Ba trường hợp đồng dạng của hai tam giác

Home Lớp 8 Toán lớp 8 - Sách Kết nối tri thức Bài 34. Ba trường hợp đồng dạng của hai tam giácBài tập nâng cao

{"common":{"save":0,"post_id":"6659","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"6898","post_id":"6659","mon_id":"1158921","chapter_id":"1158939","question":"Hai tam giác ABC và DEF \u0111\u1ed3ng d\u1ea1ng có AB = 3 cm ; AC = 4 cm ; BC = 5 cm. Chu vi c\u1ee7a tam giác DEF là 24 cm. \u0110\u1ed9 dài các c\u1ea1nh c\u1ee7a tam giác DEF theo th\u1ee9 t\u1ef1 t\u1eeb bé \u0111\u1ebfn l\u1edbn l\u1ea7n l\u01b0\u1ee3t là:<\/p>","options":["A.<\/strong> 6 cm ; 8 cm ; 10 cm","B.<\/strong> 7 cm ; 8 cm ; 9 cm","C.<\/strong> 4 cm ; 8 cm ; 12 cm","D.<\/strong> 5 cm ; 7 cm ; 12 cm"],"correct":"1","level":"3","hint":"","answer":"Chu vi c\u1ee7a tam giác DEF là 24 cm nên DE + DF + EF = 24 cm.<\/p>Vì $\\triangle ABC \\backsim \\triangle DEF$<\/span> nên $\\dfrac{AB}{DE}=\\dfrac{AC}{DF}=\\dfrac{BC}{EF}$<\/span> hay $\\dfrac{3}{DE}=\\dfrac{4}{DF}=\\dfrac{5}{EF}$<\/span> .<\/p>Áp d\u1ee5ng tính ch\u1ea5t c\u1ee7a dãy t\u1ec9 s\u1ed1 b\u1eb1ng nhau<\/p>$\\dfrac{3}{DE}=\\dfrac{4}{DF}=\\dfrac{5}{EF}=\u200b\u200b\\dfrac{3+4+5}{DE+DF+EF}=\\dfrac{12}{24}=\\dfrac{1}{2}$<\/span> .<\/p>Ta có  $\\dfrac{3}{DE}=\\dfrac{1}{2} \\Rightarrow DE = \\dfrac{3.2}{1}=6$<\/span> cm <\/p> $\\dfrac{4}{DF}=\\dfrac{1}{2} \\Rightarrow DF = \\dfrac{4.2}{1}=8$<\/span> cm<\/p> $\\dfrac{5}{EF}=\\dfrac{1}{2} \\Rightarrow EF = \\dfrac{5.2}{1}=10$<\/span> cm <\/p>\u0110áp án \u0111úng là   A.<\/strong> 6 cm ; 8 cm ; 10 cm.<\/span><\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"131","test":"0","date":"2023-09-08 21:31:03","option_type":"txt","len":2},{"id":"6900","post_id":"6659","mon_id":"1158921","chapter_id":"1158939","question":"Cho tam giác ABC và tam giác DEF có AB = 5 cm ; AC = 12 cm ; BC = 13 cm và chu vi c\u1ee7a tam giác DEF là 15 cm. \u0110\u1ed9 dài các c\u1ea1nh c\u1ee7a tam giác DEF theo th\u1ee9 t\u1ef1 t\u1eeb bé \u0111\u1ebfn l\u1edbn l\u1ea7n l\u01b0\u1ee3t là<\/p>","options":["A.<\/strong> 2 cm ; 6 cm ; 7 cm","B.<\/strong> 2,5 cm ; 6 cm ; 6,5 cm","C.<\/strong> 1 cm ; 6,5 cm ; 7,5 cm","D.<\/strong> 1,5 cm ; 6,5 cm ; 7 cm"],"correct":"2","level":"3","hint":"","answer":"Chu vi c\u1ee7a tam giác DEF là 15 cm nên DE + DF + EF = 15 cm.<\/p>Vì $\\triangle ABC \\backsim \\triangle DEF$<\/span> nên $\\dfrac{AB}{DE}=\\dfrac{AC}{DF}=\\dfrac{BC}{EF}$<\/span> hay $\\dfrac{5}{DE}=\\dfrac{12}{DF}=\\dfrac{13}{EF}$<\/span> .<\/p>Áp d\u1ee5ng tính ch\u1ea5t c\u1ee7a dãy t\u1ec9 s\u1ed1 b\u1eb1ng nhau<\/p>$\\dfrac{5}{DE}=\\dfrac{12}{DF}=\\dfrac{13}{EF}=\u200b\u200b\\dfrac{5+12+13}{DE+DF+EF}=\\dfrac{30}{15}=2$<\/span> .<\/p>Ta có  $\\dfrac{5}{DE}=2 \\Rightarrow DE = \\dfrac{5.1}{2}=2,5$<\/span> cm <\/p> $\\dfrac{12}{DF}=2 \\Rightarrow DF = \\dfrac{12.1}{2}=6$<\/span> cm<\/p> $\\dfrac{13}{EF}=2 \\Rightarrow EF = \\dfrac{13.1}{2}=6,5$<\/span> cm <\/p>\u0110áp án \u0111úng là   B.<\/strong> 2,5 cm ; 6 cm ; 6,5 cm.<\/span><\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"131","test":"0","date":"2023-09-08 21:37:18","option_type":"txt","len":2},{"id":"6902","post_id":"6659","mon_id":"1158921","chapter_id":"1158939","question":"Cho tam giác ABC. Trên c\u1ea1nh AB và AC l\u1ea7n l\u01b0\u1ee3t l\u1ea5y các \u0111i\u1ec3m M, N sao cho $\\widehat{AMN}=\\widehat{ACB}$<\/span> . K\u1ebft lu\u1eadn \u0111úng là<\/p><svg height=\"170\" width=\"190\"> <g><title><\/title><rect fill=\"#fff\" height=\"172\" id=\"canvas_background\" width=\"192\" x=\"-1\" y=\"-1\"><\/rect> <g display=\"none\" height=\"100%\" id=\"canvasGrid\" overflow=\"visible\" width=\"100%\" x=\"0\" y=\"0\"> <rect fill=\"url(#gridpattern)\" height=\"100%\" stroke-width=\"0\" width=\"100%\" x=\"0\" y=\"0\"><\/rect> <\/g> <\/g> <g><title><\/title><g id=\"svg_14\"> <g id=\"svg_8\"> <g id=\"svg_4\"> <line fill=\"none\" id=\"svg_1\" stroke=\"#000\" stroke-linecap=\"undefined\" stroke-linejoin=\"undefined\" stroke-width=\"1.5\" x1=\"72.31535\" x2=\"23.31535\" y1=\"27.52273\" y2=\"147.52273\"><\/line> <line fill=\"none\" id=\"svg_2\" stroke=\"#000\" stroke-linecap=\"undefined\" stroke-linejoin=\"undefined\" stroke-width=\"1.5\" x1=\"72.31535\" x2=\"169.31534\" y1=\"28.52273\" y2=\"147.52273\"><\/line> <line fill=\"none\" fill-opacity=\"null\" id=\"svg_3\" stroke=\"#000\" stroke-linecap=\"undefined\" stroke-linejoin=\"undefined\" stroke-opacity=\"null\" stroke-width=\"1.5\" x1=\"23.31535\" x2=\"169\" y1=\"147.52274\" y2=\"148\"><\/line> <\/g> <line fill=\"none\" fill-opacity=\"null\" id=\"svg_5\" stroke=\"#000\" stroke-linecap=\"undefined\" stroke-linejoin=\"undefined\" stroke-opacity=\"null\" stroke-width=\"1.5\" x1=\"40.31535\" x2=\"110.31535\" y1=\"105.52273\" y2=\"76.52273\"><\/line> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"20\" id=\"svg_6\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" transform=\"rotate(-45 56.64630889892574,86.7045440673828) \" x=\"53.31535\" xml:space=\"preserve\" y=\"93.52273\">)<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"20\" id=\"svg_7\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" transform=\"rotate(25 148.64631652832043,135.7045593261719) \" x=\"145.31535\" xml:space=\"preserve\" y=\"142.52273\">(<\/text> <\/g> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"20\" id=\"svg_9\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"59.31535\" xml:space=\"preserve\" y=\"24.52273\">A<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"20\" id=\"svg_10\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"8.31535\" xml:space=\"preserve\" y=\"157.52273\">B<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"20\" id=\"svg_11\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"171.31535\" xml:space=\"preserve\" y=\"159.52273\">C<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"20\" id=\"svg_12\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"19.31535\" xml:space=\"preserve\" y=\"106.52273\">M<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"20\" id=\"svg_13\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"114.31535\" xml:space=\"preserve\" y=\"75.52273\">N<\/text> <\/g> <\/g> <\/svg><\/span><\/p>","options":["A.<\/strong> \u25b3AMN$\\backsim$<\/span>\u25b3ABC","B.<\/strong> \u25b3ABC$\\backsim$<\/span>\u25b3MAN","C.<\/strong> \u25b3BCA$\\backsim$<\/span>\u25b3MNA","D.<\/strong> \u25b3AMN$\\backsim$<\/span>\u25b3ACB"],"correct":"4","level":"3","hint":"","answer":"Xét tam giác AMN và tam giác ABC có <\/p>$\\widehat{A}$<\/span> là góc chung<\/p>$\\widehat{AMN}=\\widehat{ACB}$<\/span><\/p>Suy ra  D.<\/strong> \u25b3AMN$\\backsim$<\/span>\u25b3ACB<\/span>  (g.g).<\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"131","test":"0","date":"2023-09-08 21:45:40","option_type":"math","len":0}]}