Tam giác cân. Đường trung trực của đoạn thẳng

Home Toán lớp 7 - Sách KNTT Toán lớp 7 - tập 1 Bài 16: Tam giác cân. Đường trung trực của đoạn thẳng.Bài tập nâng cao

{"common":{"save":0,"post_id":"5706","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"7569","post_id":"5706","mon_id":"1158532","chapter_id":"1158548","question":"Cho tam giác ABC cân t\u1ea1i A và các \u0111i\u1ec3m E và F l\u1ea7n l\u01b0\u1ee3t n\u1eb1m trên các c\u1ea1nh AC và AB sao cho BE $\\bot$<\/span> AC và CF $\\bot$<\/span> AB.<\/p><svg height=\"210\" width=\"200\" xmlns:xlink=\"http:\/\/www.w3.org\/1999\/xlink\"> <g><title><\/title><rect fill=\"#fff\" height=\"212\" id=\"canvas_background\" width=\"202\" 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_7\"> <image height=\"166\" id=\"svg_1\" width=\"152\" x=\"22\" 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y=\"23\"><\/image> <text fill=\"#000000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"15\" id=\"svg_2\" stroke=\"#000\" stroke-width=\"0\" text-anchor=\"start\" x=\"88\" xml:space=\"preserve\" y=\"26\">A<\/text> <text fill=\"#000000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"15\" id=\"svg_3\" stroke=\"#000\" stroke-width=\"0\" text-anchor=\"start\" x=\"19\" xml:space=\"preserve\" y=\"198\">B<\/text> <text fill=\"#000000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"15\" id=\"svg_4\" stroke=\"#000\" stroke-width=\"0\" text-anchor=\"start\" x=\"170\" xml:space=\"preserve\" y=\"194\">C<\/text> <text fill=\"#000000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"15\" id=\"svg_5\" stroke=\"#000\" stroke-width=\"0\" text-anchor=\"start\" x=\"148\" xml:space=\"preserve\" y=\"125\">E<\/text> <text fill=\"#000000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"15\" id=\"svg_6\" stroke=\"#000\" stroke-width=\"0\" text-anchor=\"start\" x=\"48\" xml:space=\"preserve\" y=\"125\">F<\/text> <\/g> <\/g> <\/svg><\/span><\/p>Trong các kh\u1eb3ng \u0111\u1ecbnh sau có bao nhiêu kh\u1eb3ng \u0111\u1ecbnh \u0111úng?<\/p>1. BE = CF          2. AE = AF           3.  $\\widehat{ACF}=\\widehat{ABE}$<\/span><\/span><\/p>","options":["0<\/b>","1<\/b>","2<\/b>","3<\/b>"],"correct":"4","level":"3","hint":"","answer":"Xét tam giác ABE (vuông t\u1ea1i E) và tam giác ACF (vuông t\u1ea1i F) có<\/p>AB = AC  (tam giác ABC cân t\u1ea1i A)<\/p>$\u200b\u200b\\widehat{A}$<\/span>   là góc chung<\/p>Suy ra $\\triangle ABE = \\triangle ACF$<\/span> (c\u1ea1nh huy\u1ec1n - góc nh\u1ecdn)<\/p>⇒ AE = AF (2 \u0111úng)<\/p> BE = CF (1 \u0111úng)<\/p> $\\widehat{ACF}=\\widehat{ABE}$<\/span>  (3 \u0111úng)<\/p>V\u1eady có 3 kh\u1eb3ng \u0111\u1ecbnh \u0111úng.<\/p>\u0110áp án \u0111úng là 3<\/span><\/strong><\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"131","test":"1","date":"2022-04-28 08:40:48","option_type":"txt","len":0},{"id":"7570","post_id":"5706","mon_id":"1158532","chapter_id":"1158548","question":"Cho hình v\u1ebd sau. Tính x trên hình?<\/p><svg height=\"160\" width=\"250\" xmlns:xlink=\"http:\/\/www.w3.org\/1999\/xlink\"> <g><title><\/title><rect fill=\"#fff\" height=\"162\" id=\"canvas_background\" width=\"252\" 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><image height=\"146\" id=\"svg_1\" width=\"242\" x=\"6\" 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y=\"4\"><\/image> <text fill=\"#000000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"14\" id=\"svg_2\" stroke=\"#000\" stroke-width=\"0\" text-anchor=\"start\" x=\"30\" xml:space=\"preserve\" y=\"125\">70<\/text> <text fill=\"#000000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"14\" id=\"svg_3\" stroke=\"#000\" stroke-width=\"0\" text-anchor=\"start\" x=\"45\" xml:space=\"preserve\" y=\"113\">o<\/text> <text fill=\"#000000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"14\" id=\"svg_4\" stroke=\"#000\" stroke-width=\"0\" text-anchor=\"start\" x=\"182\" xml:space=\"preserve\" y=\"123\">x<\/text> <\/g> <\/svg><\/span><\/p>","options":["35º","45º","55º","65º"],"correct":"1","level":"3","hint":"Tam giác cân có góc \u1edf \u0111\u1ec9nh có s\u1ed1 \u0111o b\u1eb1ng aº thì m\u1ed7i góc \u1edf \u0111áy có s\u1ed1 \u0111o b\u1eb1ng  (180º - aº ) : 2<\/p>","answer":"Xét tam giác ABC <\/span>có AB = AC (gi\u1ea3 thi\u1ebft) nên tam giác ABC cân t\u1ea1i A<\/p>⇒ $\\widehat{B}=\\widehat{ACB}$<\/span> = 70º<\/p>Ta có  $\\widehat{ACB }+\\widehat{ACD}$<\/span> = 180º  (hai góc k\u1ec1 bù)<\/p>⇒ $\\widehat{ACD}=180^0-\\widehat{ACB}$<\/span> = 180º - 70º = 110º<\/p>Xét tam giác ACD<\/span> có CA = CD (gi\u1ea3 thi\u1ebft) nên tam giác ACD cân t\u1ea1i C<\/p>⇒ $\\widehat{CAD}=\\widehat{CDA}$<\/span> = (180º - $\\widehat{ACD}$<\/span> ) : 2 = (180º - 110º) : 2 = 35º<\/p>\u0110áp án \u0111úng là 35º<\/span><\/strong><\/p>","type":"choose","extra_type":"flower","time":"0","user_id":"131","test":"1","date":"2022-04-28 08:50:30","option_type":"txt","len":0},{"id":"7571","post_id":"5706","mon_id":"1158532","chapter_id":"1158548","question":"Cho tam giác ABC có M là trung \u0111i\u1ec3m c\u1ee7a BC và AM $\\bot$<\/span> BC . K\u1ebft lu\u1eadn nào sau \u0111ây là \u0111úng?<\/p>","options":["A. <\/strong>Tam giác ABC cân t\u1ea1i A","B.<\/strong> Tam giác ABC cân t\u1ea1i B","C.<\/strong> Tam giác ABC cân t\u1ea1i C"],"correct":"1","level":"3","hint":"","answer":"Ta có hình v\u1ebd<\/p><svg height=\"170\" width=\"150\"> <g> <title><\/title> <rect fill=\"#fff\" height=\"172\" id=\"canvas_background\" width=\"152\" 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_11\"> <g id=\"svg_4\"> <path d=\"m20.675,140.86354l52.5,-123l52.5,123l-104.99999,0z\" fill=\"#fff\" id=\"svg_1\" stroke=\"#000\" stroke-width=\"1.5\"><\/path> <line fill=\"none\" fill-opacity=\"null\" id=\"svg_2\" stroke=\"#000\" stroke-linecap=\"null\" stroke-linejoin=\"null\" stroke-opacity=\"null\" stroke-width=\"1.5\" x1=\"73\" x2=\"73\" y1=\"19\" y2=\"142.14236\"><\/line> <rect fill=\"none\" fill-opacity=\"null\" height=\"9\" id=\"svg_3\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"1.5\" width=\"9\" x=\"73\" y=\"131.14236\"><\/rect> <\/g> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"15\" id=\"svg_5\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"62\" xml:space=\"preserve\" y=\"19.14236\">A<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"15\" id=\"svg_6\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"11\" xml:space=\"preserve\" y=\"150.14236\">B<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"15\" id=\"svg_7\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"126\" xml:space=\"preserve\" y=\"147.14236\">C<\/text> <text fill=\"#000000\" fill-opacity=\"null\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"15\" id=\"svg_8\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"67\" xml:space=\"preserve\" y=\"154.14236\">M<\/text> <line fill=\"none\" fill-opacity=\"null\" id=\"svg_9\" stroke=\"#000\" stroke-linecap=\"null\" stroke-linejoin=\"null\" stroke-opacity=\"null\" stroke-width=\"1.5\" x1=\"41\" x2=\"40\" y1=\"136.14236\" y2=\"145.14236\"><\/line> <line fill=\"none\" fill-opacity=\"null\" id=\"svg_10\" stroke=\"#000\" stroke-linecap=\"null\" stroke-linejoin=\"null\" stroke-opacity=\"null\" stroke-width=\"1.5\" x1=\"101\" x2=\"100\" y1=\"136.14236\" y2=\"145.14236\"><\/line> <\/g> <\/g> <\/svg><\/span><\/p>Xét tam giác AMB và tam giác AMC có<\/p>AM là c\u1ea1nh chung<\/p>$\\widehat{AMB}=\\widehat{AMC}$<\/span>   ( = 90º)<\/p>MB = MC  (M là trung \u0111i\u1ec3m c\u1ee7a BC)<\/p>Suy ra  $\\triangle AMB = \\triangle AMC$<\/span> (c.g.c)<\/p>⇒ AB = AC  (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng)<\/p>Do \u0111ó tam giác ABC cân t\u1ea1i A.<\/p>\u0110áp án \u0111úng là   A. <\/strong>Tam giác ABC cân t\u1ea1i A<\/span><\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"131","test":"1","date":"2022-04-28 09:05:03","option_type":"txt","len":2}]}