e1c vu\u00f4ng, \u0111\u01b0\u1eddng trung tuy\u1ebfn \u1ee9ng v\u1edbi c\u1ea1nh huy\u1ec1n b\u1eb1ng m\u1ed9t n\u1eeda c\u1ea1nh huy\u1ec1n. Cho tam gi\u00e1c $MNP$ vu\u00f4ng t\u1ea1i $M$ c\u00f3 hai c\u1ea1nh g\u00f3c vu\u00f4ng $MN = 7cm$; $MP = 24cm$. T\u00ednh kho\u1ea3ng c\u00e1ch t\u1eeb \u0111\u1ec9nh $M$ t\u1edbi tr\u1ecdng t\u00e2m $G$ c\u1ee7a tam gi\u00e1c. <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $MG$ = <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>$cm$ ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> D\u1ef1a v\u00e0o \u0111\u1ecbnh l\u00fd Pitago t\u00ednh $NP$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng trung tuy\u1ebfn (trung tuy\u1ebfn \u1ee9ng v\u1edbi c\u1ea1nh huy\u1ec1n b\u1eb1ng n\u1eeda c\u1ea1nh huy\u1ec1n) <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh \u0111\u1ed9 d\u00e0i $MG$ <br\/><br\/> <span class='basic_green'> B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D07.png' \/><\/center> <br\/> G\u1ecdi $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $NP$ <br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pitago cho tam gi\u00e1c vu\u00f4ng $MNP$, ta c\u00f3: <br\/> $\\begin{align} NP^2 &= MP^2 + MN^2 \\\\ &= 24^2 + 7^2 \\\\ &= 625 \\end{align}$ <br\/> $\\Rightarrow$ $NP = 25(cm)$ <br\/> $\\triangle{MNP}$ vu\u00f4ng t\u1ea1i $M$ c\u00f3 trung tuy\u1ebfn $MI$ \u1ee9ng v\u1edbi c\u1ea1nh huy\u1ec1n $NP$ <br\/> $\\Rightarrow$ $MI = \\dfrac{1}{2}NP = \\dfrac{1}{2}. 25 = 12,5(cm)$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) <br\/> V\u00ec $G$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a $\\triangle{MNP}$ n\u00ean ta c\u00f3: <br\/> $MG = \\dfrac{2}{3}MI = \\dfrac{2}{3} . 12,5 = \\dfrac{25}{3}(cm)$"}]}],"id_ques":1902},{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["CN","NC"]]],"list":[{"point":5,"width":50,"type_input":"","ques":" Cho tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$, hai \u0111\u01b0\u1eddng trung tuy\u1ebfn $BM$ v\u00e0 $CN$. <br\/> Khi \u0111\u00f3 ta c\u00f3: $BM$ = _input_ ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Ch\u1ee9ng minh $\\triangle{ABM} = \\triangle{ACN}$ <br\/><br\/> <span class='basic_green'> B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D08.png' \/><\/center> <br\/> $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$ n\u00ean $AB = AC$ (1) <br\/> $BM$ l\u00e0 trung tuy\u1ebfn n\u00ean $MA = MC = \\dfrac{AC}{2}$ (2) <br\/> $CN$ l\u00e0 trung tuy\u1ebfn n\u00ean $NA = NB = \\dfrac{AB}{2}$ (3) <br\/> T\u1eeb (1), (2), (3) $\\Rightarrow$ $MA = MC = NA = NB$ <br\/> X\u00e9t $\\triangle{ABM}$ v\u00e0 $\\triangle{ACN}$ c\u00f3: <br\/> $\\begin{cases} AN = AM (cmt) \\\\ \\widehat{A} \\hspace{0,2cm} \\text{chung} \\\\ AB = AC (cmt) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{ABM} = \\triangle{ACN}$ (c.g.c) <br\/> $\\Rightarrow$ $BM = CN$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $CN$ <\/span> <br\/> <span class='basic_green'> <i> L\u01b0u \u00fd: Ta c\u00f3 c\u00e1c k\u1ebft qu\u1ea3 <br\/> +) Trong tam gi\u00e1c c\u00e2n, hai \u0111\u01b0\u1eddng trung tuy\u1ebfn \u1ee9ng v\u1edbi hai c\u1ea1nh b\u00ean th\u00ec b\u1eb1ng nhau <br\/> +) Trong tam gi\u00e1c \u0111\u1ec1u c\u1ea1nh $a$, \u0111\u1ed9 d\u00e0i ba \u0111\u01b0\u1eddng trung tuy\u1ebfn b\u1eb1ng nhau v\u00e0 b\u1eb1ng $\\dfrac{a\\sqrt{3}}{2}$ <\/i> <\/span> "}]}],"id_ques":1903},{"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 $DEF$ c\u00f3 $DK$, $EH$ l\u00e0 c\u00e1c \u0111\u01b0\u1eddng trung tuy\u1ebfn. Bi\u1ebft $DK = EH$, khi \u0111\u00f3 tam gi\u00e1c $DEF$ c\u00e2n t\u1ea1i $F$. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>? ","select":["A. \u0110\u00daNG ","B. SAI "],"hint":"Ch\u1ee9ng minh $FD = FE$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $DH = EK$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $FD = FE$ $\\Rightarrow$ $\\triangle{DEF}$ c\u00e2n <br\/><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D09.png' \/><\/center> <br\/> G\u1ecdi $G$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c $DEF$ <br\/> Trong $\\triangle{DEF}$ c\u00f3: <br\/> $DK = EH$ (gt) (1) <br\/> $GE = \\dfrac{2}{3}EH$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) (2) <br\/> $GD = \\dfrac{2}{3}DK$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) (3) <br\/> T\u1eeb (1), (2), (3) $\\Rightarrow$ $GE = GD; GH = GK$ <br\/> X\u00e9t $\\triangle{HGD}$ v\u00e0 $\\triangle{KGE}$ c\u00f3: <br\/> $\\begin{cases} GH = GK (cmt) \\\\ \\widehat{HGD} = \\widehat{KGE} (\\text{\u0111\u1ed1i} \\hspace{0,2cm} \\text{\u0111\u1ec9nh}) \\\\ GD = GE (cmt) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{HGD} = \\triangle{KGE}$ (c.g.c) <br\/> $\\Rightarrow$ $DH = EK$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow$ $2DH = 2EK$ (4) <br\/> M\u1eb7t kh\u00e1c ta c\u00f3: $DH = HF$; $FK = KE$ (5) <br\/> T\u1eeb (4) v\u00e0 (5) $\\Rightarrow$ $FD = FE$ <br\/> Suy ra, $\\triangle{DEF}$ c\u00e2n t\u1ea1i $F$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span> ","column":2}]}],"id_ques":1904},{"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 $ABC$ c\u00e2n t\u1ea1i $A$, c\u00f3 $AB = AC = 17cm$; $BC = 16cm$. K\u1ebb trung tuy\u1ebfn $AM$ <br\/> Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. $AM \\perp BC$ ","B. $AM = \\dfrac{1}{2}BC$ ","C. $AM = 15cm$","D. A v\u00e0 C \u0111\u00fang"],"hint":"Ch\u1ee9ng minh $AM \\perp BC$ r\u1ed3i t\u00ednh $AM$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $AM \\perp BC$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> D\u1ef1a v\u00e0o \u0111\u1ecbnh l\u00fd Pitago t\u00ednh $AM$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D10.png' \/><\/center> <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{AMB}$ v\u00e0 $\\triangle{AMC}$ c\u00f3: <br\/> $\\begin{cases} AB = AC (gt) \\\\ AM \\hspace{0,2cm} \\text{chung} \\\\ BM = MC (gt) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{AMB} = \\triangle{AMC}$ (c.c.c) <br\/> $\\Rightarrow$ $\\widehat{AMB} = \\widehat{AMC} = \\dfrac{\\widehat{BMC}}{2} = \\dfrac{180^{o}}{2} = 90^{o}$ <br\/> $\\Rightarrow$ $AM \\perp BC$ <br\/> $\\blacktriangleright$ Ta c\u00f3: $MB = MC = \\dfrac{BC}{2} = \\dfrac{16}{2} = 8(cm)$ <br\/> $\\triangle{AMB}$ vu\u00f4ng t\u1ea1i $M$, theo \u0111\u1ecbnh l\u00fd Pitago ta c\u00f3: <br\/> $\\begin{align} AM^2 + MB^2 &= AB^2 \\\\ \\Rightarrow AM^2 &= AB^2 - MB^2 \\\\ &= 17^2 - 8^2 \\\\ &= 289 - 64 \\\\ &= 225(cm) \\end{align}$ <br\/> $\\Rightarrow$ $AM = 15cm$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D <\/span> ","column":2}]}],"id_ques":1905},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$, trung tuy\u1ebfn $AM$ <br\/> Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. $AM = BC$ ","B. $AM = AB$ ","C. $AM = \\dfrac{1}{2}BC$ ","D. $AM = AC$"],"hint":"Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $MA$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $MA = MD$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $AB = CD$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $AD = BC$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> Ch\u1ee9ng minh $AM = \\dfrac{1}{2}BC$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D14.png' \/><\/center> <br\/> Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $MA$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $MA = MD$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{AMB}$ v\u00e0 $\\triangle{CMD}$ c\u00f3: <br\/> $\\begin{cases} MB = MC (gt) \\\\ \\widehat{M_{1}} = \\widehat{M_{2}} (\\text{\u0111\u1ed1i} \\hspace{0,2cm} \\text{\u0111\u1ec9nh}) \\\\ MA = MD (\\text{c\u00e1ch} \\hspace{0,2cm} \\text{l\u1ea5y} \\hspace{0,2cm} \\text{\u0111i\u1ec3m} D) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{AMB} = \\triangle{DMC}$ (c.g.c) <br\/> $\\Rightarrow$ $AB = CD$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\widehat{ABM} = \\widehat{DCM}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> Hai g\u00f3c n\u00e0y \u1edf v\u1ecb tr\u00ed so le trong n\u00ean $AB \/\/ CD$ <br\/> M\u00e0 $AB \\perp AC$ $\\Rightarrow$ $CD \\perp AC$ (t\u1eeb vu\u00f4ng g\u00f3c \u0111\u1ebfn song song) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABC}$ v\u00e0 $\\triangle{CDA}$ c\u00f3: <br\/> $\\begin{cases} AB = CD (cmt) \\\\ \\widehat{BAC} = \\widehat{DCA} = 90^{o} (cmt) \\\\ AC \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{ABC} = \\triangle{DCA}$ (c.g.c) <br\/> $\\Rightarrow$ $AD = BC$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow$ $AM = \\dfrac{1}{2}AD = \\dfrac{1}{2}BC$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C <\/span> <br\/> <span class='basic_green'><i> Nh\u1eadn x\u00e9t: Trong tam gi\u00e1c vu\u00f4ng trung tuy\u1ebfn \u1ee9ng v\u1edbi c\u1ea1nh huy\u1ec1n b\u1eb1ng n\u1eeda c\u1ea1nh huy\u1ec1n <\/i> <\/span> ","column":2}]}],"id_ques":1906},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D16.png' \/><\/center> <br\/> Cho tam gi\u00e1c $ABC$, trung tuy\u1ebfn $AM$, bi\u1ebft $AM = \\dfrac{1}{2}BC$. <br\/> Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$ ","B. $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $B$ ","C. $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $C$ "],"hint":"Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $MA$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $MA = MD$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $\\widehat{A_{1}} = \\widehat{B}$; $\\widehat{A_{2}} = \\widehat{C}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ee9ng minh $\\widehat{A} = \\widehat{B} + \\widehat{C} = 90^{o}$ <br\/> <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D16.png' \/><\/center> <br\/> V\u00ec $AM = \\dfrac{1}{2}BC$ $\\Rightarrow$ $AM = MB = MC$ <br\/> $\\Rightarrow$ $\\triangle{MAB}$ v\u00e0 $\\triangle{MAC}$ c\u00e2n t\u1ea1i $M$ (\u0111\u1ecbnh ngh\u0129a tam gi\u00e1c c\u00e2n) <br\/> $\\Rightarrow$ $\\widehat{C} = \\widehat{A_{2}}$ v\u00e0 $\\widehat{B} = \\widehat{A_{1}}$ <br\/> $\\Rightarrow$ $\\widehat{A_{1}} + \\widehat{A_{2}} = \\widehat{B} + \\widehat{C}$ <br\/> $\\Leftrightarrow$ $\\widehat{A} = \\widehat{B} + \\widehat{C} = \\dfrac{180^{o}}{2} = 90^{o}$ <br\/> Do \u0111\u00f3, $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t, ta v\u1eeba ch\u1ee9ng minh m\u1ed9t \u0111\u1ecbnh l\u00fd th\u01b0\u1eddng d\u00f9ng: N\u1ebfu m\u1ed9t tam gi\u00e1c c\u00f3 \u0111\u01b0\u1eddng trung tuy\u1ebfn \u1ee9ng v\u1edbi m\u1ed9t c\u1ea1nh b\u1eb1ng n\u1eeda c\u1ea1nh \u1ea5y th\u00ec tam gi\u00e1c \u0111\u00f3 l\u00e0 tam gi\u00e1c vu\u00f4ng <\/i> <\/span> ","column":3}]}],"id_ques":1907},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 $AB < AC$, hai trung tuy\u1ebfn $BE$ v\u00e0 $CF$, $G$ l\u00e0 tr\u1ecdng t\u00e2m. H\u00e3y so s\u00e1nh $BF$ v\u00e0 $CF$. ","select":["A. $BE = CF$ ","B. $BE > CF$ ","C. $BE < CF$ "],"hint":"D\u1ef1a v\u00e0o \u0111\u1ecbnh l\u00fd b\u1ed5 sung v\u1ec1 quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong hai tam gi\u00e1c c\u00f3 hai c\u1eb7p c\u1ea1nh b\u1eb1ng nhau nh\u01b0ng c\u1eb7p g\u00f3c xen gi\u1eefa kh\u00f4ng b\u1eb1ng nhau","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> So s\u00e1nh $\\widehat{ADB}$ v\u00e0 $\\widehat{ADC}$ d\u1ef1a v\u00e0o vi\u1ec7c x\u00e9t $\\triangle{ADB}$ v\u00e0 $\\triangle{ADC}$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> So s\u00e1nh $GB$ v\u00e0 $GC$ d\u1ef1a v\u00e0o vi\u1ec7c x\u00e9t $\\triangle{GBD}$ v\u00e0 $\\triangle{GCD}$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> So s\u00e1nh $BE$ v\u00e0 $CF$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D15.png' \/><\/center> <br\/>V\u1ebd trung tuy\u1ebfn $AD$ $\\Rightarrow$ $G \\in AD$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ADB}$ v\u00e0 $\\triangle{ADB}$ c\u00f3: <br\/> $AD$ : C\u1ea1nh chung <br\/> $AB < AC$ (gt) <br\/> $DB = DC$ (v\u00ec $AD$ l\u00e0 trung tuy\u1ebfn) <br\/> $\\Rightarrow$ $\\widehat{ADB} < \\widehat{ADC}$ (\u0111\u1ed1i di\u1ec7n v\u1edbi c\u1ea1nh nh\u1ecf h\u01a1n - \u0111\u1ecbnh l\u00fd b\u1ed5 sung) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{GDB}$ v\u00e0 $\\triangle{GDC}$ c\u00f3: <br\/> $GD$: C\u1ea1nh chung <br\/> $\\widehat{GDB} < \\widehat{GDC}$ (cmt) <br\/> $DB = DC$ (gt) <br\/> $\\Rightarrow$ $GB < GC$ (\u0111\u1ed1i di\u1ec7n v\u1edbi g\u00f3c nh\u1ecf h\u01a1n - \u0111\u1ecbnh l\u00fd b\u1ed5 sung) (1) <br\/> M\u1eb7t kh\u00e1c ta c\u00f3: $GC = \\dfrac{2}{3}CF$; $GB = \\dfrac{2}{3}BE$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $BE < CF$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eafc l\u1ea1i: \u0110\u1ecbnh l\u00fd b\u1ed5 sung v\u1ec1 quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n <br\/> N\u1ebfu hai tam gi\u00e1c c\u00f3 hai c\u1eb7p c\u1ea1nh b\u1eb1ng nhau nh\u01b0ng c\u1eb7p g\u00f3c xen gi\u1eefa kh\u00f4ng b\u1eb1ng nhau th\u00ec c\u1ea1nh \u0111\u1ed1i di\u1ec7n v\u1edbi n\u00f3 c\u0169ng kh\u00f4ng b\u1eb1ng nhau. C\u1ee5 th\u1ec3: C\u1ea1nh \u0111\u1ed1i di\u1ec7n v\u1edbi g\u00f3c l\u1edbn h\u01a1n l\u00e0 c\u1ea1nh l\u1edbn h\u01a1n v\u00e0 ng\u01b0\u1ee3c l\u1ea1i <\/i> <\/span> ","column":3}]}],"id_ques":1908},{"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 $MNP$ \u0111\u1ec1u v\u00e0 c\u00f3 $G$ l\u00e0 tr\u1ecdng t\u00e2m. H\u00e3y so s\u00e1nh $GM$; $GP$ v\u00e0 $GN$. ","select":["A. $GM = GN = GP$ ","B. $GM < GN = GP$ ","C. $GM > GN > GP$ ","D. $GM = GN > GP $ "],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> So s\u00e1nh $MD; NE; PF$ l\u00e0 c\u00e1c \u0111\u01b0\u1eddng trung tuy\u1ebfn c\u1ee7a tam gi\u00e1c <br\/> <b> B\u01b0\u1edbc 2: <\/b> D\u1ef1a v\u00e0o t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c, so s\u00e1nh $GM; GN$ v\u00e0 $GP$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D17.png' \/><\/center> <br\/> G\u1ecdi $D, E, F$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 giao \u0111i\u1ec3m c\u1ee7a c\u00e1c tia $MG; NG; PG$ v\u1edbi $NP; MP$ v\u00e0 $MN$ <br\/> $\\Rightarrow$ $D; E; F$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u00e1c c\u1ea1nh $NP; MP$ v\u00e0 $MN$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{NEP}$ v\u00e0 $\\triangle{PFN}$ c\u00f3: <br\/> $\\begin{cases} NP \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{EPN} = \\widehat{FNP} = 60^{o} (gt) \\\\ FN = EP (\\text{v\u00ec} \\hspace{0,2cm} MP = MN) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{NEP} = \\triangle{PNF}$ (c.g.c) <br\/> $\\Rightarrow$ $NE = PF$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{MEN}$ v\u00e0 $\\triangle{MDN}$ c\u00f3: <br\/> $\\begin{cases} MN \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{EMN} = \\widehat{DNM} = 60^{o} (gt) \\\\ ME = ND (\\text{v\u00ec} NP = MP) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{MEN} = \\triangle{MDN}$ (c.g.c) <br\/> $\\Rightarrow$ $MD = NE$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (2) <br\/> $\\blacktriangleright$ M\u1eb7t kh\u00e1c ta c\u00f3: $GM = \\dfrac{2}{3} MD$; $GN = \\dfrac{2}{3}NE$; $GP = \\dfrac{2}{3}PF$ (t\u00ednh ch\u1ea5t tr\u1ecdng t\u00e2m) (3) <br\/> T\u1eeb (1), (2), (3) $\\Rightarrow$ $GM = GN = GP$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: Trong tam gi\u00e1c \u0111\u1ec1u \u0111\u1ed9 d\u00e0i ba \u0111\u01b0\u1eddng trung tuy\u1ebfn b\u1eb1ng nhau v\u00e0 tr\u1ecdng t\u00e2m c\u00e1ch \u0111\u1ec1u ba \u0111\u1ec9nh c\u1ee7a tam gi\u00e1c <\/i> <\/span> ","column":2}]}],"id_ques":1909},{"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 $ABC$ vu\u00f4ng \u1edf $A$ c\u00f3 $AB = 5cm$; $BC = 13cm$. Ba \u0111\u01b0\u1eddng trung tuy\u1ebfn $AM, BN, CE$ c\u1eaft nhau t\u1ea1i $O$. T\u00ednh $AM; BN; CE$. ","select":[" A. $AM = 6,5cm; BN \\approx 7,9cm ; CE \\approx 12,3cm $ ","B. $AM = 6,5cm; BN \\approx 7,5cm ; CE \\approx 12,8cm $ ","C. $AM = 2,5cm; BN \\approx 7,9cm ; CE \\approx 12,3cm $ ","D. $AM = 6cm; BN \\approx 7,9cm ; CE \\approx 12,3cm $ "],"hint":"\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd trong tam gi\u00e1c vu\u00f4ng trung tuy\u1ebfn \u1ee9ng v\u1edbi c\u1ea1nh huy\u1ec1n b\u1eb1ng n\u1eeda c\u1ea1nh huy\u1ec1n \u0111\u1ec3 t\u00ednh $AM$.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> T\u00ednh $AM$ (trung tuy\u1ebfn \u1ee9ng v\u1edbi c\u1ea1nh huy\u1ec1n b\u1eb1ng n\u1eeda c\u1ea1nh huy\u1ec1n) <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $AC$ (\u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pitago cho tam gi\u00e1c vu\u00f4ng $ABC$) <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh $BN$ v\u00e0 $CE$ (\u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pitago cho c\u00e1c tam gi\u00e1c vu\u00f4ng $ACE$ v\u00e0 $ABN$) <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai20/lv1/img\/H7C3B20_D18.png' \/><\/center> <br\/> V\u00ec $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$, $AM, BN, CE$ l\u00e0 c\u00e1c \u0111\u01b0\u1eddng trung tuy\u1ebfn <br\/> $\\Rightarrow$ $AM = \\dfrac{1}{2}BC = \\dfrac{13}{2} = 6,5(cm)$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) <br\/> $AE = \\dfrac{1}{2}AB = \\dfrac{5}{2} = 2,5cm$ (v\u00ec $CE$ l\u00e0 trung tuy\u1ebfn) <br\/> $\\blacktriangleright$ \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pitago cho tam gi\u00e1c vu\u00f4ng $ABC$ ta c\u00f3: <br\/> $\\begin{align} AB^2 + AC^2 & = BC^2 \\\\ \\Rightarrow AC^2 &= BC^2 - AB^2 \\\\ &= 13^2 - 5^2 \\\\ &= 169 - 25 \\\\ &= 144 \\end{align}$ <br\/> $\\Rightarrow$ $AC = 12(cm)$ <br\/> $\\blacktriangleright$ \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pitago cho tam gi\u00e1c vu\u00f4ng $AEC$, ta c\u00f3: <br\/> $\\begin{align} EC^2 &= AE^2 + AC^2 \\\\ &= 2,5^2 + 12^2 \\\\ &= 6,25 + 144 \\\\ &= 150,25 \\end{align}$ <br\/> $\\Rightarrow$ $EC \\approx 12,3(cm)$ <br\/> $\\Rightarrow$ $AN = \\dfrac{1}{2}AC = \\dfrac{12,3}{2} = 6,15(cm)$ (v\u00ec $BN$ l\u00e0 trung tuy\u1ebfn) <br\/> $\\blacktriangleright$ \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pitago cho tam gi\u00e1c vu\u00f4ng $ABN$, ta c\u00f3: <br\/> $\\begin{align} BN^2 &= AB^2 + AN^2 \\\\ &= 5^2 + 6,15^2 \\\\ & \\approx 62,8 \\end{align}$ <br\/> $\\Rightarrow$ $BN \\approx 7,9(cm)$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A <\/span> ","column":1}]}],"id_ques":1910}],"lesson":{"save":0,"level":1}}