CHUYEN DE HINH HOC NANG CAO-HDG

["Website: tailieumontoan.com ( )SDMN = SABCD \u2212 SADM \u2212 SBMN \u2212 SCND = 216 \u2212 36 \u2212 36 \u2212 54 = 90 cm2 B\u00e0i 54. Cho tam gi\u00e1c ABC c\u00f3 M n\u1eb1m tr\u00ean BC v\u00e0 MC = 1 BC , BK l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c ABC , 4 MH l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c AMC . T\u00ednh t\u1ec9 s\u1ed1 MH . BK L\u1eddi gi\u1ea3i K\u1ebb \u0111\u01b0\u1eddng cao AI c\u1ee7a tam gi\u00e1c ABC . A Ta c\u00f3: K Hai tam gi\u00e1c ABC v\u00e0 tam gi\u00e1c AMC c\u00f3 c\u00f9ng chi\u1ec1u cao h\u1ea1 t\u1eeb A v\u00e0 MC = 1 BC n\u00ean H M 4 SMAC = 1 S ABC B 4 I S ABC = BK \u00d7 AC 2 SMAC = MH \u00d7 AC 2 SMAC = MH SBAC BK Do \u0111\u00f3 MH = 1 . BK 4 B\u00e0i 55. Cho tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A , c\u1ea1nh AB d\u00e0i 30 c=m , AC 4=0 cm, BC 50 cm . Ta c\u1eaft m\u1ed9t \u0111o\u1ea1n th\u1eb3ng song song v\u1edbi c\u1ea1nh BC v\u00e0 c\u00e1ch c\u1ea1nh BC 3cm . \u0110o\u1ea1n th\u1eb3ng \u0111\u00f3 c\u1eaft AB t\u1ea1i M , c\u1eaft AC t\u1ea1i N . T\u00ednh di\u1ec7n t\u00edch h\u00ecnh MNCB . L\u1eddi gi\u1ea3i N\u1ed1i C v\u1edbi M ta c\u00f3 : B M 3=\u00d7 50 2 A ( )=SBMC 75 cm2 =BM 2\u00d7=SBMC 2=\u00d7 75 3, 75(cm) AC 40 AM =AB \u2212 BM =30 \u2212 3,75 =26, 25(cm) C N N\u1ed1i B v\u1edbi N ta c\u00f3 : 3=\u00d7 50 2 ( )=SBNC 75 cm2 Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com =CN 2\u00d7=SBNC 2=\u00d7 75 5(cm) AB 30 AN = AC \u2212 NC = 40 \u2212 5 = 35(cm) AB=\u00d7 AC 3=0\u00d7 40 2 2 ( )=SABC 600 cm2 A=M \u00d7 AN 2 ( )=SAMN 459,375 cm2 SB=MNC SABC \u2212 SAMN ( )=600 \u2212 459,375 =140, 625 cm2 B\u00e0i 56. Cho h\u00ecnh tam gi\u00e1c ABC c\u00f3 \u0111i\u1ec3m N l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ea1nh AC . Tr\u00ean \u0111\u00f3 c\u00f3 h\u00ecnh thang BMNE nh\u01b0 h\u00ecnh v\u1ebd b\u00ean . N\u1ed1i B v\u1edbi N , n\u1ed1i E v\u1edbi M , hai \u0111o\u1ea1n th\u1eb3ng n\u00e0y g\u1eb7p nhau \u1edf \u0111i\u1ec3m O. a.So s\u00e1nh di\u1ec7n t\u00edch hai h\u00ecnh tam gi\u00e1c OBM v\u00e0 OEN . b. So s\u00e1nh di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c EMC v\u1edbi di\u1ec7n t\u00edch h\u00ecnh AEMB . L\u1eddi gi\u1ea3i a) Do BMNE l\u00e0 h\u00ecnh thang n\u00ean SMBE = SNBE ( c\u00f3 A chung \u0111\u00e1y BE v\u00e0 c\u00f3 \u0111\u01b0\u1eddng cao l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a E h\u00ecnh thang ) O Hai tam gi\u00e1c n\u00e0y c\u00f3 ph\u1ea7n chung l\u00e0 tam gi\u00e1c OBE M N n\u00ean SOBM = SONE b) Hai tam gi\u00e1c ABN v\u00e0 tam gi\u00e1c CBN c\u00f3 c\u00f9ngchi\u1ec1u cao h\u1ea1 t\u1eeb B v\u00e0 \u0111\u00e1y NA = NC n\u00ean 1 B C 2 S=ABN S=CBN S ABC Ta c\u00f3 : SEMC = SCBN \u2212 SOBM + SOEN M\u00e0 SOEN = SOBM (theo c\u00e2u a) n\u00ean : SEMC = SCBN L\u1ea1i c\u00f3 : SCBN = 1 do \u0111\u00f3 SEMC = 1 S ABC 2 SABC 2 N\u00ean SEMC = SABME . Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com CHUY\u00caN \u0110\u1ec0. TO\u00c1N H\u00ccNH H\u1eccC B\u00c0I TO\u00c1N V\u1ec0 CHU VI - DI\u1ec6N T\u00cdCH B\u00e0i 57. Cho h\u00ecnh tam gi\u00e1c ABC. G\u1ecdi D l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa \u0111o\u1ea1n BC . L\u1ea5y \u0111i\u1ec3m E tr\u00ean c\u1ea1nh AC sao cho AE = 1 AC . N\u1ed1i DE k\u00e9o d\u00e0i c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng AB t\u1ea1i M ; N\u1ed1i M v\u1edbi C . Bi\u1ebft di\u1ec7n t\u00edch 5 tam gi\u00e1c AME b\u1eb1ng 20cm2. H\u00e3y t\u00ednh: a. Di\u1ec7n t\u00edch tam gi\u00e1c MEC . b. Di\u1ec7n t\u00edch tam gi\u00e1c ABC . L\u1eddi gi\u1ea3i M A E BC D a) Ta c\u00f3: SMEC= 4\u00d7 SMAE (v\u00ec chung chi\u1ec1u cao h\u1ea1 t\u1eeb M xu\u1ed1ng AC , EC= 4\u00d7 AE ). N\u00ean SMEC =4\u00d7 20 =80 (cm2) b) N\u1ed1i B v\u1edbi E Ta c\u00f3: SMBD = SMCD (v\u00ec chung chi\u1ec1u cao h\u1ea1 t\u1eeb M xu\u1ed1ng BC , BD \uf03d DC ) SEBD \uf03d SECD (v\u00ec chung chi\u1ec1u cao h\u1ea1 t\u1eeb E xu\u1ed1ng BC , c\u1ea1nh \u0111\u00e1y BD \uf03d DC ) M\u00e0 SMBD \uf03d SMBE \uf02b SEBD ; SMCD \uf03d SMEC \uf02b SECD . V\u1eady SMBE \uf03d SMEC \uf03d 80 (cm2). ( )\u21d2 S=ABE 80 \u2013 =20 60 cm2 M\u00e0 S ABE = \u00a01 S ABC (V\u00ec AE = 1 AC v\u00e0 chung chi\u1ec1u cao h\u1ea1 t\u1eeb B) 5 5 V\u1eady SABC = 60 :\u00a01 \u00a0= 300 cm2 5 \u0110\u00e1p s\u1ed1: 300 cm2. Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com B\u00e0i 58. Cho h\u00ecnh thang ABCD c\u00f3 hai \u0111\u00e1y l\u00e0 AB v\u00e0 CD . Hai \u0111\u01b0\u1eddng ch\u00e9o AC v\u00e0 BD c\u1eaft nhau t\u1ea1i \u0111i\u1ec3m E . H\u00e3y n\u00eau t\u00ean c\u00e1c c\u1eb7p h\u00ecnh tam gi\u00e1c c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau. L\u1eddi gi\u1ea3i SACD = SBCD (v\u00ec chi\u1ec1u cao c\u00f9ng l\u00e0 chi\u1ec1u cao c\u1ee7a h\u00ecnh thang ABCD , chung c\u1ea1nh \u0111\u00e1y DC ) SABC = SBAD (v\u00ec chi\u1ec1u cao c\u00f9ng l\u00e0 chi\u1ec1u cao c\u1ee7a h\u00ecnh thang ABCD , chung c\u1ea1nh \u0111\u00e1y AB ) V\u00ec SABC = SBAD m\u00e0 hai tam gi\u00e1c n\u00e0y l\u1ea1i c\u00f3 ph\u1ea7n chung l\u00e0 SABE n\u00ean SADE = SBCE V\u1eady c\u00f3 3 c\u1eb7p tam gi\u00e1c b\u1eb1ng nhau : SABC = SBAD ; SACD = SBCD v\u00e0 SADE = SBCE B\u00e0i 59. Cho h\u00ecnh ch\u1eef nh\u1eadt ABCD . Tr\u00ean c\u1ea1nh AB l\u1ea5y \u0111i\u1ec3m M v\u00e0 N sao cho A=M M=N NB . T\u00ednh S AMCD S NBCD L\u1eddi gi\u1ea3i Ta c\u00f3: SAMCD = ( AM + CD) \u00d7 AD : 2 v\u00e0 SNBCD = (NB + CD) \u00d7 BC : 2 . M\u00e0 AM = NB (\u0111\u1ec1 b\u00e0i cho); AD = BC (c\u1eb7p c\u1ea1nh \u0111\u1ed1i di\u1ec7n song song, b\u1eb1ng nhau) V\u1eady S AMCD = SNBCD hay S AMCD =1. S NBCD \u0110\u00e1p s\u1ed1: 1 B\u00e0i 60. ( Em m\u1ea1nh d\u1ea1n thay \u0111\u1ed5i c\u00e2u h\u1ecfi c\u1ee7a \u0111\u1ec1 b\u00e0i, c\u00e1c th\u1ea7y c\u00f4 xem gi\u00fap em \u0111\u01b0\u1ee3c kh\u00f4ng \u1ea1?) Cho h\u00ecnh thang ABCD c\u00f3 hai \u0111\u00e1y l\u00e0 AB v\u00e0 CD . Tr\u00ean AB l\u1ea5y \u0111i\u1ec3m M b\u1ea5t k\u00ec. N\u1ed1i M v\u1edbi D v\u00e0 C . Tr\u00ean DC l\u1ea5y \u0111i\u1ec3m N b\u1ea5t k\u00ec. N\u1ed1i A v\u1edbi N c\u1eaft MD t\u1ea1i E. N\u1ed1i B v\u1edbi N c\u1eaft MC \u1edf G . Bi\u1ebft di\u1ec7n t\u00edch c\u00e1c h\u00ecnh tam gi\u00e1c AED v\u00e0 BGC l\u1ea7n l\u01b0\u1ee3t l\u00e0 1, 2cm2 v\u00e0 3, 4cm2 . T\u00ednh di\u1ec7n t\u00edch h\u00ecnh t\u1ee9 gi\u00e1c MGNE . L\u1eddi gi\u1ea3i Ta c\u00f3: SADM = SANM (v\u00ec chi\u1ec1u cao c\u00f9ng l\u00e0 chi\u1ec1u cao c\u1ee7a h\u00ecnh thang ABCD , chung c\u1ea1nh \u0111\u00e1y AM ) M\u00e0 S=ADM S ADE + S AEM v\u00e0 S=ANM SMEN + S AEM N\u00ean S=ADE S=MEN 1, 2 (cm2). Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com T\u01b0\u01a1ng t\u1ef1: SMNB = SMCB n\u00ean S=MNG S=BGC 3, 4 (cm2). V\u1eady SMGNE = SMEN + SMNG =1, 2 + 3, 4 = 4, 6 (cm2). \u0110\u00e1p s\u1ed1: 4,6cm2. B\u00e0i 61. Cho h\u00ecnh thang ABCD c\u00f3 \u0111\u00e1y AB = 9cm v\u00e0 \u0111\u00e1y DC = 15cm . N\u1ebfu k\u00e9o d\u00e0i m\u1ed9t \u0111\u00e1y th\u00eam 3cm th\u00ec \u0111\u01b0\u1ee3c h\u00ecnh thang m\u1edbi c\u00f3 di\u1ec7n t\u00edch l\u1edbn h\u01a1n di\u1ec7n t\u00edch h\u00ecnh thang ABCD l\u00e0 7,5cm \u00b2 . T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang ABCD . L\u1eddi gi\u1ea3i: Gi\u1ea3 s\u1eed k\u00e9o d\u00e0i \u0111\u00e1y DC v\u1ec1 ph\u00eda C \u0111o\u1ea1n CE \uf03d 3 cm. Chi\u1ec1u cao c\u1ee7a tam gi\u00e1c BCE l\u00e0: 7,5\u00d7 2 : 3 =5 (cm). B\u00e0i 62. Chi\u1ec1u cao h\u00ecnh thang ABCD : 5cm. Di\u1ec7n t\u00edch h\u00ecnh thang ABCD l\u00e0: (15 + 9)\u00d7 5 : 2 =60(cm2 ) . \u0110\u00e1p s\u1ed1: 60cm2. Cho h\u00ecnh thang ABCD c\u00f3 \u0111\u00e1y AB = 18cm v\u00e0 \u0111\u00e1y DC = 24cm . N\u1ebfu k\u00e9o d\u00e0i \u0111\u00e1y b\u00e9 th\u00eam BE = 9cm v\u00e0 \u0111\u00e1y l\u1edbn th\u00eam CG = 12cm (v\u1ec1 c\u00f9ng m\u1ed9t ph\u00eda) th\u00ec \u0111\u01b0\u1ee3c h\u00ecnh thang BEGC c\u00f3 di\u1ec7n t\u00edch l\u00e0 157,5cm2 . T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang ABCD . L\u1eddi gi\u1ea3i: Gi\u1ea3 s\u1eed k\u00e9o d\u00e0i c\u1ea1nh \u0111\u00e1y AB v\u1ec1 ph\u00eda B m\u1ed9t \u0111o\u1ea1n BE = 9cm ; k\u00e9o d\u00e0i c\u1ea1nh \u0111\u00e1y DC v\u1ec1 ph\u00eda C m\u1ed9t \u0111o\u1ea1n CG = 12cm . Chi\u1ec1u cao c\u1ee7a h\u00ecnh thang BEGC l\u00e0: 157,5\uf0b42 :\uf0289 \uf02b12\uf029 \uf03d15 (cm). Chi\u1ec1u cao c\u1ee7a h\u00ecnh thang ABCD b\u1eb1ng15cm . Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com Di\u1ec7n t\u00edch h\u00ecnh thang ABCD l\u00e0: \uf02818 \uf02b 24\uf029\uf0b415 : 2 \uf03d 315 (cm2). \u0110\u00e1p s\u1ed1: 315cm2. B\u00e0i 63. Cho h\u00ecnh thang ABCD c\u00f3 \u0111\u00e1y AB = 18cm v\u00e0 \u0111\u00e1y DC = 36cm . Hai \u0111\u01b0\u1eddng ch\u00e9o AC v\u00e0 BD c\u1eaft nhau \u1edf E . Bi\u1ebft di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c EAB l\u00e0 5cm2 . T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang ABCD . L\u1eddi gi\u1ea3i: Ta c\u00f3: AB \uf03d18cm, DC \uf03d 36cm n\u00ean AB \uf03d\u00a0\u00a01 DC . 2 S ABD =\u00a01 SBDC (v\u00ec c\u00f3 c\u00f9ng chi\u1ec1u cao b\u1eb1ng v\u1edbi chi\u1ec1u cao h\u00ecnh thang v\u00e0 \u0111\u00e1y AB \uf03d\u00a0\u00a01 DC ) 2 2 \uf0de AH \uf03d\u00a01 CK (hai tam gi\u00e1c ABD v\u00e0 BDC c\u00f3 c\u00f9ng \u0111\u00e1y BD , S ABD \uf03d\u00a01 SBDC ) 2 2 SAEB \uf03d\u00a0SCEB (hai tam gi\u00e1c c\u00f3 chung \u0111\u00e1y EB , chi\u1ec1u cao AH =\u00a01 CK ; S ABD =\u00a01 SBDC 2 2 SABC \uf03d SAEB \uf02b SCEB \uf03d 5 \uf02b10 \uf03d 15 (cm2) (1). S ADC \uf03d 2SABC (v\u00ec c\u00f3 chung chi\u1ec1u cao v\u00e0 \u0111\u00e1y AB \uf03d\u00a0\u00a01 DC ) 2 SADC \uf03d 15\uf0b42 \uf03d 30\u00a0(cm2) (2). SABCD \uf03d SABC \uf02b SADC \uf03d 15 \uf02b 30 \uf03d 45 (cm2). V\u1eady di\u1ec7n t\u00edch h\u00ecnh thang ABCD l\u00e0 45cm2. B\u00e0i 64. Cho h\u00ecnh thang ABCD c\u00f3 \u0111\u00e1y AB = 15cm v\u00e0 \u0111\u00e1y DC = 45cm . Hai \u0111\u01b0\u1eddng ch\u00e9o c\u1eaft nhau \u1edf E . Bi\u1ebft di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c EBC l\u00e0 30cm2 \u00a0. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang ABCD . L\u1eddi gi\u1ea3i Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com Theo \u0111\u1ec1 ra, ta c\u00f3: AB \uf03d 15 \uf03d 1 . DC 45 3 S ABC = \u00a01 SBCD (v\u00ec hai tam gi\u00e1c c\u00f3 chung \u0111\u01b0\u1eddng cao l\u00e0 \u0111\u01b0\u1eddng cao h\u00ecnh thang; \u0111\u00e1y AB =1) 3 DC 3 AH \uf03d\u00a01 CK (v\u00ec hai tam gi\u00e1c ABC v\u00e0 BDC c\u00f3 chung \u0111\u00e1y BC ; S ABC = \u00a01 SBCD ) 3 3 S ABE \uf03d\u00a01 SCBE (hai tam gi\u00e1c c\u00f3 chung \u0111\u00e1y BE v\u00e0 \u0111\u01b0\u1eddng cao AH \uf03d\u00a01 CK ). 3 3 SABE \uf03d 30 \uf0b8 3\u00a0\uf03d 10 (cm2). SABC = SABE + SBCE = 10 + 30 = 40 (cm\u00b2) M\u00e0 S ABC = 1 (v\u00ec AB = 1 DC v\u00e0 chi\u1ec1u cao t\u01b0\u01a1ng \u1ee9ng \u0111\u1ec1u l\u00e0 chi\u1ec1u cao h\u00ecnh thang) 3 SACD 3 \u21d2 SADC \u00a0= 40 :\u00a01 = 120 cm2 3 V\u1eady SABCD \u00a0= 40 + 120 =160 cm2 B\u00e0i 65. Cho h\u00ecnh thang ABCD c\u00f3 \u0111\u00e1y b\u00e9 AB v\u00e0 \u0111\u00e1y l\u1edbn DC . Hai \u0111\u01b0\u1eddng ch\u00e9o c\u1eaft nhau \u1edf E . Bi\u1ebft di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c EAB l\u00e0 2,5cm2 v\u00e0 di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c EAD l\u00e0 7,5cm2 . T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang ABCD . L\u1eddi gi\u1ea3i AB 2,5cm2 7,5cm2 E DC Hai tam gi\u00e1c ADC v\u00e0 BDC c\u00f3 chung c\u1ea1nh \u0111\u00e1y DC v\u00e0 chi\u1ec1u cao h\u1ea1 t\u1eeb A xu\u1ed1ng DC b\u1eb1ng chi\u1ec1u cao h\u1ea1 t\u1eeb B xu\u1ed1ng DC n\u00ean: SADC = SBDC . M\u00e0 hai tam gi\u00e1c ADC v\u00e0 BDC c\u00f3 chung ph\u1ea7n di\u1ec7n t\u00edch tam gi\u00e1c EDC , suy ra: SBEC \uf03d SEAD \uf03d 7, 5 (cm2). Hai tam gi\u00e1c EBC v\u00e0 EAB c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb B xu\u1ed1ng AC do \u0111\u00f3: E=C SB=EC 7=, 5 3 hay EC \uf03d 3EA . EA SEAB 2, 5 Hai tam gi\u00e1c DEC v\u00e0 EAD c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb D xu\u1ed1ng AC m\u00e0 EC = 3EA n\u00ean suy ra: Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com SDEC \uf03d 3SEAD \uf03d 7, 5\uf0b43 \uf03d 22, 5 (cm2). V\u1eady di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh thang ABCD l\u00e0: 2,5 \uf02b 7,5\uf0b42 \uf02b 22,5 \uf03d 40 (cm2). \u0110\u00e1p s\u1ed1: SABCD = 40 (cm2). B\u00e0i 66. Cho h\u00ecnh thang ABCD c\u00f3 \u0111\u00e1y AB v\u00e0 DC , trong \u0111\u00f3 AB = 1 DC . Hai \u0111\u01b0\u1eddng ch\u00e9o AC v\u00e0 2 BD c\u1eaft nhau \u1edf \u0111i\u1ec3m E . a) T\u00ecm t\u1ec9 s\u1ed1 \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng EA v\u00e0 EC . b) H\u00e3y so s\u00e1nh di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c EAB v\u00e0 di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c EDC . L\u1eddi gi\u1ea3i MA B E D HC a) X\u00e9t hai tam gi\u00e1c DAB v\u00e0 DBC c\u00f3 chi\u1ec1u cao BH b\u1eb1ng v\u1edbi chi\u1ec1u cao DM (c\u00f9ng b\u1eb1ng chi\u1ec1u cao c\u1ee7a h\u00ecnh thang) v\u00e0 \u0111\u00e1y AB = 1 DC n\u00ean . 2 M\u00e0 hai tam gi\u00e1c DAB v\u00e0 DBC c\u00f3 chung \u0111\u00e1y BD n\u00ean suy ra chi\u1ec1u cao h\u1ea1 t\u1eeb A xu\u1ed1ng BD b\u1eb1ng 1 chi\u1ec1u cao h\u1ea1 t\u1eeb C xu\u1ed1ng BD (*). 2 Hai tam gi\u00e1c ADE v\u00e0 EDC c\u00f3 chung c\u1ea1nh \u0111\u00e1y DE , k\u1ebft h\u1ee3p v\u1edbi (*) \u1edf tr\u00ean ta suy ra: S ADE = 1 S EDC (**) 2 M\u1eb7t kh\u00e1c, hai tam gi\u00e1c ADE v\u00e0 EDC l\u1ea1i c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb D xu\u1ed1ng AC k\u1ebft h\u1ee3p v\u1edbi (**) ta suy ra \u0111\u01b0\u1ee3c: AE = 1 EC hay AE = 1 . 2 EC 2 b) Hai tam gi\u00e1c ADC v\u00e0 BDC c\u00f3 chung c\u1ea1nh \u0111\u00e1y DC v\u00e0 chi\u1ec1u cao h\u1ea1 t\u1eeb A xu\u1ed1ng DC b\u1eb1ng chi\u1ec1u cao h\u1ea1 t\u1eeb B xu\u1ed1ng DC n\u00ean: SADC = SBDC . M\u00e0 hai tam gi\u00e1c ADC v\u00e0 BDC c\u00f3 chung ph\u1ea7n di\u1ec7n t\u00edch tam gi\u00e1c EDC , suy ra: SBEC = SAED . X\u00e9t hai tam gi\u00e1c AEB v\u00e0 BEC c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb B xu\u1ed1ng AC v\u00e0 \u0111\u00e1y AE = 1 EC 2 n\u00ean suy ra: SEAB \uf03d 1 S BEC \uf03d 1 S AED (1). 2 2 M\u00e0 theo c\u00e2u a) ta c\u00f3: S ADE \uf03d 1 S EDC (2). 2 T\u1eeb (1) v\u00e0 (2) ta suy ra: S AEB \uf03d 1 \uf0b4 1 SEDC \uf03d 1 SEDC . 2 2 4 Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com \u0110\u00e1p s\u1ed1: a) AE = 1 . b) S AEB = 1 SEDC EC 2 4 B\u00e0i 67. Cho h\u00ecnh v\u1ebd b\u00ean, bi\u1ebft AM \uf03d MB , CN \uf03d 2BN , AC \uf03d 4CP v\u00e0 SMNP \uf03d 90 cm2. T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c ABC . L\u1eddi gi\u1ea3i A A MM C N C P P BN B S AMP = 1 S ABP (1) (hai tam gi\u00e1c c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb P xu\u1ed1ng AB v\u00e0 \u0111\u00e1y AM = 1 AB ) 2 2 SBNP = 1 SBCP (hai tam gi\u00e1c c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb P xu\u1ed1ng BC v\u00e0 \u0111\u00e1y BN = 1 BC ) 3 3 M\u00e0 SBCP = 1 S ABP (hai tam gi\u00e1c c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb B xu\u1ed1ng AC v\u00e0 \u0111\u00e1y CP = 1 AP ) 5 5 Suy ra: SBNP =1 \u00d7 1 S ABP =115 SABP (2). 35 SBMN = 1 SBMC (hai tam gi\u00e1c c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb M xu\u1ed1ng BC v\u00e0 \u0111\u00e1y BN = 1 BC ) 3 3 SBMC = 1 S ABC (hai tam gi\u00e1c c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb C xu\u1ed1ng AB v\u00e0 \u0111\u00e1y BM = 1 AB ) 2 2 S ABC = 4 (hai tam gi\u00e1c c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb B xu\u1ed1ng AC v\u00e0 \u0111\u00e1y AC = 4 AP ) 5 SABP 5 Suy ra: SBMN \uf03d 1 \uf0b4 1 \uf0b4 4 S ABP \uf03d 2 S ABP (3). 3 2 5 15 T\u1eeb (1), (2) v\u00e0 (3) ta t\u00ednh \u0111\u01b0\u1ee3c: SMNP \uf03d \uf0e8\uf0e7\uf0e7\uf0e7\uf0e61\uf02d 1 \uf02d 1 \uf02d 2 \uf0f8\uf0f6\uf0f7\uf0f7\uf0f7 S ABP \uf03d3 SABP . 2 15 15 10 T\u1eeb \u0111\u00f3 ta t\u00ednh \u0111\u01b0\u1ee3c di\u1ec7n t\u00edch tam gi\u00e1c ABP l\u00e0: 90 : 3 \uf03d 300 (cm2). 10 Di\u1ec7n t\u00edch tam gi\u00e1c ABC l\u00e0: 300\uf0b4 4 \uf03d 240 (cm2). 5 \u0110\u00e1p s\u1ed1: SABC = 240 (cm2). Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","B\u00e0i 68. Website: tailieumontoan.com Cho tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch 270cm2 . Tr\u00ean c\u1ea1nh AB l\u1ea5y \u0111i\u1ec3m M sao cho AM = 1 MB . 2 Tr\u00ean AC l\u1ea5y N v\u00e0 P sao cho A=N C=P 1 AC . Tr\u00ean BC l\u1ea5y \u0111i\u1ec3m Q sao cho 3 CQ= 2\u00d7 BQ . T\u00ednh di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh MNPQB . L\u1eddi gi\u1ea3i A MN P BQ C S AMC = 1 S ABC (hai tam gi\u00e1c c\u00f3 c\u00f9ng chi\u1ec1u cao h\u1ea1 t\u1eeb \u0111\u1ec9nh C v\u00e0 \u0111\u00e1y AM = 1 AB ) 3 3 M\u00e0 S AMN = 1 S AMC (hai tam gi\u00e1c c\u00f3 c\u00f9ng chi\u1ec1u cao h\u1ea1 t\u1eeb \u0111\u1ec9nh M v\u00e0 \u0111\u00e1y AN = 1 AC ) 3 3 Suy ra: S AMN \uf03d 1 \uf0b4 1 S ABC \uf03d 1 S ABC (1). 3 3 9 S AQC = 2 S ABC (hai tam gi\u00e1c c\u00f3 c\u00f9ng chi\u1ec1u cao h\u1ea1 t\u1eeb \u0111\u1ec9nh A v\u00e0 \u0111\u00e1y QC = 2 BC ) 3 3 M\u00e0 SPQC = 1 (hai tam gi\u00e1c c\u00f3 c\u00f9ng chi\u1ec1u cao h\u1ea1 t\u1eeb \u0111\u1ec9nh Q v\u00e0 \u0111\u00e1y PC = 1 AC ) 3 SAQC 3 Suy ra: S AMN \uf03d 1 \uf0b4 2 S ABC \uf03d 2 (2). 3 3 9 SABC T\u1eeb (1) v\u00e0 (2) ta suy ra: SMNPQB = \uf8ed\uf8eb\uf8ec1 \u2212 1 \u2212 2 \uf8f6 S ABC = 2 S ABC 9 9 \uf8f8\uf8f7 3 V\u1eady di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh MNPQB l\u00e0: 270\uf0b4 2 \uf03d 180 (cm2). 3 \u0110\u00e1p s\u1ed1: SMNPQB = 180cm2 B\u00e0i 69. Cho tam gi\u00e1c ABC , tr\u00ean c\u1ea1nh AB l\u1ea5y M sao cho AM \uf03d 2BM , tr\u00ean c\u1ea1nh AC l\u1ea5y N sao cho AN \uf03d\u00a03 NC . Bi\u1ebft di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c MNCB l\u1edbn h\u01a1n di\u1ec7n t\u00edch tam gi\u00e1c AMN l\u00e0 24cm2 . 2 T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c ABC . L\u1eddi gi\u1ea3i Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com Ta c\u00f3: AM \uf03d 2BM hay AM \uf03d\u00a02 AB 3 AN \uf03d 2 NC hay AN \uf03d\u00a03 AC . 35 S.ABN\u00a0\uf03d 3 S.ABC (chung chi\u1ec1u cao h\u1ea1 t\u1eeb B xu\u1ed1ng AC , \u0111\u00e1y AN \uf03d\u00a03 AC ). 55 S.A\u00a0MN \uf03d 2 S.ABN (chung chi\u1ec1u cao h\u1ea1 t\u1eeb N xu\u1ed1ng AB , \u0111\u00e1y AM \uf03d\u00a02 AB ). 33 Do \u0111\u00f3, S.AMN b\u1eb1ng: 2\uf0b4 3 \uf03d\u00a02 S.ABC . 35 5 Suy ra S.BMNC b\u1eb1ng S.A\u00a0BC \uf02d 2 S.ABC \uf03d 3 S.ABC . 55 V\u1eady 24cm2 \u1ee9ng v\u1edbi: 3 \u2212 2 =1 S .ABC 5 5 5 V\u1eady S.ABC b\u1eb1ng 24 : 1 = 120cm2 5 V\u1eady S.ABC b\u1eb1ng 24 : 1 \u00a0\uf03d120 (cm2). 5 B\u00e0i 70. Cho tam gi\u00e1c ABC (h\u00ecnh v\u1ebd b\u00ean); EA \uf03d EB; DC \uf03d 2BD . T\u00ednh di\u1ec7n t\u00edch ph\u1ea7n t\u00f4 \u0111\u1eadm bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c ABC b\u1eb1ng 60cm2. L\u1eddi gi\u1ea3i Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com Ta c\u00f3 AE \uf03d BE \uf03d\u00a0 1 AB; DC \uf03d\u00a02BD \uf03d 2 BC . 23 Suy ra BD \uf03d 1 BC . 3 S.EBD \uf03d 1\uf0b4\u00a01 \uf03d\u00a01 S.\u00a0ABC . (1) 23 6 S.AED \uf03d 1 \uf0b4\u00a01 \uf03d\u00a01 S.\u00a0ABC . 23 6 S.ADC \uf03d 2 S.\u00a0ABC (chung chi\u1ec1u cao h\u1ea1 t\u1eeb A xu\u1ed1ng BC , \u0111\u00e1y DC \uf03d 2 BC . 33 T\u1ec9 s\u1ed1 di\u1ec7n t\u00edch c\u1ee7a AED v\u00e0 ADC b\u1eb1ng: 1 \u00a0:\u00a02 \uf03d\u00a01 . 63 4 V\u00ec AED v\u00e0 ADC chung \u0111\u00e1y AD n\u00ean chi\u1ec1u cao h\u1ea1 t\u1eeb E xu\u1ed1ng AD b\u1eb1ng 1 chi\u1ec1u c\u1ea1o h\u1ea1 t\u1eeb C 4 xu\u1ed1ng AD . \u0110\u00e2y c\u0169ng l\u00e0 t\u1ec9 l\u1ec7 hai chi\u1ec1u cao t\u01b0\u01a1ng \u1ee9ng c\u1ee7a hai tam gi\u00e1c chung \u0111\u00e1y AF l\u00e0 AEF v\u00e0 ACF . Do \u0111\u00f3 S.AEF \uf03d 1 S.ACF hay S.AEF \uf03d 1 S.AEC . 45 L\u1ea1i c\u00f3 S.AEC \uf03d 1 S.ABC (chung chi\u1ec1u cao h\u1ea1 t\u1eeb C xu\u1ed1ng AB , \u0111\u00e1y AE \uf03d 1 AB 22 Suy ra S.AEF\u00a0\uf03d 1\uf0b4 1 \uf03d 1 S.ABC . 5 2 10 T\u1eeb \u0111\u00e2y ta c\u00f3: S. EDF \uf03d 1 \uf02d\u00a01 \uf03d 1 S.ABC . (2) 6 10 15 S. AFC =\u00a01 \u2212\u00a01 \u00a0=2 S.ABC (3) 2 10 5 T\u1eeb (1), (2) v\u00e0 (3) ta c\u00f3 di\u1ec7n t\u00edch ph\u1ea7n t\u00f4 \u0111\u1eadm b\u1eb1ng: 1 +\u00a01 +\u00a02 =\u00a019 S.ABC 6 15 5 30 V\u1eady di\u1ec7n t\u00edch ph\u1ea7n t\u00f4 \u0111\u1eadm b\u1eb1ng: 60 \u00d7 19 =38 cm2 30 B\u00e0i 72. Cho h\u00ecnh tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng 875cm2. Q v\u00e0 K l\u1ea7n l\u01b0\u1ee3t n\u1eb1m tr\u00ean c\u00e1c c\u1ea1nh AB v\u00e0 AC sao cho QB \uf03d 3 QA v\u00e0 KC \uf03d 2 \u00a0KA . N\u1ed1i QK , t\u00ednh di\u1ec7n t\u00edch t\u1ee9u gi\u00e1c QBKC . 43 Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","L\u1eddi gi\u1ea3i Website: tailieumontoan.com . KC \uf03d 2 \u00a0KA hay KC \uf03d 2 \u00a0AC 35 QB \uf03d 3 QA hay QB \uf03d 3 \u00a0AB 47 Ta c\u00f3: S. BKC \uf03d 2 \u00a0S.ABC (chung chi\u1ec1u cao h\u1ea1 t\u1eeb B xu\u1ed1ng AC , \u0111\u00e1y CK \uf03d\u00a02 AC ) 55 Suy ra S.ABK b\u1eb1ng: S.ABC\u00a0\uf02d 2 S.A\u00a0BC \uf03d 3 S.ABC . 55 V\u00e0: S.BKQ\u00a0\uf03d 3 S.ABK (chung chi\u1ec1u cao h\u1ea1 t\u1eeb K xu\u1ed1ng AB , \u0111\u00e1y BQ \uf03d\u00a03 AB ) 77 Do \u0111\u00f3 S.BKQ\u00a0\uf03d\u00a0\u00a03\uf0b43 \uf03d 9 S.ABC . 7 5 35 Nh\u01b0 v\u1eady, S.BCKQ\u00a0b\u1eb1ng: 2 S.ABC \uf02b 9 S.ABC \uf03d 23 S.ABC 5 35 35 V\u1eady di\u1ec7n t\u00edch BCKQ b\u1eb1ng 875\uf0b4 23 \uf03d 575 (cm2). 35 B\u00e0i 73. Trong h\u00ecnh b\u00ean, AM \uf03d MC, BD \uf03d DE \uf03d EM , di\u1ec7n t\u00edch tam gi\u00e1c ACD l\u00e0 2020 cm2. T\u00ednh t\u1ed5ng di\u1ec7n t\u00edch c\u00e1c tam gi\u00e1c trong h\u00ecnh nh\u1eadn \u0111i\u1ec3m B l\u00e0m \u0111\u1ec9nh. L\u1eddi gi\u1ea3i C\u00e1c tam gi\u00e1c c\u00f3 \u0111\u1ec9nh B l\u00e0: BAD; BAE; BAM ; BCD; BCE; BCM ; ABD Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com Ta c\u00f3: S=ABD S=ADE SAEM \u00a0(v\u00ec 3 tam gi\u00e1c n\u00e0y c\u00f3 BD \uf03d DE \uf03d EM v\u00e0 c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb A xu\u1ed1ng BM) (1). S=CEM S=CED SCDB (v\u00ec ba tam gi\u00e1c n\u00e0y c\u00f3 BD \uf03d DE \uf03d EM v\u00e0 c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb C xu\u1ed1ng BM) (2). SAEM = SCEM \u00a0(v\u00ec 2 tam gi\u00e1c n\u00e0y c\u00f3 AM \uf03d MC v\u00e0 c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb E xu\u1ed1ng AC) (3) T\u1eeb (1); (2); (3) suy ra S=ABD S=ADE S=AEM \u00a0\u00a0S=CEM S=CED SCDB \u00a0. \u00a0SACD = SADE + SAEM + SCEM +\u00a0SCED = 4 SAEM \u00a0\u00a0su=y ra SAEM =2020 : 4 505 (cm2). L\u1ea1i c\u00f3: SABC = SABD + SADE + SAEM +\u00a0SCEM + SCED \u00a0+ S\u00a0CD\u00a0B = 6 SAEM \u00a0 S=BCE S=BAE 2 SAEM S=BCM S=BAM 3 SAEM \u00a0S=BAD S=BCD \u00a0SAEM V\u1eady t\u1ed5ng di\u1ec7n t\u00edch c\u00e1c tam gi\u00e1c \u0111\u1ec9nh B l\u00e0: SBAD \uf02b SBAE \uf02b SBAM \uf02b SBCD \uf02b\u00a0SBCE \uf02b\u00a0SBCM\u00a0\uf02b SABC \u00a0\uf03d SAEM \uf02b 2SAEM \uf02b 3SA\u00a0EM \uf02bSAEM \uf02b 2SAEM \uf02b 3SA\u00a0EM \uf02b6SA\u00a0EM \uf03d18SAEM \uf03d 505\uf0b418 \uf03d 9090 (cm2). \u0110\u00e1p s\u1ed1: 9090cm2. B\u00e0i 74. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh sau: L\u1eddi gi\u1ea3i \u0110\u00e1p s\u1ed1: 1950cm2. Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng ABCD l\u00e0: 30\u00d7 30 =900 (cm2). Di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c BMN l\u00e0: 30\uf0b4\uf02840 \uf02b 30\uf029: 2 \uf03d 30\uf0b435 \uf03d1050 (cm2). Di\u1ec7n t\u00edch h\u00ecnh \u0111\u00e3 cho l\u00e0: 900 \uf02b105 \uf03d1950 (cm2). B\u00e0i 75. Bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c t\u00f4 \u0111\u1eadm l\u00e0 2cm2. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng l\u1edbn. Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com L\u1eddi gi\u1ea3i Ta \u0111\u1eb7t t\u00ean cho c\u00e1c \u0111i\u1ec3m nh\u01b0 h\u00ecnh v\u1ebd. Di\u1ec7n t\u00edch tam gi\u00e1c A1A2B1 \u00a0l\u00e0 2 cm2. N\u1ed1i A1C2,D1B2 . Ta c\u00f3 di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng A1B1C1D1 g\u1ea5p 4 l\u1ea7n di\u1ec7n t\u00edch tam gi\u00e1c A1A2B1 . V\u1eady di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng A1B1C1D1 l\u00e0: 2\u00d7 4 =8 (cm2). Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng A2B2C2D2 g\u1ea5p \u0111\u00f4i di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng A1B1C1D1 n\u00ean di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng A2B2C2D2 l\u00e0: 8\u00d7 2 =16 (cm2). Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng A3B3C3D3 g\u1ea5p \u0111\u00f4i di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng A2B2C2D2 n\u00ean di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng A3B3C3D3 l\u00e0: 16 \u00d7 2 =32 (cm2). Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng A4B4C4D4 g\u1ea5p \u0111\u00f4i di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng A3B3C3D3 n\u00ean di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng A4B4C4D4 l\u00e0: 32 \u00d7 2 =64 (cm2). V\u1eady di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng l\u1edbn nh\u1ea5t trong h\u00ecnh v\u1ebd l\u00e0 64 cm2. \u0110\u00e1p s\u1ed1: 64cm2. B\u00e0i 76. Cho h\u00ecnh ch\u1eef nh\u1eadt ABCD c\u00f3 AB = 18 cm v\u00e0 BC = 12 cm. Tr\u00ean AB l\u1ea5y \u0111i\u1ec3m M sao cho AM = 1 AB v\u00e0 tr\u00ean BC l\u1ea5y \u0111i\u1ec3m N sao cho BN = 1 BC . T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c DMN 32 L\u1eddi gi\u1ea3i Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com Ta c\u00f3: AM \uf03d 1\uf0b4 AB \uf03d 1\uf0b418 \uf03d 6 (cm); BM \uf03d AB \uf02d AM \uf03d18\uf02d6 \uf03d12 (cm). 33 BN \uf03d CN \uf03d 1 \uf0b4BC \uf03d\u00a01 \uf0b412 \uf03d 6 (cm). 22 T\u1eeb \u0111\u00f3 ta t\u00ednh \u0111\u01b0\u1ee3c: S DAM = 1 \u00a0\u00d7DA\u00d7 AM = 1 \u00a0\u00d712 \u00d7 6 = 36 (cm2). 2 2 \u00a0SMBN \uf03d 12\u00a0\uf0b4MB\uf0b4B\u00a0N \uf03d 12 \uf0b412\uf0b46 \uf03d 36 (cm2). S DCN =\u00a01 \u00d7 DC \u00d7 CN =\u00a01 \u00d718\u00d7 6 = 54 (cm2). 2 2 SABCD =AB \u00d7 BC =1\u00a0 8\u00d712 =216 (cm2). V\u1eady: SDMN \uf03d SABCD \u2013 \uf028SDAM \uf02b SMBN \uf02b SDCN \uf029 \uf03d 216\uf02d\uf02836 \uf02b 36 \uf02b 54\uf029 \uf03d 90 (cm2). \u0110\u00e1p S\u1ed1: 90 (cm2). B\u00e0i 77. Cho h\u00ecnh b\u00ecnh h\u00e0nh ABCD . E l\u00e0 m\u1ed9t \u0111i\u1ec3m tr\u00ean AB . N\u1ed1i E v\u1edbi C v\u00e0 B v\u1edbi D . G\u1ecdi \u0111i\u1ec3m giao nhau gi\u1eefa EC v\u00e0 BD l\u00e0 F . Di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c EFB l\u00e0 20cm2 v\u00e0 BFC l\u00e0 50cm2 . T\u00ednh di\u1ec7n t\u00edch h\u00ecnh b\u00ecnh h\u00e0nh ABCD . L\u1eddi gi\u1ea3i Hai tam gi\u00e1c EDC v\u00e0 BDC c\u00f3 chung c\u1ea1nh \u0111\u00e1y DC v\u00e0 chi\u1ec1u cao h\u1ea1 t\u1eeb A xu\u1ed1ng DC b\u1eb1ng chi\u1ec1u cao h\u1ea1 t\u1eeb E xu\u1ed1ng DC n\u00ean: SEDC .= SBDC M\u00e0 hai tam gi\u00e1c EDC v\u00e0 BDC c\u00f3 chung ph\u1ea7n di\u1ec7n t\u00edch tam gi\u00e1c DFC , suy ra: S=BFC S=DEF 50 (cm2). Hai tam gi\u00e1c BEF v\u00e0 BEC c\u00f3 chung c\u1ea1nh \u0111\u00e1y BF v\u00e0 c\u00f3 t\u1ec9 l\u1ec7 di\u1ec7n t\u00edch l\u00e0 2 n\u00ean t\u1ec9 l\u1ec7 chi\u1ec1u 5 cao h\u1ea1 t\u1eeb \u0111\u1ec9nh E xu\u1ed1ng BD v\u00e0 \u0111\u1ec9nh C xu\u1ed1ng BD l\u00e0 2 . 5 Hai tam gi\u00e1c EBF v\u00e0 DFC c\u00f3 chung c\u1ea1nh \u0111\u00e1y DF , t\u1ec9 l\u1ec7 chi\u1ec1u cao h\u1ea1 t\u1eeb \u0111\u1ec9nh E xu\u1ed1ng BD v\u00e0 \u0111\u1ec9nh C xu\u1ed1ng BD l\u00e0 2 . V\u1eady di\u1ec7n t\u00edch c\u1ee7a tam gi\u00e1c DFC l\u00e0: 50 : 2 = 125 (cm2). 55 Di\u1ec7n t\u00edch h\u00ecnh b\u00ecnh h\u00e0nh ABCD g\u1ea5p 2 l\u1ea7n di\u1ec7n t\u00edch tam gi\u00e1c BDC v\u00e0 b\u1eb1ng (50 +125) \u00d7 2 =350 (cm2). Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com B\u00e0i 78. Cho h\u00ecnh v\u1ebd b\u00ean bi\u1ebft S1 = 12cm2 . T\u00ednh S2 ? B\u00e0i gi\u1ea3i Ta c\u00f3: S AMQ = 2 ( AM = 2 AB , chung chi\u1ec1u cao h\u1ea1 t\u1eeb Q xu\u1ed1ng AB ) 3 S ABQ 3 Di\u1ec7n t\u00edch tam gi\u00e1c ABQ l\u00e0: 12 : 2 = 18 (cm2). 3 Ta l\u1ea1i c\u00f3: SAQC =2 \u00d7 SABQ =2 \u00d718 =36 (cm2). M\u00e0 S AQN = 1 S AQC ( AN = 1 AC , chung chi\u1ec1u cao h\u1ea1 t\u1eeb Q xu\u1ed1ng AC 3 3 Di\u1ec7n t\u00edch tan gi\u00e1c AQN l\u00e0 : 36 : 3 = 12 (cm2). Di\u1ec7n t\u00edch tam gi\u00e1c AMN l\u00e0: 12 +12 =24 (cm2). M\u00e0 AAMN = 2 = 2 \u00d7 1 = 2 S ABC 3 S ANB 3 3 9 Di\u1ec7n t\u00edch tam gi\u00e1c ABC l\u00e0: 24 : 2 = 108 (cm2). 9 Di\u1ec7n t\u00edch tam gi\u00e1c ABP l\u00e0 : 108 : 3 = 36 (cm2). Di\u1ec7n t\u00edch S2 l\u00e0: 108 \u2212 36 \u221212 =60 (cm2). \u0110\u00e1p s\u1ed1: 60cm2. B\u00e0i 79. Cho h\u00ecnh tam gi\u00e1c ABC . L\u1ea5y M tr\u00ean AB v\u00e0 N tr\u00ean AC sao cho AM = MB v\u00e0 NC \u00d7 2 =NA . a) T\u00ednh t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch ANM v\u00e0 BMNC . b) Cho MN c\u1eaft BC \u1edf D . So s\u00e1nh BC v\u1edbi CE L\u1eddi gi\u1ea3i Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com a) Ta c\u00f3: S ABN = 1 S ABC ( AN = 1 AC , chung chi\u1ec1u cao h\u1ea1 t\u1eeb B xu\u1ed1ng AC ) 3 3 M\u00e0 S=MNB S=AMN 1 2 S ABN Suy ra: SAMN = 1 S ABC 6 Ta c\u00f3 : S BNC = 2 S ABC ( NC = 2 AC , chung chi\u1ec1u cao h\u1ea1 t\u1eeb B xu\u1ed1ng AC ) 3 3 V\u1eady S BMNC \uf03d SMNB \uf02b SMNB \uf03d1\uf02b2 \uf03d 5 S ABC . 63 6 V\u1eady t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch c\u1ee7a ANM v\u00e0 BMNC l\u00e0 1 : 5 = 1 . 66 5 b) N\u1ed1i A v\u1edbi D , B v\u1edbi N Ta c\u00f3: SMNA = SMNB (chung chi\u1ec1u cao, AM = MB ) V\u1eady chi\u1ec1u cao h\u1ea1 t\u1eeb A xu\u1ed1ng MN v\u00e0 t\u1eeb B xu\u1ed1ng MN b\u1eb1ng nhau. Suy ra: SAND = SBND (chung \u0111\u00e1y ND ). Ta c\u00f3: S DCN = 1 S ADN (chung chi\u1ec1u cao, CN = 1 AN ). 2 2 M\u00e0 SDCN = 1 2 SBND Hay SDCN = SBCN (chung chi\u1ec1u cao) V\u1eady: BC \uf03d CD . B\u00e0i 80. Cho h\u00ecnh v\u1ebd. T\u00ednh t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch 2 tam gi\u00e1c BDF v\u00e0 AEF . L\u1eddi gi\u1ea3i: L\u1eddi gi\u1ea3i Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com SA=BF \u00a012=SAFC 2 (Chi\u1ec1u cao t\u1eeb B b\u1eb1ng 1 chi\u1ec1u cao t\u1eeb C xu\u1ed1ng \u0111\u00e1y chung AF) 4 SAFC 2 M\u00e0 S AEF = 1 S AFC n\u00ean SABE =2 S AFC + 1 SAFC =3 S AFC 4 4 4 4 M\u1eb7t kh\u00e1c SABE = 1 S ABC 4 V\u1eady S=AFC \u00a014=:\u00a034 \u00a013 SABC \u21d2\u00a0SA\u00a0EF =\u00a014\u00a0\u00d7 13\u00a0SAB\u00a0C =\u00a0112\u00a0SA\u00a0BC \u00a0(1) Ta l\u1ea1i c\u00f3: SABD =\u00a013 \u00a0SA\u00a0BC\u00a0\u00a0m\u00e0 S ABF =\u00a01 S FBC (Chi\u1ec1u cao t\u1eeb A b\u1eb1ng 1\/3 chi\u1ec1u cao t\u1eeb C xu\u1ed1ng \u0111\u00e1y chung 3 BF ) L\u1ea1i c\u00f3: SFBD =\u00a01 S FBC V\u1eady : S=FBD \u00a0S=AB\u00a0F \u00a012 S\u00a0AB\u00a0D 3 Suy ra: SBDF =\u00a012\u00a0\u00d7\u00a013\u00a0=\u00a016\u00a0S\u00a0A\u00a0BC\u00a0\u00a0(2) T\u1eeb (1) v\u00e0 (2) ta c\u00f3, t\u1ec9 s\u1ed1 2 tam gi\u00e1c AEF v\u00e0 BDF l\u00e0 : 1 :1=1 12 6 2 \u0110\u00e1p s\u1ed1: 1 2 B\u00e0i 81. Cho tam gi\u00e1c v\u1edbi c\u00e1c t\u1ef7 l\u1ec7 nh\u01b0 h\u00ecnh. Bi\u1ebft S3 \u2212 S1 =84cm2 . T\u00ednh S4 \u2212 S2 L\u1eddi gi\u1ea3i Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com Theo t\u1ef7 l\u1ec7 di\u1ec7n t\u00edch c\u00e1c h\u00ecnh tam gi\u00e1c ta c\u00f3: S1 + S4 = 2 (S3 + S2 ) \u21d2 3S1 + 3S4 = 2S3 + 2S2 3 M\u00e0 S3 \u2212 S1 =84 cm2 hay S=3 S1 + 84 V\u1eady 3S1 \uf02b 3S4 \uf03d 2\uf0b4(S1 \uf02b84) \uf02b 2S2 \uf0de S1 \uf02b S4 \uf03d168 \uf02b 2S2 (1). T\u01b0\u01a1ng t\u1ef1 ta c\u00f3: S1 + S2 = 1 ( S3 + S4 ) \u21d2 2S1 + 2S4 = S3 + S4 2 M\u00e0 S3 \u2212 S1 =84cm2 hay S=3 S1 + 84 V\u1eady 2S1 \uf02b 2S2 \uf03d (S1 \uf02b84) \uf02b S4 \uf0de S1 \uf02b 2S2 \uf03d 84 \uf02b S4 (2). Tr\u1eeb v\u1ebf v\u1edbi v\u1ebf c\u1ee7a bi\u1ec3u th\u1ee9c (1) v\u00e0 (2) ta c\u00f3: (S1 \uf02b 3S4 ) \uf02d(S1 \uf02b 2S2 ) \uf03d (168 \uf02b 2S2 ) \uf02d(84 \uf02b S4 ) \uf0de 3S4 \uf02d 2S2 \uf03d 84 \uf02b 2S2 \uf02d S4 \uf0de 4S4 \uf02d 4S2 \uf03d 84 \uf0de (S4 \uf02d S2 ) \uf03d 21(cm2). \u0110\u00e1p s\u1ed1: 21cm2. C\u00e1ch 2: Gi\u1ea3i theo d\u1ea1ng to\u00e1n Hi\u1ec7u T\u1ec9: V\u00ec S3 \u2013 S1 = 84 n\u00ean (S3 + S4) - (S1 + S4) = 84 Ta c\u00f3: (S3 + S4) = 2 S1234 3 (S1 + S4) = 2 S1234 5 V\u1eady T\u1ec9 s\u1ed1 (S3 + S4) v\u00e0 (S1 + S4) l\u00e0: 2: 2 =5 353 (S3 + S4) = 84 : (5 \u2013 3) x 5 = 210cm2 S1234 = 3 (S3 + S4) = 3 x 210 = 315 cm2 22 (S2 + S3) = 3 S1234 = 3 x 315 = 189 cm2 55 S4 \u2013 S2 = (S3 + S4) - (S2 + S3) = 210 \u2013 189 = 21cm2 \u0110\u00e1p s\u1ed1: 21cm2 B\u00e0i 82. M\u1ed9t b\u1ec3 b\u01a1i d\u00e0i 33m , r\u1ed9ng 12m v\u00e0 s\u00e2u 1m80cm . Ng\u01b0\u1eddi ta d\u00f9ng g\u1ea1ch men h\u00ecnh vu\u00f4ng c\u00f3 c\u1ea1nh 30cm \u0111\u1ec3 l\u00e1t xung quanh th\u00e0nh b\u1ec3 v\u00e0 \u0111\u00e1y b\u1ec3. T\u00ednh s\u1ed1 g\u1ea1ch \u0111\u1ec3 l\u00e1t b\u1ec3 b\u01a1i \u0111\u00f3 ? L\u1eddi gi\u1ea3i \u0110\u1ed5i: 1m80cm = 1,8m Di\u1ec7n t\u00edch xung quanh c\u1ee7a b\u1ec3 b\u01a1i l\u00e0 : (33\uf02b12)\uf0b42\uf0b41,8 \uf03d162 (cm2). Di\u1ec7n t\u00edch m\u1eb7t \u0111\u00e1y c\u1ee7a b\u1ec3 b\u01a1i l\u00e0: 33\u00d712 =396 (m2). Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com Di\u1ec7n t\u00edch ph\u1ea7n b\u1ec3 b\u01a1i \u0111\u01b0\u1ee3c l\u00e1t g\u1ea1ch l\u00e0: 162 + 396 =558 (m2). Di\u1ec7n t\u00edch m\u1ed9t v\u00ean g\u1ea1ch l\u00e1t l\u00e0: 0,3\u00d7 0,3 =0, 09 (m2). T\u1ed5ng s\u1ed1 vi\u00ean g\u1ea1ch \u0111\u1ec3 l\u00e1t b\u1ec3 b\u01a1i \u0111\u00f3 l\u00e0: 558 : 0, 09 = 6200 (vi\u00ean g\u1ea1ch). \u0110\u00e1p s\u1ed1: 6200vi\u00ean g\u1ea1ch. B\u00e0i 83. Cho tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A c\u00f3 ba c\u1ea1nh AB = 6cm , AB = 6cm v\u00e0 BC = 10m v\u1ebd chi\u1ec1u cao AH t\u1eeb \u0111\u1ec9nh A \u0111\u1ebfn c\u1ea1nh \u0111\u00e1y BC . a) T\u00ednh chi\u1ec1u cao AH . b) Tr\u00ean c\u1ea1nh BC l\u1ea5y \u0111i\u1ec3m E sao cho BE= 2\u00d7 EC . T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c AEC ? L\u1eddi gi\u1ea3i a) Di\u1ec7n t\u00edch tam gi\u00e1c ABC l\u00e0: (6\uf0b48) : 2 \uf03d 24 (cm2). Chi\u1ec1u cao AH c\u1ee7a tam gi\u00e1c ABC l\u00e0: 24\uf0b42 :10 \uf03d 4,8 (cm). b) X\u00e9t hai tam gi\u00e1c AEC v\u00e0 tam gi\u00e1c ABC c\u00f3 chung \u0111\u01b0\u1eddng cao AH v\u00e0 c\u1ea1nh \u0111\u00e1y EC = 1 BC 3 N\u00ean: S AEC \uf03d 1 S ABC \uf0de SAEC \uf03d 24 : 3 \uf03d 8 (cm2). 3 \u0110\u00e1p s\u1ed1: 8(cm2 ) B\u00e0i 84. 1. M\u1ed9t mi\u1ebfng \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chu vi b\u1eb1ng 152m, bi\u1ebft r\u1eb1ng n\u1ebfu gi\u1ea3m chi\u1ec1u d\u00e0i mi\u1ebfng \u0111\u1ea5t 5m th\u00ec di\u1ec7n t\u00edch mi\u1ebfng \u0111\u1ea5t gi\u1ea3m 170m2. T\u00ednh di\u1ec7n t\u00edch mi\u1ebfng \u0111\u1ea5t? 2. Cho tam gi\u00e1c ABC , c\u00f3 D l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a c\u1ea1nh BC . Tr\u00ean c\u1ea1nh AC l\u1ea5y \u0111i\u1ec3m M sao cho AM = 1 AC . 3 a) So s\u00e1nh di\u1ec7n t\u00edch tam gi\u00e1c ADM v\u00e0 ABC b) Tr\u00ean c\u1ea1nh AB l\u1ea5y \u0111i\u1ec3m N sao cho AN = NB .T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c DMN , n\u1ebfu bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c ABC l\u00e0 640cm2. Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com L\u1eddi gi\u1ea3i 1) N\u1eeda chu vi h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u l\u00e0: 152 : 2 \uf03d 76 (m). Chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt l\u00fac \u0111\u1ea7u l\u00e0: 170 : 5 = 34 (m). Chi\u1ec1u d\u00e0i mi\u1ebfng \u0111\u1ea5t ban \u0111\u1ea7u l\u00e0: 76 \u2212 34 =42 (m). Di\u1ec7n t\u00edch mi\u1ebfng \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt l\u00e0: 42\u00d7 34 =1428 .(m2). \u0110\u00e1p s\u1ed1: 1428m2. a) Ta c\u00f3 S ADM = 1 S ADC (hai tam gi\u00e1c ADM v\u00e0 ADC c\u00f3 chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb D v\u00e0 \u0111\u00e1y 3 AM = 1 AC ). 3 M\u00e0 SADC = 1 S ABC (hai tam gi\u00e1c ADC v\u00e0 ABC c\u00f3 chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb A v\u00e0 \u0111\u00e1y DC = 1 BC ) 2 3 \uf0de S ADM \uf03d 1 S ABC . 6 b) Ta c\u00f3 S ANM = 1 S ANC (hai tam gi\u00e1c ANM v\u00e0 ANC c\u00f3 chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb N v\u00e0 \u0111\u00e1y 3 AM = 1 AC ). 3 M\u00e0 SANC = 1 S ABC (hai tam gi\u00e1c ANC v\u00e0 ABC c\u00f3 chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb C v\u00e0 \u0111\u00e1y NA = 1 AB ) 2 2 \u21d2 S ANM = 1 6 SABC Ta l\u1ea1i c\u00f3 S ADM = 1 S ABC (theo ch\u1ee9ng minh c\u00e2u a) 6 v\u00e0 S ADN \uf03d 1 S ADB \uf03d 1 \uf0b4 1 S ABC \uf03d 1 2 2 2 4 SABC \uf0de S ANDM \uf03d SAND \uf02b SAMD \uf03d 1 S ABC \uf02b1 S ABC \uf03d5 S ABC . 4 6 12 Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com Ta c\u00f3 S NMD \uf03d S ANDM \uf02d SANM \uf03d 5 S ABC \uf02d 1 S ABC \uf03d 1 S ABC \uf03d 640 : 4 \uf03d 160 (cm2). 12 6 4 \u0110\u00e1p s\u1ed1: 160cm2. B\u00e0i 85. Cho tam gi\u00e1c ABC , l\u1ea5y \u0111i\u1ec3m M tr\u00ean c\u1ea1nh \u00a0BC sao cho BM =\u00a01 BC , l\u1ea5y \u0111i\u1ec3m N tr\u00ean c\u1ea1nh AC 2 sao cho AN = 3 AC . Bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c ABC b\u1eb1ng 60cm2. V\u1eady di\u1ec7n t\u00edch tam gi\u00e1c AMN 4 l\u00e0 \u2026..cm2 B\u00e0i gi\u1ea3i Ta c\u00f3: S tam gi\u00e1c AMB =\u00a01 S tam gi\u00e1c ABC (v\u00ec hai tam gi\u00e1c c\u00f3 chung chi\u1ec1u cao k\u1ebb t\u1eeb A xu\u1ed1ng 5 c\u1ea1nh AC , \u0110\u00e1y BM = 1 \u0111\u00e1y BC). 5 V\u1eady S tam gi\u00e1c ABM l\u00e0: 60 x 1 \u00a0= 12 (cm2) 5 \u21d2 S tam gi\u00e1c AMC l\u00e0: 60 \u201312 = 48 (cm2). Ta l\u1ea1i c\u00f3: S tam gi\u00e1c AMN =\u00a03 S tam gi\u00e1c AMC (v\u00ec hai tam gi\u00e1c c\u00f3 chung chi\u1ec1u cao k\u1ebb t\u1eeb \u0111\u1ec9nh 4 M xu\u1ed1ng \u0111\u00e1y AC, \u0110\u00e1y AN =\u00a03 \u0111\u00e1y AC ) 4 V\u1eady S tam gi\u00e1c AMN l\u00e0: 48 x 3 = 36 (cm2). 4 \u0110\u00e1p s\u1ed1: 36cm2. B\u00e0i 86. Cho h\u00ecnh v\u1ebd sau g\u1ed3m c\u00f3 m\u1ed9t h\u00ecnh vu\u00f4ng m\u00e0u cam c\u1ea1nh 21cm v\u00e0 hai n\u1eeda h\u00ecnh tr\u00f2n. B v\u00e0 C l\u00e0 t\u00e2m c\u1ee7a c\u00e1c n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n t\u01b0\u01a1ng \u1ee9ng. Di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh \u0111\u00f3 l\u00e0 \u2026cm2 Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com B\u00e0i gi\u1ea3i Ta c\u00f3: Hai n\u1eeda h\u00ecnh tr\u00f2n gh\u00e9p l\u1ea1i t\u1ea1o th\u00e0nh m\u00f4t h\u00ecnh tr\u00f2n c\u00f3 b\u00e1n k\u00ednh b\u1eb1ng c\u1ea1nh h\u00ecnh vu\u00f4ng v\u00e0 b\u1eb1ng 21 cm. Di\u1ec7n t\u00edch hai n\u1eeda h\u00ecnh tr\u00f2n l\u00e0: 21\uf0b421\uf0b43,14 \uf03d1384, 74 (cm2). Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng m\u00e0u cam l\u00e0: 21\uf0b421\uf03d 441(cm2). Di\u1ec7n t\u00edch h\u00ecnh \u0111\u00f3 l\u00e0: 1384, 74 \uf02b 441\uf03d1825, 74 (cm2). \u0110\u00e1p s\u1ed1: 1825, 74 cm2 B\u00e0i 87. Cho h\u00ecnh ch\u1eef nh\u1eadt ABCD c\u00f3 AB = 60 cm , AD = 40 cm . Tr\u00ean \u00a0AB l\u1ea5y \u0111i\u1ec3m G sao cho AG \uf03d 30 cm. Tr\u00ean AD l\u1ea5y \u0111i\u1ec3m M sao cho AM = 15 cm. a) T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c MGC . b) DG c\u1eaft MC t\u1ea1i K. T\u00ednh di\u1ec7n t\u00edch MDK . B\u00e0i gi\u1ea3i: a) \u0110\u1ed9 d\u00e0i c\u1ea1nh GB l\u00e0: 60 \u2013 30 \uf03d 30 (cm). \u0110\u1ed9 d\u00e0i c\u1ea1nh MD l\u00e0: 40 \u201315 \uf03d 25 (cm). Di\u1ec7n t\u00edch tam gi\u00e1c AGM l\u00e0: 30\uf0b415 : 2 \uf03d 225 (cm2). Di\u1ec7n t\u00edch tam gi\u00e1c MDC l\u00e0: 25\uf0b460 : 2 \uf03d 750 (cm2). Di\u1ec7n t\u00edch tam gi\u00e1c GBC l\u00e0: 30\uf0b440 : 2 \uf03d 600 (cm2). Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ABCD l\u00e0: 60\uf0b440 \uf03d 2400 (cm2). Di\u1ec7n t\u00edch tam gi\u00e1c MGC l\u00e0: 2400\uf02d\uf028225 \uf02b 750 \uf02b 600\uf029 \uf03d 825 (cm2). b) Ta c\u00f3: S\uf044MDC \uf03d 750 \uf03d 10 . S\uf044MGC 825 11 M\u00e0 hai tam gi\u00e1c n\u00e0y c\u00f3 chung \u0111\u00e1y MC \u21d2 chi\u1ec1u cao k\u1ebb t\u1eeb \u0111\u1ec9nh \u00a0D xu\u1ed1ng \u0111\u00e1y MC = 10 chi\u1ec1u cao 11 k\u1ebb t\u1eeb \u0111\u1ec9nh \u00a0G xu\u1ed1ng \u0111\u00e1y MC . M\u00e0 chi\u1ec1u cao k\u1ebb \u0111\u1ec9nh D xu\u1ed1ng \u0111\u00e1y MC v\u00e0 chi\u1ec1u cao k\u1ebb t\u1eeb \u0111\u1ec9nh G xu\u1ed1ng \u0111\u00e1y MC c\u0169ng ch\u00ednh l\u00e0 chi\u1ec1u cao c\u1ee7a tam gi\u00e1c MDK v\u00e0 GMK . M\u1eb7t kh\u00e1c hai Tam gi\u00e1c n\u00e0y c\u00f3 chung \u0111\u00e1y MK \uf0de S\uf044MDK \uf03d 10 . S\uf044MGK 11 M\u00e0 t\u1ed5ng di\u1ec7n t\u00edch hai tam gi\u00e1c b\u1eb1ng di\u1ec7n t\u00edch tam gi\u00e1c MGD v\u00e0 b\u1eb1ng: 25\uf0b430 : 2 \uf03d 375 (cm2). Ta c\u00f3 s\u01a1 \u0111\u1ed3: Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com ( )Di\u1ec7n t\u00edch tam gi\u00e1c MDK l\u00e0: 375 : (10 +11) x 10 =\u00a03750 cm2 11 \u0110\u00e1p s\u1ed1: a) 825 cm2 b) 3750 (cm2 ) 11 B\u00e0i 88: Trong h\u00ecnh v\u1ebd sau cho tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng 100cm2. Tr\u00ean AC l\u1ea5y F sao cho AF \uf03d\u00a01 AC . L\u1ea5y G l\u00e0 trung \u0111i\u1ec3m BF . N\u1ed1i AG c\u1eaft BC t\u1ea1i E . 3 a) T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c MGC . b) T\u00ednh t\u1ec9 s\u1ed1 BE . BC a) Ta c\u00f3: S tam gi\u00e1c ABF \uf03d\u00a01 S tam gi\u00e1c ABC (v\u00ec 2 tam gi\u00e1c c\u00f3 chung chi\u1ec1u cao k\u1ebb t\u1eeb \u0111\u1ec9nh B 3 xu\u1ed1ng \u0111\u00e1y AC , \u0111\u00e1y AF \uf03d\u00a01 AC ). 3 V\u1eady S tam gi\u00e1c ABF l\u00e0: 100x 1\u00a0\uf03d 100 (cm2). 33 b) Ta c\u00f3: SAGF = SAGB (hai tam gi\u00e1c c\u00f3 chung chi\u1ec1u cao k\u1ebb t\u1eeb \u0111\u1ec9nh A xu\u1ed1ng c\u1ea1nh BF , \u0111\u00e1y BG = GF ). M\u00e0: S AGF = 1 S AGC (hai tam gi\u00e1c c\u00f3 chung chi\u1ec1u cao k\u1ebb t\u1eeb \u0111\u1ec9nh G xu\u1ed1ng c\u1ea1nh AC , \u0111\u00e1y 3 AF =\u00a01 AC )\u21d2 S AGB =13 SAGC . 3 M\u1eb7t kh\u00e1c hai tam gi\u00e1c n\u00e0y c\u00f3 chung \u0111\u00e1y AG \u21d2 Chi\u1ec1u cao k\u1ebb t\u1eeb \u0111\u1ec9nh B xu\u1ed1ng \u0111\u00e1y AG =\u00a01 3 chi\u1ec1u cao k\u1ebb t\u1eeb \u0111\u1ec9nh C xu\u1ed1ng \u0111\u00e1y AG . M\u00e0 chi\u1ec1u cao k\u1ebb t\u1eeb \u0111\u1ec9nh B xu\u1ed1ng \u0111\u00e1y AG v\u00e0 chi\u1ec1u cao k\u1ebb t\u1eeb \u0111\u1ec9nh C xu\u1ed1ng \u0111\u00e1y AG l\u00e0 chi\u1ec1u cao c\u1ee7a hai tam gi\u00e1c BGE v\u00e0 CGE . H\u01a1n n\u1eefa hai tam gi\u00e1c BGE v\u00e0 CGE c\u00f3 chung \u0111\u00e1y EG \u21d2 SBEG =13 SCGE hay SBGE = 1 SBGC . 4 M\u00e0 hai tam gi\u00e1c n\u00e0y l\u1ea1i c\u00f3 chung chi\u1ec1u cao k\u1ebb t\u1eeb \u0111\u1ec9nh G xu\u1ed1ng c\u1ea1nh BC \u21d2 BE =1 BC hay 4 BE \uf03d 1 . BC 4 Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com B\u00e0i 89. Cho t\u1ee9 gi\u00e1c ABCD nh\u01b0 h\u00ecnh v\u1ebd c\u00f3 M , N, P,Q l\u1ea7n l\u01b0\u1ee3t n\u1eb1m tr\u00ean AB, BC,CD, DA sao cho: =MA M=B; NB N=C; PC P=D;QA QD . a) N\u1ebfu di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c ABCD = 90cm2 th\u00ec di\u1ec7n t\u00edch c\u1ee7a t\u1ee9 gi\u00e1c MNPA b\u1eb1ng bao nhi\u00eau? b) N\u1ed1i CM , AP, BQ, DN v\u00e0 t\u00f4 m\u00e0u nh\u01b0 h\u00ecnh v\u1ebd. H\u00e3y ch\u1ee9ng minh r\u1eb1ng di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c t\u00f4 m\u00e0u \u0111\u1ecf b\u1eb1ng t\u1ed5ng di\u1ec7n t\u00edch 4 tam gi\u00e1c t\u00f4 xanh. L\u1eddi gi\u1ea3i C\u00e2u a ph\u1ea3i l\u00e0 MNPQ th\u00ec chu\u1ea9n. B\u00e0i 90. Cho h\u00ecnh ch\u1eef nh\u1eadt ABCD c\u00f3 c\u1ea1nh AB d\u00e0i 36 cm, c\u1ea1nh AD d\u00e0i 18 cm. G\u1ecdi M l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa BC , N l\u00e0 \u0111i\u1ec3m tr\u00ean c\u1ea1nh CD sao cho DN g\u1ea5p 2 l\u1ea7n CN . a) T\u00ednh di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c AMCN . b) T\u00ecm \u0111i\u1ec3m E tr\u00ean c\u1ea1nh CD \u0111\u1ec3 di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c AMCN b\u1eb1ng m\u1ed9t n\u1eeda di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ABCD ? L\u1eddi gi\u1ea3i Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com a. H\u00ecnh ch\u1eef nh\u1eadt ABCD c\u00f3: A=B C=D 36cm n\u00ean CN =1 DC =1 \u00d7 36 =12cm 33 A=D C=B 18cm n\u00ean MB =1 BC =1 \u00d718 =9cm 33 Ta c\u00f3: SAMCN = SANC + SAMC S ANC = 1 \u00d7 NC \u00d7 AD = 1 \u00d712\u00d718 =108cm2 2 2 S AMB = 1 \u00d7 MC \u00d7 AB = 1 \u00d7 9\u00d7 36 = 162cm2 2 2 V\u1eady di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c AMNC l\u00e0: ( )SAMCN = SANC + SAMC =108 + 162 =270 cm2 b. Ta c\u00f3: SAECN = SAEC + SAMC B\u00e0i 91. \u0110\u1ec3 S AECN = 1 S ABCD th\u00ec SADE + SABM = S AEC + S AMC S=ABM S AEC + S AMC 2 Hay 1 \u00d718\u00d7 DE +162 = 1 \u00d718\u00d7 EC +162 22 Hay ED = EC Hay E l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa D v\u00e0 C . Cho tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch l\u00e0 160 cm2. G\u1ecdi M , N theo th\u1ee9 t\u1ef1 l\u00e0 \u0111i\u1ec3m thu\u1ed9c c\u00e1c c\u1ea1nh AB, AC sao =cho AM 1=AB, AN 1 AC 44 a) T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c ABN t\u1eeb \u0111\u00f3 suy ra di\u1ec7n t\u00edch tam gi\u00e1c AMN . b) MN c\u00f3 song song v\u1edbi BC kh\u00f4ng? T\u1ea1i sao? L\u1eddi gi\u1ea3i Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com a. Di\u1ec7n t\u00edch tam gi\u00e1c ABC v\u00e0 ABN AN= 1 \u00d7 AC 4 Chung chi\u1ec1u cao h\u1ea1 t\u1eeb B xu\u1ed1ng AC N\u00ean S ABN =1 \u00d7 S ABC =1 \u00d7160 =40(cm2 ) 4 4 Di\u1ec7n t\u00edch tam gi\u00e1c AMN v\u00e0 ABN c\u00f3: AM= 1 \u00d7 AB 4 Chung chi\u1ec1u cao h\u1ea1 t\u1eeb N xu\u1ed1ng AB N\u00ean SAMN =1 \u00d7 S ABN =1 \u00d7 40 =10(cm2 ) 4 4 b. Tam gi\u00e1c ABC c\u00f3: AM= 1 \u00d7 AB 4 AN= 1 \u00d7 AC 4 ( )N\u00ean S=ABN S=ACM \u00a040 cm2 SABN \u00a0\u2212SAMN =SACM \u2212 SAMN N\u00ean SABN \u00a0= SACM M\u00e0 di\u1ec7n t\u00edch tam gi\u00e1c BMN v\u00e0 CMN c\u00f3 chung c\u1ea1nh \u0111\u00e1y MN n\u00ean chi\u1ec1u cao h\u1ea1 t\u1eeb B xu\u1ed1ng MN b\u1eb1ng chi\u1ec1u cao h\u1ea1 t\u1eeb C xu\u1ed1ng MN . B\u00e0i 92. V\u1eady MN song song v\u1edbi BC . M\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chi\u1ec1u d\u00e0i g\u1ea5p 3 chi\u1ec1u r\u1ed9ng. N\u1ebfu t\u0103ng chi\u1ec1u r\u1ed9ng th\u00eam 3 dm v\u00e0 gi\u1ea3m chi\u1ec1u d\u00e0i \u0111i 3 dm th\u00ec di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt m\u1edbi t\u0103ng th\u00eam 153dm2 so v\u1edbi h\u00ecnh l\u00fac \u0111\u1ea7u. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt l\u00fac \u0111\u1ea7u ? L\u1eddi gi\u1ea3i Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com G\u1ecdi chi\u1ec1u r\u1ed9ng l\u00e0: a Di\u1ec7n t\u00edch ph\u1ea7n m\u1ea5t \u0111i (Ph\u1ea7n m\u00e0u xanh) l\u00e0 : 3\u00d7 a Di\u1ec7n t\u00edch ph\u1ea7n th\u00eam (Ph\u1ea7n m\u00e0u \u0111\u1ecf) l\u00e0: (3\u00d7 a)\u00d7 2 + (3\u00d7 a \u2212 9) = 9\u00d7 a \u2212 9 Di\u1ec7n t\u00edch t\u0103ng l\u00e0: (9\u00d7 a \u2212 9) \u2212 (3\u00d7 a) =153 (9\u00d7 a \u2212 3\u00d7 a) \u2212 9 =153 6\u00d7 a= 153 + 9 6\u00d7 a =162 a = 162 : 6 a=7 V\u1eady chi\u1ec1u d\u00e0i ban \u0111\u1ea7u l\u00e0: 27 \u00d7 3 =81(dm) Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u : 27 \u00d781 =2187(dm2 ) ( )\u0110\u00e1p s\u1ed1: 2187 dm2 B\u1ed5 sung c\u00e1ch 2 cho h\u1ecdc sinh. Theo b\u00e0i ra ta c\u00f3 : Chi\u1ec1u r\u1ed9ng = 1\/3 chi\u1ec1u d\u00e0i (1). Di\u1ec7n t\u00edch t\u0103ng th\u00eam s\u1ebd l\u00e0 : (CD-3) x 3 \u2013 (CRx3) = 153(dm2) 3 x CD \u2013 9 \u2013 CR x 3 = 153(dm2) 3 x CD \u2013 CR x 3 = 153 + 9 3 x (CD \u2013 CR) = 162 CD \u2013 CR = 162 : 3 CD \u2013 CR = 54 (2) T\u1eeb (1) v\u00e0 (2) ta \u0111\u01b0a b\u00e0i to\u00e1n v\u1ec1 d\u1ea1ng t\u00ecm 2 s\u1ed1 khi bi\u1ebft hi\u1ec7u v\u00e0 t\u1ec9 s\u1ed1 v\u00e0 ta c\u00f3: Chi\u1ec1u r\u1ed9ng ban \u0111\u1ea7u : 54 : (3 -1) x 1 = 27( cm) Chi\u1ec1u d\u00e0i ban \u0111\u1ea7u : 54 : (3 \u2013 1) x3 = 81(cm) Di\u1ec7n t\u00edch ban \u0111\u1ea7u : 27 x 81 = 2187 (dm2) B\u00e0i 93. Cho h\u00ecnh ch\u1eef nh\u1eadt ABCD c\u00f3 chu vi 100 cm ( AB > BC ) . L\u1ea5y \u0111i\u1ec3m M tr\u00ean c\u1ea1nh AB , \u0111i\u1ec3m N tr\u00ean c\u1ea1nh CD sao cho AMND l\u00e0 h\u00ecnh vu\u00f4ng c\u00f2n BMNC l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chu vi 60 cm a) T\u00ecm \u0111\u1ed9 d\u00e0i c\u00e1c c\u1ea1nh c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt ABCD . b) T\u00ednh di\u1ec7n t\u00edch c\u00e1c tam gi\u00e1c DMC L\u1eddi gi\u1ea3i Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com B\u00e0i 94. a. V\u00ec AMND l\u00e0 h\u00ecnh vu\u00f4ng c\u00f3 c\u1ea1nh b\u1eb1ng chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt ABCD n\u00ean chu vi h\u00ecnh ch\u1eef nh\u1eadt BMNC ch\u00ednh b\u1eb1ng 2 l\u1ea7n chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt ABCD . Chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef Nh\u1eadt ABCD l\u00e0 60 : 2 = 30(cm) Chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef Nh\u1eadt ABCD l\u00e0 100 : 2 \u2212 30 =20(cm2 ) b. Di\u1ec7n t\u00edch tam gi\u00e1c DMC l\u00e0 20\u00d7 30 : 2 =300(cm2 ) \u0110\u00e1p s\u1ed1: a. Chi\u1ec1u d\u00e0i: 30 cm; chi\u1ec1u r\u1ed9ng: 20 cm b. Di\u1ec7n t\u00edch 300 cm2 Cho tam gi\u00e1c ABC c\u00f3 M l\u00e0 \u0111i\u1ec3m n\u1eb1m ch\u00ednh gi\u1eefa c\u1ee7a BC , N l\u00e0 \u0111i\u1ec3m n\u1eb1m ch\u00ednh gi\u1eefa AC . Hai \u0111o\u1ea1n AM v\u00e0 BN c\u1eaft nhau t\u1ea1i O . Bi\u00eat r\u0103ng BO = 2 BN . T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c ABC 3 bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c MON l\u00e0 20 cm2 L\u1eddi gi\u1ea3i V\u00ec BO = 2 BN n\u00ean ON = 1 BN 33 Ta c\u00f3: SOMN = 1 SBMN (V\u00ec hai tam gi\u00e1c c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb \u0111\u1ec9nh M xu\u1ed1ng BN v\u00e0 c\u00f3 3 ON = 1 BN ) 3 ( )N\u00ean SBNM = 20\u00d7 3 = 60 cm2 Ta c\u00f3: SBMN = 1 SBNC (V\u00ec hai tam gi\u00e1c c\u00f3 chung \u0111\u1ec9nh, chi\u1ec1u cao h\u1ea1 t\u1eeb \u0111\u1ec9nh N xu\u1ed1ng \u0111\u00e1y BC 2 v\u00e0 BM = MC ) Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com N\u00ean SBNC = 60 \u00d7 2 = 120(cm2 ) M\u00e0 : SBNC = 1 (V\u00ec hai tam gi\u00e1c c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb \u0111\u1ec9nh B xu\u1ed1ng \u0111\u00e1y AC v\u00e0 c\u00f3 2 S ABC \u0111\u00e1y AN = NC ) ( )V\u1eady SABC= 120\u00d7 2= 240 cm2 ( )\u0110\u00e1p s\u1ed1: SABC = 240 cm2 B\u00e0i 95. Cho tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch 217,5 cm2. Tr\u00ean c\u1ea1nh AB l\u1ea5y \u0111i\u1ec3m D , tr\u00ean c\u1ea1nh BC l\u1ea5y \u0111i\u1ec3m E , bi\u1ebft AD = 8cm , BE = 1 BC v\u00e0 di\u1ec7n t\u00edch tam gi\u00e1c BDE l\u00e0 14,55cm2 . T\u00ednh \u0111\u1ed9 d\u00e0i 3 AB . L\u1eddi gi\u1ea3i N\u1ed1i D v\u1edbi C ta c\u00f3: SBDE = 1 SDBC (V\u00ec hai tam gi\u00e1c c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb \u0111\u1ec9nh D xu\u1ed1ng BC v\u00e0 \u0111\u00e1y 3 BE = 1 BC ) 3 ( )SD=BC 14,55\u00d7=3 43, 65 cm2 SADC = SABC \u2212 SDBC = 217, 5 \u2212 43, 65 =173,85(cm2 ) Chi\u1ec1u cao tam gi\u00e1c ADC l\u00e0: 173,85\u00d7 2 : 8 =43, 4265(cm2 ) Chi\u1ec1u cao tam gi\u00e1c ADC c\u0169ng ch\u00ednh l\u00e0 chi\u1ec1u cao c\u1ee7a tam gi\u00e1c ABC n\u00ean: \u0110\u1ed9 d\u00e0i \u0111\u00e1y AB l\u00e0: 217,5\u00d7 2 : 43, 4625 =10, 0086(cm) \u0110\u00e1p s\u1ed1: 10, 0086 cm B\u00e0i 96. Cho h\u00ecnh v\u1ebd b\u00ean. B=D D=E EC , =AI I=D; AK KE , Di\u1ec7n t\u00edch tam gi\u00e1c AIK l\u00e0 20cm2 . T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c ABC . L\u1eddi gi\u1ea3i Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com X\u00e9t \u2206AEI v\u00e0 \u2206KEI chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb \u0111\u1ec9nh I xu\u1ed1ng AE V\u00e0 AK = KE n\u00ean SAIK = SIKE X\u00e9t \u2206AEI v\u00e0 \u2206IDE chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb \u0111\u1ec9nh E xu\u1ed1ng AD V\u00e0 IA = ID n\u00ean SAIE = SIDE M\u00e0 SAED = SAEI + SIED = 2SAEI = 2.2SAIK = 4.20 = 80(cm2 ) X\u00e9t \u2206AED v\u00e0 \u2206ABC chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb \u0111\u1ec9nh A xu\u1ed1ng BC V\u00e0 DE = 1 BC n\u00ean S AED = 1 S ABC 3 3 Suy ra S=ABC 3S=AED 3=.80 240(cm2 ) B\u00e0i 97. Cho h\u00ecnh v\u1ebd b\u00ean. Bi\u1ebft A l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa BM , B l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa CN , C l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa PA , di\u1ec7n t\u00edch tam gi\u00e1c ABC l\u00e0 97cm2 . T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c MNP . L\u1eddi gi\u1ea3i B\u00e0i 98. V\u00ec A l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a MB n\u00ean SAMC = SABC V\u00ec A l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a MB n\u00ean SABN = SAMN V\u00ec C l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a AP n\u00ean SAMC = SMPC V\u00ec B l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a CN n\u00ean SPCB = SPBN Suy ra S=ABC S=ACM S=MCP S=PCB S=PBN S=ABN SMAN Ta l\u1ea1i c\u00f3 SABC + SACM + SMCP + SPCB + SPBN + SABN + SMAN =SMNP N\u00ean S=MNP 7.S=ABC 7=.97 679(cm2 ) . Cho tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng 420cm2 . \u0110i\u1ec3m N l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ea1nh AC . Tr\u00ean c\u1ea1nh AB l\u1ea5y \u0111i\u1ec3m P sao cho AP = 3PB . \u0110\u01b0\u1eddng th\u1eb3ng BN v\u00e0 CP c\u1eaft nhau t\u1ea1i K . T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c BKC . L\u1eddi gi\u1ea3i Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com X\u00e9t N l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a AC \u21d2 AN = NC = AC 2 Suy ra S=ABN S=BNC 1 S=ABC 210(cm2 ) 2 \u21d2 SNKC + SBKC =210 Ta l\u1ea1i c\u00f3 =AP 3PB \u21d2 \uf8f1 SPAC = 3SPBC \uf8f2 SKAP = 3SKBP \uf8f3 ( )Suy ra SPAC \u2212 SKAP = 3SBKC \u21d2 SBKC = 1 3. SPBC \u2212 SKBP \u21d2 SAKC = 3 S AKC V\u00ec N l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a AC n\u00ean S=AKN S=NKC 1 S AKC 2 \u21d2 SNKC + SBKC = 1 S AKC + 1 S AKC 2 3 \u21d2 210= 5 S AKC \u21d2 S AKC= 210.6= 252 6 5 \u21d2 SBKC =1 S AKC = 252 =84(cm2 ) 3 3 B\u00e0i 99. Cho tam gi\u00e1c ABC c\u00f3 \u0111i\u1ec3m N l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ea1nh AC , trong \u0111\u00f3 c\u00f3 h\u00ecnh thang BMNE nh\u01b0 h\u00ecnh v\u1ebd. N\u1ed1i B v\u1edbi N , n\u1ed1i E v\u1edbi M . Hai \u0111o\u1ea1n th\u1eb3ng n\u00e0y g\u1eb7p nhau \u1edf \u0111i\u1ec3m O . a. So s\u00e1nh di\u1ec7n t\u00edch hai tam gi\u00e1c OBM v\u00e0 ONE b. So s\u00e1nh di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c EMC v\u1edbi di\u1ec7n t\u00edch h\u00ecnh AEMB . L\u1eddi gi\u1ea3i a. Trong h\u00ecnh thang BENM , c\u00e1c \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb M v\u00e0 N xu\u1ed1ng \u0111\u00e1y l\u1edbn BE l\u00e0 nh\u01b0 nhau Do \u0111\u00f3 SEBM = SEBN (chung \u0111\u00e1y BC v\u00e0 hai \u0111\u01b0\u1eddng cao b\u1eb1ng nhau) M\u00e0 \uf8f1\uf8f2\uf8f3SS==EEBBMN SEON + SEOB \u21d2 SEON =SBOM SOBM + SEOB b. V\u00ec N l\u00e0 trung \u0111i\u1ec3m c\u1ea1nh AC n\u00ean Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com ( )SABN = SBCN 1 M\u00e0 SEM=C SOMCN + S=1 SOMCN + S2 (2) S =AEMB SABN + S2 \u2212 S1= SABN (3) T\u1eeb (1), (2) v\u00e0 (3) suy ra SAEMB = SEMC . B\u00e0i 100. Cho tam gi\u00e1c ABC c\u00f3 \u0111i\u1ec3m D l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ea1nh AC , \u0111i\u1ec3m E l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ea1nh AB . Hai \u0111o\u1ea1n th\u1eb3ng BD v\u00e0 CE c\u1eaft nhau t\u1ea1i G (nh\u01b0 h\u00ecnh v\u1ebd) a. So s\u00e1nh di\u1ec7n t\u00edch hai tam gi\u00e1c GEB v\u00e0 GCD b. So s\u00e1nh di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c GAB,GBC,GCA . c. K\u00e9o d\u00e0i AG c\u1eaft BC \u1edf \u0111i\u1ec3m M . So s\u00e1nh hai \u0111o\u1ea1n th\u1eb3ng MB v\u00e0 MC L\u1eddi gi\u1ea3i a. Hai tam gi\u00e1c BDA v\u00e0 BDC c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb B v\u00e0 \u0111\u00e1y DA = DC . Do v\u1eady SBDA = SBDC Suy ra SSBD = 1 2 S ABC T\u01b0\u01a1ng t\u1ef1 ta c\u00f3 EA = EB . V\u1eady S EAC = SEBC Suy ra SBEC = 1 S ABC 2 V\u1eady SCBD = SBEC M\u00e0 S=CBD SCGB + SCDG S=CBE SCGB + SEBG N\u00ean SCDG = SEGB Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com b. Ta c\u00f3 EA = EB , v\u1eady S1 = S2 (hai tam gi\u00e1c c\u00f3 c\u00f9ng \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb G ) T\u01b0\u01a1ng t\u1ef1 DA = DC , suy ra S3 = S4 Theo c\u00e2u a th\u00ec S1 = S4 V\u1eady S=1 S=2 S=3 S4 Suy ra S1 + S2 = S3 + S4 V\u1eady SGAB = SGAC C\u0169ng theo c\u00e2u a ta c\u00f3 SABD = SCBD v\u00e0 S3 = S4 n\u00ean SABD \u2212 S3= SCBD \u2212 S4 Hay SGAB = SGBC Hay S=GAB S=GBC SGAC c. Theo c\u00e2u b ta c\u00f3 SGAB = SGAC Hai tam gi\u00e1c SGAB ; SGAC c\u00f3 chung \u0111\u00e1y GA n\u00ean c\u00e1c \u0111\u01b0\u1eddng cao BH ,CK b\u1eb1ng nhau Suy ra SBGM = SCGM (Hai tam gi\u00e1c c\u00f3 chung \u0111\u00e1y GM ) Song hai tam gi\u00e1c n\u00e0y l\u1ea1i c\u00f3 chung \u0111\u01b0\u1eddng cao t\u1eeb G n\u00ean BM = CM B\u00e0i 101. Cho h\u00ecnh v\u1ebd ABCD l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt, AB = 4cm . C\u00e1c \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m A v\u00e0 t\u00e2m D c\u00f9ng b\u00e1n k\u00ednh r = AB c\u1eaft c\u1ea1nh AD t\u1ea1i G v\u00e0 E . Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com a. So s\u00e1nh di\u1ec7n t\u00edch h\u00ecnh 1 v\u00e0 h\u00ecnh 2 n\u1ebfu di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt b\u1eb1ng n\u1eeda di\u1ec7n t\u00edch h\u00ecnh tr\u00f2n t\u00e2m A b\u00e1n k\u00ednh r . b. T\u00ednh \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng EG . L\u1eddi gi\u1ea3i a. Ta c\u00f3 SABCD= AB.AD= 4 \u00d7 AD Di\u1ec7n t\u00edch h\u00ecnh tr\u00f2n t\u00e2m A b\u00e1n k\u00ednh=r A=B 4cm l\u00e0 S =\u03c0 \u00d7 r2 =16\u03c0 M\u00e0 S ABCD = 1S \u21d2 4\u00d7 AD = 1 \u00d716\u00d7\u03c0 2 2 Suy ra 4\u00d7 AD =8\u03c0 \u21d2 AD =2\u03c0 Ta c\u00f3 S ABCD = S1 + S2 + S3 + S4 = 1S (1) 2 \uf8f1 S2 + S3 =1 S =1 S (2) \uf8f4\uf8f4 + S4 4 2 M\u00e0 \uf8f2 \u21d2 S2 + S3 + S2 + S4 =1 S \uf8f3\uf8f4\uf8f4S2 4 T\u1eeb (1)&(2) S2 + S3 + S2 + S4 = S1 + S3 + S2 + S4 \u21d2 S1 = S2 b. Ta c\u00f3 AD= 2r \u2212 EG Hay 2\u03c0 = 2\u00d7 4 \u2212 EG \u21d2 EG = 8 \u2212 2\u03c0 Suy ra EG = 8 \u2212 2\u00d7 3,14 = 1, 72 B\u00e0i 102. Cho h\u00ecnh thang vu\u00f4ng ABCD c\u00f3 g\u00f3c A v\u00e0 D vu\u00f4ng. V\u1ebd \u0111\u01b0\u1eddng cao BH . AC c\u1eaft BH t\u1ea1i G . H\u00e3y so s\u00e1nh di\u1ec7n t\u00edch tam gi\u00e1c DGH v\u00e0 di\u1ec7n t\u00edch tam gi\u00e1c GBC . Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com L\u1eddi gi\u1ea3i N\u1ed1i A v\u1edbi H ta c\u00f3 hai tam gi\u00e1c AHC, BHC c\u00f3 chung c\u1ea1nh \u0111\u00e1y HC . M\u00e0 hai \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb A v\u00e0 B xu\u1ed1ng HC b\u1eb1ng nhau (c\u00f9ng b\u1eb1ng \u0111\u01b0\u1eddng cao c\u1ee7a h\u00ecnh thang ABCD ) V\u00ec v\u1eady SAHC = SBHC Do \u0111\u00f3 SAGH = SBGC (c\u00f9ng b\u1edbt ph\u1ea7n GHC )(1) M\u00e0 hai tam gi\u00e1c AGH v\u00e0 tam gi\u00e1c HDG c\u00f3 chung nhau \u0111\u00e1y GH v\u00e0 hai \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb A v\u00e0 D xu\u1ed1ng GH b\u1eb1ng nhau ( )Do \u0111\u00f3 SGHD = SAGH 2 T\u1eeb (1) v\u00e0 (2) suy ra SHDG = SBGC B\u00e0i 103. Tr\u00ean m\u1ed9t h\u00ecnh vu\u00f4ng trang tr\u00ed m\u1ed9t h\u00ecnh hoa b\u1ed1n c\u00e1nh l\u00e0 b\u1ed1n tam gi\u00e1c b\u1eb1ng nhau (H\u00ecnh v\u1ebd). Cho bi\u1ebft hi\u1ec7u s\u1ed1 \u0111o hai c\u1ea1nh g\u00f3c vu\u00f4ng OB v\u00e0 OI l\u00e0 7cm , t\u1ed5ng di\u1ec7n t\u00edch ph\u1ea7n c\u00f2n l\u1ea1i c\u1ee7a h\u00ecnh vu\u00f4ng ( Ph\u1ea7n g\u1ea1ch ch\u00e9o) l\u00e0 140cm2 . T\u00ednh di\u1ec7n t\u00edch h\u00ecnh hoa? L\u1eddi gi\u1ea3i Ta c\u00f3 OB \u2212 OI =7 (1) Ta c\u00f3 AB = OB \u00d7 2 \u21d2 SABCD = AB2 = 2OB2 =Shoa OB \u00d7 OI \u00d7 4 2 =Scheo S ABCD \u2212 Shoa Hay 140 =2OB2 \u2212 OB \u00d7 OI \u00d7 4 2 K\u1ebft h\u1ee3p v\u1edbi (1) Suy ra \uf8f1OB = 10 \u21d2 Shoa = 2OI \u00d7 OB = 60(cm2 ) \uf8f2 =3 \uf8f3 OI C\u00e1ch gi\u1ea3i tr\u00ean kh\u00f4ng ph\u00f9 h\u1ee3p v\u1edbi h\u1ecdc sinh Ti\u1ec3u h\u1ecdc,n\u00ean c\u00f3 th\u1ec3 gi\u1ea3i l\u1ea1i nh\u01b0 sau: Theo b\u00e0i ra ta c\u00f3: 4 c\u00e1nh hoa c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau. (1) Di\u1ec7n t\u00edch tam gi\u00e1c OAB = OBC = OCD = ODA ( \u0110\u1ec1u = \u00bc ABCD ) (2) Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com T\u1eeb (1) v\u00e0 (2) => Di\u1ec7n t\u00edch m\u1ed7i ph\u1ea7n g\u1ea1ch ch\u00e9o l\u00e0 140 : 4 = 35 Cm2 OA = OB (\u0110\u1ec1u b\u1eb1ng \u00bd \u0111\u01b0\u1eddng ch\u00e9o h\u00ecnh vu\u00f4ng) M\u00e0 OB \u2013 OI = 7 => OA \u2013OI = 7 => IA = 7. Tam gi\u00e1c IAB c\u00f3 \u0111\u00e1y l\u00e0 AI, => \u0111\u01b0\u1eddng cao OB = 35x2:7=10. \uf0f0 OI = 7-10 = 3. SIOB = 10x3:2 = 15Cm2 =>SHoa = 15x4 = 60 Cm2. B\u00e0i 104. M\u1ed9t s\u00e2n ch\u01a1i h\u00ecnh ch\u1eef nh\u1eadt ABCD chu vi 120m . ng\u01b0\u1eddi ta d\u1ef1 ki\u1ebfn m\u1edf r\u1ed9ng s\u00e2n ch\u01a1i \u0111\u00f3 theo s\u01a1 \u0111\u1ed3 \u1edf d\u01b0\u1edbi, th\u00e0nh h\u00ecnh ch\u1eef nh\u1eadt MPGI r\u1ed9ng h\u01a1n. T\u00ednh di\u1ec7n t\u00edch ph\u1ea7n m\u1edbi m\u1edf th\u00eam? L\u1eddi gi\u1ea3i Ta \u0111\u01b0a s\u00e2n b\u00f3ng h\u00ecnh ch\u1eef nh\u1eadt ABCD v\u1ec1 1 g\u1ed1c nh\u01b0 h\u00ecnh tr\u00ean, k\u00e9o d\u00e0i MP m\u1ed9t \u0111o\u1ea1n PE = BC , k\u00e9o d\u00e0i AK m\u1ed9t \u0111o\u1ea1n KH = BC , n\u1ed1i EH ta \u0111\u01b0\u1ee3c h\u00ecnh ch\u1eef nh\u1eadt PEHK = BCGK . V\u00ec c\u00f3 chi\u1ec1u r\u1ed9ng KH = KG v\u00e0 chi\u1ec1u d\u00e0i P=K C=G 10m . Di\u1ec7n t\u00edch ph\u1ea7n r\u1ed9ng b\u0103gng di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt AEMH . Chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt M=E 120 : 2 +1=0 20(m) Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt AEMH ( di\u1ec7n t\u00edch ph\u1ea7n m\u1edbi m\u1edf th\u00eam) l\u00e0: 70\u00d710 =700(m2 ) Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com \u0110\u00e1p s\u1ed1: 700(m2 ) B\u00e0i 105. Cho t\u1ee9 gi\u00e1c ABCD . G\u1ecdi M , N, P,Q l\u1ea7n l\u01b0\u01a1t l\u00e0 c\u00e1c \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a c\u00e1c c\u1ea1nh AB, BC,CD, DA ( H\u00ecnh v\u1ebd ). H\u00e3y so s\u00e1nh di\u1ec7n t\u00edch v\u1ee7a t\u1ee9 gi\u00e1c MNPQ v\u00e0 di\u1ec7n t\u00edch c\u1ee7a t\u1ee9 gi\u00e1c ABCD . L\u1eddi gi\u1ea3i N\u1ed1i QB ta c\u00f3: SAQM = SBQM (\u0110\u00e1y AM = BM v\u00e0 c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb Q ) . Suy ra S AQM = 1 S ABQ (1) 2 N\u1ed1i DB ta c\u00f3: SAQB = SBQD (C\u00f3 \u0111\u00e1y AQ = QD V\u00e0 c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb B ) . Suy ra S AQB = 1 S ABD (2). 2 T\u1eeb (1) V\u00e0 (2) suy ra: S AQM = 1 S ADB 4 N\u1ed1i AN v\u00e0 AC ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u00f3 SBMN = 1 S ABC . 4 N\u1ed1i DN ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u00f3 SCPN = 1 SBDC 4 N\u1ed1i CQ ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u00f3 SDPQ = 1 S ADC 4 Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com Ta c\u00f3: SAQM + SBMN + SCPN + SDQP= 1 S ABC + 1 S ABD + 1 SBCD + 1 S ADC 4 4 4 4 ( )SAQM + SBMN + SCPN + SDQP= 1 4 S ABC + S ABD + SBCD + S ADC =1 2S 4 ( )SAQM + SBMN + SCPN + SDQP ABCD S AQM + SBMN + SCPN + SDQP =12 S ABCD Suy ra SMNPQ = 1 2 S ABCD B\u00e0i 106. G\u1ecdi ABC l\u00e0 tam gi\u00e1c th\u1ee9 nh\u1ea5t. N\u1ed1i \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u00e1c c\u1ea1nh c\u1ee7a tam gi\u00e1c ABC ta \u0111\u01b0\u1ee3c tam gi\u00e1c th\u1ee9 2. N\u1ed1i \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u00e1c c\u1ea1nh c\u1ee7a tam gi\u00e1c th\u1ee9 hai ta \u0111\u01b0\u1ee3c tam gi\u00e1c th\u1ee9 3. v\u00e0 c\u1ee9 ti\u1ebfp t\u1ee5c v\u1ebd nh\u01b0 v\u1eady m\u00e3i.H\u1ecfi: a) C\u00f3 t\u1ea5t c\u1ea3 bao nhi\u00eau tam gi\u00e1c tr\u00ean h\u00ecnh khi ta v\u1ebd nh\u01b0 v\u1eady \u0111\u1ebfn tam gi\u00e1c th\u1ee9 10? b) Bi\u1ebft di\u1ec7n t\u00edch c\u1ee7a tam gi\u00e1c th\u1ee9 3 la 15 cm2. T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c th\u1ee9 nh\u00e2t? L\u1eddi gi\u1ea3i Khi n\u1ed1i \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u00e1c c\u1ea1nh c\u1ee7a tam gi\u00e1c ABC ta \u0111\u01b0\u1ee3c tam gi\u00e1c th\u1ee9 2. Ta c\u00f3 s\u1ed1 tam gi\u00e1c l\u00e0: 1+ 4\u00d71 =5 (h\u00ecnh) Khi n\u1ed1i \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u00e1c c\u1ea1nh c\u1ee7a tam gi\u00e1c th\u1ee9 hai ta \u0111\u01b0\u1ee3c tam gi\u00e1c th\u1ee9 3. Ta c\u00f3 s\u1ed1 tam gi\u00e1c l\u00e0: 1+ 4\u00d7 2 =9 ( h\u00ecnh) Khi n\u1ed1i \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u00e1c c\u1ea1nh c\u1ee7a tam gi\u00e1c th\u1ee9 10 ta \u0111\u01b0\u1ee3c tam gi\u00e1c th\u1ee9 11. Ta c\u00f3 s\u1ed1 tam gi\u00e1c l\u00e0: 1+ 4\u00d7 9 =37 ( h\u00ecnh) N\u1ed1i AP ta c\u00f3: S=BMP S=CNP 1 (C\u00f3 \u0111\u00e1y B=P C=P 1 BC v\u00e0 c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb 2 S APB 2 A ) (1) M\u1eb7t kh\u00e1c S APB = 1 S ABC (C\u00f3 \u0111\u00e1y BP = 1 BC v\u00e0 c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb A) (2). 2 2 N\u1ed1i BN Ta c\u00f3 SAMN = SCNP (C\u00f3 \u0111\u00e1y CN = AN CN = AN v\u00e0 c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb B ) (3) Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com T\u1eeb (1) , (2) v\u00e0 (3) suy ra: S=BMP S=CNP S=AMN 1 S ABC . 4 V\u1eady SMNP = 1 S ABC . 4 Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u00f3 di\u1ec7n t\u00edch tam gi\u00e1c th\u1ee9 3 b\u1eb1ng 1 SMNP . 4 Di\u1ec7n t\u00edch tam gi\u00e1c MNP l\u00e0: 15\u00d7 4 =60(cm2 ) B\u00e0i 107. Di\u1ec7n t\u00edch tam gi\u00e1c ABC l\u00e0: 60\u00d7 4 =240(cm2 ) \u0110\u00e1p s\u1ed1: : a. 37 h\u00ecnh; b. 240 cm2 Cho tam gi\u00e1c c\u00e2n ABC c\u1ea1nh AB b\u1eb1ng c\u1ea1nh AC . V\u1ebd \u0111\u01b0\u1eddng cao BH v\u00e0 CK . Tr\u00ean c\u1ea1nh AB l\u1ea5y \u0111i\u1ec3m M . Tr\u00ean AC k\u00e9o d\u00e0i v\u1ec1 ph\u00eda C l\u1ea5y \u0111i\u1ec3m N sao cho CN b\u1eb1ng BM . N\u1ed1i M v\u1edbi N , \u0111o\u1ea1n MN c\u1eaft \u0111\u00e1y BC t\u1ea1i I (h\u00ecnh v\u1ebd). a) So s\u00e1nh \u0111\u1ed9 d\u00e0i hai \u0111o\u1ea1n BH v\u00e0 CK ? b) So s\u00e1nh di\u1ec7n t\u00edch tam gi\u00e1c MIC v\u00e0 di\u1ec7n t\u00edch tam gi\u00e1c NIC ? c) So s\u00e1nh \u0111\u1ed9 d\u00e0i hai \u0111o\u1ea1n IM v\u00e0 IN ? L\u1eddi gi\u1ea3i Tam gi\u00e1c ABC c\u00f3: SABC =AC \u00d7 BH : 2 =AB \u00d7 CK : 2 m\u00e0 AB = AC suy ra BH = CK N\u1ed1i BN v\u00e0 CM ta c\u00f3 SBNC = SBMC ( \u0111\u00e1y BM = CN v\u00e0 chi\u1ec1u cao BH = CK ) m\u00e0 hai tam gi\u00e1c n\u00e0y c\u00f3 chung \u0111\u00e1y BC n\u00ean chi\u1ec1u cao h\u1ea1 t\u1eeb N xu\u1ed1ng \u0111\u00e1y BC b\u1eb1ng chi\u1ec1u cao h\u1ea1 t\u1eeb M xu\u1ed1ng \u0111\u00e1y BC(NE = ME) Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com SBDI = SBNI ( c\u00f9ng \u0111\u00e1y BI v\u00e0 chi\u1ec1u cao NE = MF ) V\u1eady SMIC = SNIC Ta c\u00f3 SMIC = SNIC m\u00e0 hai tam gi\u00e1c n\u00e0y c\u00f3 chung chi\u1ec1u cao h\u1ea1 t\u1eeb C xu\u1ed1ng \u0111\u00e1y MN suy ra MI = NI B\u00e0i 108. Cho h\u00ecnh thang ABCD c\u00f3 hai \u0111\u00e1y l\u00e0 AB v\u00e0 CD . \u0110o\u1ea1n th\u1eb3ng AC c\u1eaft \u0111o\u1ea1n th\u1eb3ng BD t\u1ea1i O (h\u00ecnh v\u1ebd): a. So s\u00e1nh di\u1ec7n t\u00edch hai h\u00ecnh tam gi\u00e1c DAO v\u00e0 BCO . b. Bi\u1ebft di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c BAO b\u1eb1ng 1cm2 v\u00e0 di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c DCO b\u1eb1ng 4cm2 . T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang ABCD . c. T\u00ednh t\u1ec9 s\u1ed1 hai \u0111\u00e1y c\u1ee7a h\u00ecnh thang AB ? CD L\u1eddi gi\u1ea3i: a) Ta c\u00f3 SADC = SBDC (chung \u0111\u00e1y DC v\u00e0 chi\u1ec1u cao t\u1eeb \u0111\u1ec9nh A b\u1eb1ng chi\u1ec1u cao t\u1eeb \u0111\u1ec9nh B ) M\u00e0 S=ADC SODC + SDAO ; S=BDC SODC + SBCO n\u00ean SDAO = SBCO b) Ta c\u00f3: SDAO = DO (Hai tam gi\u00e1c SABO OB chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb \u0111\u1ec9nh A ) Ta l\u1ea1i c\u00f3: SODC = DO (Hai tam gi\u00e1c chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb \u0111\u1ec9nh C ) SBCO OB B\u1edfi v\u1eady S=DAO S=ODC DO S ABO SBCO OB Coi di\u1ec7n t\u00edch hai h\u00ecnh tam gi\u00e1c DAO v\u00e0 BCO l\u00e0 a th\u00ec ta c\u00f3: a = 4 hay a\u00d7 a =4 1a Do 4= 2\u00d7 2 n\u00ean S=DAO S=BCO 2 cm2 SABCD =SABO + SODC + SDAO + SBOC =1cm2 + 4cm2 + 2cm2 + 2cm2 =9cm2 c) T\u1eeb ph\u1ea7n b) ta c\u00f3: SABD =1cm2 + 2cm2 =3cm2 v\u00e0 SBCD = 2cm2 + 4cm2 = 6 cm2 M\u00e0 hai tam gi\u00e1c n\u00e0y c\u00f3 chi\u1ec1u cao h\u1ea1 xu\u1ed1ng c\u00e1c \u0111\u00e1y AB v\u00e0 DC l\u00e0 b\u1eb1ng nhau n\u00ean SABD = AB V\u1eady AB= 3= 1 . SBCD DC DC 6 2 Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","B\u00e0i 109. Website: tailieumontoan.com Cho tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A . Tr\u00ean c\u1ea1nh AB l\u1ea5y \u0111i\u1ec3m M sao cho BM = 1 AB . Tr\u00ean 3 c\u1ea1nh AC l\u1ea5y \u0111i\u1ec3m N sao cho AN = 1 AC . Tr\u00ean c\u1ea1nh BC l\u1ea5y \u0111i\u1ec3m E sao cho E l\u00e0 \u0111i\u1ec3m 4 ch\u00ednh gi\u1eefa c\u1ea1nh BC . a) Ch\u1ee9ng t\u1ecf r\u1eb1ng SMNCB = 5 S ABC 6 b) Ch\u1ee9ng t\u1ecf r\u1eb1ng SAMN = SEMB c) Bi\u1ebft SABC = 24cm2 . T\u00ednh SEMN L\u1eddi gi\u1ea3i: a) Ta c\u00f3 S ABN = 1 S ABC (V\u00ec AN = 1 AC v\u00e0 4 4 chung\u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb \u0111\u1ec9nh B) Ta l\u1ea1i c\u00f3 BM = 1 AB n\u00ean AM = 2 AB 33 N\u00ean S AMN = 2 S ABN (Chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb \u0111\u1ec9nh N) 3 V\u1eady S AMN =2 \u00d7 1 S ABC =16 SABC (1) 34 N\u00ean SMNCB =S ABC \u2212 SAMN =S ABC \u2212 1 S ABC = 5 S ABC 6 6 b) Ta c\u00f3 S ABE = 1 S ABC (Chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb \u0111\u1ec9nh 2 A v\u00e0 \u0111\u00e1y BE = 1 BC ) 2 SBME = 1 S ABE (Chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb \u0111\u1ec9nh E v\u00e0 \u0111\u00e1y BM =1 AB ) 3 3 N\u00ean SBME =1 \u00d7 1 S ABC =16 SABC (2) 23 T\u1eeb (1) v\u00e0 (2) ta c\u00f3 SAMN = SEMB c) T\u01b0\u01a1ng t\u1ef1 nh\u01b0 tr\u00ean ta c\u00f3 SENC = 3 S ABC 8 SEMN = S ABC \u2212 SBME \u2212 SAMN \u2212 SENC SEMN =S ABC \u2212 1 S ABC \u2212 1 S ABC \u2212 3 S ABC =7 S ABC 6 6 8 24 SEM=N 24 cm2 \u00d7 =7 7 cm2 24 B\u00e0i 110. Cho tam gi\u00e1c ABC . Tr\u00ean AB l\u1ea5y D v\u00e0 E sao cho A=D D=E EB . Tr\u00ean AC l\u1ea5y H v\u00e0 K sao cho A=H H=K KC . Tr\u00ean BC l\u1ea5y M v\u00e0 N sao cho B=M M=N NC . (h\u00ecnh v\u1ebd) Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com a) So s\u00e1nh di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c EBM v\u00e0 ADH . b) Bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c ABC b\u1eb1ng 360 cm2 . T\u00ednh di\u1ec7n t\u00edch h\u00ecnh DEMNKH . L\u1eddi gi\u1ea3i: a) Ta c\u00f3 S ABM = 1 S ABC (Chung \u0111\u01b0\u1eddng cao h\u1ea1 3 t\u1eeb \u0111\u1ec9nh M v\u00e0 \u0111\u00e1y BM = 1 BC ) 3 Ta l\u1ea1i c\u00f3 SEBM = 1 S ABM (Chung \u0111\u01b0\u1eddng cao 3 h\u1ea1 t\u1eeb \u0111\u1ec9nh M v\u00e0 \u0111\u00e1y BE = 1 AB ) 3 V\u1eady n\u00ean: SEBM =1 \u00d7 1 S ABC =19 SABC (1) 33 T\u01b0\u01a1ng t\u1ef1: Ta c\u00f3 S ABH = 1 S ABC (Chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb \u0111\u1ec9nh B v\u00e0 \u0111\u00e1y AH = 1 AC ) 3 3 Ta l\u1ea1i c\u00f3 S ADH = 1 S ABH (Chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb \u0111\u1ec9nh H v\u00e0 \u0111\u00e1y AD = 1 AB ) 3 3 V\u1eady n\u00ean: S ADH =1 \u00d7 1 S ABC =19 SABC (2) 33 T\u1eeb (1) v\u00e0 (2) ta c\u00f3 SADH = SEBM b) Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 nh\u01b0 v\u1eady ta c\u0169ng c\u00f3 S=CKN S=ADH S=EBM 1 S=ABC 360cm=2 : 9 40cm2 9 B\u00e0i 111. SDEMNKH =SABC \u2212 (SADH + SEBM + SCKN ) S =DEMNKH 360cm2 \u2212 40cm=2 \u00d73 240cm2 Cho h\u00ecnh ch\u1eef nh\u1eadt ABCD . I l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ea1nh AB . N\u1ed1i D v\u1edbi I , \u0111o\u1ea1n th\u1eb3ng DB c\u1eaft \u0111o\u1ea1n IC t\u1ea1i K (h\u00ecnh v\u1ebd). a) Ch\u1ee9ng t\u1ecf r\u1eb1ng SDIB = 1 SDBC 2 b) K\u1ebb IP vu\u00f4ng g\u00f3c v\u1edbi DB ; k\u1ebb CQ vu\u00f4ng g\u00f3c v\u1edbi DB. Ch\u1ee9ng t\u1ecf r\u1eb1ng SDIC = 3 SDIK c) Bi\u1ebft SDIK =8cm2 . T\u00ednh di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ABCD L\u1eddi gi\u1ea3i: Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com a) X\u00e9t hai tam gi\u00e1c ABD v\u00e0 DIB c\u00f3: - Chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb \u0111\u1ec9nh D . Q - \u0110\u00e1y IB = 1 AB P 2 N\u00ean SDIB = 1 S ABD 2 M\u00e0 S ABD = SCDB n\u00ean SDIB = 1 SDBC (1) 2 b) T\u1eeb (1) ta c\u00f3 IP = 1 CQ (\u0110\u01b0\u1eddng cao c\u1ee7a hai tam gi\u00e1c DIB v\u00e0 DBC c\u00f9ng h\u1ea1 xu\u1ed1ng \u0111\u00e1y BD) 2 X\u00e9t hai tam gi\u00e1c DIK v\u00e0 CKD c\u00f3: - Chung \u0111\u00e1y DK . - \u0110\u01b0\u1eddng cao IP = 1 CQ . 2 N\u00ean SDIK = 1 SCKD hay S=DIC SDIK \u00d7 3 (2) 2 c) T\u1eeb (2) ta c\u00f3 SDI=C 8cm2 \u00d7=3 24cm2 S =ABCD 24cm2 \u00d7=2 48 cm2 B\u00e0i 112. Cho tam gi\u00e1c ABC , M l\u00e0 m\u1ed9t \u0111i\u1ec3m tr\u00ean c\u1ea1nh AB sao cho BM = 1 BC . N\u1ed1i AM . K l\u00e0 m\u1ed9t 3 \u0111i\u1ec3m tr\u00ean \u0111o\u1ea1n th\u1eb3ng AM sao cho AK = 1 AM . N\u1ed1i BK,CK . 4 a) T\u00ednh t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch tam gi\u00e1c MKC v\u00e0 tam gi\u00e1c BKC . b) T\u00ednh t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch tam gi\u00e1c MKC v\u00e0 tam gi\u00e1c AKC . c) K\u00e9o d\u00e0i CK c\u1eaft AB t\u1ea1i H . T\u00ednh t\u1ec9 s\u1ed1 AH . BH L\u1eddi gi\u1ea3i a) T\u00ednh t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch tam gi\u00e1c MKC v\u00e0 tam gi\u00e1c BKC . Th\u1ea5y SMKC = 2 SBKC ( v\u00ec chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb K xu\u1ed1ng v\u00e0 \u0111\u00e1y MC = 2 BC ). (1) 3 3 (2) b) T\u00ednh t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch tam gi\u00e1c MKC v\u00e0 tam gi\u00e1c AKC . Th\u1ea5y SAKC = 1 (v\u00ec chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb C xu\u1ed1ng \u0111\u00e1y, m\u00e0 \u0111\u00e1y AK = 1 MK ). 3 SMKC 3 Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com c) K\u00e9o d\u00e0i CK c\u1eaft AB t\u1ea1i H . T\u00ednh t\u1ec9 s\u1ed1 AH . BH T\u1eeb (1) v\u00e0 (2), ta c\u00f3: S AKC = 2 SBMK 9 X\u00e9t tam gi\u00e1c AHC v\u00e0 tam gi\u00e1c BHC c\u00f3 chung \u0111\u00e1y HC , \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb A = 2 \u0111\u01b0\u1eddng cao 9 h\u1ea1 t\u1eeb B n\u00ean: S AHC = 2 SBHC . 9 Ta l\u1ea1i c\u00f3: Tam gi\u00e1c AHC v\u00e0 tam gi\u00e1c BHC c\u00f3 chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb C n\u00ean: AH = 2 . BH 9 B\u00e0i 113. M\u1ed9t m\u1ea3nh v\u01b0\u1eddn h\u00ecnh t\u1ee9 gi\u00e1c ABCD , ng\u01b0\u1eddi ta m\u1edf r\u1ed9ng v\u01b0\u1eddn v\u1ec1 c\u00e1c ph\u00eda b\u1eb1ng c\u00e1ch k\u00e9o d\u00e0i c\u1ea1nh AB (v\u1ec1 ph\u00eda B ), c\u1ea1nh BC (v\u1ec1 ph\u00eda C ), c\u1ea1nh CD (v\u1ec1 ph\u00eda D ), c\u1ea1nh DA (v\u1ec1 ph\u00eda A ) v\u00e0 tr\u00ean c\u00e1c \u0111\u01b0\u1eddng k\u00e9o d\u00e0i \u1ea5y l\u1ea7n l\u01b0\u1ee3t l\u1ea5y c\u00e1c \u0111i\u1ec3m M , N , P , Q sao cho: BM = AB , DP = CD CN = CB , DP = CD , AQ = DA . N\u1ed1i C v\u1edbi A , C v\u1edbi M (xem h\u00ecnh). a) Ch\u1ee9ng minh hai tam gi\u00e1c MBC v\u00e0 tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau. b) T\u00ednh t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch tam gi\u00e1c BMN v\u00e0 tam gi\u00e1c ABC . c) T\u00ednh di\u1ec7n t\u00edch m\u1ea3nh v\u01b0\u1eddn MNPQ bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c ABCD l\u00e0 50 m2 . L\u1eddi gi\u1ea3i a) Ch\u1ee9ng minh hai tam gi\u00e1c MBC v\u00e0 tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau. Th\u1ea5y SBMC = SABC (chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb C , \u0111\u00e1y AB = BM ). (1) b) T\u00ednh t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch tam gi\u00e1c BMN v\u00e0 tam gi\u00e1c ABC Th\u1ea5y SBMC = SMCN (chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb M , \u0111\u00e1y BC = CN ) (2) T\u1eeb (1) v\u00e0 (2), ta c\u00f3: SBMN = 2 \u00d7 SABC (3) c) T\u00ednh di\u1ec7n t\u00edch m\u1ea3nh v\u01b0\u1eddn MNPQ bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c ABCD l\u00e0 50m2 N\u1ed1i P v\u1edbi A Th\u1ea5y \uf8f4\uf8f1 S ADC = SADP V\u1eady SADP + SAPQ =2 \u00d7 SADC \u21d2 SPQD= 2 \u00d7 SADC (4) \uf8f2 S ADP = SAPQ \uf8f4\uf8f3 T\u01b0\u01a1ng t\u1ef1 : SAQM = 2 \u00d7 SABC (5) SPCN = 2 \u00d7 SBDC (6) T\u1eeb (1),(3), (4), (5), (6), ta c\u00f3: Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com SMNPQ = S ABCD + SAQM + SPDQ + SBMN + SPCN SMN=PQ SABCD + 2 \u00d7 SABD + 2 \u00d7 SADC + 2 \u00d7 SABC + 2 \u00d7 SBCD S =MNPQ SABCD + 2 \u00d7 SABCD + 2 \u00d7 SABCD S= 5 \u00d7 SABCD MNPQ SMNPQ = 5\u00d7 50 SMNPQ = 250m2 B\u00e0i 114. Cho tam gi\u00e1c ABC . D l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a c\u1ea1nh BC , E l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a c\u1ea1nh CA , AD c\u1eaft BE t\u1ea1i G . H\u00e3y ch\u1ee9ng t\u1ecf AG g\u1ea5p \u0111\u00f4i GD . L\u1eddi gi\u1ea3i N\u1ed1i C v\u1edbi G c\u1eaft AB t\u1ea1i F . Ta c\u00f3: S=ADC S=BCE 1 2 SABC . M\u00e0 SADC v\u00e0 SBCE c\u00f3 chung SECDG n\u00ean SAGE = SBDG (1) Th\u1ea5y SAGE = SGEC (chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb G , \u0111\u00e1y AE = EC ). (2) SBGD = SGDC (chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb G , \u0111\u00e1y DC = BD ). (3) T\u1eeb (1),(2),(3), ta c\u00f3: SBGD= 1 (S BGD + SGDC + SGDC ) 3 SBGD = 1 SBEC m\u00e0 SBEC = 1 SBGD = 1 3 2 SABC n\u00ean 6 SABC Th\u1ea5y S ABD = 1 S ABC n\u00ean SBDG = 1 S ABC . 2 3 X\u00e9t SBDG v\u00e0 SABD (c\u00f3 chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb B, \u0111\u00e1y DG = 1 DA ). 3 V\u1eady AG= 2\u00d7 GD. B\u00e0i 115. M\u1ed9t h\u00ecnh tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch 2010 m2 . Tr\u00ean c\u1ea1nh AB l\u1ea5y hai \u0111i\u1ec3m M , N sao cho A=M M=N NB . Tr\u00ean c\u1ea1nh AC l\u1ea5y \u0111i\u1ec3m P,Q sao cho A=P P=Q QC . T\u00ednh di\u1ec7n t\u00edch MNPQ . L\u1eddi gi\u1ea3i Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038","Website: tailieumontoan.com N\u1ed1i N v\u1edbi P ; P v\u1edbi B Th\u1ea5y S APM = 1 S ABP (chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb P , \u0111\u00e1y AM = 1 AB ) (1) 3 3 S ABP = 1 S ABC (chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb B , \u0111\u00e1y AP = 1 AC ) (2) 3 3 T\u1eeb (1) v\u00e0 (2), ta c\u00f3 : S AMP = 1 9 SABC Th\u1ea5y \uf8f4\uf8f1S AMP = SMPN \u21d2 S AMP = 1 S ANP \uf8f4\uf8f3\uf8f2S APN = SNPQ 2 Ta l\u1ea1i c\u00f3: S =MNPQ SMNP + SNPQ S =MNPQ S AMP + S APN S =MNPQ 1 S ABC + 2 \u00d7 SAMP 9 S=MNPQ 1 S ABC + 2 S ABC 9 9 SMNPQ = 1 S ABC 3 V\u1eady SMNPQ =1 \u00d7 2010 =670m2 3 B\u00e0i 116. Cho h\u00ecnh thang ABCD c\u00f3 hai \u0111\u00e1y l\u00e0 AB v\u00e0 CD . Bi\u1ebft AB = 15 cm, CD = 20 cm, chi\u1ec1u cao h\u00ecnh thang l\u00e0 14 cm. Hai \u0111\u01b0\u1eddng ch\u00e9o AC v\u00e0 BD c\u1eaft nhau \u1edf E . a) T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang ABCD . b) T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c CED . c) Ch\u1ee9ng minh hai tam gi\u00e1c AED v\u00e0 BEC c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau. L\u1eddi gi\u1ea3i a) T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang ABCD . Di\u1ec7n t\u00edch h\u00ecnh thang ABCD l\u00e0: \uf8f0\uf8ee(15 + 20)\u00d714\uf8f9\uf8fb : 2 =245 ( cm2 ). Li\u00ean h\u1ec7 t\u00e0i li\u1ec7u word m\u00f4n to\u00e1n zalo: 039.373.2038"]