{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Tr\u1ef1c t\u00e2m c\u1ee7a m\u1ed9t tam gi\u00e1c l\u00e0 \u0111i\u1ec3m c\u1eaft nhau c\u1ee7a: ","select":["A. Ba \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a tam gi\u00e1c ","B. Ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c","C. Ba \u0111\u01b0\u1eddng trung tuy\u1ebfn c\u1ee7a tam gi\u00e1c","D. Ba \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c"],"hint":"","explain":" Tr\u1ef1c t\u00e2m c\u1ee7a tam gi\u00e1c l\u00e0 giao \u0111i\u1ec3m c\u1ee7a ba \u0111\u01b0\u1eddng cao <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D <\/span> ","column":2}]}],"id_ques":1981},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Ba \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c c\u1eaft nhau t\u1ea1i m\u1ed9t \u0111i\u1ec3m g\u1ecdi l\u00e0: ","select":["A. Tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c","B. Tr\u1ef1c t\u00e2m c\u1ee7a tam gi\u00e1c","C. T\u00e2m c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp","D. T\u00e2m c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp"],"hint":"","explain":" Ba \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c c\u00f9ng \u0111i qua m\u1ed9t \u0111i\u1ec3m, \u0111i\u1ec3m n\u00e0y l\u00e0 tr\u1ef1c t\u00e2m c\u1ee7a tam gi\u00e1c <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: B <\/span> ","column":2}]}],"id_ques":1982},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Cho tam gi\u00e1c vu\u00f4ng $MNP$ nh\u01b0 h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv1/img\/H7C3B23_D01.png' \/><\/center> <br\/> Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y \u0111\u00fang? ","select":["A. Tr\u1ef1c t\u00e2m c\u1ee7a tam gi\u00e1c n\u1eb1m b\u00ean trong tam gi\u00e1c","B. Tr\u1ef1c t\u00e2m c\u1ee7a tam gi\u00e1c n\u1eb1m b\u00ean ngo\u00e0i tam gi\u00e1c","C. Tr\u1ef1c t\u00e2m c\u1ee7a tam gi\u00e1c l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u1ea1nh huy\u1ec1n $MN$ ","D. Tr\u1ef1c t\u00e2m c\u1ee7a tam gi\u00e1c tr\u00f9ng v\u1edbi \u0111i\u1ec3m $N$ "],"hint":"","explain":" <span class='basic_left'> V\u00ec $\\triangle{MNP}$ vu\u00f4ng t\u1ea1i $N$ <br\/> N\u00ean $NM \\perp NP$ t\u1ea1i $N$ <br\/> M\u00e0 ba \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c c\u00f9ng \u0111i qua m\u1ed9t \u0111i\u1ec3m n\u00ean giao \u0111i\u1ec3m ba \u0111\u01b0\u1eddng cao ch\u00ednh l\u00e0 $N$ <br\/> Hay tr\u1ef1c t\u00e2m c\u1ee7a tam gi\u00e1c tr\u00f9ng v\u1edbi \u0111i\u1ec3m $N$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: N\u1ebfu tam gi\u00e1c nh\u1ecdn th\u00ec tr\u1ef1c t\u00e2m \u1edf b\u00ean trong \u0111\u01b0\u1eddng tr\u00f2n <br\/> N\u1ebfu tam gi\u00e1c vu\u00f4ng th\u00ec tr\u1ef1c t\u00e2m tr\u00f9ng v\u1edbi \u0111\u1ec9nh tam gi\u00e1c vu\u00f4ng <br\/> N\u1ebfu tam gi\u00e1c t\u00f9 th\u00ec tr\u1ef1c t\u00e2m n\u1eb1m b\u00ean ngo\u00e0i tam gi\u00e1c <\/i> <\/span> ","column":1}]}],"id_ques":1983},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Cho tam gi\u00e1c $ABC$, $Ax$ l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BAC}$ v\u00e0 $AD$ l\u00e0 \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c v\u1edbi c\u1ea1nh $BC$. N\u1ebfu $\\widehat{B} = 40^{o}$ v\u00e0 $\\widehat{C} = 80^{o}$ th\u00ec: ","select":["A. $\\widehat{DAx} = 60^{o} $","B. $\\widehat{DAx} = 30^{o} $","C. $\\widehat{DAx} = 20^{o} $ ","D. $\\widehat{DAx} = 10^{o} $ "],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> T\u00ednh $\\widehat{BAC}$ r\u1ed3i t\u00ednh $\\widehat{CAx}$ d\u1ef1a v\u00e0o t\u00ednh ch\u1ea5t tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\widehat{CAD}$ d\u1ef1a v\u00e0o t\u1ed5ng ba g\u00f3c trong $\\triangle{CAD}$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh $\\widehat{DAx}$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv1/img\/H7C3B23_D02.png' \/><\/center> <br\/> $\\blacktriangleright$ Trong $\\triangle{ABC}$ c\u00f3: <br\/> $\\widehat{B} + \\widehat{BAC} + \\widehat{C} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong m\u1ed9t tam gi\u00e1c) <br\/> $\\Rightarrow$ $\\widehat{BAC} = 180^{o} - (40^{o} + 80^{o}) = 60^{o}$ <br\/> V\u00ec $Ax$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BAC}$ n\u00ean: <br\/> $\\widehat{CAx} = \\widehat{BAx} = \\dfrac{\\widehat{BAC}}{2} = \\dfrac{60^{o}}{2} = 30^{o}$ (\u0111\u1ecbnh ngh\u0129a) <br\/> $\\blacktriangleright$ Trong tam gi\u00e1c vu\u00f4ng $ADC$ c\u00f3: <br\/> $\\widehat{ADC} + \\widehat{C} + \\widehat{CAD} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong m\u1ed9t tam gi\u00e1c) <br\/> $ \\begin{align} \\Rightarrow \\widehat{CAD} &= 180^{o} - (\\widehat{ADC} + \\widehat{C}) \\\\ &= 180^{o} - (90^{o} + 80^{o}) \\\\ &= 180^{o} - 170^{o} \\\\ &= 10^{o} \\end{align}$ <br\/> $\\blacktriangleright$ $\\widehat{DAx} = \\widehat{CAx} - \\widehat{CAD} = 30^{o} - 10^{o} = 20^{o}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C <\/span>","column":2}]}],"id_ques":1984},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho tam gi\u00e1c $ABC$ vu\u00f4ng c\u00e2n t\u1ea1i $A$. Tr\u00ean c\u1ea1nh $AB$ l\u1ea5y \u0111i\u1ec3m $D$, tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $AC$ l\u1ea5y \u0111i\u1ec3m $E$ sao cho $AE = AD$. Khi \u0111\u00f3 $CD$ vu\u00f4ng g\u00f3c v\u1edbi $BE$. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>? ","select":["A. \u0110\u00daNG","B. SAI"],"hint":"Ch\u1ee9ng minh $CD$ l\u00e0 \u0111\u01b0\u1eddng cao trong tam gi\u00e1c $EBC$","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> G\u1ecdi $K$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $ED$ v\u00e0 $BC$. Ch\u1ee9ng minh $EK \\perp BC$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u1eeb t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c c\u00f9ng c\u1eaft nhau t\u1ea1i m\u1ed9t \u0111i\u1ec3m suy ra \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua 1 \u0111\u1ec9nh v\u00e0 tr\u1ef1c t\u00e2m c\u0169ng l\u00e0 \u0111\u01b0\u1eddng cao. <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv1/img\/H7C3B23_D03.png' \/><\/center> <br\/> $\\blacktriangleright$ G\u1ecdi $K$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $ED$ v\u00e0 $BC$ <br\/> $\\triangle{ABC}$ vu\u00f4ng c\u00e2n t\u1ea1i $A$ n\u00ean $\\widehat{B} = \\widehat{C} = 45^{o}$ <br\/> $\\triangle{EAD}$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 $AE = AD$ n\u00ean l\u00e0 tam gi\u00e1c vu\u00f4ng c\u00e2n <br\/> $\\Rightarrow$ $\\widehat{AED} = \\widehat{ADE} = 45^{o}$ <br\/> $\\triangle{EKC}$ c\u00f3 $\\widehat{CEK} + \\widehat{KCE} + \\widehat{EKC} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong m\u1ed9t tam gi\u00e1c) <br\/> $\\Rightarrow$ $\\widehat{EKC} = 180^{o} - (45^{o} + 45^{o}) = 90^{o}$ <br\/> Hay $EK \\perp BC$ <br\/>$\\blacktriangleright$ $\\triangle{BEC}$ c\u00f3 $BA; EK$ l\u00e0 hai \u0111\u01b0\u1eddng cao ch\u00fang c\u1eaft nhau \u1edf $D$ <br\/> N\u00ean $CD$ c\u0169ng l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a $\\triangle{BEC}$ hay $CD \\perp BE$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span> <br\/> <span class='basic_green'> <i> Nh\u1eadn x\u00e9t: T\u1eeb t\u00ednh ch\u1ea5t ba \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c c\u00f9ng \u0111i qua 1 \u0111i\u1ec3m (tr\u1ef1c t\u00e2m) ta suy ra: Trong m\u1ed9t tam gi\u00e1c, \u0111\u01b0\u1eddng th\u1eb3ng \u0111i qua m\u1ed9t \u0111\u1ec9nh v\u00e0 tr\u1ef1c t\u00e2m c\u0169ng l\u00e0 m\u1ed9t \u0111\u01b0\u1eddng cao. \u0110\u00e2y l\u00e0 m\u1ed9t c\u00e1ch m\u1edbi \u0111\u1ec3 ch\u1ee9ng minh hai \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c. <\/i> <\/span> ","column":2}]}],"id_ques":1985},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["6"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"T\u00ednh \u0111\u01b0\u1eddng cao \u1ee9ng v\u1edbi c\u1ea1nh \u0111\u00e1y c\u1ee7a m\u1ed9t tam gi\u00e1c c\u00e2n c\u00f3 \u0111\u00e1y $5cm$, c\u1ea1nh b\u00ean $6,5cm$. <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> _input_ $cm$ ","hint":"Trong tam gi\u00e1c c\u00e2n \u0111\u01b0\u1eddng cao \u1ee9ng v\u1edbi c\u1ea1nh \u0111\u00e1y \u0111\u1ed3ng th\u1eddi l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c, \u0111\u01b0\u1eddng trung tuy\u1ebfn, \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a tam gi\u00e1c \u1ea5y","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv1/img\/H7C3B23_D04.png' \/><\/center> <br\/> Gi\u1ea3 s\u1eed $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$ c\u00f3 $AC = AB = 6,5cm; BC = 5cm$ <br\/> $AH$ l\u00e0 \u0111\u01b0\u1eddng cao \u1ee9ng v\u1edbi c\u1ea1nh $BC$ n\u00ean $AH$ c\u0169ng \u0111\u1ed3ng th\u1eddi l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> $\\Rightarrow$ $HB = HC = \\dfrac{BC}{2} = \\dfrac{5}{2} = 2,5(cm)$ <br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pitago cho tam gi\u00e1c vu\u00f4ng $AHC$ ta c\u00f3: <br\/> $AH^2 + HC^2 = AC^2$ <br\/> $\\Rightarrow$ $AH^2 = AC^2 - HC^2 = 6,5^2 - 2,5^2 = 36$ <br\/> $\\Rightarrow$ $AH = 6(cm)$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $6$ <\/span>"}]}],"id_ques":1986},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["15"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"Cho tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $B$ c\u00f3 chu vi $50cm$. K\u1ebb \u0111\u01b0\u1eddng cao $BH$. Bi\u1ebft chu vi tam gi\u00e1c $ABH$ b\u1eb1ng $40cm$. T\u00ednh \u0111\u1ed9 d\u00e0i $BH$. <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $BH = $ _input_ $cm$ ","hint":"Trong tam gi\u00e1c c\u00e2n \u0111\u01b0\u1eddng cao \u1ee9ng v\u1edbi c\u1ea1nh \u0111\u00e1y \u0111\u1ed3ng th\u1eddi l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c, \u0111\u01b0\u1eddng trung tuy\u1ebfn, \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a tam gi\u00e1c \u1ea5y","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv1/img\/H7C3B23_D05.png' \/><\/center> <br\/> $\\triangle{ABC}$ c\u00e2n t\u1ea1i $B$ n\u00ean $BA = BC$, \u0111\u01b0\u1eddng cao $BH$ \u0111\u1ed3ng th\u1eddi l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> $\\Rightarrow$ $HA = HC$ <br\/> Chu vi tam gi\u00e1c $ABC$ l\u00e0: <br\/> $AB + BC + AC = 2AB + HA + HC = 2AB + 2HA$ <br\/> V\u00ec chu vi tam gi\u00e1c $ABC$ l\u00e0 $50cm$ n\u00ean: $2AB + 2HA = 50$ hay $AB + HA = 25$ (1)<br\/> Chu vi tam gi\u00e1c $ABH$ l\u00e0: $AB + HA + BH$ <br\/> V\u00ec chu vi tam gi\u00e1c $ABH$ b\u1eb1ng $40cm$ n\u00ean: $AB + HA + BH = 40$ (2) <br\/> L\u1ea5y (2) tr\u1eeb (1) theo t\u1eebng v\u1ebf ta \u0111\u01b0\u1ee3c: $BH = 40 - 25 = 15(cm)$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $15$ <\/span>"}]}],"id_ques":1987},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$ c\u00f3 c\u00e1c \u0111\u01b0\u1eddng cao $BD, CE$ c\u1eaft nhau t\u1ea1i $I$. Bi\u1ebft $\\widehat{BIC} = 110^{o}$ th\u00ec c\u00e1c g\u00f3c c\u1ee7a tam gi\u00e1c $ABC$ l\u00e0: ","select":["A. $\\widehat{A} = 80^{o}; \\widehat{B} = \\widehat{C} = 50^{o} $","B. $\\widehat{A} = 70^{o}; \\widehat{B} = \\widehat{C} = 55^{o} $","C. $\\widehat{A} = 90^{o}; \\widehat{B} = \\widehat{C} = 45^{o} $ ","D. $\\widehat{A} = 110^{o}; \\widehat{B} = \\widehat{C} = 35^{o} $ "],"hint":"","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> <b> B\u01b0\u1edbc 1: <\/b> T\u00ednh $\\widehat{CID}$ r\u1ed3i t\u00ednh $\\widehat{A}$ d\u1ef1a v\u00e0o t\u00ednh ch\u1ea5t hai g\u00f3c k\u1ec1 b\u00f9 v\u00e0 hai g\u00f3c ph\u1ee5 nhau <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\widehat{B}$ v\u00e0 $\\widehat{C}$ d\u1ef1a v\u00e0o t\u1ed5ng ba g\u00f3c trong m\u1ed9t tam gi\u00e1c v\u00e0 t\u00ednh ch\u1ea5t tam gi\u00e1c 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/bai23/lv1/img\/H7C3B23_D06.png' \/><\/center> <br\/> $\\blacktriangleright$ $\\widehat{CID} = 180^{o} - \\widehat{BIC} = 180^{o} - 110^{o} = 70^{o}$ (hai g\u00f3c k\u1ec1 b\u00f9) <br\/> Ta c\u00f3: $\\widehat{A} = \\widehat{CID} = 70^{o}$ (c\u00f9ng ph\u1ee5 v\u1edbi $\\widehat{ACE}$) <br\/> $\\blacktriangleright$ $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$ n\u00ean $\\widehat{B} = \\widehat{C}$ (t\u00ednh ch\u1ea5t) <br\/> $\\widehat{A} + \\widehat{B} + \\widehat{C} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong m\u1ed9t tam gi\u00e1c) <br\/> $\\Rightarrow$ $2\\widehat{C} = 180^{o} - \\widehat{A} = 180^{o} - 70^{o} = 110^{o}$ <br\/> $\\Rightarrow$ $\\widehat{B} = \\widehat{C} = \\dfrac{110^{o}}{2} = 55^{o}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: B <\/span>","column":2}]}],"id_ques":1988},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["14,4"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"Tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 $BC = 26cm$, \u0111\u01b0\u1eddng cao $AH = 12cm$. T\u00ednh \u0111\u1ed9 d\u00e0i $AB$ (k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t) <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $AB = $ _input_ $cm$ ","hint":"","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv1/img\/H7C3B23_D07.png' \/><\/center> <br\/> $\\blacktriangleright$ $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 \u0111\u01b0\u1eddng trung tuy\u1ebfn $AM$ <br\/> N\u00ean $AM = BM = MC = \\dfrac{BC}{2} = \\dfrac{26}{2} = 13(cm)$ (t\u00ednh ch\u1ea5t trung tuy\u1ebfn trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\blacktriangleright$ Trong tam gi\u00e1c vu\u00f4ng $HAM$ c\u00f3: <br\/> $HM^2 = AM^2 - AH^2 = 13^2 - 12^2 = 25$ (\u0111\u1ecbnh l\u00fd Pitago) <br\/> $\\Rightarrow$ $HM = 5(cm)$ <br\/> Ta c\u00f3: $BH = BM - HM = 13 - 5 = 8(cm)$ <br\/> $\\blacktriangleright$ Trong $\\triangle{AHB}$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: <br\/> $AB^2 = AH^2 + BH^2 = 12^{2} + 8^{2} = 208$ (\u0111\u1ecbnh l\u00fd Pitago) <br\/> $\\Rightarrow$ $AB = \\sqrt{208} \\approx 14,4 (cm) $ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $14,4$ <\/span>"}]}],"id_ques":1989},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{A} = 45^{o}$ v\u00e0 $AC < BC$. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $CE$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $EB = ED$. Khi \u0111\u00f3 $BC$ vu\u00f4ng g\u00f3c v\u1edbi $AD$. <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 $AC \\perp BD$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> Ch\u1ec9 ra $BC$ c\u0169ng l\u00e0 m\u1ed9t \u0111\u01b0\u1eddng cao c\u1ee7a $\\triangle{ABD}$ hay $BC \\perp AD$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv1/img\/H7C3B23_D08.png' \/><\/center> <br\/> $\\blacktriangleright$ X\u00e9t tam gi\u00e1c $BED$ vu\u00f4ng t\u1ea1i $E$, ta c\u00f3 $ED = EB$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow$ $\\triangle{BED}$ vu\u00f4ng c\u00e2n t\u1ea1i $E$ (\u0111\u1ecbnh ngh\u0129a) <br\/> $\\Rightarrow$ $\\widehat{EDB} = \\widehat{EBD} = 45^{o}$ (t\u00ednh ch\u1ea5t) <br\/> Suy ra: $\\widehat{CAE} + \\widehat{EBD} = 45^{o} + 45^{o} = 90^{o}$ <br\/> N\u00ean $AC \\perp BD$ <br\/> $\\blacktriangleright$ $\\triangle{ABD}$ c\u00f3 $AC \\perp BD$; $DE \\perp AB$ suy ra $C$ l\u00e0 tr\u1ef1c t\u00e2m $\\triangle{ABD}$ <br\/> $\\Rightarrow$ $BC$ l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a $\\triangle{ABD}$ hay $BC \\perp AD$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span> ","column":2}]}],"id_ques":1990},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["120"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"Cho tam gi\u00e1c nh\u1ecdn $MNP$, $MN < NP$, $\\widehat{N} = 60^{o}$. Hai \u0111\u01b0\u1eddng cao $MD$ v\u00e0 $PE$ c\u1eaft nhau \u1edf $F$. T\u00ednh $\\widehat{DFE}$ <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $\\widehat{DFE} $ _input_ $^{o}$ ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b> B\u01b0\u1edbc 1: <\/b> T\u00ednh $\\widehat{EPN}$ d\u1ef1a v\u00e0o tam gi\u00e1c vu\u00f4ng $EPN$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\widehat{EFD}$ d\u1ef1a v\u00e0o g\u00f3c ngo\u00e0i c\u1ee7a $\\triangle{DFP}$ <br\/><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv1/img\/H7C3B23_D09.png' \/><\/center> <br\/> $\\blacktriangleright$ $\\triangle{ENP}$ vu\u00f4ng t\u1ea1i $E$ c\u00f3 $\\widehat{N} = 60^{o}$ n\u00ean: <br\/> $\\widehat{EPN} = 90^{o} - \\widehat{N} = 90^{o} - 60^{o} = 30^{o}$ (Trong tam gi\u00e1c vu\u00f4ng hai g\u00f3c nh\u1ecdn ph\u1ee5 nhau) <br\/> $\\blacktriangleright$ $\\widehat{DFE}$ l\u00e0 g\u00f3c ngo\u00e0i t\u1ea1i \u0111\u1ec9nh $F$ c\u1ee7a $\\triangle{DFP}$ n\u00ean: <br\/> $\\widehat{DFE} = \\widehat{FPD} + \\widehat{FDP} = 30^{o} + 90^{o} = 120^{o}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $120$ <\/span>"}]}],"id_ques":1991},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Trong m\u1ed9t tam gi\u00e1c, \u0111\u01b0\u1eddng cao \u1ee9ng v\u1edbi c\u1ea1nh nh\u1ecf h\u01a1n l\u00e0 \u0111\u01b0\u1eddng cao l\u1edbn h\u01a1n. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>? ","select":["A. \u0110\u00daNG","B. SAI"],"hint":"","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv1/img\/H7C3B23_D10.png' \/><\/center> <br\/> $\\blacktriangleright$ Gi\u1ea3 s\u1eed $\\triangle{ABC}$ c\u00f3 $AB < AC$, hai \u0111\u01b0\u1eddng cao $BE; CF$ <br\/> Ta c\u00f3 $S_{\\triangle{ABC}} = \\dfrac{1}{2}CF. AB = \\dfrac{1}{2}BE. AC$ (1) <br\/> M\u1eb7t kh\u00e1c $AB < AC$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $CF > BE$ hay \u0111\u01b0\u1eddng cao \u1ee9ng v\u1edbi c\u1ea1nh nh\u1ecf h\u01a1n th\u00ec l\u1edbn h\u01a1n <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'> \u0110\u00daNG <\/span> ","column":2}]}],"id_ques":1992},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Cho tam gi\u00e1c $ABC$ c\u00e2n \u1edf $B$, tr\u1ef1c t\u00e2m $H$. Th\u00eam \u0111i\u1ec1u ki\u1ec7n g\u00ec \u0111\u1ec3 $H$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a tam gi\u00e1c n\u00e0y? ","select":["A. $AB > AC$ ","B. $AB \\perp AC $","C. $\\widehat{A} = 60^{o}$ ","D. $\\widehat{B} = 90^{o}$ "],"hint":"","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv1/img\/H7C3B23_D12.png' \/><\/center> <br\/> $\\blacktriangleright$ $\\triangle{ABC}$ c\u00e2n \u1edf $B$ c\u00f3 tr\u1ef1c t\u00e2m $H$ v\u00e0 tr\u1ecdng t\u00e2m $G$ <br\/> \u0110i\u1ec1u ki\u1ec7n \u0111\u1ec3 $H$ v\u00e0 $G$ tr\u00f9ng nhau l\u00e0 $\\triangle{ABC}$ \u0111\u1ec1u hay $\\widehat{A} = 60^{o}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: C. $\\widehat{A} = 60^{o}$ <\/span> ","column":2}]}],"id_ques":1993},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["f","t","t","t"]],"list":[{"point":5,"image":"","col_name":["Kh\u1eb3ng \u0111\u1ecbnh","\u0110\u00fang","Sai"],"arr_ques":[" Tr\u1ef1c t\u00e2m c\u1ee7a m\u1ed9t tam gi\u00e1c l\u00e0 giao ba \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a tam gi\u00e1c"," Trong tam gi\u00e1c c\u00e2n, hai \u0111\u01b0\u1eddng cao \u1ee9ng v\u1edbi hai c\u1ea1nh b\u00ean b\u1eb1ng nhau "," Tr\u1ef1c t\u00e2m c\u1ee7a m\u1ed9t tam gi\u00e1c \u0111\u1ec1u c\u00e1ch \u0111\u1ec1u ba c\u1ea1nh c\u1ee7a tam gi\u00e1c \u0111\u00f3 "," Tr\u1ef1c t\u00e2m c\u1ee7a tam gi\u00e1c vu\u00f4ng ch\u00ednh l\u00e0 \u0111\u1ec9nh g\u00f3c vu\u00f4ng c\u1ee7a tam gi\u00e1c \u0111\u00f3 "],"hint":"","explain":["SAI: Tr\u1ef1c t\u00e2m c\u1ee7a tam gi\u00e1c l\u00e0 giao ba \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c","\u0110\u00daNG ","\u0110\u00daNG v\u00ec trong tam gi\u00e1c \u0111\u1ec1u tr\u1ef1c t\u00e2m v\u00e0 \u0111i\u1ec3m c\u00e1ch \u0111\u1ec1u ba \u0111\u1ec9nh tr\u00f9ng nhau","\u0110\u00daNG "]}]}],"id_ques":1994},{"time":24,"part":[{"title":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u","title_trans":"Cho tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$, c\u00f3 $\\widehat{A} = 45^{o}$, \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $AD$. \u0110\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AB$ c\u1eaft $AC$ t\u1ea1i $M$. Tr\u00ean c\u1ea1nh $AB$ l\u1ea5y m\u1ed9t \u0111i\u1ec3m $N$ sao cho $BN = CM$. H\u00e3y s\u1eafp x\u1ebfp c\u00e1c c\u00e2u sau \u0111\u1ec3 ch\u1ee9ng minh r\u1eb1ng ba \u0111\u01b0\u1eddng th\u1eb3ng $AD, BM, CN$ \u0111\u1ed3ng quy. ","temp":"sequence","correct":[[[2],[4],[1],[3]]],"list":[{"point":5,"image":"","left":["$\\triangle{NBC} = \\triangle{MCB}$ (c.g.c) v\u00ec: $\\begin{cases} BN = CM (gt) \\\\ \\widehat{NBC} = \\widehat{MCB} (\\triangle{ABC} \\hspace{0,2cm} \\text{c\u00e2n}) \\\\ BC \\hspace{0,2cm} \\text{chung} \\end{cases}$"," Do $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$ n\u00ean $AD$ l\u00e0 ph\u00e2n gi\u00e1c \u0111\u1ed3ng th\u1eddi l\u00e0 \u0111\u01b0\u1eddng cao <br\/> X\u00e9t $\\triangle{ABC}$ v\u00ec $AD, BM, CN$ l\u00e0 ba \u0111\u01b0\u1eddng cao n\u00ean ch\u00fang \u0111\u1ed3ng quy"," V\u00ec $M$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a c\u1ea1nh $AB$ n\u00ean $MA = MB$ (t\u00ednh ch\u1ea5t) v\u00e0 $\\widehat{BAC} = 45^{o}$ <br\/> N\u00ean $\\triangle{AMB}$ vu\u00f4ng c\u00e2n t\u1ea1i $M$ $\\Rightarrow$ $BM \\perp AC$ ","$\\Rightarrow$ $\\widehat{BNC} = \\widehat{BMC}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow$ $\\widehat{BNC} = 90^{o}$ $\\Rightarrow$ $CN \\perp AB$"],"top":100,"hint":"","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv1/img\/H7C3B23_D11.png' \/><\/center> $\\blacktriangleright$ V\u00ec $M$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a c\u1ea1nh $AB$ n\u00ean $MA = MB$ (t\u00ednh ch\u1ea5t) v\u00e0 $\\widehat{BAC} = 45^{o}$ <br\/> N\u00ean $\\triangle{AMB}$ vu\u00f4ng c\u00e2n t\u1ea1i $M$ $\\Rightarrow$ $BM \\perp AC$ <br\/> $\\blacktriangleright$ $\\triangle{NBC} = \\triangle{MCB}$ (c.g.c) v\u00ec: <br\/> $\\begin{cases} BN = CM (gt) \\\\ \\widehat{NBC} = \\widehat{MCB} (\\triangle{ABC} \\hspace{0,2cm} \\text{c\u00e2n}) \\\\ BC \\hspace{0,2cm} \\text{chung} \\end{cases}$ $\\Rightarrow$ $\\widehat{BNC} = \\widehat{BMC}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow$ $\\widehat{BNC} = 90^{o}$ $\\Rightarrow$ $CN \\perp AB$ <br\/> Do $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$ n\u00ean $AD$ l\u00e0 ph\u00e2n gi\u00e1c \u0111\u1ed3ng th\u1eddi l\u00e0 \u0111\u01b0\u1eddng cao <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{ABC}$ v\u00ec $AD, BM, CN$ l\u00e0 ba \u0111\u01b0\u1eddng cao n\u00ean ch\u00fang \u0111\u1ed3ng quy. "}]}],"id_ques":1995},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Cho tam gi\u00e1c $MNP$ vu\u00f4ng t\u1ea1i $M$, \u0111\u01b0\u1eddng cao $MH$. C\u00e1c tr\u1ef1c t\u00e2m c\u1ee7a tam gi\u00e1c $MNP, MHN, MHP$ l\u00e0 \u0111i\u1ec3m n\u00e0o? ","select":["<span class='basic_left'> A. C\u1ea3 ba tr\u1ef1c t\u00e2m tam gi\u00e1c $MNP, MHN, MHP$ l\u00e0 \u0111i\u1ec3m $M$ <\/span> ","<span class='basic_left'> B. Tr\u1ef1c t\u00e2m tam gi\u00e1c $MNP$ l\u00e0 \u0111i\u1ec3m $H$, tr\u1ef1c t\u00e2m tam gi\u00e1c $MHN, MHP$ l\u00e0 \u0111i\u1ec3m $M$ <\/span>","<span class='basic_left'> C. C\u1ea3 ba tr\u1ef1c t\u00e2m tam gi\u00e1c $MNP, MHN, MHP$ l\u00e0 \u0111i\u1ec3m $H$ <\/span> ","<span class='basic_left'> D. Tr\u1ef1c t\u00e2m tam gi\u00e1c $MNP$ l\u00e0 $M$, tr\u1ef1c t\u00e2m tam gi\u00e1c $MHN, MHP$ l\u00e0 $H$ <\/span> "],"hint":"","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv1/img\/H7C3B23_D18.png' \/><\/center> <br\/> $\\triangle{MNP}$ vu\u00f4ng t\u1ea1i $M$ n\u00ean tr\u1ef1c t\u00e2m tam gi\u00e1c ch\u00ednh l\u00e0 \u0111i\u1ec3m $M$ <br\/> $\\triangle{MHN}$ v\u00e0 $\\triangle{MHP}$ vu\u00f4ng t\u1ea1i $H$ n\u00ean c\u00f3 tr\u1ef1c t\u00e2m l\u00e0 \u0111i\u1ec3m $H$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: D <\/span> ","column":1}]}],"id_ques":1996},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Cho h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv1/img\/H7C3B23_D16.png' \/><\/center> <br\/> Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y <b> sai <\/b>?","select":["A. $I$ l\u00e0 tr\u1ef1c t\u00e2m $\\triangle{ABC}$ ","B. $CI \\perp AB$ ","C. $\\widehat{BID} = 42^{o}$ ","D. $\\widehat{DIE} = 132^{o}$ "],"hint":"","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv1/img\/H7C3B23_D16.png' \/><\/center> <br\/> $\\blacktriangleright$ $\\triangle{ABC}$ c\u00f3 $AD, BE$ l\u00e0 c\u00e1c \u0111\u01b0\u1eddng cao ch\u00fang c\u1eaft nhau t\u1ea1i $I$ <br\/> N\u00ean $I$ l\u00e0 tr\u1ef1c t\u00e2m $\\triangle{ABC}$ <b> (\u0110\u00e1p \u00e1n A \u0111\u00fang) <\/b> <br\/> $\\blacktriangleright$ $I$ l\u00e0 tr\u1ef1c t\u00e2m $\\triangle{ABC}$ $\\Rightarrow$ $CI$ l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a $\\triangle{ABC}$ (t\u00ednh ch\u1ea5t) <br\/> $\\Rightarrow$ $CI \\perp AB$ <b> (\u0110\u00e1p \u00e1n B \u0111\u00fang) <\/b> <br\/> $\\blacktriangleright$ $\\widehat{BID} = \\widehat{ECB} = 48^{o}$ (c\u00f9ng ph\u1ee5 v\u1edbi $\\widehat{EBC}$) <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n C sai <\/b> <br\/> $\\blacktriangleright$ $\\widehat{DIE}$ v\u00e0 $\\widehat{BID}$ l\u00e0 hai g\u00f3c k\u1ec1 b\u00f9 <br\/> $\\Rightarrow$ $\\widehat{DIE} = 180^{o} - \\widehat{BID} = 180^{o} - 48^{o} = 132^{o}$ <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n D \u0111\u00fang <\/b> <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n sai l\u00e0: C <\/span> ","column":2}]}],"id_ques":1997},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 c\u00e1c \u0111\u01b0\u1eddng cao $AD, BE, CF$. Bi\u1ebft $AD = BE = CF$, khi \u0111\u00f3 tam gi\u00e1c $ABC$ l\u00e0 tam gi\u00e1c \u0111\u1ec1u. <b> \u0110\u00fang <\/b> hay <b> sai <\/b>?","select":["A. \u0110\u00daNG ","B. SAI "],"hint":"","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv1/img\/H7C3B23_D13.png' \/><\/center> <br\/> $\\blacktriangleright$ X\u00e9t hai tam gi\u00e1c vu\u00f4ng $BEC$ v\u00e0 $CFB$ c\u00f3: <br\/> $\\begin{cases} BE = CF (gt) \\\\ BC \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{BEC} = \\triangle{CFB}$ (c\u1ea1nh huy\u1ec1n - c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> $\\Rightarrow$ $\\widehat{FBC} = \\widehat{ECB}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> $\\blacktriangleright$ Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta \u0111\u01b0\u1ee3c $\\triangle{BFC} = \\triangle{AEB}$ (c\u1ea1nh huy\u1ec1n - c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> $\\Rightarrow$ $\\widehat{FBC} = \\widehat{EAB}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $\\widehat{FBC} = \\widehat{ECB} = \\widehat{EAB}$ hay $\\widehat{ABC} = \\widehat{ACB} = \\widehat{BAC}$ <br\/> $\\Rightarrow$ $\\triangle{ABC}$ l\u00e0 tam gi\u00e1c \u0111\u1ec1u (t\u00ednh ch\u1ea5t) <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <span class='basic_pink'>\u0110\u00daNG <\/span> ","column":2}]}],"id_ques":1998},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["8,7"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"T\u00ednh \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c \u0111\u1ec1u c\u00f3 c\u1ea1nh b\u1eb1ng $10cm$ (k\u1ebft qu\u1ea3 l\u1ea5y \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t). <br\/> <b> \u0110\u00e1p \u00e1n l\u00e0: <\/b> _input_ $cm$ ","hint":"Trong tam gi\u00e1c \u0111\u1ec1u \u0111\u01b0\u1eddng cao c\u0169ng l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv1/img\/H7C3B23_D14.png' \/><\/center> $\\blacktriangleright$ Gi\u1ea3 s\u1eed $\\triangle{ABC}$ \u0111\u1ec1u v\u00e0 c\u00f3 \u0111\u01b0\u1eddng cao $AM$ <br\/> V\u00ec $\\triangle{ABC}$ \u0111\u1ec1u n\u00ean \u0111\u01b0\u1eddng cao $AM$ c\u0169ng l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn <br\/> $\\Rightarrow$ $MB = MC = \\dfrac{1}{2}BC = \\dfrac{10}{2} = 5(cm) $ (t\u00ednh ch\u1ea5t) <br\/> $\\blacktriangleright$ $\\triangle{AMB}$ vu\u00f4ng t\u1ea1i $M$ c\u00f3: <br\/> $AM^2 + MB^2 = AB^2$ (\u0111\u1ecbnh l\u00fd Pitago) <br\/> $\\begin{align} \\Rightarrow AM^2 &= AB^2 - MB^2 \\\\ &= 10^2 - 5^2 \\\\ &= 75 \\end{align}$ <br\/> $\\Rightarrow$ $AM \\approx 8,7(cm)$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $8,7$ <\/span>"}]}],"id_ques":1999},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Cho h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv1/img\/H7C3B23_D17.png' \/><\/center> <br\/> Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y <b> sai <\/b>?","select":["A. $PI \\perp MN$ ","B. $\\widehat{MPN} = 70^{o} $ ","C. $\\widehat{PME} = 30^{o}$ ","D. $\\widehat{MIN} = 110^{o}$ "],"hint":"","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai23/lv1/img\/H7C3B23_D17.png' \/><\/center> <br\/> $\\blacktriangleright$ $\\triangle{MNP}$ c\u00f3 $NH, ME$ l\u00e0 c\u00e1c \u0111\u01b0\u1eddng cao v\u00e0 ch\u00fang c\u1eaft nhau t\u1ea1i $I$ <br\/> N\u00ean $I$ l\u00e0 tr\u1ef1c t\u00e2m $\\triangle{MNP}$ $\\Rightarrow$ $PI$ c\u0169ng l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a $\\triangle{MNP}$ <br\/> Hay $PI \\perp MN$ <b> (\u0110\u00e1p \u00e1n A \u0111\u00fang) <\/b> <br\/> $\\blacktriangleright$ $\\widehat{MPN}$ v\u00e0 $\\widehat{NPx}$ l\u00e0 hai g\u00f3c k\u1ec1 b\u00f9 n\u00ean: <br\/> $\\widehat{MPN} = 180^{o} - \\widehat{NPx} = 180^{o} - 110^{o} = 70^{o}$ <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n B \u0111\u00fang <\/b> <br\/> $\\blacktriangleright$ $\\triangle{PME}$ vu\u00f4ng t\u1ea1i $E$ n\u00ean $\\widehat{PME}$ v\u00e0 $\\widehat{EPM}$ l\u00e0 hai g\u00f3c ph\u1ee5 nhau <br\/> $\\Rightarrow$ $\\widehat{PME} = 90^{o} - \\widehat{EPM} = 90^{o} - 70^{o} = 20^{o}$ <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n C sai <\/b> <br\/> $\\blacktriangleright$ Ta c\u00f3 $\\widehat{MIH} = \\widehat{MPN} = 70^{o}$ (c\u00f9ng ph\u1ee5 v\u1edbi $\\widehat{PME}$) <br\/> M\u00e0 $\\widehat{MIH}$ v\u00e0 $\\widehat{MIN}$ l\u00e0 hai g\u00f3c k\u1ec1 b\u00f9 n\u00ean: <br\/> $\\widehat{MIN} = 180^{o} - \\widehat{MIH} = 180^{o} - 70^{o} = 110^{o}$ <br\/> $\\Rightarrow$ <b> \u0110\u00e1p \u00e1n D \u0111\u00fang <\/b> <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n sai l\u00e0: C <\/span> ","column":2}]}],"id_ques":2000}],"lesson":{"save":0,"level":1}}