{"segment":[{"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":" T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ee7a $x$ trong h\u00ecnh sau, bi\u1ebft $AB \/\/ FG$: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai19/lv3/img\/H8C3B6_D101.png' \/><\/center>","select":[" A. $8cm$ "," B. $9,6cm$","C. $10cm$","D. $12cm$"],"hint":"S\u1eed d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t.","explain":"<span class='basic_left'>X\u00e9t $\\triangle EFG$ c\u00f3:<br\/>$\\left.\\begin{array}{l} AB \/\/ FG \\text{(gi\u1ea3 thi\u1ebft)}\\\\ A \\in EG; B \\in EF \\end{array} \\right\\}$ $\\Rightarrow \\dfrac{AE}{EG} = \\dfrac{BE}{EF}$ (h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00fd Ta-l\u00e9t)<br\/> $\\Rightarrow \\dfrac{4}{x} = \\dfrac{5}{12,5}$<br\/>$\\Rightarrow x = \\dfrac{4.12,5}{5} = 10 \\text{(cm)}$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'>C. $10cm$<\/span><\/span> ","column":2}],"id_ques":1800},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<span class='basic_left'>T\u00ecm \u0111\u1ed9 d\u00e0i c\u1ee7a $x$ trong h\u00ecnh v\u1ebd sau: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai19/lv3/img\/H8C3B6_D102.png' \/><\/center><\/span>","select":["A. $x = 6$","B. $x = 6,5$","C. $x = 8$","D. $x = 12,6$"],"hint":"Ch\u1ee9ng minh $EF \/\/ BC$. T\u1eeb \u0111\u00f3 \u00e1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t \u0111\u1ec3 t\u00ecm $x$.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> T\u00ednh t\u1ec9 s\u1ed1 $\\dfrac{EA}{EB}$ v\u00e0 $\\dfrac{FA}{FC}$<br\/><b>B\u01b0\u1edbc 2:<\/b> T\u1eeb t\u1ec9 s\u1ed1 t\u00ecm \u0111\u01b0\u1ee3c \u1edf b\u01b0\u1edbc 1, \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Ta-l\u00e9t \u0111\u1ea3o ch\u1ee9ng minh \u0111\u01b0\u1ee3c $EF$ \/\/ $BC$<br\/><b>B\u01b0\u1edbc 3:<\/b> \u00c1p d\u1ee5ng h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t t\u00ecm $x$.<\/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/bai19/lv3/img\/H8C3B6_D102.png' \/><\/center><br\/>Ta c\u00f3:<br\/>$\\left.\\begin{array}{l} \\dfrac{EA}{EB} = \\dfrac{4}{7} \\\\ \\dfrac{FA}{FC} = \\dfrac{5}{8,75} = \\dfrac{4}{7} \\end{array} \\right\\}$ $\\Rightarrow \\dfrac{EA}{EB} = \\dfrac{FA}{FC}$ <br\/> $\\Rightarrow EF \/\/ BC$ (\u0111\u1ecbnh l\u00ed Ta-l\u00e9t \u0111\u1ea3o)<br\/>X\u00e9t $\\triangle ABC$ c\u00f3: $EF \/\/ BC$ (ch\u1ee9ng minh tr\u00ean)<br\/> $\\Rightarrow \\dfrac{AE}{AB}=\\dfrac{EF}{BC} $ (h\u1ec7 qu\u1ea3 c\u1ee7a \u0111\u1ecbnh l\u00ed Ta-l\u00e9t)<br\/>$\\Rightarrow \\dfrac{AE}{AE + EB}=\\dfrac{EF}{BC} $<br\/> $\\Rightarrow$ $\\dfrac{4}{4 + 7}=\\dfrac{x}{22} $<br\/> $\\Rightarrow$ $\\dfrac{4}{11}=\\dfrac{x}{22} $<br\/> $\\Rightarrow$ $x = \\dfrac{4.22}{11} = 8$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: <span class='basic_pink'>C. $x = 8$<\/span>","column":2}],"id_ques":1801},{"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 ph\u00e2n gi\u00e1c $AD$. Bi\u1ebft $DB = 9cm, DC = 12cm$.<br\/><b>H\u00e3y ch\u1ecdn nh\u1eefng \u0111\u00e1p \u00e1n \u0111\u00fang<\/b><\/span>","hint":"","column":2,"number_true":2,"select":["A. $AC = 16,8cm$","B. $AB = 12,6 cm$","C. $\\dfrac{AB}{AC} = \\dfrac{3}{4}$","D. $\\dfrac{AC}{AB} = \\dfrac{DB}{DC}$"],"explain":"<span class='basic_left'><span class='basic_green'>Ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i:<\/span><br\/>+ \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Py-ta-go \u0111\u1ec3 \u0111\u01b0\u1ee3c: $AB^2 + AC^2 = m^2$ ($m$ l\u00e0 h\u1eb1ng s\u1ed1)<br\/>+ S\u1eed d\u1ee5ng t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong tam gi\u00e1c l\u00e0m xu\u1ea5t hi\u1ec7n t\u1ec9 l\u1ec7 th\u1ee9c $\\dfrac{AB}{AC} = \\dfrac{a}{b}$ ($a + b = m$, $a, b$ l\u00e0 c\u00e1c h\u1eb1ng s\u1ed1) t\u1eeb \u0111\u00f3 bi\u1ec3n \u0111\u1ed5i \u0111\u1ec3 xu\u1ea5t hi\u1ec7n t\u1ec9 l\u1ec7 th\u1ee9c c\u00f3 d\u1ea1ng $\\dfrac{AB^2}{a^2} = \\dfrac{AC^2}{b^2}$<br\/>+ \u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau \u0111\u1ec3 t\u00ednh $AB^2, AC^2$ t\u1eeb \u0111\u00f3 suy ra $AB, AC$.<\/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/bai19/lv3/img\/H8C3B6_D103.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\/> $\\Rightarrow AB^2 + AC^2 = (DB + DC)^2 = (9 + 12)^2 = 21^2 = 441$<br\/> $\\triangle ABC$ c\u00f3:<br\/> $AD$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BAC}$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow \\dfrac{AB}{AC} = \\dfrac{DB}{DC}$ (t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c)<br\/>$\\Rightarrow \\dfrac{AB}{AC} = \\dfrac{9}{12} = \\dfrac{3}{4}$<br\/> $\\Rightarrow \\dfrac{AB}{3} = \\dfrac{AC}{4 }$<br\/> $\\Rightarrow \\dfrac{AB^2}{9} = \\dfrac{AC^2}{16}$<br\/>\u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau, ta c\u00f3:<br\/>$\\dfrac{AB^2}{9} = \\dfrac{AC^2}{16} = \\dfrac{AB^2 + AC^2}{9 + 16} = \\dfrac{441}{25} = 17,64 $ <br\/>$\\Rightarrow$ $\\begin{cases}AC^2 = 16.17,64 = 282,24 \\\\ AB^2 = 9.17,64 = 158,76\\end{cases}$<br\/>$\\Rightarrow$ $\\begin{cases}AC = 16,8 \\text{(cm)} \\\\ AB = 12,6 \\text{(cm)} \\end{cases}$<br\/>V\u1eady nh\u1eefng \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'> A, B v\u00e0 C<\/span><br\/> "}],"id_ques":1802},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["27"],["30"],["36"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'>Cho tam gi\u00e1c $ABC$, c\u00e1c \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c $BD$, $CE$. Bi\u1ebft $\\dfrac{AD}{DC} = \\dfrac{3}{4}$, $\\dfrac{AE}{EB} = \\dfrac{5}{6}$. T\u00ednh \u0111\u1ed9 d\u00e0i c\u00e1c c\u1ea1nh $AB$, $AC$, $BC$, bi\u1ebft chu vi tam gi\u00e1c $ABC$ b\u1eb1ng $93cm$.<br\/> <b>\u0110\u00e1p \u00e1n:<\/b> $AB$ = _input_ ($cm$); $AC$ = _input_ ($cm$); $BC$ = _input_ ($cm$)<\/span> ","hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai19/lv3/img\/H8C3B6_D104.png' \/><\/center><br\/>$\\triangle ABC$ c\u00f3: $BD$ v\u00e0 $CE$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{ABC}$ v\u00e0 $\\widehat{ACB}$(gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow \\begin{cases}\\dfrac{AC}{BC} = \\dfrac{AE}{EB} = \\dfrac{5}{6} \\\\ \\dfrac{AB}{BC} = \\dfrac{AD}{DC} = \\dfrac{3}{4} \\end{cases}$ (\u0111\u1ecbnh l\u00ed t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c)<br\/>$\\Rightarrow \\begin{cases}\\dfrac{AC}{5} = \\dfrac{BC}{6} \\Rightarrow \\dfrac{AC}{10} = \\dfrac{BC}{12} \\\\ \\dfrac{AB}{3} = \\dfrac{BC}{4} \\Rightarrow \\dfrac{AB}{9} = \\dfrac{BC}{12} \\end{cases}$ <br\/>$\\Rightarrow \\dfrac{AB}{9} = \\dfrac{AC}{10} = \\dfrac{BC}{12}$<br\/>\u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau, ta c\u00f3:<br\/>$\\dfrac{AB}{9} = \\dfrac{AC}{10} = \\dfrac{BC}{12} = \\dfrac{AB + AC + BC}{9 + 10 + 12} = \\dfrac{93}{31} = 3$<br\/>Do \u0111\u00f3<br\/>$AC = 3.10 = 30 \\text{(cm)}$ <br\/>$AB = 3.9 = 27 \\text{(cm)}$<br\/>$BC = 3.12 = 36 \\text{(cm)}$<br\/>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 <span class='basic_pink'>27; 30; 36<\/span> <br\/> <span class='basic_left'><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span> \u0110\u1ed1i v\u1edbi d\u1ea1ng b\u00e0i cho bi\u1ebft c\u00e1c \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c v\u00e0 c\u00e1c t\u1ec9 s\u1ed1 \u0111\u1ed9 d\u00e0i hai \u0111o\u1ea1n th\u1eb3ng ta \u0111i l\u1eadp c\u00e1c \u0111o\u1ea1n th\u1eb3ng t\u1ec9 l\u1ec7 t\u1eeb t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c \u0111\u1ec3 gi\u1ea3i.<\/span>"}],"id_ques":1803},{"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 $ABC$ v\u00e0 \u0111i\u1ec3m $O$ n\u1eb1m trong tam gi\u00e1c $ABC$. G\u1ecdi $M, N, P$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $OA, OB, OC$. Bi\u1ebft chu vi tam gi\u00e1c $ABC$ b\u1eb1ng $30cm$. <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{ABC} \\backsim \\triangle{MNP} $","B. Chu vi $\\triangle{MNP}$ $= 20cm$","C. $C_{\\triangle{MNP}} = \\dfrac{ C_{\\triangle{ABC}}}{2}$","D. $ \\dfrac{MN}{AB} = \\dfrac{NP}{BC} = \\dfrac{MP}{AC}$"],"explain":"<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/bai19/lv3/img\/H8C3B6_D105.png' \/><\/center><br\/> X\u00e9t $\\triangle{OAB}$ c\u00f3:<br\/> $M, N$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $OA,OB$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow MN$ l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a $\\triangle{OAB}$<br\/>$\\Rightarrow MN = \\dfrac{1}{2}AB$ hay $\\dfrac{MN}{AB} = \\dfrac{1}{2}$<br\/>T\u01b0\u01a1ng t\u1ef1 ta ch\u1ee9ng minh \u0111\u01b0\u1ee3c: $\\dfrac{NP}{BC} = \\dfrac{1}{2}$ ; $\\dfrac{MP}{AC} = \\dfrac{1}{2}$<br\/>$\\Rightarrow \\dfrac{MN}{AB} = \\dfrac{NP}{BC} = \\dfrac{MP}{AC} = \\dfrac{1}{2}$<br\/>$\\Rightarrow $ $\\triangle{ABC} \\backsim \\triangle{MNP} $ (c - c - c) v\u00e0 t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng l\u00e0 $\\dfrac{1}{2}$<br\/>L\u1ea1i c\u00f3: $ \\dfrac{MN}{AB} = \\dfrac{NP}{BC} = \\dfrac{MP}{AC} = \\dfrac{1}{2} = \\dfrac{MN + NP + MP }{AB + BC + AC} = \\dfrac{ C_{\\triangle_{MNP}}}{C_{\\triangle_{ABC}}}$<br\/>$\\Rightarrow C_{\\triangle{MNP}} = \\dfrac{ C_{\\triangle{ABC}}}{2} = \\dfrac{30}{2} = 15 \\text{(cm)}$<br\/>V\u1eady nh\u1eefng \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'>A, C, D<\/span> "}],"id_ques":1804},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["7,5"]]],"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/bai19/lv3/img\/H8C3B6_D106.png' \/><\/center><br\/>T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh $DE$ (k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t)<br\/><b>\u0110\u00e1p \u00e1n:<\/b> $DE =$ _input_ ($cm$)<\/span>","hint":"S\u1eed d\u1ee5ng tr\u01b0\u1eddng h\u1ee3p \u0111\u1ed3ng d\u1ea1ng th\u1ee9 hai: c\u1ea1nh-g\u00f3c-c\u1ea1nh. ","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 $\\triangle{ADE} \\backsim \\triangle{ACB}$ t\u1eeb \u0111\u00f3 suy ra c\u00e1c \u0111o\u1ea1n th\u1eb3ng t\u1ec9 l\u1ec7<br\/><b>B\u01b0\u1edbc 2:<\/b> T\u1eeb b\u01b0\u1edbc 1 suy ra \u0111\u1ed9 d\u00e0i c\u1ea1nh $DE$<\/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/bai19/lv3/img\/H8C3B6_D106.png' \/><\/center><br\/>X\u00e9t $\\triangle{ADE}$ v\u00e0 $\\triangle{ACB}$ c\u00f3:<br\/>$\\widehat{A}$ chung<br\/>$\\dfrac{AE}{EB} = \\dfrac{AD}{AC}$ (v\u00ec $\\dfrac{4}{8} = \\dfrac{6}{12}$)<br\/> $\\Rightarrow$ $\\triangle{ADE} \\backsim \\triangle{ACB}$ (c\u1ea1nh-g\u00f3c-c\u1ea1nh)<br\/>$\\Rightarrow \\dfrac{AD}{AC} = \\dfrac{DE}{BC}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/>$\\Rightarrow DE = \\dfrac{AD.BC}{AC} = \\dfrac{6.15}{12} = 7,5 \\text{(cm)}$ <br\/>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 <span class='basic_pink'>7,5<\/span>"}],"id_ques":1805},{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["4,5"],["3,5"]]],"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/bai19/lv3/img\/H8C3B6_D107.png' \/><\/center><br\/>Bi\u1ebft $AB = 6cm; AC = 8cm$, $\\widehat{ABD} = \\widehat{ACB}$. T\u00ednh $AD, DC$ (k\u1ebft qu\u1ea3 vi\u1ebft \u1edf d\u1ea1ng s\u1ed1 th\u1eadp ph\u00e2n)<br\/><b>\u0110\u00e1p \u00e1n:<\/b> $AD =$ _input_ ($cm$); $DC = $ _input_ ($cm$) <\/span>","hint":"","explain":"<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/bai19/lv3/img\/H8C3B6_D107.png' \/><\/center><br\/>X\u00e9t $\\triangle{ABD}$ v\u00e0 $\\triangle{ACB}$ c\u00f3:<br\/>$\\widehat{ABD} = \\widehat{ACB}$ (gi\u1ea3 thi\u1ebft)<br\/>$\\widehat{A} $ chung<br\/> $\\Rightarrow$ $\\triangle{ABD} \\backsim \\triangle{ACB}$ (g\u00f3c - g\u00f3c)<br\/>$\\Rightarrow \\dfrac{AB}{AC} = \\dfrac{AD}{AB} $<br\/>$\\Rightarrow \\dfrac{6}{8} = \\dfrac{AD}{6} $ <br\/>$\\Rightarrow AD = \\dfrac{6.6}{8} = 4,5 \\text{(cm)} $<br\/>L\u1ea1i c\u00f3: $AD + DC = AC$<br\/>$\\Rightarrow DC = AC - AD = 8 - 4,5 = 3,5 \\text{cm}$<br\/>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 <span class='basic_pink'>$4,5; 3,5$<\/span>"}],"id_ques":1806},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["60"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'>Cho tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{BAC} = 80^o$, $\\dfrac{AB}{BC} = \\dfrac{BC}{AB + AC}$. T\u00ednh $\\widehat{ABC}$<br\/><b>\u0110\u00e1p \u00e1n:<\/b> $\\widehat{B} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^o$<\/span>","hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai19/lv3/img\/H8C3B6_D108.png' \/><\/center><br\/>+) V\u1ebd $AD$ l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BAC}$, $D \\in BC$<br\/>$\\Rightarrow$ $\\dfrac{BD}{DC} = \\dfrac{AB}{AC}$ (t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c)<br\/>$\\Rightarrow$ $\\dfrac{BD}{AB} = \\dfrac{DC}{AC}$<br\/>\u00c1p d\u1ee5ng t\u00ednh ch\u1ea5t d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau, ta c\u00f3:<br\/>$\\dfrac{BD}{AB} = \\dfrac{DC}{AC} = \\dfrac{BD + DC}{AB + AC} = \\dfrac{BC}{AB + AC}$ <br\/>M\u00e0 $\\dfrac{AB}{BC} = \\dfrac{BC}{AB + AC}$ (gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow$ $\\dfrac{AB}{BC} = \\dfrac{DB}{AB}$ <br\/>+) X\u00e9t $\\triangle{BAD}$ v\u00e0 $\\triangle{BCA}$ c\u00f3:<br\/>$\\widehat{B}$ chung<br\/>$\\dfrac{AB}{BC} = \\dfrac{DB}{AB}$ (ch\u1ee9ng minh tr\u00ean) <br\/>$\\Rightarrow$ $\\triangle{BAD}$ $\\backsim$ $\\triangle{BCA}$ (c\u1ea1nh-g\u00f3c-c\u1ea1nh)<br\/>$\\Rightarrow$ $\\widehat{BAD} = \\widehat{BCA}$ (c\u1eb7p g\u00f3c t\u01b0\u01a1ng \u1ee9ng)<br\/> L\u1ea1i c\u00f3: $\\widehat{BAD} = \\dfrac{\\widehat{BAC}}{2} = \\dfrac{\\widehat{80}}{2} = 40^o$ (v\u00ec $AD$ l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BAC}$)<br\/>$\\Rightarrow \\widehat{BCA} = 40^o$<br\/> $\\triangle{ABC}$ c\u00f3: $\\widehat{ABC} + \\widehat{BCA} + \\widehat{BAC} = 180^o$ (\u0111\u1ecbnh l\u00ed t\u1ed5ng ba g\u00f3c trong m\u1ed9t tam gi\u00e1c) <br\/>$\\Rightarrow \\widehat{ABC} = 180^o - (\\widehat{BAC} + \\widehat{BCA}) = 180^o - (80^o + 40^o) = 60^o$<br\/>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 <span class='basic_pink'>$60$<\/span><br\/><span class='basic_left'><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/> Khi b\u00e0i to\u00e1n cho c\u00e1c t\u1ec9 s\u1ed1 b\u1eb1ng nhau v\u00e0 c\u00f3 xu\u1ea5t hi\u1ec7n d\u1ea1ng t\u1ec9 s\u1ed1 $\\dfrac{BC}{AB + AC}$ th\u00ec ch\u00fang ta c\u1ea7n ngh\u0129 \u0111\u1ebfn vi\u1ec7c v\u1ebd th\u00eam $AD$ l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BAC}$<\/span> <br\/><br\/> "}],"id_ques":1807},{"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 $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$. Tr\u00ean c\u1ea1nh $BC$ l\u1ea5y \u0111i\u1ec3m $K$, t\u1eeb $K$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi $AB$ t\u1ea1i $I$. Bi\u1ebft $BI = 4cm, AB = 9cm$ v\u00e0 $S_{\\triangle{ABC}} = 162m^2$. T\u00ednh $S_{\\triangle{IBK}}$<\/span>","select":[" A. $S_{\\triangle{IBK}} = 28cm^2$"," B. $S_{\\triangle{IBK}} = 25cm^2$","C. $S_{\\triangle{IBK}} = 30cm^2$","D. $S_{\\triangle{IBK}} = 32cm^2$"],"hint":"Ch\u1ee9ng minh $\\triangle{BIK} \\backsim \\triangle{BAC}$, t\u1eeb \u0111\u00f3 suy ra $\\dfrac{S_{\\triangle{BIK}}}{S_{\\triangle{BAC}}}$","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai19/lv3/img\/H8C3B6_D109.png' \/><\/center><br\/>X\u00e9t $\\triangle$ vu\u00f4ng $BIK$ v\u00e0 $\\triangle$ vu\u00f4ng $BAC$ c\u00f3:<br\/>$\\widehat{BIK} = \\widehat{BAC}$ (c\u00f9ng $ = 90^o$)<br\/>$\\widehat{B}$ chung<br\/>$\\Rightarrow$ $\\triangle$ vu\u00f4ng $BIK$ $\\backsim$ $\\triangle$ vu\u00f4ng $BAC$ (g\u00f3c - g\u00f3c)<br\/>$\\Rightarrow \\left(\\dfrac{BI}{BA}\\right)^2 = \\dfrac{S_{\\triangle{BIK}}}{S_{\\triangle{BAC}}} $ hay $ \\dfrac{S_{\\triangle{BIK}}}{S_{\\triangle{BAC}}} = \\left(\\dfrac{4}{9}\\right)^2 = \\dfrac{16}{81}$ <br\/>$\\Rightarrow S_{\\triangle{BIK}} = \\dfrac{16.S_{\\triangle{BAC}}}{81} = \\dfrac{16.162}{81} = 32 (\\text{cm}^2)$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 <span class='basic_pink'>D. $S_{\\triangle{IBK}} = 32 (\\text{cm}^2)$<\/span><br\/><span class='basic_left'><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/>T\u1ec9 s\u1ed1 di\u1ec7n t\u00edch c\u1ee7a hai tam gi\u00e1c \u0111\u1ed3ng d\u1ea1ng b\u1eb1ng t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng<\/span> <br\/><br\/>","column":2}],"id_ques":1808},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["4,5"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'>T\u00ecm $x$ trong h\u00ecnh v\u1ebd sau:<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai19/lv3/img\/H8C3B6_D110.png' \/><\/center><br\/><b>\u0110\u00e1p \u00e1n:<\/b> $x = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ (k\u1ebft qu\u1ea3 vi\u1ebft d\u01b0\u1edbi d\u1ea1ng s\u1ed1 th\u1eadp ph\u00e2n)<\/span>","hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai19/lv3/img\/H8C3B6_D110.png' \/><\/center><br\/> X\u00e9t $\\triangle{ABC}$ v\u00e0 $\\triangle{EBD}$ c\u00f3:<br\/>$\\widehat{ABC} = \\widehat{EBD}$ (hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh)<br\/>$\\widehat{CAB} = \\widehat{BED}$ (c\u00f9ng = $90^o$)<br\/>$\\Rightarrow$ $\\triangle{ABC}$ $\\backsim$ $\\triangle{EBD}$ (g\u00f3c - g\u00f3c)<br\/>$\\Rightarrow$ $\\dfrac{AB}{EB} = \\dfrac{AC}{ED}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7)<br\/>$\\Rightarrow$ $\\dfrac{4}{6} = \\dfrac{3}{x} \\Rightarrow x = \\dfrac{3.6}{4} = 4,5 \\text{(cm)} $ <br\/>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 <span class='basic_pink'>$4,5$<\/span> "}],"id_ques":1809}],"id_ques":0}],"lesson":{"save":1,"level":3,"time":44}}