CHUYEN DE HINH HOC NANG CAO-HDG

["Website: tailieumontoan.com b) T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c CED . Ta c\u00f3 tam gi\u00e1c ABC v\u00e0 tam gi\u00e1c ADC l\u00e0 hai tam gi\u00e1c c\u00f3 chung \u0111\u01b0\u1eddng cao ch\u00ednh l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a h\u00ecnh thang ABCD n\u00ean t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch c\u1ee7a ch\u00fang b\u1eb1ng t\u1ec9 s\u1ed1 c\u1ee7a c\u1ea1nh AB v\u00e0 c\u1ea1nh CD . V\u1eady SAB=C A=B 1=5 3 . SADC CD 20 4 Hai tam gi\u00e1c ABC v\u00e0 tam gi\u00e1c ADC l\u1ea1i c\u00f3 chung c\u1ea1nh \u0111\u00e1y AC n\u00ean 3 l\u00e0 t\u1ec9 l\u1ec7 chi\u1ec1u cao c\u1ee7a 4 hai tam gi\u00e1c v\u00e0 l\u00e0 t\u1ec9 l\u1ec7 di\u1ec7n t\u00edch tam gi\u00e1c BEC v\u00e0 tam gi\u00e1c DEC . Di\u1ec7n t\u00edch tam gi\u00e1c BCD l\u00e0: 14\u00d7 20 : 2 =140 ( cm2 ). Di\u1ec7n t\u00edch tam gi\u00e1c DEC l\u00e0: 140 : (3 + 4)\u00d7 4 =80 ( cm2 ). B\u00e0i 117. c) Ch\u1ee9ng minh hai tam gi\u00e1c AED v\u00e0 BEC c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau. Ta c\u00f3: SACD = SBCD (hai tam gi\u00e1c c\u00f3 chung c\u1ea1nh \u0111\u00e1y CD v\u00e0 chung chi\u1ec1u cao). M\u00e0: S=ACD SDEC + SAED v\u00e0 S=BCD SDEC + SBEC . N\u00ean SAED = SBEC . Cho tam gi\u00e1c ABC , \u0111i\u1ec3m M n\u1eb1m tr\u00ean c\u1ea1nh BC sao cho BM = 2\u00d7 MC , \u0111i\u1ec3m N tr\u00ean c\u1ea1nh CA sao cho CN = 3\u00d7 NA . G\u1ecdi D l\u00e0 giao \u0111i\u1ec3m c\u1ee7a AM v\u00e0 BN . T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c ABC n\u1ebfu bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c AND b\u1eb1ng 10 cm2 . L\u1eddi gi\u1ea3i Ta c\u00f3: CN = 3\u00d7 NA hay CA= 4\u00d7 NA . S AND= 1 \u00d7 S ADC (tam gi\u00e1c AND v\u00e0 tam gi\u00e1c ADC c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb D v\u00e0 c\u00f3 4 CA= 4\u00d7 NA ) V\u1eady SADC =4 \u00d7 SAND =4 \u00d710 =40 ( cm2 ). Ta l\u1ea1i c\u00f3: SAMC= 1 \u00d7 S AMB (v\u00ec BM = 2\u00d7 MC v\u00e0 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A ). M\u00e0 hai tam 2 gi\u00e1c AMC v\u00e0 tam gi\u00e1c AMB c\u00f3 chung c\u1ea1nh AM n\u00ean \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B g\u1ea5p 2 l\u1ea7n chi\u1ec1u cao k\u1ebb t\u1eeb C xu\u1ed1ng AM . Hai \u0111\u01b0\u1eddng cao n\u00e0y c\u0169ng l\u00e0 hai \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c ADB v\u00e0 tam gi\u00e1c ADC . Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com S ADC= 1 \u00d7 S ADB \u21d2 S ADB =2 \u00d7 SADC = 40\u00d7 2 = 80 ( cm2 ). 2 S=ANB SAND + SADB =10 + 80 = 90 ( cm2 ). M\u00e0 SANB= 1 \u00d7 S ABC (hai tam gi\u00e1c n\u00e0y c\u00f3 CA= 4\u00d7 NA , chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B ). 4 V\u1eady SABC = 90\u00d7 4 = 360 ( cm2 ). B\u00e0i 118. Cho tam gi\u00e1c ABC . Tr\u00ean c\u1ea1nh AB l\u1ea5y \u0111i\u1ec3m E sao cho AE= 2 \u00d7 AB . Tr\u00ean c\u1ea1nh AC l\u1ea5y 3 \u0111i\u1ec3m D sao cho AD= 1 \u00d7 AC . 3 a) N\u1ed1i D v\u1edbi B . T\u00ednh t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch c\u1ee7a hai tam gi\u00e1c ABD v\u00e0 ABC . b) N\u1ed1i E v\u1edbi D . T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c ABC bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c AED l\u00e0 4 cm2 . c) N\u1ed1i C v\u1edbi E , CE c\u1eaft BD t\u1ea1i G . T\u00ednh t\u1ec9 s\u1ed1 \u0111\u1ed9 d\u00e0i hai \u0111o\u1ea1n th\u1eb3ng EG v\u00e0 CG . L\u1eddi gi\u1ea3i a) N\u1ed1i D v\u1edbi B . T\u00ednh t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch c\u1ee7a hai tam gi\u00e1c ABD v\u00e0 ABC . X\u00e9t tam gi\u00e1c ABD v\u00e0 tam gi\u00e1c ABC c\u00f3: AD= 1 \u00d7 AC v\u00e0 chung chi\u1ec1u cao h\u1ea1 t\u1eeb B xu\u1ed1ng 3 c\u1ea1nh AC n\u00ean SABD= 1 \u00d7 S ABC . 3 b) N\u1ed1i E v\u1edbi D . T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c ABC bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c AED l\u00e0 4 cm2 . X\u00e9t tam gi\u00e1c AED v\u00e0 tam gi\u00e1c AEC c\u00f3: AD= 1 \u00d7 AC v\u00e0 chung chi\u1ec1u cao h\u1ea1 t\u1eeb E xu\u1ed1ng 3 c\u1ea1nh AC n\u00ean SAED= 1 \u00d7 S AEC . V\u1eady di\u1ec7n t\u00edch tam gi\u00e1c AEC l\u00e0: 3 4 : 1 = 12 ( cm2 ). 3 X\u00e9t tam gi\u00e1c ABC v\u00e0 tam gi\u00e1c AEC c\u00f3: AE= 2 \u00d7 AB v\u00e0 chung chi\u1ec1u cao h\u1ea1 t\u1eeb C xu\u1ed1ng 3 AB n\u00ean SAEC= 2 \u00d7 S ABC . V\u1eady di\u1ec7n t\u00edch tam gi\u00e1c ABC l\u00e0: 3 12 : 1 = 36 ( cm2 ). 3 V\u1eady di\u1ec7n t\u00edch tam gi\u00e1c ABC l\u00e0 36 cm2 . c) N\u1ed1i C v\u1edbi E , CE c\u1eaft BD t\u1ea1i G . T\u00ednh t\u1ec9 s\u1ed1 \u0111\u1ed9 d\u00e0i hai \u0111o\u1ea1n th\u1eb3ng EG v\u00e0 CG . Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com Ta c\u00f3 AE= 2 \u00d7 AB hay EB= 1 \u00d7 AB n\u00ean SBDE= 1 \u00d7 SBDA . 3 3 3 V\u00e0 S ABD= 1 \u00d7 S ABC hay SABD= 1 \u00d7 SCBD . 3 2 V\u1eady SBDE =\uf8ec\uf8ed\uf8eb 1 \u00d7 1 \uf8f6 \u00d7 SBDC =1 \u00d7 S BDC . 2 3 \uf8f7\uf8f8 6 Hai tam gi\u00e1c BDE v\u00e0 BDC c\u00f3 chung c\u1ea1nh \u0111\u00e1y BD n\u00ean \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb E b\u1eb1ng 1 \u0111\u01b0\u1eddng 6 cao k\u1ebb t\u1eeb C xu\u1ed1ng BD . X\u00e9t hai tam gi\u00e1c DGE v\u00e0 tam gi\u00e1c DGC c\u00f3 c\u1ea1nh \u0111\u00e1y chung l\u00e0 DG v\u00e0 \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb E b\u1eb1ng 1 \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb C xu\u1ed1ng DG n\u00ean SDGE= 1 \u00d7 SDGC . M\u1eb7t kh\u00e1c hai tam gi\u00e1c n\u00e0y c\u00f3 6 6 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb D n\u00ean EG= 1 \u00d7 CG . 6 V\u1eady EG= 1 \u00d7 CG . 6 B\u00e0i 119. Cho tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch 64 cm2 . Tr\u00ean c\u1ea1nh AB l\u1ea5y \u0111i\u1ec3m M sao cho AM= 1 \u00d7 AB . 4 Tr\u00ean c\u1ea1nh AC l\u1ea5y \u0111i\u1ec3m N sao cho AN= 1 \u00d7 AC . N\u1ed1i B v\u1edbi N . 4 a) T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c BNC . b) T\u00ednh t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch hai tam gi\u00e1c AMN v\u00e0 tam gi\u00e1c ABC . c) Qua A v\u1ebd m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng c\u1eaft MN \u1edf K v\u00e0 c\u1eaft BC \u1edf E . T\u00ednh t\u1ec9 s\u1ed1 KE . AK L\u1eddi gi\u1ea3i a) T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c BNC . V\u00ec AN= 1 \u00d7 AC n\u00ean CN= 3 \u00d7 AC . 44 V\u1eady SBNC= 3 \u00d7 S ABC (v\u00ec c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb B xu\u1ed1ng AC v\u00e0 \u0111\u00e1y CN= 3 \u00d7 AC ). 4 4 V\u1eady SBNC =3 \u00d7 64 =48 ( cm2 ). 4 b) T\u00ednh t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch hai tam gi\u00e1c AMN v\u00e0 tam gi\u00e1c ABC . Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com S AMN= 1 \u00d7 S ABN (v\u00ec chung chi\u1ec1u cao h\u1ea1 t\u1eeb N xu\u1ed1ng AB v\u00e0 \u0111\u00e1y AM= 1 \u00d7 AB ). 4 4 S ABN= 1 \u00d7 S ABC (v\u00ec chung chi\u1ec1u cao h\u1ea1 t\u1eeb B xu\u1ed1ng AC v\u00e0 \u0111\u00e1y AN= 1 \u00d7 AC ). 4 4 V\u1eady S AMN = 1 \u00d7 1 \u00d7 S AB=C 1 \u00d7 S ABC hay KE =3 SAMN = 1 . 4 4 16 KA SABC 16 c) Qua A v\u1ebd m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng c\u1eaft MN \u1edf K v\u00e0 c\u1eaft BC \u1edf E . T\u00ednh t\u1ec9 s\u1ed1 KE . AK N\u1ed1i EM , EN . S AEM= 1 \u00d7 S ABE (v\u00ec chung chi\u1ec1u cao h\u1ea1 t\u1eeb E xu\u1ed1ng AB v\u00e0 \u0111\u00e1y AM= 1 \u00d7 AB ). 4 4 S ANE= 1 \u00d7 S ACE (v\u00ec chung chi\u1ec1u cao h\u1ea1 t\u1eeb E xu\u1ed1ng AC v\u00e0 \u0111\u00e1y AN= 1 \u00d7 AC ). 4 4 V\u1eady S =AMEN 1 \u00d7 S ABC . 4 Ta c\u00f3 SAMN = 1 . SABC 16 V\u1eady SMEN =S AMEN \u2212 SAMN =3 \u00d7 S ABC . 16 Suy ra SMEN = 3\u00d7 SAMN . Hai tam gi\u00e1c MEN v\u00e0 AMN l\u1ea1i chung \u0111\u00e1y MN n\u00ean chi\u1ec1u cao s\u1ebd g\u1ea5p 3 l\u1ea7n AH . V\u1eady SEMK = 3\u00d7 SAMK (chung \u0111\u00e1y MK ). Tam gi\u00e1c EMK v\u00e0 tam gi\u00e1c AMK l\u1ea1i c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb M xu\u1ed1ng AE n\u00ean KE= 3\u00d7 AK hay KE = 3. KA Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038"]