{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho $\\triangle{ABC}$ v\u00e0 c\u00e1c \u0111\u01b0\u1eddng cao $BD$, $CE$. T\u00ednh $\\widehat{AED}$ bi\u1ebft $\\widehat{ACB} = 60^o$.<br\/><\/span>","select":[" A. $\\widehat{AED} = 30^o$"," B. $\\widehat{AED} = 60^o$","C. $\\widehat{AED} = 120^o$","D. $\\widehat{AED} = 150^o$"],"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 $\\triangle$ vu\u00f4ng $ABD$ $\\backsim$ $\\triangle$ vu\u00f4ng $ACE$<br\/><b>B\u01b0\u1edbc 2:<\/b> Ch\u1ee9ng minh $\\triangle{ADE}$ $\\backsim$ $\\triangle{ABC}$<\/span><br\/> <span class='basic_left'><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai17/lv3/img\/H8C3B4_K101.png' \/><\/center><br\/> +) $BD$, $CE$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a $\\triangle{ABC}$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow \\widehat{AEC} = \\widehat{ADB} = 90^o$ <br\/>+) X\u00e9t$\\triangle$ vu\u00f4ng $ABD$ v\u00e0 $\\triangle$ vu\u00f4ng $ACE$ c\u00f3: <br\/>$\\widehat{AEC} = \\widehat{ADB} = 90^o$ (ch\u1ee9ng minh tr\u00ean)<br\/>$\\widehat{A} $ chung<br\/>$\\Rightarrow$ $\\triangle$ vu\u00f4ng $ABD$ $\\backsim$ $\\triangle$ vu\u00f4ng $ACE$ (g\u00f3c - g\u00f3c)<br\/>$\\Rightarrow \\dfrac{AD}{AE} = \\dfrac{AB}{AC}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7)<br\/>$\\Rightarrow \\dfrac{AD}{AB} = \\dfrac{AE}{AC}$<br\/> +) X\u00e9t $\\triangle{ADE}$ v\u00e0 $\\triangle{ABC}$ c\u00f3: <br\/>$\\widehat{A} $ chung<br\/> $\\dfrac{AD}{AB} = \\dfrac{AE}{AC}$ (ch\u1ee9ng minh tr\u00ean) <br\/>$\\Rightarrow \\triangle{ADE}$ $\\backsim$ $\\triangle{ABC}$ (c\u1ea1nh-g\u00f3c-c\u1ea1nh)<br\/>$\\Rightarrow \\widehat{AED} = \\widehat{ACB}$ (c\u1eb7p g\u00f3c t\u01b0\u01a1ng \u1ee9ng)<br\/> M\u00e0 $\\widehat{ACB} = 60^o$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow \\widehat{AED} = 60^o$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0<span class='basic_pink'> B. $\\widehat{AED} = 60^o$<\/span> <\/span> <br\/><\/span> ","column":2}]}],"id_ques":1760},{"time":24,"part":[{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["1","3","4"]],"list":[{"point":10,"img":"","ques":"<span class='basic_left'>Cho tam gi\u00e1c $ABC$, \u0111\u01b0\u1eddng cao $AH$ ($H$ $\\in$ $BC$). Bi\u1ebft $AH = 6cm$, $BH = 3cm$, $HC = 12cm$<br\/><b>H\u00e3y ch\u1ecdn nh\u1eefng \u0111\u00e1p \u00e1n \u0111\u00fang<\/b><\/span><br\/>","hint":"","column":2,"number_true":2,"select":["A. $\\triangle{ABH} \\backsim \\triangle{CAH} $","B. $\\triangle{ABH} \\backsim \\triangle{ACH} $","C. $\\widehat{HAB} = \\widehat{HCA}$","D. $\\widehat{BAC} = 90^o$"],"explain":"<span class='basic_left'><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai17/lv3/img\/H8C3B4_K102.png' \/><\/center><br\/> X\u00e9t $\\triangle$ vu\u00f4ng $AHB$ v\u00e0 $\\triangle$ vu\u00f4ng $CHA$ c\u00f3:<br\/>$\\widehat{AHB} = \\widehat{AHC}$ (c\u00f9ng $= 90^o$)<br\/> $\\dfrac{HB}{HA} = \\dfrac{HA}{HC}$ (v\u00ec $\\dfrac{3}{6} = \\dfrac{6}{12}$ ) <br\/>$\\Rightarrow$ $\\triangle$ vu\u00f4ng $AHB$ $\\backsim$ $\\triangle$ vu\u00f4ng $CHA$ (c\u1ea1nh-g\u00f3c-c\u1ea1nh) <b>(\u0111\u00e1p \u00e1n A sai, \u0111\u00e1p \u00e1n B \u0111\u00fang)<\/b><br\/>$\\Rightarrow$ $\\widehat{HAB} = \\widehat{HCA}$ (c\u1eb7p g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <b>(\u0111\u00e1p \u00e1n C \u0111\u00fang)<\/b><br\/>L\u1ea1i c\u00f3: $\\widehat{HAC} + \\widehat{HCA} = 90^o$ (hai g\u00f3c ph\u1ee5 nhau trong $\\triangle$ vu\u00f4ng $AHC$)<br\/> $\\Rightarrow$ $\\widehat{HAB} + \\widehat{HAC} = 90^o$ hay $\\widehat{BAC} = 90^o$ <b>(\u0111\u00e1p \u00e1n D \u0111\u00fang)<\/b><br\/> <br\/><br\/>V\u1eady nh\u1eefng \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'>A, C, D<\/span><br\/> "}]}],"id_ques":1761},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" <span class='basic_left'> Tam gi\u00e1c $ABC$ c\u00f3 \u0111\u1ed9 d\u00e0i c\u00e1c c\u1ea1nh l\u00e0 $AB = 3cm$, $AC = 4cm$, $BC = 5cm$. Bi\u1ebft $\\triangle{ABC}$ $\\backsim$ $\\triangle{A'B'C'}$ v\u00e0 $S_{\\triangle{A'B'C'}} = 54 \\text{cm}^2$. T\u00ednh chu vi $\\triangle{A'B'C'}$ <br\/><\/span>","select":[" A. $28cm$"," B. $30cm$","C. $32cm$","D. $35cm$"],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3: $3^2 + 4^2 = 9 + 16 = 25 = 5^2$<br\/> $\\Rightarrow AB^2 + AC^2 = BC^2$<br\/>$\\Rightarrow \\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$ (\u0111\u1ecbnh l\u00ed Py-ta-go \u0111\u1ea3o)<br\/>$\\Rightarrow \\widehat{BAC} = 90^o$<br\/>L\u1ea1i c\u00f3: $\\triangle{ABC}$ $\\backsim$ $\\triangle{A'B'C'}$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow \\widehat{BAC} = \\widehat{B'A'C'} = 90^o$ (c\u1eb7p g\u00f3c t\u01b0\u01a1ng \u1ee9ng)<br\/>$\\Rightarrow \\triangle{A'B'C'}$ vu\u00f4ng t\u1ea1i $A'$ <br\/>$\\Rightarrow S_{\\triangle{A'B'C'}} = \\dfrac{A'B'. A'C'}{2}$ (1)<br\/>G\u1ecdi $k$ l\u00e0 h\u1ec7 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng c\u1ee7a $\\triangle{ABC}$ v\u00e0 $\\triangle{A'B'C'}$ ($k > 0$) <br\/>$\\Rightarrow A'B' = 3k; A'C' = 4k; B'C' = 5k$ (2)<br\/>T\u1eeb (1) v\u00e0 (2) $\\Rightarrow S_{\\triangle{A'B'C'}} = \\dfrac{3k.4k}{2} = 6k^2$ <br\/>L\u1ea1i c\u00f3: $S_{\\triangle{A'B'C'}} = 54 \\text{cm}^2$<br\/>$\\Rightarrow 6k^2 = 54 \\Rightarrow k^2 = 9 \\Rightarrow k = 3$ <br\/>$\\Rightarrow A'B' = 3.3 = 9 \\text{(cm)}; A'C' = 4.3 = 12 \\text{(cm)}; B'C' = 5.3 = 15 \\text{(cm)}$ <br\/>Chu vi $\\triangle{A'B'C'}$ l\u00e0: $9 + 12 + 15 = 36$ ($cm$)<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'>C. $36cm$<\/span> <\/span> ","column":2}]}],"id_ques":1762},{"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":" <span class='basic_left'> Cho tam gi\u00e1c nh\u1ecdn $ABC$, \u0111\u01b0\u1eddng cao $AH$. Qua $H$ k\u1ebb $HK$ song song v\u1edbi $AB$ ($K \\in AC$). Bi\u1ebft $S_{\\triangle{AHK}} = \\dfrac{3}{16}S_{\\triangle{ABC}}$, h\u00e3y t\u00ednh t\u1ec9 s\u1ed1 $\\dfrac{AK}{KC}$. <br\/><\/span>","select":[" A. $\\dfrac{AK}{KC} = \\dfrac{2}{3}$ "," B. $\\dfrac{AK}{KC} = \\dfrac{1}{3}$ ","C. $\\dfrac{AK}{KC} = \\dfrac{2}{3}$ ho\u1eb7c $\\dfrac{AK}{KC} = 3 $","D. $\\dfrac{AK}{KC} = \\dfrac{1}{3}$ ho\u1eb7c $\\dfrac{AK}{KC} = 3 $"],"hint":"$ \\dfrac{S_{\\triangle{AHK}}}{S_{\\triangle{ABC}}} = \\dfrac{S_{\\triangle{AHK}}}{S_{\\triangle{CHK}}}.\\dfrac{S_{\\triangle{CHK}}}{S_{\\triangle{ABC}}}$.","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai17/lv3/img\/H8C3B4_K103.png' \/><\/center><br\/>K\u1ebb $HI \\perp AC$ ($I \\in AC$)<br\/>\u0110\u1eb7t $AK = x, KC = y$<br\/> $ \\dfrac{S_{\\triangle{AHK}}}{S_{\\triangle{CHK}}} = \\dfrac{\\dfrac{HI.AK}{2}}{\\dfrac{HI.KC}{2}} = \\dfrac{AK}{KC} = \\dfrac{x}{y}$<br\/> $\\triangle{ABC}$ c\u00f3: $HK \/\/ AB$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow$ $\\triangle{CHK}$ $\\backsim$ $\\triangle{CBA}$<br\/>$\\Rightarrow \\dfrac{S_{\\triangle{CHK}}}{S_{\\triangle{CBA}}} = \\left(\\dfrac{CK}{CA}\\right)^2 = \\left(\\dfrac{y}{x + y}\\right)^2$<br\/> $\\dfrac{S_{\\triangle{AHK}}}{S_{\\triangle{ABC}}} = \\dfrac{S_{\\triangle{AHK}}}{S_{\\triangle{CHK}}}.\\dfrac{S_{\\triangle{CHK}}}{S_{\\triangle{ABC}}} = \\dfrac{x}{y}. \\left(\\dfrac{y}{x + y}\\right)^2 = \\dfrac{xy}{(x+y)^2}$ (1)<br\/>M\u00e0 $S_{\\triangle{AHK}} = \\dfrac{3}{16}S_{\\triangle{ABC}}$ (gi\u1ea3 thi\u1ebft)<br\/> $\\Rightarrow \\dfrac{S_{\\triangle{AHK}}}{S_{\\triangle{ABC}}} = \\dfrac{3}{16}$ (2)<br\/>T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\dfrac{xy}{(x+y)^2} = \\dfrac{3}{16}$ <br\/>$\\Leftrightarrow 3(x + y)^2 = 16xy$<br\/>$\\Leftrightarrow 3x^2 + 3y^2 - 10xy = 0$<br\/>$\\Leftrightarrow 3x^2 - 9xy - xy + 3y^2 = 0$<br\/> $\\Leftrightarrow 3x(x - 3y) - y(x - 3y) = 0$<br\/> $\\Leftrightarrow (3x - y)(x - 3y) = 0$ <br\/>$\\Leftrightarrow \\left[\\begin{array}{l} 3x - y = 0\\\\ x - 3y = 0 \\end{array} \\right.$<br\/> $\\Leftrightarrow \\left[\\begin{array}{l} \\dfrac{x}{y} = \\dfrac{1}{3}\\\\ \\dfrac{x}{y} = 3 \\end{array} \\right.$<br\/>V\u1eady $\\dfrac{AK}{KC} = \\dfrac{1}{3}$ ho\u1eb7c $\\dfrac{AK}{KC} = 3$<br\/>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'>D. $\\dfrac{AK}{KC} = \\dfrac{1}{3}$ ho\u1eb7c $\\dfrac{AK}{KC} = 3$<\/span><\/span>","column":2}]}],"id_ques":1763},{"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":[[["150"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'> Cho h\u00ecnh v\u1ebd nh\u01b0 sau: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai17/lv3/img\/H8C3B4_K105.png' \/><\/center><br\/>T\u00ednh $S_{\\triangle{ABC}}$ <br\/><b>\u0110\u00e1p \u00e1n:<\/b> $S_{\\triangle{ABC}}$ = _input_ ($cm^2$)<\/span>","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 $\\triangle{ABC}$ $\\backsim$ $\\triangle{HAC}$ $\\rightarrow$ T\u00ednh $AC$<br\/><b>B\u01b0\u1edbc 2:<\/b> Ch\u1ee9ng minh $\\triangle{ABC}$ $\\backsim$ $\\triangle{HBA}$ $\\rightarrow$ T\u00ednh $AB$<br\/><b>B\u01b0\u1edbc 3:<\/b> T\u00ednh $S_{\\triangle{ABC}}$.<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai17/lv3/img\/H8C3B4_K105.png' \/><\/center><br\/>Ta c\u00f3: $BC = BH + HC = 9 + 16 = 25$ ($cm$)<br\/>X\u00e9t $\\triangle$ vu\u00f4ng $ABC$ v\u00e0 $\\triangle$ vu\u00f4ng $HAC$ c\u00f3:<br\/>$\\widehat{C} $ chung<br\/> $\\Rightarrow$ $\\triangle$ vu\u00f4ng $ABC$ $\\backsim$ $\\triangle$ vu\u00f4ng $HAC$ (g\u00f3c-g\u00f3c)<br\/>$\\Rightarrow \\dfrac{AC}{HC} = \\dfrac{BC}{AC}$ <br\/>$\\Rightarrow AC^2 = BC.HC = 25.16 \\Rightarrow AC = 20$<br\/>T\u01b0\u01a1ng t\u1ef1 ch\u1ee9ng minh \u0111\u01b0\u1ee3c: $\\triangle{ABC}$ $\\backsim$ $\\triangle{HBA}$ (g\u00f3c-g\u00f3c)<br\/>$\\Rightarrow \\dfrac{AB}{HB} = \\dfrac{BC}{BA}$ <br\/>$\\Rightarrow AB^2 = BC.HB = 25.9 \\Rightarrow AB = 15$<br\/> $S_{\\triangle{ABC}} = \\dfrac{AB.AC}{2} = \\dfrac{15.20}{2} = 150 $ ($cm^2$)<br\/>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 <span class='basic_pink'>150<\/span> <\/span> "}]}],"id_ques":1764},{"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":[[["="]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$ c\u00f3 hai g\u00f3c $B$ v\u00e0 $C$ th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n $\\widehat{B} - \\widehat{C} = 90^o$. K\u1ebb \u0111\u01b0\u1eddng cao $AH$. So s\u00e1nh $AH^2$ v\u00e0 $BH.CH$<br\/><b>\u0110\u00e1p \u00e1n:<\/b> $AH^2$ _input_ $BH.CH$<\/span>","hint":"","explain":"<span class='basic_left'><span class='basic_green'>Ph\u00e2n t\u00edch b\u00e0i to\u00e1n theo s\u01a1 \u0111\u1ed3 ng\u01b0\u1ee3c:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai17/lv3/img\/H8C3B4_K108.png' \/><\/center><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai17/lv3/img\/H8C3B4_K106.png' \/><\/center><br\/>+) $\\triangle{ABH}$ c\u00f3: <br\/>$\\widehat{ABC} = \\widehat{BAH} + \\widehat{AHB}$ (t\u00ednh ch\u1ea5t g\u00f3c ngo\u00e0i tam gi\u00e1c)<br\/>$\\Rightarrow \\widehat{ABC} = \\widehat{BAH} + 90^o$ <b>(1)<\/b><br\/>L\u1ea1i c\u00f3: $\\widehat{ABC} - \\widehat{ACB} = 90^o$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow \\widehat{ABC} = \\widehat{ACB} + 90^o$ <b>(2)<\/b><br\/>T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\widehat{BAH} = \\widehat{ACB}$ hay $\\widehat{BAH} = \\widehat{ACH}$<br\/>X\u00e9t $\\triangle$ vu\u00f4ng $ABH$ v\u00e0 $\\triangle$ vu\u00f4ng $CAH$ c\u00f3:<br\/>$\\widehat{BAH} = \\widehat{ACH}$ (ch\u1ee9ng minh tr\u00ean)<br\/> $\\Rightarrow$ $\\triangle$ vu\u00f4ng $ABH$ $\\backsim$ $\\triangle$ vu\u00f4ng $CAH$ (g\u00f3c-g\u00f3c) <br\/>$\\Rightarrow \\dfrac{AH}{CH} = \\dfrac{BH}{AH}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7)<br\/>$\\Rightarrow AH^2 = BH.CH$<br\/>V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 <span class='basic_pink'> \"= \"<\/span> "}]}],"id_ques":1765},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$, \u0111\u01b0\u1eddng cao $AH$. Bi\u1ebft $\\dfrac{AB}{AC} = \\dfrac{2}{3}$. T\u00ednh t\u1ec9 s\u1ed1 $\\dfrac{HB}{HC}$. <br\/><\/span>","select":[" A. $\\dfrac{HB}{HC} = \\dfrac{2}{3}$ "," B. $\\dfrac{HB}{HC} = \\dfrac{4}{9}$ ","C. $\\dfrac{HB}{HC} = \\dfrac{1}{3}$ ","D. $\\dfrac{HB}{HC} = \\dfrac{3}{2}$ "],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>S\u01a1 \u0111\u1ed3 ng\u01b0\u1ee3c:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai17/lv3/img\/H8C3B4_K109.png' \/><\/center><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai17/lv3/img\/H8C3B4_K107.png' \/><\/center><br\/> Ta c\u00f3: $\\dfrac{S_{\\triangle{AHB}}}{S_{\\triangle{CHA}}} = \\dfrac{ \\dfrac{AH.HB}{2}}{ \\dfrac{AH.HC}{2}} = \\dfrac{HB}{HC}$ <b>(1)<\/b><br\/> X\u00e9t $\\triangle$ vu\u00f4ng $AHB$ v\u00e0 $\\triangle$ vu\u00f4ng $CHA$ c\u00f3:<br\/>$\\widehat{ABH} = \\widehat{HAC}$ (c\u00f9ng ph\u1ee5 $\\widehat{BAH}$ )<br\/> $\\Rightarrow$ $\\triangle$ vu\u00f4ng $AHB$ $\\backsim$ $\\triangle$ vu\u00f4ng $CHA$ (g\u00f3c-g\u00f3c)<br\/>$\\Rightarrow \\dfrac{S_{\\triangle{AHB}}}{S_{\\triangle{CHA}}} = \\left(\\dfrac{AB}{AC}\\right)^2 = \\left(\\dfrac{2}{3}\\right)^2 = \\dfrac{4}{9}$ <b>(2)<\/b><br\/>T\u1eeb (1) v\u00e0 (2) $\\Rightarrow \\dfrac{HB}{HC} = \\dfrac{4}{9}$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'>B. $\\dfrac{HB}{HC} = \\dfrac{4}{9}$<\/span> ","column":2}]}],"id_ques":1766},{"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":[[["5,4"],["9,6"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'> Cho h\u00ecnh v\u1ebd nh\u01b0 sau: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai17/lv3/img\/H8C3B4_K110.png' \/><\/center><br\/>T\u00ednh $x, y$ <br\/><b>\u0110\u00e1p \u00e1n:<\/b> $x$ = _input_ ($cm$); $y$ = _input_ ($cm$)<\/span>","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 $BC$<br\/><b>B\u01b0\u1edbc 2:<\/b> Ch\u1ee9ng minh $\\triangle{AHB}$ $\\backsim$ $\\triangle{CAB}$ <br\/><b>B\u01b0\u1edbc 3:<\/b> T\u00ednh $x, y$.<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai17/lv3/img\/H8C3B4_K110.png' \/><\/center><br\/>$\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow AB^2 + AC^2 = BC^2$ (\u0111\u1ecbnh l\u00ed Py-ta-go)<br\/>$\\Rightarrow BC^2 = 9^2 + 12^2 = 225 \\Rightarrow BC = 15 \\text{(cm)} $ <br\/>X\u00e9t $\\triangle$ vu\u00f4ng $AHB$ v\u00e0 $\\triangle$ vu\u00f4ng $CBA$ c\u00f3:<br\/>$\\widehat{ABH} = \\widehat{CAH}$ (c\u00f9ng ph\u1ee5 $\\widehat{HAB}$<br\/> $\\Rightarrow$ $\\triangle$ vu\u00f4ng $AHB$ $\\backsim$ $\\triangle$ vu\u00f4ng $CAB$ (g\u00f3c-g\u00f3c)<br\/>$\\Rightarrow \\dfrac{AB}{CB} = \\dfrac{HB}{AB}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7)<br\/>$\\Rightarrow HB = \\dfrac{AB^2}{CB} = \\dfrac{9^2}{15} = 5,4 \\text{(cm)} $ hay $x = 5,4cm$<br\/>L\u1ea1i c\u00f3: $x + y = BC = 15 \\Rightarrow y = 15 - x = 15 - 5,4 = 9,6 \\text{(cm)}$ <br\/>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 <span class='basic_pink'>5,4 ; 9,6<\/span> <\/span> "}]}],"id_ques":1767},{"time":24,"part":[{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["1","2","3"]],"list":[{"point":10,"img":"","ques":"<span class='basic_left'>Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$, \u0111\u01b0\u1eddng cao $AH$. G\u1ecdi $I$ v\u00e0 $K$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a \u0111i\u1ec3m $H$ l\u00ean $AB, AC$. Bi\u1ebft $BC = 10cm, AH = 4cm$.<br\/><b>H\u00e3y ch\u1ecdn nh\u1eefng \u0111\u00e1p \u00e1n \u0111\u00fang<\/b><\/span><br\/>","hint":"","column":2,"number_true":2,"select":["A. $\\triangle{AIK} \\backsim \\triangle{ACB} $","B. $S_{\\triangle{AIK}} = 3,2 \\text{cm}^2$","C. $\\widehat{AIK} = \\widehat{ACB}$","D. $S_{\\triangle{AIK}} = 40 \\text{cm}^2$"],"explain":"<span class='basic_left'><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai17/lv3/img\/H8C3B4_K111.png' \/><\/center><br\/>T\u1ee9 gi\u00e1c $AIHK$ c\u00f3:<br\/> $\\widehat{A} = \\widehat{I} = \\widehat{K} = 90^o$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow$ T\u1ee9 gi\u00e1c $AIHK$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft)<br\/>G\u1ecdi $O$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AH$ v\u00e0 $IK$<br\/>$\\Rightarrow$ $OA = OH = OI = OK$ (t\u00ednh ch\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt)<br\/>$\\Rightarrow$ Tam gi\u00e1c $AOI$ c\u00e2n t\u1ea1i $O$ (t\u00ednh ch\u1ea5t)<br\/>$\\Rightarrow$ $\\widehat{OIA} = \\widehat{OAI}$ (\u0111\u1ecbnh ngh\u0129a) (1)<br\/> L\u1ea1i c\u00f3: $\\widehat{OAI} + \\widehat{HAC} = 90^o$ v\u00e0 $\\widehat{HAC} + \\widehat{HCA} = 90^o$<br\/>$\\Rightarrow$ $\\widehat{OAI} = \\widehat{HCA}$ (2)<br\/>T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $\\widehat{OIA} = \\widehat{HCA}$ hay $\\widehat{AIK} = \\widehat{ACB}$ <b>(\u0111\u00e1p \u00e1n C \u0111\u00fang)<\/b><br\/> X\u00e9t $\\triangle$ vu\u00f4ng $AIK$ v\u00e0 $\\triangle$ vu\u00f4ng $ACB$ c\u00f3:<br\/>$\\widehat{AIK} = \\widehat{ACB}$ (ch\u1ee9ng minh tr\u00ean)<br\/> $\\Rightarrow$ $\\triangle$ vu\u00f4ng $AIK$ $\\backsim$ $\\triangle$ vu\u00f4ng $ACB$ (g\u00f3c - g\u00f3c) <b>(\u0111\u00e1p \u00e1n A \u0111\u00fang)<\/b><br\/>$\\Rightarrow$ $\\dfrac{S_{\\triangle{AIK}}}{S_{\\triangle{ACB}}} = \\left(\\dfrac{IK}{BC}\\right)^2 = \\left(\\dfrac{AH}{BC}\\right)^2 $<br\/> $\\Rightarrow$ $S_{\\triangle{AIK}} = \\left(\\dfrac{AH}{BC}\\right)^2.S_{\\triangle{ACB}} $<br\/> L\u1ea1i c\u00f3: $S_{\\triangle{ACB}} = \\dfrac{1}{2}.BC.AH = \\dfrac{1}{2}.10.4 = 20 (\\text{cm}^2)$ <b>(\u0111\u00e1p \u00e1n D sai)<\/b><br\/> $\\Rightarrow$ $S_{\\triangle{AIK}} = \\left(\\dfrac{4}{10}\\right)^2.20 = 3,2 (\\text{cm}^2)$ <b>(\u0111\u00e1p \u00e1n B \u0111\u00fang) <\/b><br\/>V\u1eady nh\u1eefng \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'>A, B, C<\/span><br\/> "}]}],"id_ques":1768},{"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":[[["15"],["20"],["25"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$, \u0111\u01b0\u1eddng cao $AH$ ($H \\in BC$). Bi\u1ebft $BH = 9cm, CH = 16cm$. T\u00ednh c\u00e1c c\u1ea1nh c\u1ee7a tam gi\u00e1c $ABC$. <br\/><b>\u0110\u00e1p \u00e1n:<\/b> $AB$ = _input_ ($cm$); $AC$ = _input_ ($cm$), $BC$ = _input_ ($cm$)<\/span>","hint":"","explain":"<span class='basic_left'><span class='basic_green'>Ph\u00e2n t\u00edch b\u00e0i to\u00e1n theo s\u01a1 \u0111\u1ed3 ng\u01b0\u1ee3c, ta c\u00f3:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai17/lv3/img\/H8C3B4_K112.png' \/><\/center><br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai17/lv3/img\/H8C3B4_K113.png' \/><\/center><br\/>X\u00e9t $\\triangle$ vu\u00f4ng $HAB$ v\u00e0 $\\triangle$ vu\u00f4ng $HCA$ c\u00f3:<br\/>$\\widehat{ABH} = \\widehat{CAH}$ (c\u00f9ng ph\u1ee5 $\\widehat{HAB}$)<br\/> $\\Rightarrow$ $\\triangle$ vu\u00f4ng $HAB$ $\\backsim$ $\\triangle$ vu\u00f4ng $HCA$ (g\u00f3c-g\u00f3c)<br\/>$\\Rightarrow \\dfrac{HA}{HC} = \\dfrac{HB}{HA}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7)<br\/>$\\Rightarrow HA^2 = HB.HC = 9.16 = 144 \\Rightarrow HA = 12 \\text{(cm)}$ <br\/>$\\triangle{ABH}$ vu\u00f4ng t\u1ea1i $H$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow AH^2 + BH^2 = AB^2$ (\u0111\u1ecbnh l\u00ed Py-ta-go)<br\/>$\\Rightarrow AB^2 = 12^2 + 9^2 = 225 \\Rightarrow AB = 15 \\text{(cm)}$<br\/> $\\triangle{ACH}$ vu\u00f4ng t\u1ea1i $H$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow AH^2 + CH^2 = AC^2$ (\u0111\u1ecbnh l\u00ed Py-ta-go)<br\/>$\\Rightarrow AC^2 = 12^2 + 16^2 = 400 \\Rightarrow AC = 20 \\text{(cm)}$<br\/>$\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow AB^2 + AC^2 = BC^2$ (\u0111\u1ecbnh l\u00ed Py-ta-go)<br\/>$\\Rightarrow BC^2 = 15^2 + 20^2 = 625 \\Rightarrow BC = 25 \\text{(cm)}$ <br\/>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 <span class='basic_pink'>15; 20; 25<\/span> <\/span> "}]}],"id_ques":1769}],"lesson":{"save":0,"level":3}}