Sự đồng quy của ba đường trung trực, ba đường cao trong một tam giác

Home Lớp 7 Toán lớp 7 - Sách kết nối tri thức Bài 35: Sự đồng quy của ba đường trung trực, ba đường cao trong một tam giácBài tập nâng cao

{"common":{"save":0,"post_id":"5844","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"8721","post_id":"5844","mon_id":"1158532","chapter_id":"1158662","question":"G\u1ecdi H là tr\u1ef1c tâm c\u1ee7a tam giác ABC không vuông. Tr\u1ef1c tâm c\u1ee7a tam giác HAB là <\/p>","options":["A. <\/strong>\u0111i\u1ec3m C","B.<\/strong> \u0111i\u1ec3m A","C. <\/strong>\u0111i\u1ec3m B","D. <\/strong>\u0111i\u1ec3m H"],"correct":"1","level":"3","hint":"","answer":"Ta có hình v\u1ebd:<\/p><svg height=\"150\" width=\"200\" xmlns:xlink=\"http:\/\/www.w3.org\/1999\/xlink\"> <g> <title><\/title> <rect fill=\"#fff\" height=\"152\" 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> <image height=\"130\" id=\"svg_1\" width=\"184\" x=\"6\" 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y=\"9\"><\/image> <ellipse cx=\"114.82639\" cy=\"44\" fill=\"#fff\" id=\"svg_2\" rx=\"9\" ry=\"9\" stroke-width=\"1.5\"><\/ellipse> <ellipse cx=\"76.82639\" cy=\"133\" fill=\"#fff\" id=\"svg_3\" rx=\"9\" ry=\"9\" stroke-width=\"1.5\"><\/ellipse> <rect fill=\"#fff\" fill-opacity=\"null\" height=\"10\" id=\"svg_4\" stroke=\"#000000\" stroke-width=\"1.5\" transform=\"rotate(24.443946838378906 55.82638549804694,71.99999999999999) \" width=\"10\" x=\"50.82639\" y=\"67\"><\/rect> <\/g> <\/svg><\/span><\/p>Vì H là tr\u1ef1c tâm tam giác ABC nên<\/p>  CH $\\bot$<\/span> AB ⇒ CH là \u0111\u01b0\u1eddng cao \u1ee9ng v\u1edbi c\u1ea1nh AB c\u1ee7a tam giác HAB<\/p>  CB $\\bot$<\/span> HA ⇒ CB là \u0111\u01b0\u1eddng cao \u1ee9ng v\u1edbi c\u1ea1nh HA c\u1ee7a tam giác HAB<\/p>  CA $\\bot$<\/span> HB ⇒ CA là \u0111\u01b0\u1eddng cao \u1ee9ng v\u1edbi c\u1ea1nh HB c\u1ee7a tam giác HAB<\/p>Mà CH, CB, CA \u0111\u1ed3ng quy t\u1ea1i C ⇒   C là giao \u0111i\u1ec3m c\u1ee7a \u0111\u01b0\u1eddng cao c\u1ee7a tam giác HAB<\/p>Do \u0111ó tr\u1ef1c tâm c\u1ee7a tam giác HAB là \u0111i\u1ec3m C.<\/p>\u0110áp án \u0111úng là    A. <\/strong>\u0111i\u1ec3m C<\/span><\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"131","test":"3","date":"2022-06-04 09:47:14","option_type":"txt","len":0},{"id":"8724","post_id":"5844","mon_id":"1158532","chapter_id":"1158662","question":"Trên m\u1ed9t b\u1ea3n \u0111\u1ed3, ba khu dân c\u01b0 \u0111\u01b0\u1ee3c quy ho\u1ea1ch t\u1ea1i ba \u0111\u1ecba \u0111i\u1ec3m M, N và P không th\u1eb3ng hàng. Xác \u0111\u1ecbnh \u0111i\u1ec3m O trên b\u1ea3n \u0111\u1ed3 sao cho O cách \u0111\u1ec1u ba \u0111i\u1ec3m M, N, P \u0111\u1ec3 quy ho\u1ea1ch m\u1ed9t b\u1ec7nh vi\u1ec7n (hình v\u1ebd).<\/p><svg height=\"200\" width=\"260\"> <g stroke=\"null\"><title><\/title><rect fill=\"#fff\" height=\"202\" id=\"canvas_background\" stroke=\"null\" width=\"262\" x=\"-1\" y=\"-1\"><\/rect> <g display=\"none\" id=\"canvasGrid\" stroke=\"null\"> <rect fill=\"url(#gridpattern)\" height=\"100%\" id=\"svg_15\" stroke-width=\"0\" width=\"100%\" x=\"0\" y=\"0\"><\/rect> <\/g> <\/g> <g stroke=\"null\"><title><\/title><line fill=\"none\" id=\"svg_1\" stroke=\"#000\" stroke-linecap=\"undefined\" stroke-linejoin=\"undefined\" stroke-width=\"1.5\" x1=\"18.82639\" x2=\"205.82639\" y1=\"102.02083\" y2=\"29.02083\"><\/line> <line fill=\"none\" fill-opacity=\"null\" id=\"svg_2\" stroke=\"#000\" stroke-linecap=\"undefined\" stroke-linejoin=\"undefined\" stroke-opacity=\"null\" stroke-width=\"1.5\" x1=\"17.82639\" x2=\"228.82639\" y1=\"102.02083\" y2=\"172.02083\"><\/line> <ellipse cx=\"21.32639\" cy=\"102.02083\" fill=\"#ff5656\" id=\"svg_3\" rx=\"3\" ry=\"3\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"1.5\"><\/ellipse> <ellipse cx=\"140.32639\" cy=\"97.02083\" fill=\"#ff5656\" id=\"svg_6\" rx=\"3\" ry=\"3\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"1.5\"><\/ellipse> <line fill=\"none\" fill-opacity=\"null\" id=\"svg_7\" stroke=\"#000\" stroke-linecap=\"undefined\" stroke-linejoin=\"undefined\" stroke-opacity=\"null\" stroke-width=\"1.5\" x1=\"203.82639\" x2=\"230.82639\" y1=\"28.02083\" y2=\"172.02083\"><\/line> <ellipse cx=\"204.32639\" cy=\"30.02083\" fill=\"#ff5656\" id=\"svg_4\" rx=\"3\" ry=\"3\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"1.5\"><\/ellipse> <ellipse cx=\"232.32635\" cy=\"172.52089\" fill=\"#ff5656\" id=\"svg_5\" rx=\"3\" ry=\"3\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"1.5\"><\/ellipse> <text fill=\"#000000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"15\" id=\"svg_8\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"10.82639\" xml:space=\"preserve\" y=\"119.02083\">M<\/text> <text fill=\"#000000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"15\" id=\"svg_9\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"210.82639\" xml:space=\"preserve\" y=\"28.02083\">N<\/text> <text fill=\"#000000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"15\" id=\"svg_10\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"236.82639\" xml:space=\"preserve\" y=\"180.02083\">P<\/text> <text fill=\"#000000\" font-family=\"Helvetica, Arial, sans-serif\" font-size=\"15\" id=\"svg_11\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"0\" text-anchor=\"start\" x=\"130.82639\" xml:space=\"preserve\" y=\"87.02083\">O<\/text> <path d=\"m27.812,88.66539l-2.72189,-2.19758l0,-5.24849l-2.33343,0l0,3.36455l-5.44467,-4.39587l-10.5,8.47739l2.49966,-0.04809l0,8.57069l15.88956,0l0,-8.52261l2.61078,0l-0.00001,0.00001zm-10.5,7.29679l-5.18511,0l0,-5.02385l5.18511,0l0,5.02385zm5.03755,-2.15308l-3.40725,0l0,-2.90092l3.40725,0l0,2.90092z\" fill=\"#56ff56\" id=\"svg_12\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"1.5\"><\/path> <path d=\"m206.86325,14.99241l-2.33305,-2.84392l0,-6.79216l-2.00008,0l0,4.35412l-4.66686,-5.68877l-8.99999,10.97074l2.14257,-0.06223l0,11.09148l13.61962,0l0,-11.02926l2.23781,0l-0.00002,0zm-8.99999,9.4429l-4.44438,0l0,-6.50146l4.44438,0l0,6.50146zm4.31789,-2.78634l-2.9205,0l0,-3.75413l2.9205,0l0,3.75413z\" fill=\"#ffff00\" id=\"svg_13\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"1.5\"><\/path> <path d=\"m254.812,156.1614l-3.062,-2.58538l0,-6.1747l-2.625,0l0,3.95829l-6.125,-5.17161l-11.812,9.9734l2.812,-0.05657l0,10.08317l17.875,0l0,-10.0266l2.937,0zm-11.812,8.58446l-5.833,0l0,-5.91042l5.833,0l0,5.91042zm5.667,-2.53304l-3.833,0l0,-3.41284l3.833,0l0,3.41284z\" fill=\"#005fbf\" id=\"svg_14\" stroke=\"#000\" stroke-opacity=\"null\" stroke-width=\"1.5\"><\/path> <\/g> <\/svg><\/span><\/p>","options":["A.<\/strong> O là giao \u0111i\u1ec3m c\u1ee7a ba \u0111\u01b0\u1eddng cao c\u1ee7a tam giác MNP","B. <\/strong>O là giao \u0111i\u1ec3m c\u1ee7a ba \u0111\u01b0\u1eddng trung tuy\u1ebfn c\u1ee7a tam giác MNP","C.<\/strong> O là giao \u0111i\u1ec3m c\u1ee7a ba \u0111\u01b0\u1eddng phân giác c\u1ee7a tam giác MNP","D. <\/strong>O là giao \u0111i\u1ec3m c\u1ee7a ba \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a tam giác MNP"],"correct":"4","level":"3","hint":"","answer":"Vì ba \u0111i\u1ec3m M, N, P không th\u1eb3ng hàng nên M, N, P l\u1eadp thành tam giác MNP.<\/p>Vì \u0111i\u1ec3m O cách \u0111\u1ec1u ba \u0111i\u1ec3m M, N, P nên O là giao \u0111i\u1ec3m c\u1ee7a ba \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a tam giác MNP.<\/p>\u0110áp án \u0111úng là   D. <\/strong>O là giao \u0111i\u1ec3m c\u1ee7a ba \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a tam giác MNP<\/span><\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"131","test":"1","date":"2022-06-04 10:08:12","option_type":"txt","len":3},{"id":"8730","post_id":"5844","mon_id":"1158532","chapter_id":"1158662","question":"Cho tam giác ABC có góc A b\u1eb1ng 110°. G\u1ecdi H là tr\u1ef1c tâm tam giác ABC. Tính s\u1ed1 \u0111o góc BHC.<\/p>","options":["A.<\/strong> 80°","B.<\/strong> 90°","C.<\/strong> 70°","D.<\/strong> 60°"],"correct":"3","level":"3","hint":"","answer":"G\u1ecdi D, E l\u1ea7n l\u01b0\u1ee3t là chân \u0111\u01b0\u1eddng cao xu\u1ea5t phát t\u1eeb \u0111\u1ec9nh B và \u0111\u1ec9nh C c\u1ee7a tam giác ABC (hình v\u1ebd).<\/p><svg height=\"180\" width=\"260\" xmlns:xlink=\"http:\/\/www.w3.org\/1999\/xlink\"> <g> <title><\/title> <rect fill=\"#fff\" height=\"182\" id=\"canvas_background\" width=\"262\" 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=\"164\" id=\"svg_1\" width=\"258\" x=\"6\" 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y=\"8\"><\/image> <\/g> <\/svg><\/span><\/p>Ta có  $\\widehat{BAC}=\\widehat{DBA}+\\widehat{ADB}$<\/span>  (góc ngoài t\u1ea1i \u0111\u1ec9nh A c\u1ee7a tam giác ABD)<\/p>⇒ $\\widehat{DBA } = \\widehat{BAC} - \\widehat{ADB}$<\/span>  = 110° - 90° = 20°<\/p>Xét tam giác  BHE vuông t\u1ea1i E có  $\\widehat{BHE }+ \\widehat{ABD}=90^0$<\/span><\/p>⇒ $\\widehat{BHE}=90^0 - \\widehat{ABD}$<\/span> = 90° - 20° = 70° ⇒ $\\widehat{BHC}$<\/span> = 70°.<\/p>V\u1eady góc BHC có s\u1ed1 \u0111o b\u1eb1ng  70°.<\/p>\u0110áp án \u0111úng là  C.<\/strong> 70°<\/span><\/p>","type":"choose","extra_type":"classic","time":"0","user_id":"131","test":"1","date":"2022-06-04 10:48:16","option_type":"txt","len":0}]}