["S\u0111\u1ed5i = (a-5) x (b+5) = (b+5) x (b+5) = b.b + 5b + 5b + 25 = b.b + 10b + 25 Hi\u1ec7u di\u1ec7n t\u00edch khi \u0111\u00e3 thay \u0111\u1ed5i v\u00e0 ban \u0111\u1ea7u: (b.b + 10b + 25) \u2013 (b.b + 10b) = 25 (m2) V\u1edbi m\u1ecdi a; b ta \u0111\u1ec1u c\u00f3 di\u1ec7n t\u00edch sau khi thay \u0111\u1ed5i s\u1ed1 \u0111o nh\u01b0 \u0111\u1ec1 b\u00e0i \u0111\u1ec1u l\u1edbn h\u01a1n 25 m2. (d\u00f9ng d\u1ea5u ch\u1ea5m(.) thay d\u1ea5u nh\u00e2n (x) cho d\u1ec5 nh\u00ecn m\u1ed9t ch\u00fat). B\u00e0i 42: M\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chu vi 60m. N\u1ebfu gi\u1ea3m chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt 5m v\u00e0 t\u0103ng chi\u1ec1u r\u1ed9ng l\u00ean 5m th\u00ec \u0111\u01b0\u1ee3c m\u1ed9t h\u00ecnh vu\u00f4ng. T\u00ecm di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt? N\u1eeda chu vi h\u00ecnh ch\u1eef nh\u1eadt l\u00e0: 60 : 2 = 30 (m) Chi\u1ec1u d\u00e0i h\u01a1n chi\u1ec1u r\u1ed9ng l\u00e0: 5 + 5 = 10 (m) Chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt l\u00e0: (30 \u2013 10) : 2 = 10 (m) Chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt l\u00e0: 30 \u2013 10 = 20 (m) Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt l\u00e0: 20 x 10 = 200 (m2) \u0110\u00e1p s\u1ed1: 200 m2. B\u00e0i 42: Cho tam gi\u00e1c ABC c\u00f3 BC = 9 cm. G\u1ecdi D l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ea1nh AC, k\u00e9o d\u00e0i c\u1ea1nh AB m\u1ed9t \u0111o\u1ea1n BE = AB. N\u1ed1i D v\u1edbi E, \u0111o\u1ea1n DE c\u1eaft \u0111o\u1ea1n BC t\u1ea1i G a-So s\u00e1nh di\u1ec7n t\u00edch c\u00e1c tam gi\u00e1c GBE, GBA, GAD, GDC b.T\u00ednh \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n BG a) N\u1ed1i CE. SGBE=SGBA. V\u00ec c\u00f3 AB=BE chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb G. SGAD=SGDC. V\u00ec c\u00f3 CD=DA chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb G. Ta c\u0169ng c\u00f3: SABC=SEBC => SGAC=SGEC (1) SDAE=SDCE => SGAE=SGEC (2) T\u1eeb (1) v\u00e0 (2) ta \u0111\u01b0\u1ee3c: SGAE=SGCA","V\u1eady: SGBE=SGBA= SGAD=SGDC b) Hai tam gi\u00e1c ABC v\u00e0 ABG coa chung \u0111\u01b0\u1eddng c\u00f3 k\u1ebb t\u1eeb A n\u00ean 2 c\u1ea1nh \u0111\u00e1y CB v\u00e0 GB s\u1ebd t\u1ec9 l\u1ec7 v\u1edbi di\u1ec7n t\u00edch. T\u1eeb k\u1ebft qu\u1ea3 c\u00e2u a. Suy ra: SABC = SAGB x 3 V\u1eady: CB = GB x 3 GB = 9 : 3 = 3 (cm) B\u00e0i 43: T\u00fd c\u00f3 m\u1ed9t t\u1ea5m b\u00eca h\u00ecnh vu\u00f4ng, t\u00fd c\u1eaft t\u1ea5m b\u00eca th\u00e0nh hai h\u00ecnh ch\u1eef nh\u1eadt kh\u00f4ng b\u1eb1ng nhau, chu vi c\u1ee7a hai h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 150cm. T\u00ednh di\u1ec7n t\u00edch t\u1ea5m b\u00eca h\u00ecnh vu\u00f4ng Chu vi 2 h\u00ecnh ch\u1eef nh\u1eadt b\u1eb1ng 6 l\u1ea7n c\u1ea1nh h\u00ecnh vu\u00f4ng. C\u1ea1nh h\u00ecnh vu\u00f4ng: 150 : 6 = 25 (cm) Di\u1ec7n t\u00edch t\u1ea5m b\u00eca: 25 x 25 = 625 (cm2) \u0110\u00e1p s\u1ed1: 625 cm2 B\u00e0i 44: Cho h\u00ecnh tam gi\u00e1c ABC. Tr\u00ean c\u1ea1nh AB ta l\u1ea5y \u0111i\u1ec3m E sao cho BE g\u1ea5p \u0111\u00f4i AE; tr\u00ean c\u1ea1nh AC ta l\u1ea5y \u0111i\u1ec3m D sao cho CD g\u1ea5p \u0111\u00f4i AD. N\u1ed1i E v\u1edbi D ta \u0111\u01b0\u1ee3c h\u00ecnh tam gi\u00e1c AED c\u00f3 di\u1ec7n t\u00edch 5 cm2. H\u00e3y t\u00ednh di\u1ec7n t\u00edch h\u00ecnh t\u1ee9 gi\u00e1c BCDE. Ta c\u00f3 DC=AD x 2 N\u00ean SDCE= 5 x 2 = 10 (cm2) (\u0111\u00e1y DC=2AD v\u00e0 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A). SACE = 5 + 10 = 15 (cm2) Ta l\u1ea1i c\u00f3 EB = EA x 2 N\u00ean SECB = SACE x 2 = 15 x 2 = 30 (cm2) SBCDE = SDEC+ SECB = 10 + 30 = 40 (cm2) B\u00e0i 45: Cho tam gi\u00e1c ABC tr\u00ean c\u1ea1nh AB l\u1ea5y \u0111i\u1ec3m E sao cho AE = 2\/3 AB, tr\u00ean c\u1ea1nh AC l\u1ea5y \u0111i\u1ec3m D sao cho AD= 1\/3 AC a. N\u1ed1i B v\u1edbi D . T\u00ednh t\u1ef7 s\u1ed1 di\u1ec7n t\u00edch hai tam gi\u00e1c ABD v\u00e0 ABC","b. N\u1ed1i E v\u1edbi D .Bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c AED l\u00e0 8 cm2 . T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c ABC c. N\u1ed1i C v\u1edbi E c\u1eaft BD t\u1ea1i G. T\u00ednh t\u1ef7 s\u1ed1 \u0111\u1ed9 d\u00e0i hai \u0111o\u1ea1n th\u1eb3ng EG v\u00e0 CG a).Do AD = 1\/3 AC n\u00ean SABD = 1\/3SABC. V\u00ec 2 tam gi\u00e1c n\u00e0y c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B b).T\u01b0\u01a1ng t\u1ef1 ta c\u00f3 SAED= 1\/3SAEC N\u00ean SAEC = 8 x 3 = 24 (cm2) M\u00e0 AE = 2\/3AB v\u00e0 2 tam gi\u00e1c AEC v\u00e0 EBC c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb C. N\u00ean SAEC = 2\/3SABC Di\u1ec7n t\u00edch tam gi\u00e1c ABC: 24 : 2 x 3 = 36 (cm2) c). SEBD= 1\/3 SABD = 1\/3.1\/3SABC = 4 (cm2) SEBC = 12 (cm2) \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.(1\/3 c\u1ee7a SABC) SDEC = 2\/3.24 = 16 (cm2) \u2026\u2026\u2026\u2026\u2026.(2\/3 c\u1ee7a SAEC) 2 tam gi\u00e1c BCE v\u00e0 DCE c\u00f3 chung c\u1ea1nh \u0111\u00e1y CE n\u00ean 2 \u0111\u01b0\u1eddng cao t\u1ec9 l\u1ec7 v\u1edbi di\u1ec7n t\u00edch. T\u1ec9 s\u1ed1: Bh\/Dk = 12\/16 = 3\/4 T\u01b0\u01a1ng t\u1ef1 ta c\u00f3: SEBG \/ SDEG = 3\/4 Suy ra SDEG = 4 : (4+3) x 4 = 16\/7 (cm2) SDCG = SDEC \u2013 SDEG = 16 \u2013 16\/7 = 96\/7 (cm2) T\u1ec9 s\u1ed1 c\u1ee7a EG v\u00e0 CG l\u00e0 t\u1ec9 s\u1ed1 c\u1ee7a SDEG v\u00e0 SDCG (16\/7) \/ (96\/7) = 16\/96 = 1\/6 B\u00e0i 47: Cho tam gi\u00e1c ABC \u0111i\u1ec3m N n\u1eb1m tr\u00ean AC \u0111i\u1ec3m M n\u1eb1m tr\u00ean BC sao cho AM c\u1eaft BN t\u1ea1i O di\u1ec7n t\u00edch c\u00e1c tam gi\u00e1c ANO = 2cm2 , ABO = 6cm2 , BMO = 4cm2 T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c ABC?","SABO = 3SAON ( v\u00ec 6:2=3) ==> BO = 3ON (chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A). ==> SOMN = 1\/3SOBM = 1\/3 x 4 = 4\/3 (cm2) (chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb M) X\u00e9t 2 tam gi\u00e1c ABN v\u00e0 AMN c\u00f3 chung \u0111\u00e1y AN n\u00ean Bk v\u00e0 Mh t\u1ec9 l\u1ec7 v\u1edbi di\u1ec7n t\u00edch. Bk\/Mh = (6+2)\/(2+4\/3) = 8\/(10\/3) = 24\/10 Hai tam gi\u00e1c ABC v\u00e0 AMC c\u00f3 chung \u0111\u00e1y AC n\u00ean di\u1ec7n t\u00edch t\u1ec9 l\u1ec7 v\u1edbi \u0111\u01b0\u1eddng cao. SABC\/SAMC = 24\/10 | SABC - SAMC = SABM = 6+4 = 10 (cm2) | Hi\u1ec7u v\u00e0 T\u1ec9 Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 24 - 10 = 14 (ph\u1ea7n) Di\u1ec7n t\u00edch tam gi\u00e1c ABC: 10:14x24 = 17,14286 (cm2) B\u00e0i 48: C\u00f3 m\u1ed9t mi\u1ebfng \u0111\u1ea5t h\u00ecnh thang. H\u00f9ng \u01b0\u1edbc l\u01b0\u1ee3ng \u0111\u00e1y l\u1edbn b\u1eb1ng 32m, D\u0169ng \u01b0\u1edbc l\u01b0\u1ee3ng \u0111\u00e1y l\u1edbn b\u1eb1ng 37m v\u00e0 c\u1ea3 hai \u0111\u1ec1u \u01b0\u1edbc l\u01b0\u1ee3ng sai. N\u1ebfu \u01b0\u1edbc l\u01b0\u1ee3ng nh\u01b0 H\u00f9ng th\u00ec di\u1ec7n t\u00edch mi\u1ebfng \u0111\u1ea5t gi\u1ea3m 36m2, c\u00f2n n\u1ebfu \u01b0\u1edbc l\u01b0\u1ee3ng nh\u01b0 D\u0169ng th\u00ec t\u0103ng 24m2. H\u1ecfi \u0111\u00e1y l\u1edbn c\u1ee7a mi\u1ebfng \u0111\u1ea5t d\u00e0i bao nhi\u00eau m? \u01af\u1edbc l\u01b0\u1ee3ng v\u1ec1 \u0111\u00e1y l\u1edbn \u1edf hai b\u1ea1n l\u1ec7ch nhau: 37 - 32 = 5 (m) T\u1eeb sai l\u00each v\u1ec1 \u0111\u1ea5y l\u1edbn gi\u1eefa hai b\u1ea1n n\u00ean di\u1ec7n t\u00edch c\u0169ng l\u1ec7ch theo: 36 + 24 = 60 (m2) Chi\u1ec1u cao h\u00ecnh thang: 60 x 2 : 5 = 24 (m) Theo H\u00f9ng th\u00ec \u0111\u00e1y l\u1edbn c\u00f2n thi\u1ebfu: 36 x 2 : 24 = 3 (m) \u0110\u1ed9 d\u00e0i c\u1ee7a \u0111\u00e1y l\u1edbn mi\u1ebfng \u0111\u00e2t: 32 + 3 = 35 (m) \u0110\u00e1p s\u1ed1: 35m B\u00e0i 49: Cho tam gi\u00e1c ABC .Tr\u00ean c\u1ea1nh AB l\u1ea5y \u0111i\u1ec3m M sao cho AM = 1\/3 AB.Tr\u00ean c\u1ea1nh AC l\u1ea5y \u0111i\u1ec3m N sao cho AN b\u1eb1ng 1\/3 AC.N\u1ed1i B v\u1edbi N, n\u1ed1i C v\u1edbi M; BN c\u1eaft CM t\u1ea1i I. T\u00ednh di\u1ec7n t\u00edcg tam gi\u00e1c ABC, bi\u1ebft di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c AMIN b\u1eb1ng 90cm2. SABN = SACM (b\u1eb1ng 1\/3 SABC) M\u00e0 2 tam gi\u00e1c n\u00e0y c\u00f3 ph\u1ea7n chung AMIN n\u00ean SMBI = SNIC. N\u1ed1i AI ta c\u00f3: SABI = 3\/2 SMBI (AB = 3\/2MB). T\u01b0\u01a1ng t\u1ef1: SAIC = 3\/2 SNIC Suy ra SABI = SAIC ==> SAMI = SAIN = 90\/2 = 45 (cm2) V\u1eady SMBI = 45 x 2 = 90 (cm2) ==> SABN = SMBI + SAMIN = 90+90 = 180 (cm2) Do \u0111\u00f3: SABC = 180 x 3 = 540 (cm2) B\u00e0i 50:","Cho h\u00ecnh vu\u00f4ng ABCD, g\u1ecdi M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u1ea1nh AD. \u0110o\u1ea1n th\u1eb3ng AC c\u1eaft BM t\u1ea1i N. a, Di\u1ec7n t\u00edch tam gi\u00e1c BMC g\u1ea5p m\u1ea5y l\u1ea7n Di\u1ec7n t\u00edch tam gi\u00e1c AMB? b, Di\u1ec7n t\u00edch tam gi\u00e1c BNC g\u1ea5p m\u1ea5y l\u1ea7n di\u1ec7n t\u00edch tam gi\u00e1c ANB ? T\u00ednh di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng ABCD bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c ANB b\u1eb1ng 1,5 dm2. a\/ 2 tam gi\u00e1c BMC v\u00e0 AMB c\u00f3 \u0111\u00e1y BC=2AM, 2 \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B xu\u1ed1ng AM v\u00e0 t\u1eeb M xu\u1ed1ng BC b\u1eb1ng nhau b\u1eb1ng c\u1ea1nh h\u00ecnh vu\u00f4ng. N\u00ean SBMC = 2 SAMB. b\/ T\u01b0\u01a1ng t\u1ef1 nh\u01b0 tr\u00ean. Ta c\u00f3 SABC = 2SAMC Suy ra: BH = 2 MK (c\u0169ng l\u00e0 2 \u0111\u01b0\u1eddng cao c\u1ee7a 2 tam gi\u00e1c BNC v\u00e0 MNC c\u00f3 chung \u0111\u00e1y NC) N\u00ean SBNC = 2SMNC (1) (do SABM = SACM v\u00e0 2 tam gi\u00e1c n\u00e0y c\u00f3 ph\u1ea7n chung l\u00e0 M\u00e0 SMNC = SANB (2) SANM) T\u1eeb (1) v\u00e0 (2). Ta \u0111\u01b0\u1ee3c: SBNC = 2 SBNA. SABC = SABN + SBNC = 1,5 + 1,5 x 2 = 4,5 (dm2) Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng ABC: 4,5 x 2 = 9 (dm2) B\u00e0i 51 M\u1ed9t b\u1ec3 n\u01b0\u1edbc d\u1ea1ng h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt c\u00f3 chi\u1ec1u d\u00e0i 2,7m, chi\u1ec1u r\u1ed9ng 2m. Hi\u1ec7n b\u1ec3 \u0111ang ch\u1ee9a 6480 l\u00edt n\u01b0\u1edbc th\u00ec m\u1ef1c n\u01b0\u1edbc trong b\u1ec3 b\u1eb1ng 3\/4 chi\u1ec1u cao c\u1ee7a b\u1ec3. T\u00ednh chi\u1ec1u cao c\u1ee7a b\u1ec3 bi\u1ebft 1 l\u00edt = 1 dm3 Di\u1ec7n t\u00edch \u0111\u00e1y b\u1ec3 HHCN: 2,7 x 2 = 5,4 (m2) \u0110\u1ed5i: 6480 l\u00edt = 6480 dm3 = 6,48 m3. Chi\u1ec1u cao m\u1ef1c n\u01b0\u1edbc trong b\u1ec3: 6,48 : 5,4 = 1,2 (m) Chi\u1ec1u cao c\u1ee7a b\u1ec3 : 1,2 : 3 x 4 = 1,6 (m) \u0110\u00e1p s\u1ed1: 1,6m B\u00e0i 52: Tr\u00ean 1 \u0111\u01b0\u1eddng tr\u00f2n ta l\u1ea5y 10 \u0111i\u1ec3m, n\u1ed1i 2 \u0111i\u1ec3m kh\u00f4ng li\u1ec1n k\u1ec1 v\u1edbi nhau th\u00ec ta \u0111\u01b0\u1ee3c 1 \u0111o\u1ea1n th\u1eb3ng. H\u1ecfi t\u1eeb 10 \u0111i\u1ec3m tr\u00ean ta n\u1ed1i \u0111\u01b0\u1ee3c bao nhi\u00eau \u0111o\u1ea1n th\u1eb3ng? - M\u1ed7i \u0111i\u1ec3m b\u1ea5t k\u00ec s\u1ebd n\u1ed1i v\u1edbi 10 - 3 = 7 \u0111i\u1ec3m c\u00f2n l\u1ea1i. - C\u00f3 10 \u0111i\u1ec3m s\u1ebd n\u1ed1i \u0111\u01b0\u1ee3c s\u1ed1 \u0111o\u1ea1n th\u1eb3ng l\u00e0: 7 x 10 : 2 = 35 \u0111o\u1ea1n th\u1eb3ng. \u0110\u00e1p s\u1ed1: 35 \u0111o\u1ea1n B\u00e0i 53:","Ng\u01b0\u1eddi ta ng\u0103n th\u1eeda \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt th\u00e0nh 2 m\u1ea3nh, m\u1ed9t m\u1ea3nh h\u00ecnh vu\u00f4ng, m\u1ed9t m\u1ea3nh h\u00ecnh ch\u1eef nh\u1eadt. Bi\u1ebft chu vi ban \u0111\u1ea7u h\u01a1n chu vi m\u1ea3nh \u0111\u1ea5t h\u00ecnh vu\u00f4ng l\u00e0 28 m. Di\u1ec7n t\u00edch c\u1ee7a th\u1eeda \u0111\u1ea5t ban \u0111\u1ea7u h\u01a1n di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng l\u00e0 224 m2. T\u00ednh di\u1ec7n t\u00edch th\u1eeda \u0111\u1ea5t ban \u0111\u1ea7u. N\u1eeda chu vi h\u00ecnh ABCD h\u01a1n n\u1eeda chu vi h\u00ecnh AMND l\u00e0 : 28 : 2 = 14 (m). N\u1eeda chu vi h\u00ecnh ABCD l\u00e0 AD + AB. N\u1eeda chu vi h\u00ecnh AMND l\u00e0 AD + AM. Do \u0111\u00f3 : MB = AB - AM = 14 (m). Chi\u1ec1u r\u1ed9ng BC c\u1ee7a h\u00ecnh ABCD l\u00e0 : 224 : 14 = 16 (m) Chi\u1ec1u d\u00e0i AB c\u1ee7a h\u00ecnh ABCD l\u00e0 : 16 + 14 = 30 (m) Di\u1ec7n t\u00edch h\u00ecnh ABCD l\u00e0 : 30 x 16 = 480 (m2). \u0110\u00e1p s\u1ed1: 480 m2. B\u00e0i 54: Cho h\u00ecnh thang ABCD, c\u00f3 BC=5cm. Tr\u00ean BC l\u1ea5y 1 \u0111i\u1ec3m E sao cho BE = 1cm. T\u00ednh t\u1ef7 s\u1ed1 \u0111\u1ed9 d\u00e0i hai c\u1ea1nh \u0111\u00e1y CD v\u00e0 AB, bi\u1ebft di\u1ec7n t\u00edch c\u1ee7a tam gi\u00e1c ABE bang 1\/6 di\u1ec7n t\u00edch t\u01b0 gi\u00e1c AECD. * Ta c\u00f3: S_ABE = 1\/4 S_ACE (\u0110\u00e1y BE = 1\/4 \u0111\u00e1y CE; Chi\u1ec1u cao \u0111\u1ec9nh A chung). \u0110\u1ec3 S_ABE = 1\/6 S_ADCE. N\u1ebfu coi S_ABE b\u1eb1ng 1 ph\u1ea7n di\u1ec7n t\u00edch th\u00ec S_ADC = 2 ph\u1ea7n di\u1ec7n t\u00edch. => S_ABC = 5 ph\u1ea7n di\u1ec7n t\u00edch. => S_ADC = 2\/5 S_ABC Hai tam gi\u00e1c n\u00e0y c\u00f3 chi\u1ec1u cao b\u1eb1ng nhau n\u00ean \u0111\u00e1y DC = 2\/5 AB \u0110\u00e1p s\u1ed1: CD = 2\/5 AB B\u00e0i 55: M\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chi\u1ec1u d\u00e0i g\u1ea5p 4 l\u1ea7n chi\u1ec1u r\u1ed9ng. N\u1ebfu t\u0103ng chi\u1ec1u r\u1ed9ng th\u00eam 45 m th\u00ec \u0111\u01b0\u1ee3c h\u00ecnh ch\u1eef nh\u1eadt m\u1edbi c\u00f3 chi\u1ec1u d\u00e0i v\u1eabn g\u1ea5p 4 l\u1ea7n chi\u1ec1u r\u1ed9ng. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u. Khi t\u0103ng chi\u1ec1u r\u1ed9ng th\u00eam 45 m th\u00ec khi \u0111\u00f3 chi\u1ec1u r\u1ed9ng s\u1ebd tr\u1edf th\u00e0nh chi\u1ec1u d\u00e0i c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt m\u1edbi, c\u00f2n chi\u1ec1u d\u00e0i ban \u0111\u1ea7u s\u1ebd tr\u1edf th\u00e0nh chi\u1ec1u r\u1ed9ng c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt m\u1edbi. Theo \u0111\u1ec1 b\u00e0i ta c\u00f3 s\u01a1 \u0111\u1ed3 :","Do \u0111\u00f3 45 m \u1ee9ng v\u1edbi s\u1ed1 ph\u1ea7n l\u00e0 : 16 - 1 = 15 (ph\u1ea7n) Chi\u1ec1u r\u1ed9ng ban \u0111\u1ea7u l\u00e0 : 45 : 15 = 3 (m) Chi\u1ec1u d\u00e0i ban \u0111\u1ea7u l\u00e0 : 3 x 4 = 12 (m) Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u l\u00e0 : 3 x 12 = 36 (m2) \u0110\u00e1p s\u1ed1: 36 m2. B\u00e0i 56: Cho h\u00ecnh vu\u00f4ng ABCD, g\u1ecdi M l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u1ea1nh AD. \u0110o\u1ea1n th\u1eb3ng AC c\u1eaft BM t\u1ea1i N. a, Di\u1ec7n t\u00edch tam gi\u00e1c BMC g\u1ea5p m\u1ea5y l\u1ea7n Di\u1ec7n t\u00edch tam gi\u00e1c AMB? b, Di\u1ec7n t\u00edch tam gi\u00e1c BNC g\u1ea5p m\u1ea5y l\u1ea7n di\u1ec7n t\u00edch tam gi\u00e1c ANB ? T\u00ednh di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng ABCD bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c ANB b\u1eb1ng 1,5 dm2. a\/.Hai tam gi\u00e1c BMC v\u00e0 AMB c\u00f3 c\u1ea1nh \u0111\u00e1y BC = 2.AM, c\u00f3 2 \u0111\u01b0\u1eddng cao t\u01b0\u01a1ng \u1ee9ng b\u1eb1ng nhau (t\u1eeb B xu\u1ed1ng AM v\u00e0 t\u1eeb M xu\u1ed1ng BC (c\u1ea1nh h\u00ecnh vu\u00f4ng)). N\u00ean SBMC = 2.SAMB . b\/.T\u1eeb SBMC = 2.SAMB v\u00e0 2 tam gi\u00e1c n\u00e0y c\u00f3 chung \u0111\u00e1y MB. N\u00ean \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb C xu\u1ed1ng MB g\u1ea5p 2 l\u1ea7n \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A xu\u1ed1ng MB. Hai \u0111\u01b0\u1eddng cao c\u1ee7a 2 tam gi\u00e1c n\u00e0y c\u0169ng ch\u00ednh l\u00e0 2 \u0111\u01b0\u1eddng cao c\u1ee7a 2 tam gi\u00e1c CNB v\u00e0 ANB. M\u1eb7t kh\u00e1c 2 tam gi\u00e1c CNB v\u00e0 ANB c\u00f3 chung c\u1ea1nh \u0111\u00e1y NB. N\u00ean SBNC = 2.SANB. SBNC = 1,5 x 2 = 3 (dm2) SABC = 1,5 + 3 = 4,5 (dm2) Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng ABCD: 4,5 x 2 = 9 (dm2) B\u00e0i 57: M\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt, n\u1ebfu t\u0103ng chi\u1ec1u r\u1ed9ng \u0111\u1ec3 b\u1eb1ng chi\u1ec1u d\u00e0i c\u1ee7a n\u00f3 th\u00ec di\u1ec7n t\u00edch t\u0103ng th\u00eam 20m2, c\u00f2n khi gi\u1ea3m chi\u1ec1u d\u00e0i cho b\u1eb1ng chi\u1ec1u r\u1ed9ng th\u00ec di\u1ec7n t\u00edch gi\u1ea3m 16m2. T\u00ednh di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt","H\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u l\u00e0 ABCD. Theo \u0111\u1ec1 b\u00e0i ta c\u00f3: MD=DC chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt BC=ME chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt (c\u1ea1nh h\u00ecnh vu\u00f4ng nh\u1ecf) MA=KB hi\u1ec7u c\u1ee7a chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng Suy ra: SMEKA=SKBCP=16m2 SENBK=20-16=4(m2) C\u1ea1nh h\u00ecnh vu\u00f4ng ENBK l\u00e0 2m (2x2=4) Chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt: 16 : 2 = 8 (m) Chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt: 8 + 2 = 10 (m) Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt: 10 x 8 = 80 (m2) \u0110\u00e1p s\u1ed1: 80 m2. B\u00e0i 58: M\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 t\u1ed5ng chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng g\u1ea5p 5 l\u1ea7n hi\u1ec7u chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng. Di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 600m2. T\u00ednh chi\u1ec1u d\u00e0i, chi\u1ec1u r\u1ed9ng? Xem hi\u1ec7u c\u1ee7a 2 c\u1ea1nh l\u00e0 1 ph\u1ea7n, ta c\u00f3 s\u01a1 \u0111\u1ed3: Hi\u1ec7u 2 c\u1ea1nh: |-----| T\u1ed5ng 2 c\u1ea1nh: |-----|-----|-----|-----|-----| Chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt l\u00e0: ( 1+ 5) : 2 = 3 ( ph\u1ea7n). Chi\u1ec1u r\u1ed9ng nh\u1eadt l\u00e0: 5 \u2013 3 = 2 (ph\u1ea7n). Ta c\u00f3 h\u00ecnh v\u1ebd: S\u1ed1 h\u00ecnh vu\u00f4ng c\u00f3 l\u00e0: 2 x 3 = 6 (h\u00ecnh). Di\u1ec7n t\u00edch m\u1ed9t h\u00ecnh vu\u00f4ng l\u00e0: 600 : 6 = 100 (m2). C\u1ea1nh h\u00ecnh vu\u00f4ng l\u00e0 10 m (10 x 10 = 100). Chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 10 x 3 = 30 (m). Chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 10 x 2 = 30 (m).","Chu vi h\u00ecnh ch\u1eef nh\u1eadt l\u00e0: (30 + 20) x 2 = 100 (m). \u0110\u00e1p s\u1ed1: 100m B\u00e0i 59: Cho h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chu vi 142m. N\u1ebfu gi\u1ea3m chi\u1ec1u d\u00e0i 15m v\u00e0 t\u0103ng chi\u1ec1u r\u1ed9ng 15m th\u00ec di\u1ec7n t\u00edch kh\u00f4ng \u0111\u1ed5i. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt \u0111\u00f3 ? Khi gi\u1ea3m chi\u1ec1u d\u00e0i 15m v\u00e0 t\u0103ng chi\u1ec1u r\u1ed9ng 15m th\u00ec di\u1ec7n t\u00edch kh\u00f4ng \u0111\u1ed5i, l\u00fac n\u00e0y chi\u1ec1u r\u1ed9ng tr\u1edf th\u00e0nh chi\u1ec1u d\u00e0i m\u1edbi v\u00e0 chi\u1ec1u d\u00e0i l\u1ea1i tr\u1edf th\u00e0nh chi\u1ec1u r\u1ed9ng m\u1edbi. Nh\u01b0 v\u1eady chi\u1ec1u d\u00e0i h\u01a1n chi\u1ec1u r\u1ed9ng 15m. N\u1eeda chu vi h\u00ecnh ch\u1eef nh\u1eadt: 142 : 2 = 71 (m) Chi\u1ec1u r\u1ed9ng l\u00e0: (71 \u2013 15) : 2 = 28 (m) Chi\u1ec1u d\u00e0i l\u00e0: 71 \u2013 28 = 43 (m) Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt l\u00e0: 43 x 28 = 1204 (m2) \u0110\u00e1p s\u1ed1: 1204 m2. B\u00e0i 60: L\u1edbp 5A v\u00e0 l\u1edbp 5B nh\u1eadn ch\u0103m s\u00f3c hai th\u1eeda ru\u1ed9ng c\u00f3 di\u1ec7n t\u00edch t\u1ed5ng c\u1ed9ng l\u00e0 1560 m2. N\u1ebfu l\u1ea5y \u00bc di\u1ec7n t\u00edch th\u1eeda ru\u1ed9ng c\u1ee7a l\u1edbp 5A chuy\u1ec3n sang cho l\u1edbp 5B ch\u0103m s\u00f3c th\u00ec di\u1ec7n t\u00edch ch\u0103m s\u00f3c c\u1ee7a hai l\u1edbp b\u1eb1ng nhau. T\u00ednh di\u1ec7n t\u00edch c\u1ee7a m\u1ed7i th\u1eeda ru\u1ed9ng. Ph\u00e2n s\u1ed1 ch\u1ec9 di\u1ec7n t\u00edch th\u1eeda ru\u1ed9ng l\u1edbp 5A c\u00f2n l\u1ea1i: 1 \u2013 \u00bc = \u00be (ru\u1ed9ng 5A) \u00be di\u1ec7n t\u00edch th\u1eeda ru\u1ed9ng l\u1edbp 5A l\u00e0: 1560 : 2 = 780 (m2) Di\u1ec7n t\u00edch th\u1eeda ru\u1ed9ng l\u1edbp 5A l\u00e0: 780 : 3 x 4 = 1040 (m2) Di\u1ec7n t\u00edch th\u1eeda ru\u1ed9ng l\u1edbp 5B l\u00e0: 1560 \u2013 1040 = 520 (m2) \u0110\u00e1p s\u1ed1: 5A 1040 m2 ; 5B 520 m2. B\u00e0i 60: Cho 8 \u0111i\u1ec3m n\u1eb1m tr\u00ean m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n s\u1ed1 tam gi\u00e1c \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh c\u00f3 c\u00e1c \u0111\u1ec9nh n\u1eb1m tr\u00ean 8 \u0111i\u1ec3m thu\u1ed9c \u0111\u01b0\u1eddng tr\u00f2n l\u00e0 \u2026\u2026\u2026\u2026.","Tr\u01b0\u1edbc ti\u00ean ta l\u1ea5y \u0111i\u1ec3m A l\u00e0m chu\u1ea9n, s\u1ebd c\u00f3 c\u00e1c tam gi\u00e1c: ABC;ABD;ABE;ABF;ABG;ABH (6) \u0110\u1ebfn AC, ta c\u00f3: ACD;ACE;ACF;ACG;ACH (5) \u0110\u1ebfn AD, ta c\u00f3: ADE;ADF;ADG;ADH (4) \u0110\u1ebfn AE, ta c\u00f3: AEF;AEG;AEH (3) \u0110\u1ebfn AF, ta c\u00f3: AFG;AFH (2) \u0110\u1ebfn AG, ta c\u00f3: AGH (1) Ta c\u00f3: 1+2+3+4+5+6 = 21 (h\u00ecnh tam gi\u00e1c) c\u00f3 \u0111\u00ecnh t\u1eeb 8 \u0111i\u1ec3m tr\u00ean 1 \u0111\u01b0\u1eddng tr\u00f2n. T\u01b0\u01a1ng t\u1ef1, ta l\u1ea5y: *.B l\u00e0m chu\u1ea9n, l\u00fac n\u00e0y kh\u00f4ng k\u1ec3 \u0111i\u1ec3m A.Ta c\u00f3: 5+4+3+2+1 = 15 (tam gi\u00e1c) *.C l\u00e0m chu\u1ea9n, ta kh\u00f4ng k\u1ec3 \u0111\u1ebfn A; v\u00e0 B: C\u00f3 4+3+2+1=10 (tam gi\u00e1c) *.D l\u00e0m chu\u1ea9n, ta kh\u00f4ng k\u1ec3 \u0111\u1ebfn A; B v\u00e0 C: C\u00f3 3+2+1=6 (tam gi\u00e1c) *.E l\u00e0m chu\u1ea9n, ta kh\u00f4ng k\u1ec3 \u2026\u2026\u2026\u2026\u2026\u2026: C\u00f3 2+1= 3 (tam gi\u00e1c) *.F l\u00e0m chu\u1ea9n, ta kh\u00f4ng k\u1ec3\u2026\u2026\u2026\u2026\u2026\u2026: C\u00f3 1 (tam gi\u00e1c) T\u1ea5t c\u1ea3 c\u00e1c tam gi\u00e1c l\u00e0: 1+3+6+10+15+21= 56 (tam gi\u00e1c) B\u00e0i 61: C\u1eaft 1 mi\u1ebfng b\u00eca h\u00ecnh vu\u00f4ng th\u00e0nh 2 mi\u1ebfng b\u00eca h\u00ecnh ch\u1eef nh\u1eadt.Bi\u1ebft t\u1ed5ng chu vi 2 mi\u1ebfng b\u00eca h\u00ecnh ch\u1eef nh\u1eadt \u0111\u00f3 l\u00e0 192cm v\u00e0 hi\u1ec7u chu vi b\u1eb1ng 16cm.T\u00ednh di\u00ean t\u00edch mi\u1ebfng b\u00eca h\u00ecnh ch\u1eef nh\u1eadt l\u1edbn T\u1ed5ng chu vi 2 h\u00ecnh ch\u1eef nh\u1eadt b\u1eb1ng 6 l\u1ea7n c\u1ea1nh h\u00ecnh vu\u00f4ng (do l\u1eb1n c\u1eaft t\u1ea1o th\u00eam 2 c\u1ea1nh h\u00ecnh vu\u00f4ng) C\u1ea1nh h\u00ecnh vu\u00f4ng c\u0169ng l\u00e0 t\u1ed5ng 2 chi\u1ec1u r\u1ed9ng c\u1ee7a 2 h\u00ecnh ch\u1eef nh\u1eadt l\u00e0: 192 : 6 = 32 (cm) Chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt l\u1edbn h\u01a1n chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt nh\u1ecf l\u00e0: 16 : 2 = 8 (cm) Chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt l\u1edbn l\u00e0: (32 + 8) : 2 = 20 (cm) Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt l\u1edbn l\u00e0: 32 x 20 = 640 (cm2) \u0110\u00e1p s\u1ed1: 640 cm2.","B\u00e0i 62: M\u1ed9t th\u1eeda ru\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chu vi 0,450 km. Bi\u1ebft chi\u1ec1u r\u1ed9ng b\u1eb1ng 2\/3 chi\u1ec1u d\u00e0i. T\u00ednh s\u1ed1 \u0111o chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng c\u1ee7a th\u1eeda ru\u1ed9ng \u0111\u00f3. 0,450km = 450 m N\u1eeda chu vi l\u00e0: 450 : 2 = 225 (m) T\u1ed5ng s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 2+3 = 5 (ph\u1ea7n) Chi\u1ec1u r\u1ed9ng: 225 : 5 x 2 = 90 (m) Chi\u1ec1u d\u00e0i: 225 \u2013 90 = 135 (m) \u0110\u00e1p s\u1ed1: 90 m ; 135 m B\u00e0i 63: Cho h\u00ecnh thang vu\u00f4ng ABCD c\u00f3 g\u00f3c A v\u00e0 D vu\u00f4ng. H\u1ea1 \u0111\u01b0\u1eddng cao BH, \u0111\u01b0\u1eddng cao BH c\u1eaft \u0111\u01b0\u1eddng ch\u00e9o AC t\u1ea1i I. So s\u00e1nh di\u1ec7n t\u00edch 2 tam gi\u00e1c DIH v\u00e0 BIC ABHD l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt n\u00ean AD=BH ; AB=DH SABD=SABC=SABI+SBIC (1) (2 tam gi\u00e1c ABD v\u00e0 ABC c\u00f3 chung \u0111\u00e1y AB, 2 \u0111\u01b0\u1eddng cao b\u1eb1ng \u0111\u01b0\u1eddng cao h\u00ecnh thang). SABD=SABI+SDIH (2) (Tam gi\u00e1c ABD c\u00f3 \u0111\u00e1y AD =BI+IH, 3 tam gi\u00e1c n\u00e0y (ABD, ABI, DIH) c\u00f3 \u0111\u01b0\u1eddng cao b\u1eb1ng chi\u1ec1u r\u1ed9ng (AB) h\u00ecnh ch\u1eef nh\u1eadt ABHD). T\u1eeb (1) v\u00e0 (2) suy ra SBIC = SDIH B\u00e0i 64: Huy c\u00f3 m\u1ed9t m\u1ea3nh gi\u1ea5y h\u00ecnh vu\u00f4ng c\u00f3 chu vi l\u00e0 80cm. Huy \u0111\u00e3 g\u1ea5p h\u00ecnh vu\u00f4ng \u0111\u00f3 l\u1ea1i v\u00e0 c\u1eaft \u0111\u01b0\u1ee3c m\u1ed9t h\u00ecnh tr\u00f2n (to nh\u1ea5t). a.T\u00ednh chu vi h\u00ecnh tr\u00f2n m\u00e0 Huy \u0111\u00e3 c\u1eaft \u0111\u01b0\u1ee3c b.N\u1ebfu d\u00f9ng m\u1ea3nh gi\u1ea5y h\u00ecnh tr\u00f2n \u0111\u00f3 \u0111\u1ec3 c\u1eaft m\u1ed9t m\u1ea3nh gi\u1ea5y h\u00ecnh vu\u00f4ng c\u00f3 c\u1ea1nh 16cm th\u00ec c\u00f3 c\u1eaft \u0111\u01b0\u1ee3c kh\u00f4ng?V\u00ec sao? C\u1ea1nh h\u00ecnh vu\u00f4ng: 80 : 4 = 20 (cm) Chu vi h\u00ecnh tr\u00f2n: 20 x 3,14 = 62,8 (cm) Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng l\u1edbn nh\u1ea5t c\u00f3 th\u1ec3 c\u1eaft \u0111\u01b0\u1ee3c:","(20 : 2) x ( 20 : 2 ) x 2 = 200 (cm2) N\u00ean kh\u00f4ng th\u1ec3 c\u1eaft \u0111\u01b0\u1ee3c h\u00ecnh vu\u00f4ng c\u00f3 c\u1ea1nh l\u00e0 16cm. V\u00ec 16 x 16 = 256 (cm2) B\u00e0i 65: Cho h\u00ecnh tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch l\u00e0 216m2 ,AB=AC v\u00e0 BC b\u1eb1ng 36 m. Tr\u00ean c\u1ea1nh AB l\u1ea5y \u0111i\u1ec3m M sao cho MB b\u1eb1ng 2\/3 AB ,tr\u00ean AC l\u1ea5y \u0111i\u1ec3m N sao cho NC b\u1eb1ng 2\/3 AC v\u00e0 tr\u00ean BC l\u1ea5y \u0111i\u1ec3m I sao cho BI b\u1eb1ng 2\/3 BC. N\u1ed1i M v\u1edbi N v\u1edbi I \u0111\u01b0\u1ee3c h\u00ecnh thang MNIB. T\u00ednh : a. Di\u1ec7n t\u00edch h\u00ecnh thang MNIB b. \u0110\u1ed9 d\u00e0i \u0111o\u1ea1n MN MB=2\/3AB => AM=1\/2AM=1\/3AB NC=2\/3AC => AN=1\/2NC=1\/3AC a) SBNA=1\/3SABC= 216 : 3 = 72 (m2) Chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B. T\u01b0\u01a1ng t\u1ef1: SNMB=2\/3SNBA= 72 x 2\/3 = 48 (m2) SBNC=SABC-SBNA= 216 \u2013 72 = 144 (m2) SNBI=2\/3SNBC= 144 x 2\/3 = 96 (m2) SMNIB = SMNB+SNIB = 48+96 = 144 (m2) b) Chi\u1ec1u cao k\u1ebb t\u1eeb N c\u1ee7a tam gi\u00e1c NBC 144 x 2 : 36 = 8 (m) C\u0169ng l\u00e0 \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B c\u1ee7a tam gi\u00e1c BMN. \u0110\u1ed9 d\u00e0i c\u1ea1nh MN (trong tam gi\u00e1c BMN). 48 x 2 : 8 = 12 (m) B\u00e0i 66: Cho m\u1ed9t h\u00ecnh thang c\u00f3 chu vi l\u00e0 405 cm, t\u1ed5ng hai \u0111\u00e1y( AB v\u00e0 v\u00e0 CD ) d\u00e0i h\u01a1n t\u1ed5ng hai c\u1ea1nh b\u00ean (AD v\u00e0 BC) l\u00e0 15 cm. C\u1ea1nh AB b\u1eb1ng 2 ph\u1ea7n 5 c\u1ea1nh CD v\u00e0 c\u1ea1nh BC ng\u1eafn h\u01a1n AD 15 cm. Tr\u00ean AD l\u1ea5y \u0111i\u1ec3m M sao cho \u0111o\u1ea1n th\u1eb3ng AM b\u1eb1ng 2 ph\u1ea7n 3 c\u1ea1nh AD. N\u1ed1i M v\u1edbi B v\u00e0 C. T\u00ednh : a. Di\u1ec7n t\u00edch h\u00ecnh thang ABCD bi\u1ebft chi\u1ec1u cao l\u00e0 36 cm b. C\u1ea1nh AD, BC c\u1ee7a h\u00ecnh thang ABCD c. Chi\u1ec1u cao h\u1ea1 t\u1eeb M c\u1ee7a h\u00ecnh MBC","a) T\u1ed5ng 2 \u0111\u00e1y AB v\u00e0 CD: (405+15):2 = 210 (cm) T\u1ed5ng s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 2+5 = 7 (ph\u1ea7n) C\u1ea1nh \u0111\u00e1y AB: 210 : 7 x 2 = 60 (cm) C\u1ea1nh \u0111\u00e1y DC: 210 \u2013 60 = 150 (cm) Di\u1ec7n t\u00edch h\u00ecnh thang ABCD: (60+150) x 36 : 2 = 3780 (cm2) b) T\u1ed5ng 2 c\u1ea1nh AD v\u00e0 BC: 405 \u2013 210 = 195 (cm) C\u1ea1nh AD: (195+15):2 = 105 (cm) C\u1ea1nh BC: 195 \u2013 105 = 90 (cm) c) AM=2\/3AD => DM=1\/2MA=1\/3AD N\u1ed1i AC v\u00e0 n\u1ed1i BD. *.Ta c\u00f3: SABC = 2\/5SADC T\u1ed5ng s\u1ed1 ph\u1ea7n b\u1eb1ng nhau : 2 + 5 = 7 (ph\u1ea7n) SADC = 3780 : 7 x 5 = 2700 (cm2) SCDM=1\/3SADC = 2700 : 3 = 900 (cm2) *.T\u01b0\u01a1ng t\u1ef1: SADB = 3780 :7 x 2 = 1080 (cm2) SBMA=2\/3SADB = 1080 x 2\/3 = 720 (cm2) M\u00e0: SMBC = SABCD \u2013 (SMAB+SMCD) = 3780 \u2013 (720+900) = 2160 (cm2) Chi\u1ec1u cao h\u1ea1 t\u1eeb M c\u1ee7a tam gi\u00e1c MBC 2160 x 2 : 90 = 48 (cm) B\u00e0i 67: Cho h\u00ecnh thang vu\u00f4ng ABCD, vu\u00f4ng t\u1ea1i A. C\u00f3 \u0111\u00e1y DC g\u1ea5p 2 l\u1ea7n \u0111\u00e1y AB. K\u00e9o d\u00e0i AD c\u1eaft BC t\u1ea1i G. T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c GAB. Bi\u1ebft di\u1ec7n t\u00edch h\u00ecnh thang ABCD l\u00e0 48dm2.","X\u00e9t 2 tam gi\u00e1c BDG v\u00e0 CDG c\u00f3 chung c\u1ea1nh \u0111\u00e1y DG, AB = 1\/2DC n\u00ean SBDG = 1\/2SCDG Suy ra SBDG = SBDC SDAB = 1\/2SBDC (2 \u0111\u01b0\u1eddng cao b\u1eb1ng nhau b\u1eb1ng \u0111\u01b0\u1eddng cao h\u00ecnh thang, AB=1\/2DC). Suy ra SGAB = SDAB M\u00e0 SDAB = 48 : (1+2) = 16 (dm2) B\u00e0i 68: Cho h\u00ecnh thang ABCD. \u0110\u00e1y l\u1edbn CD g\u1ea5p \u0111\u00f4i \u0111\u00e1y b\u00e9 AB. Hai \u0111\u01b0\u1eddng ch\u00e9o AC v\u00e0 BD c\u1eaft nhau t\u1ea1i G. Bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c ABG l\u00e0 34,5cm2. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang ABCD. SABC = 1\/2SADC (AB=1\/2CD, \u0111\u01b0\u1eddng cao b\u1eb1ng nhau b\u1eb1ng \u0111\u01b0\u1eddng cao h\u00ecnh thang) Suy ra \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B b\u1eb1ng 1\/2 \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb D xu\u1ed1ng AC. Hai \u0111\u01b0\u1eddng cao n\u00e0y c\u0169ng l\u00e0 2 \u0111\u01b0\u1eddng cao c\u1ee7a 2 tam gi\u00e1c ABG v\u00e0 AGD m\u00e0 hai tam gi\u00e1c n\u00e0y c\u00f3 c\u1ea1nh \u0111\u00e1y chung AG. N\u00ean SAGD = SABG x 2 = 34,5 x 2 = 69 (cm2). SABD = SABG + SAGD = 34,5 + 69 = 103,5 (cm2) T\u01b0\u01a1ng t\u1ef1: SBDC = SABD x 2 = 103,5 x 2 = 207 (cm2) M\u00e0 SABCD = SABD + SBDC = 103,5 + 207 = 310,5 (cm2) B\u00e0i 69: Cho h\u00ecnh thang ABCD c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng 600cm2 Bi\u1ebft AM=MQ=QD;BN=NP=PC. T\u00ednh di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c MNPQ","N\u1ed1i BD; BM; PD. Ta c\u00f3: SABD+SCBD= 600 cm2 (1) M\u00e0 SABM = 1\/3SABD (2) (AM=1\/3AD, chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B) T\u01b0\u01a1ng t\u1ef1: SDPC = 1\/3SCBD (3) T\u1eeb (1), (2), (3) cho ta: SABM + SDPC = 600 : 3 = 200 (cm2) Suy ra : SMBPD = 600 \u2013 200 = 400 (cm2) N\u1ed1i MP, ta \u0111\u01b0\u1ee3c : SMBP + SPMD = 400 (cm2) T\u01b0\u01a1ng t\u1ef1 nh\u01b0 tr\u00ean, ta c\u00f3 : SMBN = 1\/2 SMBP SPDQ = 1\/2 SPDM Suy ra : SPDQ + SMBN = 400 : 2 = 200 (cm2) M\u00e0 SMNPQ = SMBPD \u2013 (SPDQ + SMBN) = 400 \u2013 200 SMNPQ = 200 cm2. B\u00e0i 70: H\u00ecnh t\u1ee9 gi\u00e1c MNPQ c\u00f3 hai \u0111\u01b0\u1eddng ch\u00e9o MP v\u00e0 NQ c\u1eaft nhau t\u1ea1i O. Bi\u1ebft di\u1ec7n t\u00edch c\u00e1c h\u00ecnh tam gi\u00e1c MNO; NPO; OPQ l\u1ea7n l\u01b0\u1ee3t l\u00e0 : 670cm2; 2010cm2; 2070cm2. Di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c MNPQ l\u00e0 : \u2026\u2026\u2026.cm2. X\u00e9t 2 tam gi\u00e1c MON v\u00e0 PON c\u00f3 ON chung n\u00ean \u0111\u01b0\u1eddng cao c\u1ee7a 2 tam gi\u00e1c t\u1ec9 l\u1ec7 v\u1edbi di\u1ec7n t\u00edch. T\u1ec9 s\u1ed1 \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb P v\u00e0 \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb M xu\u1ed1ng ON l\u00e0 2010\/670 = 201\/67 2 \u0111\u01b0\u1eddng cao n\u00e0y c\u0169ng l\u00e0 2 \u0111\u01b0\u1eddng cao c\u1ee7a 2 tam gi\u00e1c PQN v\u00e0 MQN. SPQN = 2070+2010 = 4080 (cm2) Suy ra SMQN = 4080 : 201 x 67 = 1360 (cm2) SMNPQ = SPQN + SMQN = 4080 + 1360 = 5440 (cm2) B\u00e0i 71:","Cho h\u00ecnh ch\u1eef nh\u1eadt ABCD, tr\u00ean CD l\u1ea5y M, n\u1ed1i B v\u1edbi M. L\u1ea5y \u0111i\u1ec3m I l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng BM. N\u1ed1i A v\u1edbi I. Tr\u00ean \u0111o\u1ea1n th\u1eb3ng AI l\u1ea5y \u0111i\u1ec3m N sao cho AN b\u1eb1ng 2\/3 AI. N\u1ed1i M v\u1edbi N. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ABCD, bi\u1ebft di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c MNI b\u1eb1ng 15 cm2. AN = 2\/3 AI ==> NI = 1\/3 AI SAIM = SMNI x 3 (AI=NI x 3, chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb M). SAIM = 15 x 3 = 45 (cm2) SABM = SAIM x 2 (BM=IM x 2, chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A). SABM = 45 x 2 = 90 (cm2) X\u00e9t 3 tam gi\u00e1c ABM ; BMC v\u00e0 AMD. Ta th\u1ea5y AB = MD+MC (chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt), 3 tam gi\u00e1c n\u00e0y c\u00f3 3 \u0111\u01b0\u1eddng cao b\u1eb1ng nhau b\u1eb1ng chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt n\u00ean. SABM = SBMC + SAMD = 90 cm2. Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ABCD 90 x 2 = 180 (cm2) B\u00e0i 72: Cho tam gi\u00e1c ABC. \u0110i\u1ec3m M l\u00e0 di\u1ec3m ch\u00ednh gi\u1eefa c\u1ea1nh AB. Tr\u00ean AC l\u1ea5y \u0111i\u1ec3m N sao cho AN = 1\/2 NC. Hai \u0111o\u1ea1n th\u1eb3ng BN v\u00e0 CM c\u1eaft nhau t\u1ea1i K. T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c AKC bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c KAB b\u1eb1ng 42dm2 Ta c\u00f3: SABN = 1\/2SBCN (AN=1\/2NC, chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B). Hai tam gi\u00e1c n\u00e0y l\u1ea1i c\u00f3 chung c\u1ea1nh BN n\u00ean hai \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A v\u00e0 t\u1eeb C xu\u1ed1ng BN b\u1eb1ng nhau. Hai \u0111\u01b0\u1eddng cao n\u00e0y c\u0169ng l\u00e0 hai \u0111\u01b0\u1eddng cao c\u1ee7a hai tam gi\u00e1c ABK v\u00e0 CBK c\u00f3 c\u1ea1nh \u0111\u00e1y chung l\u00e0 BK. N\u00ean SABK = 1\/2SCBK. (1) T\u01b0\u01a1ng t\u1ef1 ta l\u1ea1i c\u00f3 SCBK = SACK (2) T\u1eeb (1) v\u00e0 (2) ta \u0111\u01b0\u1ee3c SABK = 1\/2SACK V\u1eady SACK = SABK x 2 = 42 x 2 = 84 (dm2)","B\u00e0i 73: Cho t\u1ee9 gi\u00e1c ABCD, \u0111\u01b0\u1eddng ch\u00e9o AC v\u00e0 BD. G\u1ecdi E l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AC, t\u1eeb E k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi BD c\u1eaft DC t\u1ea1i F. N\u1ed1i B v\u1edbi F. Ch\u1ee9ng t\u1ecf r\u1eb1ng \u0111o\u1ea1n BF chia t\u1ee9 gi\u00e1c ABCD th\u00e0nh hai ph\u1ea7n c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau. N\u1ed1i BE v\u00e0 DE c\u1eaft BF t\u1ea1i K. Trong tam gi\u00e1c ABC ta c\u00f3: SABE = 1\/2 SABC (1) (AE = 1\/2AC , chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B). T\u01b0\u01a1ng t\u1ef1 ta c\u00f3 SADE = 1\/2 SADC (2) T\u1eeb (1) v\u00e0 (2) cho ta SABED = 1\/2 SABCD H\u00ecnh thang DBEF cho ta SBFE = SDFE (chung c\u1ea1nh \u0111\u00e1y FE, hai \u0111\u01b0\u1eddng cao b\u1eb1ng nhau b\u1eb1ng chi\u1ec1u cao h\u00ecnh thang). M\u00e0 2 tam gi\u00e1c n\u00e0y c\u00f3 ph\u1ea7n chung l\u00e0 SKFE suy ra SBKE = SDKF (3) Ta th\u1ea5y: SABFD = SABED \u2013 SBKE + SDKF Theo (3) ta c\u00f3: SABFD = SABED Hay SABFD = 1\/2 SABCD V\u1eady \u0111o\u1ea1n th\u1eb3ng BF chia h\u00ecnh t\u1ee9 gi\u00e1c ABCD th\u00e0nh hai ph\u1ea7n c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau. B\u00e0i 74: Cho h\u00ecnh thang vu\u00f4ng ABCD, vu\u00f4ng g\u00f3c t\u1ea1i A v\u00e0 D, \u0111\u00e1y AB=1\/3 CD.K\u00e9o d\u00e0i DA v\u00e0 CB c\u1eaft nhau t\u1ea1i E. a) So s\u00e1nh di\u1ec7n t\u00edch hai h\u00ecnh tam gi\u00e1c ABC v\u00e0 ADC. b)Bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c ABE b\u1eb1ng 7 x\u0103ng-ti-m\u00e9t vu\u00f4ng. T\u00ecm di\u1ec7n t\u00edch h\u00ecnh thang ABCD a) X\u00e9t 2 tam gi\u00e1c ABC v\u00e0 ADC c\u00f3: AB = 1\/3DC, hai \u0111\u01b0\u1eddng cao t\u01b0\u01a1g \u1ee9ng v\u1edbi 2 c\u1ea1nh \u0111\u00e1y b\u1eb1ng nhau b\u1eb1ng chi\u1ec1u cao h\u00ecnh thang. V\u1eady SABC = 1\/3 SADC","b) N\u1ed1i BD. T\u01b0\u01a1ng t\u1ef1 ta c\u00f3 SABD = 1\/3 SBDC 2 tam gi\u00e1c EBD v\u00e0 ECD c\u00f3 chung c\u1ea1nh \u0111\u00e1y AD, 2 \u0111\u01b0\u1eddng cao c\u1ee7a 2 tam gi\u00e1c n\u00e0y AB = 1\/3DC V\u1eady: SEBD = 1\/3 SECD M\u1eb7t kh\u00e1c 2 tam gi\u00e1c n\u00e0y c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb D xu\u1ed1ng EC n\u00ean EB = 1\/3 EC hay EB = 1\/2 BC SEBD = 1\/2SBDC. Ph\u00e2n s\u1ed1 ch\u1ec9 7cm2 l\u00e0: 1\/2 \u2013 1\/3 = 1\/6 (SBDC) Di\u1ec7n t\u00edch tam gi\u00e1c BDC : 7 x 6 = 42 (cm2) Di\u1ec7n t\u00edch tam gi\u00e1c ABD: 42 : 3 = 14 (cm2) Di\u1ec7n t\u00edch h\u00ecnh thang ABCD: 42 + 14 = 56 (cm2) B\u00e0i 75: Cho h\u00ecnh thang ABCD, AB = 1\/2 CD. K\u00e9o d\u00e0i DA c\u1ec1 ph\u00eda A v\u00e0 CB v\u1ec1 ph\u00eda B c\u1eaft t\u1ea1i M. a) T\u00ec t\u1ec9 s\u1ed1 MA\/MD v\u00e0 MB\/MC b) t\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang ABCD, bi\u1ebft di\u1ec7n t\u00edch MAB = 9cm2 a)Ta c\u00f3 SABD = 1\/2 SACD = 1\/2 SBDC (1) (V\u00ec AB=1\/2CD, 2 \u0111\u01b0\u1eddng cao t\u01b0\u01a1ng \u1ee9ng b\u1eb1ng nhau b\u1eb1ng chi\u1ec1u cao h\u00ecnh thang). M\u00e0 2 tam gi\u00e1c n\u00e0y c\u00f3 AD chung. Suy ra \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb C xu\u1ed1ng AD g\u1ea5p 2 l\u1ea7n \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B xu\u1ed1ng AD. Hai \u0111\u01b0\u1eddng cao n\u00e0y c\u0169ng l\u00e0 2 \u0111\u01b0\u1eddng cao c\u1ee7a 2 tam gi\u00e1c MBD v\u00e0 MCD. Hai tam gi\u00e1c n\u00e0y c\u00f3 c\u1ea1nh \u0111\u00e1y MD chung n\u00ean SMBD = 1\/2SMCD (2) T\u1eeb (1) v\u00e0 (2) cho ta SMAB = SABD. Hai tam gi\u00e1c n\u00e0y l\u1ea1i c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B. Suy ra MA = AD hay MA = 1\/2MD => MA\/MD = 1\/2 T\u01b0\u01a1ng t\u1ef1: MB\/MC = 1\/2 b)SABCD = SABD + SBCD = 9 + 9x2 = 27 (cm2) B\u00e0i 76: M\u1ed9t s\u00e2n tr\u01b0\u1eddng c\u00f3 chu vi b\u1eb1ng 142m. N\u1ebfu t\u0103ng chi\u1ec1u r\u1ed9ng th\u00eam 15m, \u0111\u1ed3ng th\u1eddi gi\u1ea3m chi\u1ec1u d\u00e0i \u0111i 15m th\u00ec di\u1ec7n t\u00edch c\u1ee7a s\u00e2n tr\u01b0\u1eddng kh\u00f4ng thay \u0111\u1ed5i. T\u00ednh di\u1ec7n t\u00edch s\u00e2n tr\u01b0\u1eddng \u0111\u00f3?","\u0110\u1ec3 di\u1ec7n t\u00edch s\u00e2n tr\u01b0\u1eddng kh\u00f4ng \u0111\u1ed5i th\u00ec 2 h\u00ecnh ch\u1eef nh\u1eadt nh\u1ecf ph\u1ea3i c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau v\u00e0 c\u00f3 chi\u1ec1u r\u1ed9ng b\u1eb1ng nhau 15m, chi\u1ec1u d\u00e0i b\u1eb1ng chi\u1ec1u r\u1ed9ng s\u00e2n tr\u01b0\u1eddng. Cho ta th\u1ea5y s\u00e2n tr\u01b0\u1eddng c\u00f3 chi\u1ec1u d\u00e0i h\u01a1n chi\u1ec1u r\u1ed9ng 15m. N\u1eeda chu vi s\u00e2n tr\u01b0\u1eddng l\u00e0: 142 : 2 = 71 (m) Chi\u1ec1u r\u1ed9ng s\u00e2n tr\u01b0\u1eddng l\u00e0: (71 \u2013 15) : 2 = 28 (m) Chi\u1ec1u d\u00e0i s\u00e2n tr\u01b0\u1eddng l\u00e0: 71 \u2013 28 = 43 (m) Di\u1ec7n t\u00edch s\u00e2n tr\u01b0\u1eddng l\u00e0: 43 x 28 = 1204 (m2) \u0110\u00e1p s\u1ed1: 1204 m2. B\u00e0i 77: H\u00ecnh thang ABCD c\u00f3 hai \u0111\u01b0\u1eddng ch\u00e9o AC v\u00e0 BD c\u1eaft nhau t\u1ea1i O. T\u00ednh di\u1ec7n tich h\u00ecnh thang \u0111\u00f3, bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c AOB = 4cm2, di\u1ec7n t\u00edch tam gi\u00e1c COD = 8cm2 SABD\/SBDC = AB\/DC => SAOB\/SBOC = AB\/DC T\u01b0\u01a1ng t\u1ef1: SBOC\/SDOC = AB\/DC Suy ra: SAOB\/SBOC = SBOC\/SDOC SBOC x SBOC = SAOB x SDOC = 4 x 8 = 32 SBOC \u2248 5,65 (cm2) Ta l\u1ea1i c\u00f3: SAOD = SBOC (Do SABD = SABC c\u00f3 ABO l\u00e0 ph\u1ea7n chung) V\u1eady: SABCD = 8 + 4 + 5,65 + 5,65 = 23,3 (cm2) B\u00e0i 78: Cho h\u00ecnh thang ABCD; \u0111\u00e1y nh\u1ecf AB; \u0111\u00e1y l\u1edbn CD. Hai \u0111\u01b0\u1eddng ch\u00e9o AC v\u00e0 BD c\u1eaft nhau t\u1ea1i I. Bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c ABI l\u00e0 24,5 cm2; di\u1ec7n t\u00edch tam gi\u00e1c ICD l\u00e0 98 cm2. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang ABCD.","H\u00ecnh thang ABCD cho ta SAID=SBIC g\u1ecdi di\u1ec7n t\u00edch 2 h\u00ecnh tam gi\u00e1c n\u00e0y l\u00e0 n. X\u00e9t 2 tam gi\u00e1c AIB v\u00e0 AID chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A n\u00ean 2 c\u1ea1nh \u0111\u00e1y IB v\u00e0 ID t\u1ec9 l\u1ec7 v\u1edbi 2 di\u1ec7n t\u00edch: IB\/ID = 24,5\/n T\u01b0\u01a1ng t\u1ef1 v\u1edbi 2 tam gi\u00e1c CIB v\u00e0 CID ta c\u00f3 IB\/ID = n\/98 Suy ra: 24,5\/n = n\/98 n x n = 98 x 24,5 = 2401 V\u1eady n= 49 SABCD = 24,5 + 98 + 49x2 = 220,5 (cm2) B\u00e0i 79: M\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chu vi l\u00e0 18 cm. N\u1ebfu gi\u1ea3m chi\u1ec1u d\u00e0i 20% v\u00e0 t\u0103ng chi\u1ec1u r\u1ed9ng 25% thi chu vi kh\u00f4ng \u0111\u1ed5i. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt. N\u1eeda chu vi (a+b) l\u00e0: 18 : 2 = 9 (cm) Ta c\u00f3 : a + b = 80% a + 125% b => 20% a = 25% b => 1\/5. a = 1\/4 . b Cho bi\u1ebft a c\u00f3 5 ph\u1ea7n th\u00ec b c\u00f3 4 ph\u1ea7n. T\u1ed5ng s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 4 + 5 = 9 (ph\u1ea7n) Chi\u1ec1u d\u00e0i: 9 : 9 x 5 = 5 (cm) Chi\u1ec1u r\u1ed9ng: 9 \u2013 5 = 4 (cm) Di\u1ec7n t\u00edch: 5 x 4 = 20 (cm2) \u0110\u00e1p s\u1ed1: 20 (cm2) B\u00e0i 80: Cho h\u00ecnh thang ABCD c\u00f3 \u0111\u00e1y DC = AB x 2. Tr\u00ean AB l\u1ea5y \u0111i\u1ec3m M sao cho AM = BM; DM c\u1eaft AC t\u1ea1i K. a. T\u00ecm t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch 2 tam gi\u00e1c ACM v\u00e0 DMC; AKD v\u00e0 MKC. b. Bi\u1ebft di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c MCB = 15 cm2. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c AMK a). Do DC=ABx2 n\u00ean SABC = 1\/2 SDMC (1) Hai tam gi\u00e1c n\u00e0y c\u00f3 2 \u0111\u01b0\u1eddng cao b\u1eb1ng nhau b\u1eb1ng chi\u1ec1u cao h\u00ecnh thang ABCD. M\u1eb7t kh\u00e1c SACM = 1\/2 SABC (2) V\u00ec AM=BM hay AM=1\/2AB v\u00e0 hai tam gi\u00e1c n\u00e0y c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb C T\u1eeb (1) v\u00e0 (2) suy ra SACM = 1\/4 SDMC","Trong h\u00ecnh thang AMCD cho ta SAMD = SAMC (chung c\u1ea1nh \u0111\u00e1y AM v\u00e0 c\u00f3 2 \u0111\u01b0\u1eddng cao b\u1eb1ng nhau b\u1eb1ng chi\u1ec1u cao c\u1ee7a h\u00ecnh thang). M\u00e0 2 tam gi\u00e1c AMD v\u00e0 AMC c\u00f3 ph\u1ea7n chung l\u00e0 AMK n\u00ean: SAKD = SMKC b). Ta c\u00f3 SAMC = SBMC = 15 cm2 (AM=MB; chung \u0111\u01b0\u1eddng cao). Hay: SAMD = SAMC = SBMC X\u00e9t 2 tam gi\u00e1c AMD v\u00e0 CMD c\u00f3 AM=1\/4DC (t\u1eeb \u0111\u1ec1 b\u00e0i) v\u00e0 2 \u0111\u01b0\u1eddng cao b\u1eb1ng nhau b\u1eb1ng chi\u1ec1u cao h\u00ecnh thang. N\u00ean SAMD = 1\/4SCMD. Hai tam gi\u00e1c n\u00e0y l\u1ea1i c\u00f3 chung c\u1ea1nh \u0111\u00e1y MD n\u00ean \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A b\u1eb1ng 1\/4 \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb C xu\u1ed1ng MD. Hai \u0111\u01b0\u1eddng cao n\u00e0y c\u0169ng ch\u00ednh l\u00e0 2 \u0111\u01b0\u1eddng cao c\u1ee7a 2 tam gi\u00e1c AMK v\u00e0 CMK. M\u1eb7t kh\u00e1c hai tam gi\u00e1c n\u00e0y l\u1ea1i c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb M xu\u1ed1ng AC n\u00ean suy ra AK = 1\/4KC => SAMK = 1\/4SCMK => SAMK = 1\/5SAMC SAMK = 15 : 5 = 3 (cm2) B\u00e0i 81: Cho tam gi\u00e1c ABC. Tr\u00ean c\u1ea1nh BC l\u1ea5y \u0111i\u1ec3m M sao cho BM=1\/2 MC v\u00e0 tr\u00ean c\u1ea1nh CA l\u1ea5y \u0111i\u1ec3m N sao cho NC=1\/3 NA. \u0110\u01b0\u1eddng th\u1eb3ng MN c\u1eaft c\u1ea1nh AB k\u00e9o d\u00e0i t\u1ea1i \u0111i\u1ec3m K. a, \u0110\u01b0\u1eddng th\u1eb3ng MN c\u1eaft tam gi\u00e1c ABC th\u00e0nh 2 ph\u1ea7n. T\u00ednh di\u1ec7n t\u00edch c\u00e1c ph\u1ea7n \u0111\u00f3 n\u1ebfu bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c ABC b\u1eb1ng 36cm2. b, So s\u00e1nh c\u00e1c \u0111o\u1ea1n KA v\u00e0 KB. a).N\u1ed1i AM. T\u1eeb BM = 1\/2 MC => MC = 2\/3 BC ; NC = 1\/3 NA => NC = 1\/4 AC SAMC = 2\/3 SABC = 2\/3 x 36 = 24 (cm2) V\u00ec MC = 2\/3 BC, hai tam gi\u00e1c c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A. SMNC = 1\/4 SAMC = 1\/4 x 24 = 6 (cm2) SAMNB = SABC \u2013 SMNC = 36 \u2013 6 = 30 (cm2) b).N\u1ed1i BN ; KC. SKBM = 1\/2 SKCM (V\u00ec BM=1\/2MC, chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb K). Hai tam gi\u00e1c n\u00e0y c\u00f3 KM chung n\u00ean \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb C g\u1ea5p 2 l\u1ea7n \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B xu\u1ed1ng KM. M\u00e0 2 \u0111\u01b0\u1eddng cao n\u00e0y c\u0169ng l\u00e0 2 \u0111\u01b0\u1eddng cao c\u1ee7a hai tam gi\u00e1c NKC v\u00e0 NKB c\u00f3 c\u1ea1nh \u0111\u00e1y KN chung. Suy ra : SNKB = 1\/2 SNKC (1) M\u00e0 : SNKC = 1\/3 SNKA (2) (NC = 1\/3NA, chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb K). T\u1eeb (1) v\u00e0 (2) cho ta : SNKB = 1\/6 SNKA.","Hai tam gi\u00e1c n\u00e0y l\u1ea1i c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb N xu\u1ed1ng AK. Suy ra : KB = 1\/6 KA B\u00e0i 82: Trong h\u00ecnh v\u1ebd b\u00ean l\u00e0 m\u1ed9t tam gi\u00e1c \u0111\u01b0\u1ee3c t\u1ea1o n\u00ean t\u1eeb ba m\u1ea3nh gi\u1ea5y m\u00e0u kh\u00e1c nhau. Hai m\u1ea3nh gi\u1ea5y \u0111\u1ecf v\u00e0 xanh c\u00f3 h\u00ecnh tam gi\u00e1c vu\u00f4ng v\u1edbi c\u1ea1nh l\u1edbn nh\u1ea5t d\u00e0i t\u01b0\u01a1ng \u1ee9ng 10 cm v\u00e0 6 cm, c\u00f2n m\u1ea3nh gi\u1ea5y v\u00e0ng l\u00e0 h\u00ecnh vu\u00f4ng. T\u00ednh t\u1ed5ng di\u1ec7n t\u00edch c\u00e1c m\u1ea3nh gi\u1ea5y \u0111\u1ecf v\u00e0 xanh? Do ADEG l\u00e0 h\u00ecnh vu\u00f4ng n\u00ean DE=EG. G\u00f3c G c\u1ee7a tam gi\u00e1c GEC l\u00e0 g\u00f3c vu\u00f4ng n\u00ean ta c\u00f3 th\u1ec3 di chuy\u1ec3n tam gi\u00e1c GEC gh\u00e9p v\u00e0o c\u1ea1nh DE \u0111\u1ec3 c\u00f3 tam gi\u00e1c m\u1edbi DEH b\u1eb1ng v\u1edbi tam gi\u00e1c GEC. Tam gi\u00e1c vu\u00f4ng BEH (ki\u1ec3m tra b\u1eb1ng \u00ea ke) c\u00f3 2 c\u1ea1nh g\u00f3c vu\u00f4ng b\u1eb1ng 10cm v\u00e0 6cm. Di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c n\u00e0y b\u1eb1ng t\u1ed5ng di\u1ec7n t\u00edch 2 tam gi\u00e1c DBE v\u00e0 DEH. Ch\u00ednh l\u00e0 t\u1ed5ng c\u1ee7a 2 tam gi\u00e1c DBE v\u00e0 GEC. T\u1ed5ng di\u1ec7n t\u00edch 2 tam gi\u00e1c \u0111\u00f3 l\u00e0: 10 x 6 : 2 = 30 (cm2) B\u00e0i 83: Cho tam gi\u00e1c ABC .Tr\u00ean c\u1ea1nh AB l\u1ea5y \u0111i\u1ec3m M sao cho AM = 1\/3 x AB , tr\u00ean c\u1ea1nh BC l\u1ea5y \u0111i\u1ec3m N sao cho BN = 1\/2 x BC . K\u00e9o d\u00e0i NM v\u00e0 CA c\u1eaft nhau t\u1ea1i K a, So s\u00e1nh di\u1ec7n t\u00edch tam gi\u00e1c AMN v\u00e0 di\u1ec7n t\u00edch tam gi\u00e1c CMN. b. So s\u00e1nh \u0111\u1ed9 d\u00e0i c\u00e1c \u0111o\u1ea1n th\u1eb3ng KA v\u00e0 AC","a).Ta c\u00f3: SABN = SANC (BN=NC v\u00e0 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A). => SABN = 1\/2 SABC SAMN = 1\/3 SABN (AM=1\/3AB v\u00e0 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb N). => SAMN = 1\/3 x 1\/2 SABC = 1\/6 SABC (1) Ta l\u1ea1i c\u00f3 : SCMB = 2\/3SABC (MB=2\/3AB v\u00e0 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb C). SCMN = 1\/2SCMB (BN=NC v\u00e0 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb M). => SCMN = 1\/2 x 2\/3 SABC = 2\/6SABC (2) T\u1eeb (1) v\u00e0 (2) ta \u0111\u01b0\u1ee3c: SCMN = SAMN x 2 b)N\u1ed1i KB. T\u1eeb 2 tam gi\u00e1c MBN v\u00e0 MNC c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau v\u00e0 chung c\u1ea1nh \u0111\u00e1y MN n\u00ean hai \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B v\u00e0 C xu\u1ed1ng MN b\u1eb1ng nhau. Suy ra SKBM = SKMC (chung c\u1ea1nh \u0111\u00e1y KM v\u00e0 hai \u0111\u01b0\u1eddng cao t\u01b0\u01a1ng \u1ee9ng b\u1eb1ng nhau). M\u1eb7t kh\u00e1c: SKMA = 1\/2SKMB (AM=1\/2MB v\u00e0 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb K). Hay: SKMA = 1\/2SKMC => SKMA = SCMA Hay tam gi\u00e1c n\u00e0y l\u1ea1i c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb M n\u00ean: KA = AC B\u00e0i 84: Cho h\u00ecnh vu\u00f4ng ABCD c\u00f3 c\u1ea1nh 20cm. M l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ea1nh BC . N l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ea1nh CD. \u0110o\u1ea1n AM v\u00e0 \u0111o\u1ea1n BN c\u1eaft nhau t\u1ea1i O. T\u00ednh di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c AOND. AB OM DN C N\u1ed1i AN; OC. SABN = 2.SBNC (AB=2.NC, hai \u0111\u01b0\u1eddng cao b\u1eb1ng nhau b\u1eb1ng c\u1ea1nh h\u00ecnh vu\u00f4ng). Hai tam gi\u00e1c n\u00e0y c\u00f3 c\u1ea1nh BN chung n\u00ean \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A g\u1ea5p 2 l\u1ea7n \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb C xu\u1ed1ng BN.","=> SABO = 2.SBOC (c\u00f3 chung c\u1ea1ng \u0111\u00e1y BO). M\u00e0 SOBM = SOCM (MB=MC, chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb O). N\u00ean SABO = 4.SBOM Di\u1ec7n t\u00edch tam gi\u00e1c ABM: 20 x 10 : 2 = 100 (cm2) Di\u1ec7n t\u00edch tam gi\u00e1c BOM: 100 : (4+1) = 20 (cm2) Ta l\u1ea1i c\u00f3 SABM = SBNC = 100 cm2. Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng ABCD l\u00e0 : 20 x 20 = 400 (cm2) Di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c AOND l\u00e0 : 400 \u2013 (100 + 100 \u2013 20) = 220 (cm2) \u0110\u00e1p s\u1ed1 : 220 cm2. PH\u1ea6N B\u1ed4 SUNG B\u00e0i 85: M\u1ed9t h\u00ecnh tam gi\u00e1c c\u00f3 \u0111\u00e1y b\u1eb1ng 12,8cm, bi\u1ebft n\u1ebfu t\u0103ng \u0111\u00e1y th\u00eam 6,4cm th\u00ec di\u1ec7n t\u00edch t\u0103ng th\u00eam 27,2 cm2. T\u00ednh di\u1ec7n t\u00edch h\u00ednh tam gi\u00e1c \u0111\u00f3. 12,8cm 27,2cm2 6,4cm C\u00e1ch 1: Chi\u1ec1u cao h\u00ecnh tam gi\u00e1c l\u00e0: 27,2 x 2 : 6,4 = 8,5 (cm) Di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c l\u00e0: 12,8 x 8,5 : 2 = 54,4 (cm2) \u0110\u00e1p s\u1ed1: 54,4 cm2 C\u00e1ch 2: Hai tam gi\u00e1c c\u00f9ng chi\u1ec1u cao th\u00ec di\u1ec7n t\u00edch t\u1ec9 l\u1ec7 v\u1edbi c\u1ea1nh \u0111\u00e1y. 12,8cm th\u00ec g\u1ea5p 6,4cm: 12,8 : 6,4 = 2 (l\u1ea7n) Di\u1ec7n t\u00edch tam gi\u00e1c \u0111\u00f3 l\u00e0: 27,2 x 2 = 54,4 (cm2) \u0110\u00e1p s\u1ed1: 54,4 cm2 B\u00e0i 86: M\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chi\u1ec1u d\u00e0i g\u1ea5p r\u01b0\u1edfi chi\u1ec1u r\u1ed9ng. N\u1ebfu m\u1ed7i chi\u1ec1u t\u0103ng th\u00eam 1m th\u00ec di\u1ec7n t\u00edch t\u0103ng th\u00eam 31m2.T\u00ednh di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt \u0111\u00f3.","1m 1m Chi\u1ec1u d\u00e0i 3 ph\u1ea7n, chi\u1ec1u r\u1ed9ng 2 ph\u1ea7n. N\u1eeda chu vi h\u00ecnh ch\u1eef nh\u1eadt: (31 \u2013 1) : 1 = 30 (m) T\u1ed5ng s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 3 + 2 = 5 (ph\u1ea7n) Gi\u00e1 tr\u1ecb 1 ph\u1ea7n: 30 : 5 = 6 (m) Chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt l\u00e0: 6 x 2 = 12 (m) Chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt l\u00e0: 30 \u2013 12 = 18 (m) Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt l\u00e0: 18 x 12 = 216 (m2) \u0110\u00e1p s\u1ed1: 216 m2 B\u00e0i 87: Cho h\u00ecnh tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng 240m2. L\u1ea5y \u0111i\u1ec3m M tr\u00ean c\u1ea1nh AB sao cho AM b\u1eb1ng MB; tr\u00ean c\u1ea1nh AC l\u1ea5y N sao cho AN b\u1eb1ng 1\/2 NC; n\u1ed1i M v\u1edbi N. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c AMN. A N M BC AN = 1\/3 AC. N\u1ed1i NB. Di\u1ec7n t\u00edch tam gi\u00e1c ABN l\u00e0: 240 : 3 = 80 (cm2) AM = 1\/2 AB. Di\u1ec7n t\u00edch tam gi\u00e1c AMN l\u00e0: 80 : 2 = 40 (cm2) \u0110\u00e1p s\u1ed1: 40 cm2. B\u00e0i 88: M\u1ed9t m\u1ea3nh v\u01b0\u1eddn h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chi\u1ec1u d\u00e0i b\u1eb1ng 2 l\u1ea7n chi\u1ec1u r\u1ed9ng. N\u1ebfu t\u0103ng chi\u1ec1u r\u1ed9ng th\u00eam 2 m\u00e9t v\u00e0 gi\u1ea3m chi\u1ec1u d\u00e0i 2m th\u00ec di\u1ec7n t\u00edch m\u1ea3nh v\u01b0\u1eddn t\u0103ng th\u00eam 12mvu\u00f4ng. T\u00ednh di\u1ec7n t\u00edch c\u1ee7a m\u1ea3nh v\u01b0\u1eddn \u0111\u00f3.","Chi\u1ec1u h\u00ecnh ch\u1eef nh\u1eadt m\u00e0u v\u00e0ng h\u01a1n chi\u1ec1u h\u00ecnh ch\u1eef nh\u1eadt m\u00e0u xanh : 12 : 2 = 6 (m) Chi\u1ec1u d\u00e0i m\u1ea3nh v\u01b0\u1eddn h\u01a1n chi\u1ec1u r\u1ed9ng l\u00e0: 6 + 2 = 8 (m) 2m Hi\u1ec7u s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 2 \u2013 1 = 1 (ph\u1ea7n) Chi\u1ec1u r\u1ed9ng m\u1ea3nh v\u01b0\u1eddn l\u00e0: 8 : 1 = 8 (m) Chi\u1ec1u d\u00e0i m\u1ea3nh v\u01b0\u1eddn l\u00e0: 8 x 2 = 16 (m) Di\u1ec7n t\u00edch m\u1ea3nh v\u01b0\u1eddn l\u00e0: 16 x 8 = 128 (m2) \u0110\u00e1p s\u1ed1; 128m2. B\u00e0i 89: M\u1ed9t th\u1eeda ru\u1ed9ng h\u00ecnh tam gi\u00e1c c\u00f3 \u0111\u00e1y l\u00e0 c\u1ea1nh k\u1ec1 v\u1edbi g\u00f3c vu\u00f4ng v\u00e0 d\u00e0i 20m, chi\u1ec1u cao l\u00e0 24m. Nay ng\u01b0\u1eddi ta l\u1ea5y 1 ph\u1ea7n di\u1ec7n t\u00edch c\u1ee7a th\u1eeda ru\u1ed9ng l\u00e0m \u0111\u01b0\u1eddng \u0111i. \u0110\u01b0\u1eddng \u0111i vu\u00f4ng g\u00f3c v\u1edbi chi\u1ec1u cao. Do \u0111\u00f3, \u0111\u00e1y th\u1eeda ru\u1ed9ng ch\u1ec9 c\u00f2n c\u00f3 15m. H\u1ecfi: a. Di\u1ec7n t\u00edch c\u00f2n l\u1ea1i c\u1ee7a th\u1eeda ru\u1ed9ng? b. \u0110\u01b0\u1eddng m\u1edf r\u1ed9ng m\u1ea5y m\u00e9t v\u00e0o chi\u1ec1u cao c\u1ee7a th\u1eeda \u0111\u1ea5t? c. Do th\u1eeda \u0111\u1ea5t gi\u00e1p m\u1eb7t \u0111\u01b0\u1eddng n\u00ean gi\u00e1 tr\u1ecb c\u1ee7a th\u1eeda \u0111\u1ea5t t\u0103ng l\u00ean 400% gi\u00e1 tr\u1ecb tr\u01b0\u1edbc \u0111\u00e2y. H\u1ecfi ch\u1ee7 th\u1eeda \u0111\u1ea5t l\u1ee3i hay thi\u1ec7t trong vi\u1ec7c l\u00e0m \u0111\u01b0\u1eddng v\u00e0 l\u1ee3i hay thi\u1ec7t bao nhi\u00eau ph\u1ea7n tr\u0103m? SABC = 24 x 20 : 2 = 240 (m2) A 15m E SABE = 24 x 15 : 2 = 180 (m2) 20m H C SBEC = 240 \u2013 180 (m2) 24m \u0110\u01b0\u1eddng cao EH = 60 x 2 : 20 = 6 (m) D SDBE = 15 x 6 : 2 = 45 (m2) B a)Di\u1ec7n t\u00edch c\u00f2n l\u1ea1i c\u1ee7a th\u1eeda ru\u1ed9ng l\u00e0: 240 \u2013 (60 + 45) = 135 (m2) b)\u0110\u01b0\u1eddng r\u1ed9ng b\u1eb1ng \u0111\u01b0\u1eddng cao EH = 6m c)Gi\u1ea3 s\u1eed gi\u00e1 \u0111\u1ea5t tr\u01b0\u1edbc \u0111\u00e2y l\u00e0 100 \u0111\u1ed3ng\/m2 th\u00ec gi\u00e1 th\u1eeda ru\u1ed9ng l\u00e0: 100 x 240 = 24000 (\u0111\u1ed3ng) Gi\u00e1 \u0111\u1ea5t c\u00f2n l\u1ea1i: 100 x 135 x 400% = 54000 (\u0111\u1ed3ng) Ti\u1ec1n l\u00e3i l\u00e0: 54000 \u2013 24000 = 30000 (\u0111\u1ed3ng) T\u1ec9 s\u1ed1 % ti\u1ec1n l\u00e3i so v\u1edbi gi\u00e1 d\u1ea5t ban \u0111\u1ea7u l\u00e0 : 30000 : 24000 = 125% B\u00e0i 90: M\u1ed9t khu v\u01b0\u1eddn h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chu vi l\u00e0 210m. Ng\u01b0\u1eddi ta mu\u1ed1n m\u1edf r\u1ed9ng khu v\u01b0\u1eddn \u0111\u00f3 \u0111\u1ec3 \u0111\u01b0\u1ee3c khu v\u01b0\u1eddn m\u1edbi c\u00f3 di\u1ec7n t\u00edch g\u1ea5p 3 l\u1ea7n di\u1ec7n t\u00edch ban \u0111\u1ea7u. Bi\u1ebft r\u1eb1ng chi\u1ec1u r\u1ed9ng \u0111\u01b0\u1ee3c t\u0103ng l\u00ean g\u1ea5p \u0111\u00f4i v\u00e0 khu v\u01b0\u1eddn m\u00f3i l\u00e0 h\u00ecnh vu\u00f4ng. T\u00ednh di\u1ec7n t\u00edch khu v\u01b0\u1eddn ban \u0111\u1ea7u? (2) (3) (1) (4)","H\u00ecnh v\u1ebd cho ta th\u1ea5y di\u1ec7n t\u00edch h\u00ecnh (1) b\u1eb1ng di\u1ec7n t\u00edch h\u00ecnh (2), b\u1eb1ng t\u1ed5ng di\u1ec7n t\u00edch h\u00ecnh (3) v\u00e0 h\u00ecnh (4). (Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng g\u1ea5p 3 l\u1ea7n di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u). S(3) = S(4) => Chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u g\u1ea5p 2 l\u1ea7n chi\u1ec1u r\u1ed9ng h\u00ecnh (4). C\u1ea1nh h\u00ecnh vu\u00f4ng b\u1eb1ng 2 l\u1ea7n chi\u1ec1u r\u1ed9ng v\u00e0 b\u1eb1ng 1,5 l\u1ea7n chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u. G\u1ecdi r l\u00e0 r\u1ed9ng, d l\u00e0 d\u00e0i. ta \u0111\u01b0\u1ee3c : r x 2 = 1,5 x d hay rx4 = dx3 Chi\u1ec1u d\u00e0i c\u00f3 4 ph\u1ea7n, chi\u1ec1u r\u1ed9ng c\u00f3 3 ph\u1ea7n: T\u1ed5ng s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 3 + 4 = 7 (ph\u1ea7n) N\u1eeda chu vi h\u00ecnh ch\u1eef nh\u1eadt l\u00e0: 210 : 2 = 105 (m) Gi\u00e1 tr\u1ecb 1 ph\u1ea7n : 105 : 7 = 15 (m) Chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 : 15 x 4 = 60 (m) Chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 : 105 \u2013 60 = 45 (m) Di\u1ec7n t\u00edch khu v\u01b0\u1eddn ban \u0111\u1ea7u l\u00e0 : 60 x 45 = 2700 (m2) \u0110\u00e1p s\u1ed1 : 2700m2. B\u00e0i 91: Cho tam gi\u00e1c ABC c\u00f3 \u0111\u00e1y BC , k\u00e9o d\u00e0i BC th\u00eam 1 \u0111o\u1ea1n CD = 1\/2 BC th\u00ec di\u1ec7n t\u00edch tam gi\u00e1c ABC t\u0103ng th\u00ean 20 dm2 . T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c ABC . A 20dm2 B CD CD = 1\/2 BC hay BC = CD x 2 Hai tam gi\u00e1c ABC v\u00e0 ACD c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A n\u00ean di\u1ec7n t\u00edch t\u1ec9 l\u1ec7 v\u1edbi c\u1ea1nh \u0111\u00e1y t\u01b0\u01a1ng \u1ee9ng. Di\u1ec7n t\u00edch tam gi\u00e1c ABC l\u00e0: 20 x 2 = 40 (dm2) \u0110\u00e1p s\u1ed1: 40 dm2. B\u00e0i 92: T\u1ed5ng \u2013 T\u1ec9 Cho tam gi\u00e1c ABC vu\u00f4ng g\u00f3c t\u1ea1i B, chu vi l\u00e0 37dm. C\u1ea1nh AB =2\/3 AC, c\u1ea1nh BC = 4\/5AC. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c ABC. AB = 2\/3AC = 10\/15AC BC = 4\/5AC = 12\/15AC","Nh\u01b0 v\u1eady n\u1ebfu AC c\u00f3 15 ph\u1ea7n b\u1eb1ng nhau th\u00ec AB c\u00f3 10 ph\u1ea7n v\u00e0 BC c\u00f3 12 ph\u1ea7n. T\u1ed5ng s\u1ed1 ph\u1ea7n b\u1eb1ng nhau l\u00e0: 10 + 12 + 15 = 37 (ph\u1ea7n) A Gi\u00e1 tr\u1ecb 1 ph\u1ea7n l\u00e0: 37 : 37 = 1 (dm) \u0110\u1ed9 d\u00e0i c\u1ea1nh AB l\u00e0: 1 x 10 = 10 (dm) \u0110\u1ed9 d\u00e0i c\u1ea1nh BC l\u00e0 : 1 x 12 = 12 (dm) B C Di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c ABC l\u00e0 : 10 x 12 : 2 = 60 (dm2) \u0110\u00e1p s\u1ed1 : 60dm2. B\u00e0i 93: M\u1ed9t m\u1ea3nh v\u01b0\u1eddn HCN c\u00f3 chi\u1ec1u r\u1ed9ng b\u1eb1ng 1\/4 chi\u1ec1u d\u00e0i. N\u1ebfu b\u1edbt chi\u1ec1u d\u00e0i \u0111i 5 m \u0111\u1ec3 th\u00eam v\u00e0o chi\u1ec1u r\u1ed9ng th\u00ec l\u00fac \u0111\u00f3 chi\u1ec1u r\u1ed9ng b\u1eb1ng 4\/11 chi\u1ec1u d\u00e0i.T\u00ednh di\u1ec7n th\u00edch m\u1ea3nh v\u01b0\u1eddn? Ta th\u1ea5y: 1\/4 = 3\/12 Khi b\u1edbt \u0111i chi\u1ec1u d\u00e0i v\u00e0 th\u00eam chi\u1ec1u r\u1ed9ng th\u00ec n\u1eeda chu vi v\u1eabn kh\u00f4ng \u0111\u1ed5i v\u00e0 hi\u1ec7u ch\u00fang s\u1ebd gi\u1ea3m \u0111i: 5 x 2 = 10 (m) 10m \u1ee9ng v\u1edbi s\u1ed1 ph\u1ea7n b\u1eb1ng nhau l\u00e0: (12-3) \u2013 (11-4) = 2 (ph\u1ea7n) Gi\u00e1 tr\u1ecb 1 ph\u1ea7n l\u00e0: 10 : 2 = 5 (m) Chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 : 5 x 3 = 15 (m) Chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 : 5 x 12 = 60 (m) Di\u1ec7n t\u00edch m\u1ea3nh v\u01b0\u1eddn h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 : 60 x 15 = 900 (m2) \u0110\u00e1p s\u1ed1 : 900m2. B\u00e0i 94: Cho h\u00ecnh thang ABCD c\u00f3 \u0111\u00e1y AB = 15 cm,DC=45 cm .Hai \u0111\u01b0\u1eddng ch\u00e9o c\u1eaft nhau t\u1ea1i E bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c EBC l\u00e0 30 cm2.T\u00ednh AB di\u1ec7n t\u00edch h\u00ecnh thang ABCD? 30cm2 E DC AB\/DC = 15\/45 = 1\/3 SABD = 1\/3SBDC (AB\/DC=1\/3 , chi\u1ec1u cao b\u1eb1ng nhau b\u1eb1ng chi\u1ec1u cao h\u00ecnh thang). M\u00e0 2 tam gi\u00e1c n\u00e0y c\u00f3 c\u1ea1nh BD chung n\u00ean 2 \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A v\u00e0 t\u1eeb C xu\u1ed1ng BD t\u1ec9 l\u1ec7 v\u1edbi 2 di\u1ec7n t\u00edch.","Cao t\u1eeb A = 1\/3 cao t\u1eeb C (xu\u1ed1ng BD). T\u01b0\u01a1ng t\u1ef1 suy ra: SABE = 1\/3 SEBC = 30 : 3 = 10 (cm2) SABC = SABE + SEBC = 10 + 30 = 40 (cm2) T\u01b0\u01a1ng t\u1ef1: SADC = 3SABC = 40 x 3 = 120 (cm2) SABCD = SABC + SADC = 40 + 120 = 160 (cm2) \u0110\u00e1p s\u1ed1: SABCD = 160cm2. B\u00e0i 95: Cho h\u00ecnh thanh ABCD c\u00f3 \u0111\u00e1y b\u00e9 b\u1eb1ng 2\/3 \u0111\u00e1y l\u1edbn .Hai \u0111\u01b0\u1eddng ch\u00e9o AC v\u00e0 DB c\u1eaft hau t\u1ea1i M .Bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c AMB b\u1eb1ng 40 cm2 . a. So s\u00e1nh di\u1ec7n t\u00edch tam gi\u00e1c ABC v\u00e0 di\u1ec7n t\u00edch tam gi\u00e1c ADC b. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang ABCD? AB 40cm2 M DC a).Do AB=2\/3CD n\u00ean SABC=2\/3SADC. Hai chi\u1ec1u cao t\u01b0\u01a1ng \u1ee9ng 2 c\u1ea1nh \u0111\u00e1y b\u1eb1ng nhau b\u1eb1ng chi\u1ec1u cao h\u00ecnh thang. b).M\u00e0 hai tam gi\u00e1c n\u00e0y c\u00f3 AC chung n\u00ean \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B b\u1eb1ng 2\/3 \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb D xu\u1ed1ng AC. Hai \u0111\u01b0\u1eddng cao n\u00e0y c\u0169ng l\u00e0 2 \u0111\u01b0\u1eddng cao c\u1ee7a 2 tam gi\u00e1c ABM v\u00e0 ADM. hai tam gi\u00e1c n\u00e0y l\u1ea1i c\u00f3 c\u1ea1nh \u0111\u00e1y AM chung n\u00ean: SAMD = 3\/2SAMB = 40 x 3\/2 = 60 (cm2) SABD = SAMD + SAMB = 60 + 40 = 100 (cm2) T\u01b0\u01a1ng t\u1ef1 ta c\u00f3 SCDB = 3\/2SADB = 100 x 3\/2 = 150 (cm2) SABCD = SABD + SCDB = 100 + 150 = 250 (cm2) B\u00e0i 96: Cho m\u1ed9t h\u00ecnh vu\u00f4ng , bi\u1ebft n\u1ebfu t\u0103ng c\u1ea1nh h\u00ecnh vu\u00f4ng l\u00ean 5 cm th\u00ec di\u1ec7n t\u00edch t\u0103ng 185cm2 . T\u00ednh di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng \u0111\u00e3 cho 5cm 5cm Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng m\u00e0u v\u00e0ng.","5 x 5 = 25 (cm2) T\u1ed5ng di\u1ec7n t\u00edch 2 h\u00ecnh ch\u1eef nh\u1eadt m\u00e0u tr\u1eafng. 185 \u2013 25 = 160 (cm2) C\u1ea1nh h\u00ecnh vu\u00f4ng ban \u0111\u1ea7u. (160 : 5) : 2 = 16 (cm) Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng \u0111\u00e3 cho. 16 x 16 = 256 (cm2) \u0110\u00e1p s\u1ed1: 256cm2. B\u00e0i 97: Cho tam giac ABC c\u00f3 AB = 15 cm, AC = 20 cm. Tren canh AB lay diem D sao cho AD = 10 cm, tren canh AC lay diem E sao cho AE = 15cm.Noi D voi E . Tinh dien tich tam giac ABC biet dien tich tam giac ADE bang 45 cm2 A D E B C BD = 15 \u2013 10 = 5 (cm) => BD\/AD = 5\/10 = 1\/2 EC = 20 \u2013 15 = 5 (cm) => EC\/AE = 5\/15 = 1\/3 SEBD = 1\/2SEAD = 45 : 2 = 22,5 (cm2) BD = 1\/2AD, chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb E. SABE = 45 + 22,5 = 67,5 (cm2) T\u01b0\u01a1ng t\u1ef1. SBEC = 1\/3SABE = 67,5 : 3 = 22,5 (cm2) SABC = 67,5 + 22,5 = 90 (cm2) B\u00e0i 98: Cho h\u00ecnh tam gi\u00e1c ABC c\u00f3 AB=12cm, AC=15cm. K\u00e9o d\u00e0i AB v\u1ec1 ph\u00eda B, AC v\u1ec1 ph\u00eda C. L\u1ea7n l\u01b0\u1ee3t l\u1ea5y \u0111i\u1ec3m M, N sao cho AM=AN=20cm. N\u1ed1i M v\u1edbi N, bi\u1ebft di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c ABC b\u1eb1ng 45cm2. T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c AMN. A 12 15 B C 8 5 M N MB = 20 \u2013 12 = 8 (cm) NC = 20 \u2013 15 = 5 (cm)","CN\/AC = 5\/15 = 1\/3 MB\/AB = 8\/12 = 2\/3 SBCN = 1\/3 SABC = 45 : 3 = 15 (cm2) (CN=1\/3AC , chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B). SABN = 45 + 15 = 60 (cm2) T\u01b0\u01a1ng t\u1ef1 : SMBN = 2\/3SABN = 60 x 2\/3 = 40 (cm2) SAMN = SABN + SMBN = 60 + 40 = 100 (cm2) \u0110\u00e1p s\u1ed1 : 100cm2. B\u00e0i 99: Cho h\u00ecnh thang ABCD, \u0111\u00e1y nh\u1ecf AB,\u0111 \u00e1y l\u1edbn CD. Hai \u0111\u01b0\u1eddng ch\u00e9o AC v\u00e0 BD c\u1eaft nhau t\u1ea1i I . Bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c ABI b\u1eb1ng 2,5cm2 v\u00e0 di\u1ec7n t\u00edch tam gi\u00e1c IDC b\u1eb1ng 4,9cm2. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang ABCD. AB 2,5 I C 4,9 D Hai tam gi\u00e1c AIB v\u00e0 CIB c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B n\u00ean SAIB\/SCIB = IA\/IC T\u01b0\u01a1ng t\u1ef1 ta c\u00f3: SAID\/SCID = IA\/IC Suy ra: SAIB\/SCIB = SAID\/SCID = 2,5\/SCIB = SAID\/4,9 Hay: SCIB x SAID = 2,5 x 4,9 = 12,25 (cm2) M\u00e0 SCIB = SAID Do SADC=SBDC v\u00e0 2 tam gi\u00e1c n\u00e0y c\u00f3 ph\u1ea7n chung IDC. Suy ra: SCIB = SAID = 3,5 (cm2) (V\u00ec 3,5 x 3,5 = 12,25). Di\u1ec7n t\u00edch h\u00ecnh thang ABCD l\u00e0: 2,5 + 4,9 + 3,5 + 3,5 = 14,4 (cm2) \u0110\u00e1p s\u1ed1: 14,4cm2. B\u00e0i 100: Cho h\u00ecnh tam gi\u00e1c ABC, l\u1ea5y D l\u00e0 trung \u0111i\u1ec3m c\u1ea1nh BC. N\u1ed1i A v\u1edbi D l\u1ea5y I trung \u0111i\u1ec3m \u0111o\u1ea1n AD. N\u1ed1i B v\u1edbi I k\u00e9o d\u00e0i c\u1eaft AC t\u1ea1i K. T\u00ednh BK \/ IK A K I BC D Ta c\u00f3:","SABD = SACD (BD=DC, chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A) T\u01b0\u01a1ng t\u1ef1 ta c\u00f3: SABI = SBDI = SCID = SCIA = 1\/4 SABC => SABI = 1\/2SBIC Hai tam gi\u00e1c n\u00e0y c\u00f3 chung c\u1ea1nh BI n\u00ean \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb C g\u1ea5p 2 l\u1ea7n \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A. Hai \u0111\u01b0\u1eddng cao n\u00e0y c\u0169ng l\u00e0 hai \u0111\u01b0\u1eddng cao c\u1ee7a 2 tam gi\u00e1c CIK v\u00e0 AIK. => SAIK = 1\/2SCIK => SAIK = 1\/3SCIA Hay SAIK = 1\/3SABI (SABI = 3SAIK). Hai tam gi\u00e1c n\u00e0y c\u00f3 ch\u00fang \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A n\u00ean suy ra: BI = 3 x IK Hay BK = 4 x IK => BK\/IK = 4 B\u00e0i 101: Cho tam gi\u00e1c ABC.Tr\u00ean AB l\u1ea5y \u0111i\u1ec3m M sao cho AM =2\/3 AB. Tr\u00ean AC l\u1ea5y \u0111i\u1ec3m N sao cho AN=3\/4 AC.N\u1ed1i M v\u1edbi N ta \u0111\u01b0\u1ee3c h\u00ecnh t\u1ee9 gi\u00e1c BMNC c\u00f3 di\u1ec7n t\u00edch l\u00e0 120cm2.T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c ABC A MD EN B C Ta th\u1ea5y: SBNC = 1\/4SABC => SABN = 3\/4SABC SNBM = 1\/3SABN = 1\/3 x 3\/4SABC = 1\/4SABC Suy ra : SBMNC = SBNC + SNBM = 1\/4SABC + 1\/4SABC = 1\/2SABC Di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c ABC l\u00e0 : 120 : 2 = 240 (cm2) \u0110\u00e1p s\u1ed1 : 240cm2. B\u00e0i 102: M\u1ed9t c\u00e1i th\u00f9ng h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt c\u00f3 di\u1ec7n t\u00edch \u0111\u00e1y l\u00e0 35dm2. Hi\u1ec7n th\u00f9ng ch\u1ee9a 175 l\u00edt n\u01b0\u1edbc. T\u00ednh chi\u1ec1u cao c\u1ee7a th\u00f9ng, bi\u1ebft chi\u1ec1u cao m\u1ef1c n\u01b0\u1edbc b\u1eb1ng 0,4 l\u1ea7n chi\u1ec1u cao c\u1ee7a th\u00f9ng. L\u01b0\u1ee3ng n\u01b0\u1edbc n\u1ebfu ch\u1ee9a \u0111\u1ea7y th\u00f9ng l\u00e0: 175 : 0,4 = 437,5 (l\u00edt) 437,5 l\u00edt = 437,5dm3 Chi\u1ec1u cao c\u1ee7a th\u00f9ng \u0111\u00f3 l\u00e0: 437,5 : 35 = 12,5 (dm) \u0110\u00e1p s\u1ed1 : 12,5dm. B\u00e0i 103:","C\u00f3 hai b\u1ec3 c\u00e1 d\u1ea1ng h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt .B\u1ec3 l\u1edbn kh\u00f4ng ch\u1ee9a n\u01b0\u1edbc d\u00e0i 1,6m; r\u1ed9ng 0,6m ; cao 1m.B\u1ec3 nh\u1ecf ch\u1ee9a \u0111\u1ea7y n\u01b0\u1edbc d\u00e0i 1m;r\u1ed9ng 0,6m; cao 0,8m.H\u1ecfi n\u1ebfu \u0111\u1ed5 h\u1ebft n\u01b0\u1edbc t\u1eeb b\u1ec3 nh\u1ecf sang b\u1ec3 l\u1edbn th\u00ec m\u1ef1c n\u01b0\u1edbc \u1edf b\u1ec3 l\u1edbn cao bao nhi\u00eau cm ? Th\u1ec3 t\u00edch b\u1ec3 l\u1edbn: 1,6 x 0,6 x 1 = 0,96 (m3) Th\u1ec3 t\u00edch b\u1ec3 nh\u1ecf: 1 x 0,6 x 0,8 = 0,48 (m3) T\u1ec9 s\u1ed1 % c\u1ee7a TT b\u1ebb nh\u1ecf so v\u1edbi b\u1ec3 l\u1edbn l\u00e0. 0,48 : 0,96 = 50% M\u1ee9c n\u01b0\u1edbc \u1edf b\u1ec3 l\u1edbn cao: 1 x 50% = 0,5 (m) 0,5m = 50cm \u0110\u00e1p s\u1ed1: 50cm B\u00e0i 104: Cho tam gi\u00e1c ABC c\u00f3 AB = AC=20 cm.Tr\u00ean c\u1ea1nh AB l\u1ea5y \u0111i\u1ec3m M sao cho MB =8cm.Tr\u00ean c\u1ea1nh AC l\u1ea5y \u0111i\u1ec3m N sao cho AN = 5cm. N\u1ed1i M v\u1edbi N. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c AMN bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c ABC l\u00e0 100cm2. A 5 N M 8 BC SBMC = 8\/20SABC = 100 x 8\/20 = 40 (cm2) Hai tam gi\u00e1c n\u00e0y c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb C v\u00e0 MB = 8\/20AB. SAMC = SABC \u2013 SBMC = 100 \u2013 40 = 60 (cm2) T\u01b0\u01a1ng t\u1ef1: SAMN = 5\/20SAMC = 60 x 5\/20 = 15 (cm2) \u0110\u00e1p s\u1ed1: 15cm2. B\u00e0i 105: Hai c\u00e1i th\u00f9ng h\u00ecnh tr\u00f2n c\u00f3 t\u1ec9 s\u1ed1 b\u00e1n k\u00ednh l\u00e0 1,5. T\u1ed5ng kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a 2 \u0111\u00e1y th\u00f9ng l\u00e0 1,3kg. T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng m\u1ed7i \u0111\u00e1y th\u00f9ng. Bi\u1ebft ch\u00fang \u0111\u01b0\u1ee3c c\u1eaft ra t\u1eeb c\u00f9ng m\u1ed9t l\u00e1 t\u00f4n? T\u1ec9 s\u1ed1 b\u00e1n k\u00ednh l\u00e0 1,5 = 3\/2 8m 4m T\u1ec9 s\u1ed1 di\u1ec7n tich s\u1ebd l\u00e0: 3r x 3r x 3,14 \/ 2r x 2r x 3,14 = 9\/4 4m T\u1ed5ng s\u1ed1 ph\u1ea7n b\u1eb1ng nhau : 4 + 9 = 13 (ph\u1ea7n) Gi\u00e1 tr\u1ecb 1 ph\u1ea7n l\u00e0 : 1,3 : 13 = 0,1 (kg) Kh\u1ed1i l\u01b0\u1ee3ng d\u00e1y th\u00f9ng nh\u1ecf l\u00e0 : 0,1 x 4 = 0,4 (kg) Kh\u1ed1i l\u01b0\u1ee3ng \u0111\u00e1y th\u00f9ng l\u1edbn l\u00e0 :","1,3 \u2013 0,4 = 0,9 (kg) B\u00e0i 106: M\u1ed9t th\u1eeda \u0111\u1ea5t h\u00ecnh tam gi\u00e1c vu\u00f4ng c\u00f3 c\u1ea1nh \u0111\u00e1y l\u00e0 c\u1ea1nh k\u1ec1 v\u1edbi g\u00f3c vu\u00f4ng v\u00e0 d\u00e0i 20m, chi\u1ec1u cao l\u00e0 24m. Nay ng\u01b0\u1eddi ta l\u1ea5y b\u1edbt m\u1ed9t ph\u1ea7n di\u1ec7n t\u00edch c\u1ee7a th\u1eeda \u0111\u1ea5t \u0111\u1ec3 l\u00e0m \u0111\u01b0\u1eddng \u0111i. \u0110\u01b0\u1eddng \u0111i c\u1eaft d\u1ecdc theo c\u1ea1nh \u0111\u00e1y vu\u00f4ng g\u00f3c v\u1edbi chi\u1ec1u cao c\u1ee7a th\u1eeda \u0111\u1ea5t. Do \u0111\u00f3, \u0111\u00e1y th\u1eeda \u0111\u1ea5t ch\u1ec9 c\u00f2n l\u00e0 15m. T\u00ednh di\u1ec7n t\u00edch c\u00f2n l\u1ea1i c\u1ee7a th\u1eeda \u0111\u1ea5t. Di\u1ec7n t\u00edch th\u1eeda \u0111\u1ea5t l\u00e0: A 15 N 20 x 24 : 2 = 240 (m2) 20 C Hai tam ANB v\u00e0 ACB c\u00f3 chung c\u1ea1nh \u0111\u00e1y AB n\u00ean 24 di\u1ec7n t\u00edch ch\u00fang t\u1ec9 l\u1ec7 v\u1edbi \u0111\u01b0\u1eddng cao. M Di\u1ec7n t\u00edch tam gi\u00e1c ABN l\u00e0: B 240 : 20 x 15 = 180 (m2) Di\u1ec7n t\u00edch tam gi\u00e1c NBC l\u00e0: 240 \u2013 180 = 60 (m2) Chi\u1ec1u cao k\u1ebb t\u1eeb N l\u00e0: (MB) 60 x 2 : 20 = 6 (m) Chi\u1ec1u cao c\u00f2n l\u1ea1i c\u1ee7a th\u1eeda \u0111\u1ea5t l\u00e0 : (AM) 24 \u2013 6 = 18 (m) Di\u1ec7n t\u00edch c\u00f2n l\u1ea1i c\u1ee7a th\u1eeda \u0111\u1ea5t l\u00e0 : 15 x 18 : 2 = 135 (m2) \u0110\u00e1p s\u1ed1 : 135m2. B\u00e0i 107: Cho h\u00ecnh thang ABCD . \u0110\u00e1y l\u1edbn CD g\u1ea5p \u0111\u00f4i \u0111\u00e1y b\u00e9 AB.Hai \u0111\u01b0\u1eddng ch\u00e9o AC v\u00e0 BD c\u1eaft nhau t\u1ea1i G.Bi\u1ebft di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c ABG l\u00e0 37,5 cm2.T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang ABCD. Hai tam gi\u00e1c ABD v\u00e0 CBD c\u00f3 AB = 1\/2CD v\u00e0 2 \u0111\u01b0\u1eddng cao t\u01b0\u01a1ng \u1ee9ng b\u1eb1ng nhau b\u1eb1ng \u0111\u01b0\u1eddng cao h\u00ecnh thang ABCD. Suy ra: SABD = 1\/2 SCBD M\u00e0 hai tam gi\u00e1c n\u00e0y l\u1ea1i c\u00f3 c\u1ea1nh BD chung n\u00ean \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A b\u1eb1ng 1\/2 \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb C xu\u1ed1ng BD. M\u1eb7t kh\u00e1c hai \u0111\u01b0\u1eddng cao n\u00e0y c\u0169ng l\u00e0 hai \u0111\u01b0\u1eddng cao c\u1ee7a 2 tam gi\u00e1c ABG v\u00e0 CBG. hai tam gi\u00e1c n\u00e0y l\u1ea1i c\u00f3 c\u1ea1nh chung l\u00e0 BG. Suy ra: SABG = 1\/2SCBG AB C Di\u1ec7n t\u00edch tam gi\u00e1c CBG l\u00e0: 37,5 37,5 x 2 = 75 (cm2). G Di\u1ec7n t\u00edch tam gi\u00e1c ABC l\u00e0: D 37,5 + 75 = 112,5 (cm2) T\u01b0\u01a1ng t\u1ef1 ta c\u00f3 SABC = 1\/2SADC Di\u1ec7n t\u00edch tam gi\u00e1c ADC l\u00e0 : 112,5 x 2 = 225 (cm2) SABCD = SABC + SADC SABCD = 112,5 + 225 = 337,5 (cm2) \u0110\u00e1p s\u1ed1 : 337,5cm2. B\u00e0i 108:","Cho h\u00ecnh tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch 66cm2. D l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u1ea1nh AB, E l\u00e0 \u0111i\u1ec3m n\u1eb1m tr\u00ean c\u1ea1nh AC sao cho AE = 2EC. T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c ADE. AE = 2EC => EC = 1\/3AC A E SBEC = 1\/3SABC = 66 : 3 = 22 (cm2) D C B V\u00ec hai tam gi\u00e1c n\u00e0y c\u00f3 EC=1\/3AC v\u00e0 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B. Di\u1ec7n t\u00edch tam gi\u00e1c ABE l\u00e0: 66 \u2013 22 = 44 (cm2) D l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AB n\u00ean SADE = 1\/2SABE Di\u1ec7n t\u00edch tam gi\u00e1c ADE l\u00e0: 44 : 2 = 22 (cm2) \u0110\u00e1p s\u1ed1: 22cm2. B\u00e0i 109: Cho tam gi\u00e1c ABC. L\u1ea5y M l\u00e0 trung \u0111i\u1ec3m c\u1ea1nh AB; N l\u00e0 trung \u0111i\u1ec3m c\u1ea1nh AC. N\u1ed1i M v\u1edbi N ta \u0111\u01b0\u1ee3c t\u1ee9 gi\u00e1c BMNC c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng 225cm2 . T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c ABC. A MN BC SBNC = 1\/2SABC SMNB = 1\/2SABN = 1\/4SABC SBMNC = SMNB + SBNC = 3\/4SABC = 225cm2 SABC = 225 : 3 x 4 = 300cm2 B\u00e0i 110: Cho tam gi\u00e1c ABC vu\u00f4ng \u1edf A v\u00e0 c\u00f3 chu vi 120cm. Bi\u1ebft \u0111\u1ed9 d\u00e0i c\u1ea1nh AC b\u1eb1ng 75% \u0111\u1ed9 d\u00e0i c\u1ea1nh AB. \u0110\u1ed9 d\u00e0i c\u1ea1nh BC b\u1eb1ng 5\/7 t\u1ed5ng \u0111\u1ed9 d\u00e0i c\u1ee7a hai c\u1ea1nh AC v\u00e0 AB. H\u00e3y t\u00ednh chi\u1ec1u cao AH \u1ee9ng v\u1edbi c\u1ea1nh BC c\u1ee7a h\u00ecnh tam gi\u00e1c ABC. 75% = 3\/4 AC c\u00f3 3 ph\u1ea7n th\u00ec AB c\u00f3 4 ph\u1ea7n, AB+AC c\u00f3 3=4=7 (ph\u1ea7n) CB c\u00f3 5 ph\u1ea7n. T\u1ed5ng s\u1ed1 ph\u1ea7n b\u1eb1ng nhau: 3 + 4 + 5 = 12 (ph\u1ea7n) C Gi\u00e1 tr\u1ecb 1 ph\u1ea7n: 120 : 12 = 10 (cm) H \u0110\u1ed9 d\u00e0i c\u1ea1nh AC l\u00e0: 10 x 3 = 30 (cm) \u0110\u1ed9 d\u00e0i c\u1ea1nh AB l\u00e0 : 10 x 4 = 40 (cm) A B \u0110\u1ed9 d\u00e0i c\u1ea1nh BC l\u00e0 : 10 x 5 = 50 (cm) Di\u1ec7n t\u00edch tam gi\u00e1c ABC l\u00e0 :","40 x 30 : 2 = 600 (cm2) Chi\u1ec1u cao AH \u1ee9ng v\u1edbi c\u1ea1nh BC l\u00e0 : 600 x 2 : 50 = 24 (cm) \u0110\u00e1p s\u1ed1 : 24cm B\u00e0i 111: Cho tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng 360cm2. L\u1ea5y D l\u00e0 m\u1ed9t \u0111i\u1ec3m b\u1ea5t k\u00ec tr\u00ean BC. N\u1ed1i A v\u1edbi D. L\u1ea5y K l\u00e0 trung \u0111i\u1ec3m c\u1ee7a AD. N\u1ed1i K v\u1edbi B v\u00e0 C.T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c BCK. A K B C D SABC = SABD + SADC (1) Do AK = KD n\u00ean ta \u0111\u01b0\u1ee3c: SBAK = SBDK = 1\/2SABD (2) SCAK = SCDK = 1\/2SADC (3) T\u1eeb (1) , (2) v\u00e0 (3) cho ta: SBCK = SBDK + SCDK = 1\/2SABC SBCK = 360 : 2 = 180 (cm2) B\u00e0i 112: M\u1ed9t v\u01b0\u1eddn tr\u01b0\u1eddng h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chu vi g\u1ea5p 8 l\u1ea7n chi\u1ec1u r\u1ed9ng c\u1ee7a n\u00f3. N\u1ebfu t\u0103ng chi\u1ec1u r\u1ed9ng th\u00eam 2m v\u00e0 gi\u1ea3m chi\u1ec1u d\u00e0i 2m th\u00ec di\u1ec7n t\u00edch t\u0103ng th\u00eam 144 m (vu\u00f4ng). T\u00ednh di\u1ec7n t\u00edch v\u01b0\u1eddn tr\u01b0\u1eddng tr\u01b0\u1edbc khi m\u1edf r\u1ed9ng? 2m 144m2 2m N\u1eeda chu vi th\u00ec g\u1ea5p chi\u1ec1u r\u1ed9ng: 8 : 2 = 4 (l\u1ea7n) Chi\u1ec1u d\u00e0i g\u1ea5p chi\u1ec1u r\u1ed9ng: 4 \u2013 1 = 3 (l\u1ea7n) N\u1ebfu chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt m\u00e0u v\u00e0ng th\u00eam 2m s\u1ebd g\u1ea5p 2 l\u1ea7n chi\u1ec1u r\u1ed9ng ban \u0111\u1ea7u. 2 l\u1ea7n chi\u1ec1u r\u1ed9ng ban \u0111\u1ea7u l\u00e0: 144 : 2 + 2 = 74 (m) Chi\u1ec1u r\u1ed9ng ban \u0111\u1ea7u l\u00e0: 74 : 2 = 37 (m) Chi\u1ec1u d\u00e0i ban \u0111\u1ea7u l\u00e0 :","37 x 3 = 111 (m) Di\u1ec7n t\u00edch ban \u0111\u1ea7u l\u00e0 : 111 x 37 = 4107 (m2) \u0110\u00e1p s\u1ed1 : 4107m2. B\u00e0i 113: Cho tam gi\u00e1c ABC , tr\u00ean AC l\u1ea5y \u0111i\u1ec3m E sao cho CE=2\/3 CA , tr\u00ean BC l\u1ea5y \u0111i\u1ec3m D sao cho CD =1\/3 CB.AD v\u00e0 BE c\u1eaft nhau t\u1ea1i O 1. So sanh BO v\u00e0 OE 2.T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c AOE .Bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c BOD b\u1eb1ng 800 cm2 1). S_ABE = 1\/3S_ABC (AE = 1\/3AC). S_BEC = 2\/3S_ABC ; S_BED = 2\/3S_BEC => S_BED = 2\/3 x 2\/3S_ABC = 4\/9S_ABC. A T\u1ec9 s\u1ed1 S_ABE\/S_DBE = (1\/3)\/(4\/9) = 3\/4 E Hai tam gi\u00e1c n\u00e0y c\u00f3 BE chung n\u00ean \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A xu\u1ed1ng BE c\u00f3 3 ph\u1ea7n th\u00ec \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb D c\u00f3 4 ph\u1ea7n. O => S_AOE\/S_DOE = 3\/4 D S_ADC = 1\/3S_ABC ; S_ADE = 1\/3S_ADC => S_ADE = 1\/9S_ABC => S_AOE = 1\/9 : (3+4) x 3 = 1\/21S_ABC (1) S_ABO = S_ABE \u2013 S_AOE = (1\/3 \u2013 1\/21)S_ABC = 6\/21S_ABC (4) B C => T\u1eeb (1) v\u00e0 (2) cho ta S_AOE = 1\/6S_ABO. Hai tam gi\u00e1c n\u00e0y c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A n\u00ean OE = 1\/6BO 2). T\u01b0\u01a1ng t\u1ef1 ta c\u00f3 S_DOE = 1\/6S_DOB => S_DOE = 800 : 6 = 400\/3 (cm2) M\u00e0 S_AOE = 3\/4S_DOE => S_AOE = 400\/3 x 3\/4 = 100 (cm2) B\u00e0i 114: Cho tam gi\u00e1c ABC , tr\u00ean AC l\u1ea5y \u0111i\u1ec3m N sao cho AN = 1\/4 AC , tr\u00ean BC l\u1ea5y \u0111i\u1ec3m M sao cho BM = MC . K\u00e9o d\u00e0i AB v\u00e0 c\u1eaft nhau \u1edf P . a, T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c ABC , bi\u1ebft di\u1ec7n t\u00edch tam giac APN b\u1eb1ng 100 x\u0103ng - ti - m\u00e9t vu\u00f4ng b, So s\u00e1nh PN v\u00e0 NM P Vi\u1ebft t\u1eaft l\u00e0 di\u1ec7n t\u00edch. BKC = 1\/2ABC ; BMK = 1\/2BKC A N => BMK = AMK = 1\/4ABC K T\u1ee9 gi\u00e1c AKMB l\u00e0 h\u00ecnh thang. => PMK = AMK = 1\/4ABC M\u00e0 NMK = 1\/2AMK = 1\/8ABC N\u00ean : PNK = PMK \u2013 NMK = 1\/4ABC \u2013 1\/8ABC = 1\/8ABC BC M","=> PAN = PNK = 1\/8ABC = 100cm2. => ABC = 100 x 8 = 800 (cm2) b).APN = ANM = 1\/8ABC Hai tam gi\u00e1c n\u00e0y c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb A n\u00ean PN = NM B\u00e0i 115: Cho hai h\u00ecnh vu\u00f4ng ABCD v\u00e0 EGHD (nh\u01b0 h\u00ecnh v\u1ebd). H\u00e3y so s\u00e1nh hai \u0111o\u1ea1n th\u1eb3ng BK v\u00e0 DE B KC EG A DH N\u1ed1i DK, HC, GC. D\u1ec5 th\u1ea5y KCHD l\u00e0 h\u00ecnh thang. Ta c\u00f3: SDCK = SHCK ( Chung \u0111\u00e1y CK v\u00e0 chi\u1ec1u cao \u0111\u1ec1u b\u1eb1ng c\u1ea1nh h\u00ecnh vu\u00f4ng ABCD) => SDEK = SHCE (Hai tam gi\u00e1c c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau c\u00f9ng tr\u1eeb \u0111i m\u1ed9t ph\u1ea7n di\u1ec7n t\u00edch chung ECK) SHCE = SGCE ( Chung \u0111\u00e1y CE v\u00e0 chi\u1ec1u cao \u0111\u1ec1u b\u1eb1ng c\u1ea1nh h\u00ecnh vu\u00f4ng EGHD) => SDEK = SGCE (v\u00ec c\u00f9ng b\u1eb1ng SHCE) Hai tam gi\u00e1c n\u00e0y c\u00f3 \u0111\u00e1y DE b\u1eb1ng \u0111\u00e1y EG (c\u1ea1nh h\u00ecnh vu\u00f4ng EGHD) n\u00ean chi\u1ec1u cao KC=CE. Suy ra: DE = BK (Hai \u0111o\u1ea1n th\u1eb3ng b\u1eb1ng nhau c\u00f9ng tr\u1eeb \u0111i hai \u0111o\u1ea1n b\u1eb1ng nhau kh\u00e1c). B\u00e0i 116: Cho h\u00ecnh vu\u00f4ng ABCD c\u1ea1nh 12cm. E l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a c\u1ea1nh AD. G\u1ecdi M l\u00e0 giao \u0111i\u1ec3m c\u1ee7a hai \u0111o\u1ea1n th\u1eb3ng AC v\u00e0 BE. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c MAE","A B M E DC SABCD = 12 x 12 = 144 (cm2) SACE = SABE = 6 x 12 : 2 = 36 (cm2) (ED = 1\/2AD) SABC = 144 : 2 = 72 (cm2) Ta th\u1ea5y 2 tam gi\u00e1c ABC v\u00e0 AEC c\u00f3 chung c\u1ea1nh \u0111\u00e1y AC nen hai \u0111\u01b0\u1eddng cao c\u1ee7a ch\u00fang t\u1ec9 l\u1ec7 v\u1edbi hai di\u1ec7n t\u00edch b\u1eb1ng 36\/72 = 1\/2 M\u1eb7t kh\u00e1c, hai tam gi\u00e1c ABM v\u00e0 AEM c\u00f3 chung c\u1ea1nh \u0111\u00e1y AM n\u00ean hai di\u1ec7n t\u00edch t\u1ec9 l\u1ec7 v\u1edbi hai \u0111\u01b0\u1eddng cao b\u1eb1ng 1\/2. SAME = 36 : (1+2) = 12 (cm2) B\u00e0i 117: Cho tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng 48cm2, c\u1ea1nh \u0111\u00e1y BC b\u1eb1ng 12cm. a,T\u00ednh chi\u1ec1u cao c\u1ee7a tam gi\u00e1c \u0111\u00f3. b, Tr\u00ean c\u1ea1nh BC l\u1ea5y \u0111i\u1ec3m M sao cho BM=MC, N l\u00e0 \u0111i\u1ec3m tr\u00ean c\u1ea1nh AC sao cho NC=2xNA. K\u00e9o d\u00e0i MN c\u1eaft BA t\u1ea1i K. H\u00e3y so s\u00e1nh hai \u0111o\u1ea1n th\u1eb3ng AK v\u00e0 AB. a\/.Chi\u1ec1u cao c\u1ee7a tam gi\u00e1c ABC l\u00e0: K 48 x 2 : 12 = 8 (cm) b\/.Ta c\u00f3: A N SKMC = SKMB C\u00f3 BM = MC, chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb K. SNMC = SNMB (t\u01b0\u01a1ng t\u1ef1) => SKNC = SKNB (1) SKNC = SKNA x 2 (2) T\u1eeb (1) v\u00e0 (2) suy ra SKNA = SBNA Hai tam gi\u00e1c n\u00e0y l\u1ea1i c\u00f3 \u0111\u01b0\u1eddng cao chung k\u1ebb t\u1eeb A n\u00ean hai c\u1ea1nh \u0111\u00e1y AK = AB BC M","B\u00e0i 118: Cho tam gi\u00e1c MNP, di\u1ec7n t\u00edch MNP= 480(cm2), h\u00ecnh thang ONPQ, O n\u1eb1m tr\u00ean MN v\u00e0 Q n\u1eb1m tr\u00ean PM, bi\u1ebft OQ = \u00bd NP. G\u1ecdi I l\u00e0 giao \u0111i\u1ec3m c\u1ee7a QN v\u00e0 PO. a\/ Ch\u1ee9ng t\u1edf r\u1eb1ng MO = NO v\u00e0 MQ = QP b\/ di\u1ec7n t\u00edch OIQ =? c\/ MI c\u1eaft OQ t\u1ea1i K. So s\u00e1nh KM v\u00e0 KI a)X\u00e9t 2 tam gi\u00e1c OQP v\u00e0 NQP c\u00f3 OQ = 1\/2NP v\u00e0 2 \u0111\u01b0\u1eddng cao t\u01b0\u01a1ng \u1ee9ng b\u1eb1ng nhau. N\u00ean S_OQP = 1\/2S_NQP. M\u1eb7t kh\u00e1c hai tam gi\u00e1c n\u00e0y c\u00f3 c\u1ea1nh \u0111\u00e1y chung QP. Suy ra hai \u0111\u01b0\u1eddng cao OH = 1\/2NF X\u00e9t 2 tam gi\u00e1c MOP v\u00e0 MNP c\u00f3 MP chung v\u00e0 hai \u0111\u01b0\u1eddng cao OH = 1\/2NF. N\u00ean S_MOP = 1\/2S_MNP. M\u1eb7t kh\u00e1c hai tam gi\u00e1c n\u00e0y l\u1ea1i c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb P v\u1edbi 2 c\u1ea1nh \u0111\u00e1y t\u01b0\u01a1ng \u1ee9ng l\u00e0 MO v\u00e0 MN. Suy ra MO = 1\/2MN hay MO = NO T\u01b0\u01a1ng t\u1ef1: MQ = QP b)S_MOP = 1\/2S_MNP = 480 : 2 = 240 (cm2) (do MO M = 1\/2MN) O H FA T\u01b0\u01a1ng t\u1ef1: S_OQP = 1\/2 S_MOP = 240 : 2 = 120 (cm2) K Q T\u01b0\u01a1ng t\u1ef1: S_OQN = 1\/2SPQN => Hai \u0111\u01b0\u1eddng cso t\u01b0\u01a1ng \u1ee9ng OB = 1\/2PA. M\u00e0 hai \u0111\u01b0\u1eddng cao n\u00e0y c\u0169ng l\u00e0 hai \u0111\u01b0\u1eddng cao c\u1ee7a hai tam gi\u00e1c OQI v\u00e0 PQI c\u00f3 chung c\u1ea1nh \u0111\u00e1y IQ n\u00ean S_OQI = 1\/2SPQI. I P Di\u1ec7n t\u00edch OQI l\u00e0: 120 : (1+2) = 40 (cm2) B C c)Ch\u1ee9ng minh tr\u00ean ta c\u00f3 S_MOQ = S_POQ = N 1\/2S_MOP = 120 (cm2) => S_OQI\/S_MOQ = 40\/120 = 1\/3 Hai tam gi\u00e1c n\u00e0y l\u1ea1i c\u00f3 OQ chung n\u00ean hai \u0111\u01b0\u1eddng cao ch\u00fang c\u0169ng c\u00f3 t\u1ec9 l\u1ec7 l\u00e0 1\/3. Hai \u0111\u01b0\u1eddng cao n\u00e0y c\u0169ng l\u00e0 hai \u0111\u01b0\u1eddng cao c\u1ee7a hai tam gi\u00e1c MKQ v\u00e0 IKQ. M\u1eb7t kh\u00e1c hai tam gi\u00e1c MKQ v\u00e0 IKQ l\u1ea1i c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb Q. => IK = 1\/3MK B\u00e0i 119: Cho tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch l\u00e0 150 cm2. C\u00e1c \u0111i\u1ec3m K; M; N l\u1ea7n l\u01b0\u1ee3t n\u1eb1m tr\u00ean c\u00e1c c\u1ea1nh BA; AC; BC sao cho BK = 1\/4 AK; AM = 1\/4 MC; NC = 1\/4 NB. N\u1ed1i BM; KC; AN c\u1eaft nhau l\u1ea7n l\u01b0\u1ee3t t\u1ea1i O; P; Q. T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c OPQ. S_ACN = S_KCB = 1\/5S_ABC = 150 : 5 = 30 (cm2) A Q S_KNB = 4\/5S_KCB = 4\/5 x 30 = 24 (cm2) M N S_ANK = S_ANB \u2013 S_KNB = 4\/5 x 150 \u2013 24 = 96 (cm2) S_KNC = S_KCB \u2013 S_KNB = 30 \u2013 24 = 6 (cm2) P S_ANC \/ S_ANK = 30\/96 = 10\/32 KO Hai tam gi\u00e1c n\u00e0y c\u00f3 AN chung n\u00ean 2 \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb C v\u00e0 t\u1eeb B K xu\u1ed1ng AN c\u00f3 t\u1ec9 l\u1ec7 10\/32. Hai \u0111\u01b0\u1eddng cao n\u00e0y c\u0169ng l\u00e0 2 \u0111\u01b0\u1eddng cao c\u1ee7a 2 tam gi\u00e1c QNC v\u00e0 QNK. Hai tam gi\u00e1c n\u00e0y l\u1ea1i c\u00f3 c\u1ea1nh chung QN n\u00ean di\u1ec7n t\u00edch c\u1ee7a ch\u00fang c\u0169ng c\u00f3 t\u1ec9 l\u1ec7 10\/32. Di\u1ec7n t\u00edch c\u1ee7a tam gi\u00e1c QNC = 6 : (10 + 32) x 10 = 60\/42","(cm2) T\u01b0\u01a1ng t\u1ef1 ta c\u0169ng c\u00f3 S_AMP = S_KOB = S_QNC = 60\/42cm2 T\u1ed5ng di\u1ec7n t\u00edch 3 tam gi\u00e1c n\u00e0y l\u00e0: 60\/42 x 3 = 60\/14 (cm2) S_OPQ = S_ABC \u2013 (S_ANC + S_CKB + S_BMA) + (S_AMP + S_KOB + S_QNC) S_OPQ = 150 \u2013 (30 x 3) + 60\/14 = 60 + 60\/14 = 900\/14 (cm2) B\u00e0i 120: Cho tam gi\u00e1c ABC, tr\u00ean AB l\u1ea5y \u0111i\u1ec3m M sao cho AM = BM. Tr\u00ean AC l\u1ea5y \u0111i\u1ec3m N sao cho AN = 2NC. K\u00e9o d\u00e0i \u0111o\u1ea1n th\u1eb3ng MN v\u00e0 BC ch\u00fang c\u1eaft nhau t\u1ea1i D. Ch\u1ee9ng t\u1ecf r\u1eb1ng BC = CD. Ta c\u00f3: A N D M C C S_DMA = S_DMB (MA=MB, B \u0111\u01b0\u1eddng cao chung k\u1ebb t\u1eeb D) (1) S_NMA = S_NMB (2) T\u1eeb (1) v\u00e0 (2) suy ra S_DNA = S_DNB M\u00e0 S_DNA = 2 S_DNC (AN=2NC, chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb D). Suy ra S_NBC = S_NDC Hai tam gi\u00e1c n\u00e0y c\u00f3 chung \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb N n\u00ean BC = CD B\u00e0i 121: Cho h\u00ecnh thang abcd c\u00f3 \u0111\u00e1y b\u1eb1ng AB b\u1eb1ng 1\/3 CD .ACv\u00e0 BD c\u1eaft nhau t\u1ea1i O . T\u00ednh di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c AOB . Bi\u1ebft di\u1ec7n t\u00edch h\u00ecnh thang ABCD l\u00e0 96 cm2 X\u00e9t tam gi\u00e1c ABC v\u00e0 ADC c\u00f3 AB = 1\/3CD v\u00e0 hai AB \u0111\u01b0\u1eddng cao b\u1eb1ng nhau b\u1eb1ng \u0111\u01b0\u1eddng cao h\u00ecnh thang. N\u00ean S_ABC = 1\/3S_ADC O Di\u1ec7n t\u00edch tam gi\u00e1c ABC l\u00e0: 96 : (1 + 3) = 24 (cm2) D Hai tam gi\u00e1c n\u00e0y l\u1ea1i c\u00f3 c\u1ea1nh AC chung n\u00ean \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb B xu\u1ed1ng AC b\u1eb1ng 1\/3 \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb D xu\u1ed1ng BC. Hai \u0111\u01b0\u1eddng cao n\u00e0y c\u0169ng l\u00e0 hai \u0111\u01b0\u1eddng cao c\u1ee7a hai tam gi\u00e1c ABO v\u00e0 ADO. Hai tam gi\u00e1c n\u00e0y l\u1ea1i c\u00f3 AO chung n\u00ean S_ABO = 1\/3S_ADO. Ta l\u1ea1i c\u00f3 S_ABD = S_ABC = 24cm2 Di\u1ec7n t\u00edch tam gi\u00e1c AOB l\u00e0: 24 : (1+3) = 6 (cm2) B\u00e0i 122: Nh\u00e0 tr\u01b0\u1eddng m\u1edf r\u1ed9ng v\u01b0\u1eddn tr\u01b0\u1eddng h\u00ecnh vu\u00f4ng v\u1ec1 c\u1ea3 4 ph\u00eda, m\u1ed7i ph\u00eda th\u00eam 3 m\u00e9t n\u00ean di\u1ec7n t\u00edch t\u0103ng th\u00eam 336 m\u00e9t vu\u00f4ng. T\u00ednh chu vi v\u01b0\u1eddn tr\u01b0\u1eddng sau khi m\u1edf r\u1ed9ng?","T\u1ed5ng di\u1ec7n t\u00edch 4 h\u00ecnh vu\u00f4ng m\u00e0u v\u00e0ng. 3m 3m 3 x 3 x 4 = 36 (m2) 9m2 3m Di\u1ec7n t\u00edch 1 h\u00ecnh ch\u1eef nh\u1eadt m\u00e0u tr\u1eafng. (336 \u2013 36) : 4 = 75 (m2) 3m C\u1ea1nh v\u01b0\u1eddn tr\u01b0\u1eddng l\u00fac ban \u0111\u1ea7u. 75 : 3 = 25 (m) C\u1ea1nh v\u01b0\u1eddn tr\u01b0\u1eddng sau khi m\u1edf r\u1ed9ng. 25 + 3 + 3 = 31 (m) Chu vi v\u01b0\u1eddn tr\u01b0\u1eddng sau khi m\u1edf r\u1ed9ng. 31 x 4 = 124 (m) B\u00e0i 123: Qua \u0111\u1ec9nh c\u1ee7a 1 h\u00ecnh thang h\u00e3y k\u1ebb 1 \u0111\u01b0\u1eddng th\u1eb3ng chia \u0111\u00f4i di\u1ec7n t\u00edch c\u1ee7a n\u00f3 AB N DK C Xem h\u00ecnh thang ABCD, ta ch\u1ecdn \u0111\u1ec9nh B \u0111\u1ec3 k\u1ebb \u0111o\u1ea1n th\u1eb3ng chia \u0111\u00f4i di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh thang n\u00e0y. N\u1ed1i BD ta c\u00f3 3 tr\u01b0\u1eddng h\u1ee3p: -BD chia \u0111\u00f4i di\u1ec7n t\u00edch ABCD -SABD > SBDC -SABD < SBDC. *.Ch\u1ecdn \u0111i\u1ec3n h\u00ecnh 1 tr\u01b0\u1eddng h\u1ee3p l\u00e0 SABD < SBDC T\u1eeb A k\u1ec3 \u0111\u01b0\u1eddng song song v\u1edbi BD c\u1eaft CD n\u1ed1i d\u00e0i t\u1ea1i N. Ta c\u00f3 SABD = SBND (AB=ND, hai \u0111\u01b0\u1eddng cao t\u01b0\u01a1ng \u1ee9ng b\u1eb1ng chi\u1ec1u cao h\u00ecnh b\u00ecnh h\u00e0nh ABDN) => SBCN = SABCD (1) (SBCD+SBDN = SBCD+SABD) Ch\u1ecdn K l\u00e0 trung \u0111i\u1ec3m c\u1ee7a NC n\u1ed1i BK ta \u0111\u01b0\u1ee3c di\u1ec7n t\u00edch 2 h\u00ecnh tam gi\u00e1c b\u1eb1ng nhau SBCK = SBKN. Hay SBCK = 1\/2SBCN T\u1eeb (1) suy ra SBCK = 1\/2SABCD BK chia \u0111\u00f4i di\u1ec7n t\u00edch h\u00ecnh thang ABCD B\u00e0i 124: M\u1ed9t m\u1ea3nh \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chu vi 220 m\u00e9t. N\u1ebfu t\u0103ng chi\u1ec1u d\u00e0i l\u00ean 5 m\u00e9t, gi\u1ea3m chi\u1ec1u r\u1ed9ng 5 m\u00e9t th\u00ec di\u1ec7n t\u00edch m\u1ea3nh \u0111\u1ea5t gi\u1ea3m 225 m\u00e9t vu\u00f4ng. T\u00ednh chi\u1ec1u d\u00e0i, chi\u1ec1u r\u1ed9ng? H\u00ecnh v\u1ebd cho th\u1ea5y di\u1ec7n t\u00edch gi\u1ea3m \u0111i b\u1eb1ng 5m 225m2 di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt m\u00e0u xanh. 5m","Chi\u1ec1u d\u00e0i h\u01a1n chi\u1ec1u r\u1ed9ng l\u00e0: 225 : 5 \u2013 5 = 40 (m) T\u1ed5ng chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng (n\u1eeda chu vi) l\u00e0: 220 : 2 = 110 (m) Chi\u1ec1u d\u00e0i l\u00e0: (110 + 40) : 2 = 75 (m) Chi\u1ec1u r\u1ed9ng l\u00e0: 110 \u2013 75 = 35 (m) B\u00e0i 124: Cho tam gi\u00e1c ABC v\u00e0 c\u00e1c \u0111i\u1ec3m M; N; P sao cho BM = 3AB; CN = 3CB; AP = 3AC . H\u1ecfi di\u1ec7n t\u00edch tam gi\u00e1c MNP g\u1ea5p bao nhi\u00eau l\u1ea7n di\u1ec7n t\u00edch tam gi\u00e1c ABC M A N CB P Ta c\u00f3: S_ABC = 1\/3S_CBM M\u00e0 S_MCB = 1\/2S_MBN => S_ABC = 1\/3 x 1\/2S_MBN = 1\/6S_MBN Hay S_MBN = 6 S_ABC T\u01b0\u01a1ng t\u1ef1: S_MAP = 6 S_ABC S_NCP = 6 S_ABC V\u00e0 S_MNP = S_MBN + S_MAP + S_NCP + S_ABC => S_MNP = 6 S_ABC + 6 S_ABC + 6 S_ABC + S_ABC V\u1eady: S_MNP = 19 S_ABC B\u00e0i 125: Cho hai h\u00ecnh vu\u00f4ng c\u00f3 t\u1ed5ng hai chu vi l\u00e0 280m, hi\u1ec7u hai di\u1ec7n t\u00edch l\u00e0 1400m2. T\u00ecm c\u1ea1nh c\u1ee7a hai h\u00ecnh vu\u00f4ng \u0111\u00f3 Xem 2 h\u00ecnh vu\u00f4ng ABCD v\u00e0 MPND nh\u01b0 h\u00ecnh v\u00e9. A P B T\u1ed5ng 2 c\u1ea1nh h\u00ecnh vu\u00f4ng (AB+MP): N C M 280 : 4 = 70 (m) D Di\u1ec7n t\u00edch h\u00ecnh thang ABPM: 1400 : 2 = 700 (m2) Chi\u1ec1u cao AM c\u1ee7a h\u00ecnh thang ABPM c\u0169ng l\u00e0 hi\u1ec7u 2 c\u1ea1nh h\u00ecnh vu\u00f4ng: 700 x 2 : 70 = 20 (m) C\u1ea1nh h\u00ecnh vu\u00f4ng l\u1edbn l\u00e0: (70 + 20) : 2 = 45 (m) C\u1ea1nh h\u00ecnh vu\u00f4ng nh\u1ecf :","70 \u2013 45 = 25 (m) B\u00e0i 126: Cho h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chi\u1ec1u d\u00e0i g\u1ea5p 2 l\u1ea7n chi\u1ec1u r\u1ed9ng. T\u00ednh chu vi v\u00e0 di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt \u0111\u00f3, bi\u1ebft r\u1eb1ng n\u1ebfu ta t\u0103ng chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng m\u1ed7i chi\u1ec1u 3 m th\u00ec \u0111\u01b0\u1ee3c m\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt m\u1edbi c\u00f3 di\u1ec7n t\u00edch h\u01a1n h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u 72 m2. Gi\u1ea3i Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng (1) 3 x 3 = 9 (m2) 3m (a) (a) (1) Di\u1ec7n t\u00edch 1 h\u00ecnh ch\u1eef nh\u1eadt (a) (72 \u2013 9) : 3 = 21 (m) Chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u: (a) 21 : 3 = 7 (m) Chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u: 7 x 2 = 14 (m) 3m Chu vi h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u: (14 + 7) x 2 = 42 (m) Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u: 14 x 7 = 98 (m2) \u0110\u00e1p s\u1ed1: Chu vi: 42m Di\u1ec7n t\u00edch: 98m2. B\u00e0i 127: M\u1ed9t tr\u01b0\u1eddng Ti\u1ec3u h\u1ecdc c\u00f3 m\u1ed9t m\u1ea3nh \u0111\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt. N\u1ebfu chi\u1ec1u d\u00e0i t\u0103ng th\u00eam 4m, chi\u1ec1u r\u1ed9ng t\u0103ng th\u00eam 5m th\u00ec di\u1ec7n t\u00edch t\u0103ng th\u00eam 250m2. N\u1ebfu ch\u1ec9 c\u00f3 chi\u1ec1u r\u1ed9ng t\u0103ng th\u00eam 5m th\u00ec di\u1ec7n t\u00edch t\u0103ng th\u00eam 150m2. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt ban \u0111\u1ea7u? Gi\u1ea3i Di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt m\u00e0u xanh: 4m 250 \u2013 150 = 100 (m2) 5m 150m2 Chi\u1ec1u r\u1ed9ng m\u1ea3nh \u0111\u1ea5t c\u1ee7a tr\u01b0\u1eddng: 100 : 4 \u2013 5 = 20 (m) Chi\u1ec1u d\u00e0i m\u1ea3nh \u0111\u1ea5t c\u1ee7a tr\u01b0\u1eddng: 150 : 5 = 30 (m) Di\u1ec7n t\u00edch m\u1ea3nh \u0111\u1ea5t c\u1ee7a tr\u01b0\u1eddng: 30 x 20 = 600 (m2) \u0110\u00e1p s\u1ed1: 600m2. B\u00e0i 128: Tr\u00ean m\u1ed9t m\u1ea3nh \u0111\u1ea5t h\u00ecnh vu\u00f4ng, ng\u01b0\u1eddi ta \u0111\u00e0o m\u1ed9t c\u00e1i ao c\u0169ng h\u00ecnh vu\u00f4ng. C\u1ea1nh c\u1ee7a m\u00e3nh \u0111\u1ea5t h\u01a1n c\u1ea1nh ao 18m. Di\u1ec7n t\u00edch \u0111\u1ea5t c\u00f2n l\u1ea1i \u0111\u1ec3 tr\u1ed3ng rau l\u00e0 468 m2. T\u00ednh chu vi m\u1ea3nh \u0111\u1ea5t. Gi\u1ea3i Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng (1) l\u00e0: 18 x 18 = 324 (m2) Di\u1ec7n t\u00edch 1 h\u00ecnh ch\u1eef nh\u1eadt (a) l\u00e0: 18m (a) (1) (468 \u2013 324) : 2 = 72 (m2) Chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt (a) c\u0169ng l\u00e0 c\u1ea1nh c\u00e1i ao l\u00e0: 72 : 18 = 4 (m) Ao (a) C\u1ea1nh m\u1ea3nh \u0111\u1ea5t h\u00ecnh vu\u00f4ng l\u00e0: 18m 18 + 4 = 22 (m) Chu vi m\u1ea3nh \u0111\u1ea5t l\u00e0:","22 x 4 = 88 (m) \u0110\u00e1p s\u1ed1 : 88m B\u00e0i 129: M\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt, m\u1ed9t h\u00ecnh tam gi\u00e1c v\u00e0 m\u1ed9t h\u00ecnh vu\u00f4ng c\u00f3 chu vi b\u1eb1ng nhau.Chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt g\u1ea5p \u0111\u00f4i chi\u1ec1u r\u1ed9ng.Tam gi\u00e1c c\u00f3 ba c\u1ea1nh b\u1eb1ng nhau. T\u1ed5ng di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt v\u00e0 h\u00ecnh vu\u00f4ng l\u00e0 425cm2 . T\u00ecm c\u1ea1nh c\u1ee7a m\u1ed7i h\u00ecnh? Gi\u1ea3i G\u1ecdi a l\u00e0 c\u1ea1nh h\u00ecnh vu\u00f4ng, d chi\u1ec1u d\u00e0i v\u00e0 r chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt. Do chu vi h\u00ecnh ch\u1eef nh\u1eadt b\u1eb1ng chu vi h\u00ecnh vu\u00f4ng n\u00ean. a+a=d+r m\u00e0 d = r x 2 a + a = r + r + r => a = 3\/2 r Svu\u00f4ng = a x a = 3\/2 r x 3\/2 r = 9\/4 x r x r Scn = 2 r x r = 2 x r x r T\u1ec9 s\u1ed1 di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng v\u00e0 di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt l\u00e0: 9\/4 : 2 = 9\/8 Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng: 425 : (9+8) x 9 = 225 (cm2) C\u1ea1nh h\u00ecnh vu\u00f4ng l\u00e0 15cm v\u00ec 15x15=225 Chu vi c\u1ee7a m\u1ed7i h\u00ecnh l\u00e0 : 15 x 4 = 60 (cm) C\u1ea1nh h\u00ecnh tam gi\u00e1c l\u00e0 : 60 : 3 = 20 (cm) Chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 : (60 : 2) : (2 + 1) = 10 (cm) Chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt l\u00e0 : 10 x 2 = 20 (cm) \u0110\u00e1p s\u1ed1 : -C\u1ea1nh h\u00ecnh vu\u00f4ng 15cm -C\u1ea1nh h\u00ecnh tam gi\u00e1c 20cm -Chi\u1ec1u r\u1ed9ng HCN 10cm -Chi\u1ec1u d\u00e0i HCN 20cm. B\u00e0i 130: Cho h\u00ecnh ch\u1eef nh\u1eadt ABCD. G\u1ecdi M v\u00e0 N l\u00e0 hai \u0111i\u1ec3m tr\u00ean AB vag CD sao cho BM=DN, P l\u00e0 m\u1ed9t \u0111i\u1ec3m t\u00f9y \u00fd tr\u00ean c\u1ea1nh AD. \u0110o\u1ea1n th\u1eb3ng PB v\u00e0 PC l\u1ea7n l\u01b0\u1ee3t c\u1eaft MN t\u1ea1i E v\u00e0 F. H\u00e3y so s\u00e1nh di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c EBCF v\u1edbi t\u1ed5ng di\u1ec7n t\u00edch 2 t\u1ee9 gi\u00e1c AMPE v\u00e0 DNFP. Gi\u1ea3i Gi\u1ea3i th\u00edch chi ti\u1ebft th\u00ec kh\u00e1 d\u00e0i d\u00f2ng n\u00ean ch\u1ec9 ghi ng\u1eafn g\u1ecdn v\u00e0 xem c\u00e1ch ghi ABCD l\u00e0 di\u1ec7n t\u00edch h\u00ecnh ABCD. A MB Ta c\u00f3: E P AMND = MBCN = PBC = 1\/2 ABCD F M\u00e0: D MBCN = MBE + EBCF + FCN (1) N PBC = PEF + EBCF (2) T\u1eeb (1) v\u00e0 (2) suy ra: C","PEF = MBE + FCN (3) M\u1eb7t kh\u00e1c: EBCF = MBCN \u2013 (MBE + FCN) = 1\/2 ABCD \u2013 (MBE + FCN) AMEP + PFND = AMND \u2013 PEF = 1\/2 ABCD \u2013 (PEF) Do PEF = MBE + FCN (t\u1eeb (3)) N\u00ean: EBCF = AMEP + PFND 7.D\u00c3Y S\u1ed0 (quy lu\u1eadt) B\u00e0i 1: Cho d\u00e3y s\u1ed1 2; 16; 42; 80; \u2026; 560; 682; \u2026 1. H\u00e3y vi\u1ebft ti\u1ebfp ba s\u1ed1 h\u1ea1ng li\u1ec1n sau s\u1ed1 80. 2. T\u00ecm s\u1ed1 h\u1ea1ng th\u1ee9 20 c\u1ee7a d\u00e3y s\u1ed1. Gi\u1ea3i C\u00e1ch 1: Quy lu\u1eadt 1 1,Ta th\u1ea5y quy lu\u1eadt l\u00e0 S\u1ed1 th\u1ee9 nh\u1ea5t 1*1+1=2 s\u1ed1 th\u1ee9 t\u1ef1 x1+s\u1ed1 th\u1ee9 t\u1ef1 S\u1ed1 th\u1ee9 hai 2*7+2=16 s\u1ed1 th\u1ee9 t\u1ef1 *s\u1ed1 l\u1edbn h\u01a1n 6 \u0111\u01a1n v\u1ecb (1+6)+s\u1ed1 th\u1ee9 t\u1ef1 S\u1ed1 th\u1ee9 ba 3*13+3=42 s\u1ed1 th\u1ee9 t\u1ef1 * s\u1ed1 l\u1edbn h\u01a1n 6 \u0111v(7+6)* s\u1ed1 th\u1ee9 t\u1ef1 S\u1ed1 th\u1ee9 t\u01b0 4*19+4=80 t\u01b0\u01a1ng t\u1ef1 nh\u01b0 tr\u00ean V\u1eady 3 s\u1ed1 ti\u1ebfp theo l\u00e0 5*25+5=130 6*31+6=192 7*37+7= 266 2, t\u1eeb s\u1ed1 th\u1ee9 nh\u1ea5t \u0111\u1ebfn s\u1ed1 th\u1ee9 20 c\u00f3 hai m\u01b0\u01a1i s\u1ed1 suy ra c\u00f3 19 kho\u1ea3ng c\u00e1ch m\u00e0 m\u1ed7i kho\u1ea3ng c\u00e1ch l\u00e0 6 \u0111\u01a1n v\u1ecb v\u1eady s\u1ed1 th\u1ee9 20 l\u00e0: 20*(19*6+1)+20=2320 C\u00e1ch 2: Quy lu\u1eadt 2 s\u1ed1 h\u1ea1ng th\u1ee9 nh\u1ea5t : 2 = 2x1 + 10x0 th\u1ee9 hai : 16 = 2x(1+2) + 10x1 th\u1ee9 ba : 42 = 2x(1+2+3) + 10x(1+2) th\u1ee9 t\u01b0 : 80 = 2x(1+2+3+4) +10x(1+2+3) N\u1ebfu g\u1ecdi n l\u00e0 s\u1ed1 h\u1ea1ng th\u1ee9 n c\u1ee7a d\u1ea3y s\u1ed1 ta c\u00f3 c\u00f4ng th\u1ee9c t\u1ed5ng qu\u00e1t c\u1ee7a s\u1ed1 h\u1ea1ng th\u1ee9 n : s\u1ed1 Th\u1ee9 n = 2xnx(n+1)\/2 + 10xnx(n-1) = 6xnxn - 4xn = 2xn(3xn - 2) V\u1eady 3 s\u1ed1 h\u1ea1ng li\u1ec1n sau s\u1ed1 80 l\u00e0: TH\u1ee9 5: = 2x5x(3x5 - 2) = 10x13 = 130 TH\u1ee9 6: = 2x6x(3x6 - 2) = 12x16 = 192 Th\u1ee9 7: = 2x7x(3x7 - 2) = 14x19 = 266 => s\u1ed1 h\u1ea1ng th\u1ee9 20: = 2x20x(3x20 - 2) = 40x 58 = 2320 B\u00e0i 2: Cho d\u00e3y s\u1ed1 2, 17, 47, 92, 152, \u2026 T\u00ecm s\u1ed1 h\u1ea1ng th\u1ee9 120 c\u1ee7a d\u00e3y. D\u00e3y s\u1ed1 tr\u00ean c\u00f3 quy lu\u1eadt k\u1ec3 t\u1eeb s\u1ed1 h\u1ea1ng th\u1ee9 2 nh\u01b0 sau: S\u1ed1 h\u1ea1ng th\u1ee9 2 l\u00e0 17 = 2 + 15 x (1) S\u1ed1 h\u1ea1ng th\u1ee9 3 l\u00e0 47 = 2 + 15 x (1+2) S\u1ed1 h\u1ea1ng th\u1ee9 2 l\u00e0 92 = 2 + 15 x (1+2+3) S\u1ed1 h\u1ea1ng th\u1ee9 2 l\u00e0 152 = 2 + 15 x (1+2+3+4) \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.","S\u1ed1 h\u1ea1ng th\u1ee9 120 l\u00e0 : 2 + 15 x (1+2+3+4+\u2026\u2026.+117+118+119) = 2 + 15 x [(1+119) x 119] : 2 = 2 + 15 x 7140 = 107102 \u0110\u00e1p s\u1ed1: 107102 B\u00e0i 3: Cho m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean c\u00f3 2014 ch\u1eef s\u1ed1. Bi\u1ebft r\u1eb1ng v\u1edbi hai ch\u1eef s\u1ed1 li\u00ean ti\u1ebfp theo th\u1ee9 t\u1ef1 \u0111\u00e3 vi\u1ebft th\u00ec t\u1ea1o th\u00e0nh s\u1ed1 c\u00f3 hai ch\u1eef s\u1ed1 chia h\u1ebft cho 17 ho\u1eb7c chia h\u1ebft cho 23. N\u1ebfu ch\u1eef s\u1ed1 cu\u1ed1i c\u00f9ng c\u1ee7a s\u1ed1 \u0111\u00f3 l\u00e0 ch\u1eef s\u1ed1 1 th\u00ec ch\u1eef s\u1ed1 \u0111\u1ea7u ti\u00ean l\u00e0 ch\u1eef s\u1ed1 n\u00e0o? Ta suy lu\u1eadn d\u1ea7n th\u1ebf n\u00e0y: Ch\u1eef s\u1ed1 cu\u1ed1i c\u00f9ng l\u00e0 1 th\u00ec ch\u1eef s\u1ed1 li\u1ec1n tr\u01b0\u1edbc ph\u1ea3i l\u00e0 5 \u0111\u1ec3 51 chia h\u1ebft cho 17 (51:17=3) \u2026\u2026\u202651 Tr\u01b0\u1edbc ch\u1eef s\u1ed1 5 ph\u1ea3i l\u00e0 8 \u0111\u1ec3 c\u00f3 85:17=5 \u2026\u2026\u2026.851 Tr\u01b0\u1edbc ch\u1eef s\u1ed1 8 ph\u1ea3i l\u00e0 6 \u0111\u1ec3 c\u00f3 68:17=4 \u2026\u2026\u2026.6851 Tr\u01b0\u1edbc ch\u1eef s\u1ed1 6 ph\u1ea3i l\u00e0 4 \u0111\u1ec3 c\u00f3 46:23=2 \u2026\u2026\u2026.46851 Tr\u01b0\u1edbc ch\u1eef s\u1ed1 4 ph\u1ea3i l\u00e0 3 \u0111\u1ec3 c\u00f3 34:17=2 \u2026\u2026\u2026.346851 Tr\u01b0\u1edbc ch\u1eef s\u1ed1 3 ph\u1ea3i l\u00e0 2 \u0111\u1ec3 c\u00f3 23:23=1 \u2026\u2026\u2026.2346851 Tr\u01b0\u1edbc ch\u1eef s\u1ed1 2 ph\u1ea3i l\u00e0 9 \u0111\u1ec3 c\u00f3 92:23=4 \u2026\u2026\u2026.92346851 Tr\u01b0\u1edbc ch\u1eef s\u1ed1 9 ph\u1ea3i l\u00e0 6 \u0111\u1ec3 c\u00f3 69:23=3 \u2026\u2026\u2026.692346851 \u2026\u2026\u2026\u2026.. Ti\u1ebfp t\u1ee5c: \u2026\u2026\u2026\u2026.92346 92346 92346 851 Ta th\u1ea5y quy lu\u1eadt \u0111\u01b0\u1ee3c l\u1eadp l\u1ea1i nhi\u1ec1u l\u1ea7n 5 ch\u1eef s\u1ed1: 9;2;3;4;6 cu\u1ed1i c\u00f9ng l\u00e0 851. B\u1ecf 3 ch\u1eef s\u1ed1 cu\u1ed1i c\u00f9ng (851) th\u00ec c\u00f2n l\u1ea1i: 2014-3=2011 (ch\u1eef s\u1ed1) Chia nh\u00f3m 5 th\u00ec \u0111\u01b0\u1ee3c: 2011 : 5 = 402 (nh\u00f3m 5) d\u01b0 1. Ch\u1eef s\u1ed1 cu\u1ed1i c\u00f9ng trong nh\u00f3m 5 l\u00e0 ch\u1eef s\u1ed1 6. V\u1eady ch\u1eef s\u1ed1 \u0111\u1ea7u ti\u00ean trong d\u00e3y s\u1ed1 tr\u00ean l\u00e0 6 (6 92346 92346 92346 \u2026\u2026\u2026\u2026. 92346 92346 851) B\u00e0i 4 Cho d\u00e3y s\u1ed1 4; 6; 9; 13; 18; \u2026\u2026\u2026. T\u00ecm s\u1ed1 h\u1ea1ng th\u1ee9 2014 c\u1ee7a d\u00e3y s\u1ed1 tr\u00ean. D\u00e3y s\u1ed1 tr\u00ean c\u00f3 quy lu\u1eadt t\u1eeb s\u1ed1 th\u1ee9 hai v\u1ec1 sau b\u1eb1ng s\u1ed1 li\u1ec1n tr\u01b0\u1edbc c\u1ed9ng th\u00eam th\u1ee9 t\u1ef1 c\u1ee7a n\u00f3. S\u1ed1 h\u1ea1ng th\u1ee9 2: 6 = 4+2 S\u1ed1 h\u1ea1ng th\u1ee9 3: 9 = 4+2+3 S\u1ed1 h\u1ea1ng th\u1ee9 4: 13 = 4+2+3+4 \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 S\u1ed1 h\u1ea1ng th\u1ee9 2014: ? = 4+2+3+4+\u2026.+2014 D\u00e3y s\u1ed1 t\u1ef1 nhi\u00ean t\u1eeb 2 \u0111\u1ebfn 2014 c\u00f3: 2014-2+1= 2013 (s\u1ed1 h\u1ea1ng) S\u1ed1 h\u1ea1ng th\u1ee9 2014 l\u00e0: 4 + (2+2014) x 2013 : 2 = 2 029 108 B\u00e0i 5 S\u1ed1 th\u1ee9 100 trong d\u00e3y sau l\u00e0 s\u1ed1 n\u00e0o? 1\/1 ; 2\/1 ; 1\/2 ; 3\/1 ; 2\/2 ; 1\/3 ; 4\/1 ; 3\/2 ; 2\/3 ; 1\/4 ; 5\/1 ; 4\/2 ; 3\/3 ; 2\/4 ; 1\/5 ;... 1\/1 ; 2\/1 ; 1\/2 ; 3\/1 ; 2\/2 ; 1\/3 ; 4\/1 ; 3\/2 ; 2\/3 ; 1\/4 ; 5\/1 ; 4\/2 ; 3\/3 ; 2\/4 ; 1\/5;... Ta c\u00f3 th\u1ec3 chia d\u00e3y s\u1ed1 tr\u00ean theo c\u00e1c nh\u00f3m c\u00f3 t\u1ed5ng c\u1ee7a t\u1eed s\u1ed1 v\u00e0 m\u1eabu s\u1ed1 b\u1eb1ng nhau nh\u01b0 sau:","(1\/1) ; (2\/1 ; 1\/2) ; (3\/1 ; 2\/2 ; 1\/3) ; (4\/1 ; 3\/2 ; 2\/3 ; 1\/4) ; (5\/1 ; 4\/2 ; 3\/3 ; 2\/4 ; 1\/5) ; ... B\u1eb1ng 2: c\u00f3 1 ps. 1\/1 B\u1eb1ng 3: c\u00f3 2 ps. 2\/1 ; 1\/2 B\u1eb1ng 4: c\u00f3 3 ps. 3\/1 ; 2\/2 ; 1\/3 B\u1eb1ng 5: c\u00f3 4 ps. 4\/1 ; 3\/2 ; 2\/3 ; 4\/1 B\u1eb1ng 6: c\u00f3 5 ps. 5\/1 ; 4\/2 ; 3\/3 ; 2\/4 ; 1\/5 \u2026\u2026\u2026\u2026\u2026\u2026. Ph\u00e2n s\u1ed1 \u0111\u1ea7u ti\u00ean trong nh\u00f3m c\u00f3 m\u1eabu s\u1ed1 b\u1eb1ng 1 r\u1ed3i t\u0103ng d\u1ea7n \u0111\u1ebfn ph\u00e2n s\u1ed1 cu\u1ed1i c\u00f9ng c\u00f3 t\u1eed s\u1ed1 b\u1eb1ng 1. Ph\u00e2n s\u1ed1 th\u1ee9 100 thu\u1ed9c nh\u00f3m s\u1ed1 h\u1ea1ng cua d\u00e3y s\u1ed1: 1+2+3+4+5+ \u2026. c\u00f3 t\u1ed5ng l\u00e0 100 Ta th\u1ea5y: 1+2+3+4+\u2026.+13+14 = (14+1)x14:2 = 105 Nh\u01b0 v\u1eady ph\u00e2n s\u1ed1 th\u1ee9 100 n\u1eb1m trong nh\u00f3m c\u00f3 14 ph\u00e2n s\u1ed1 c\u00f3 t\u1ed5ng c\u1ee7a t\u1eed s\u1ed1 v\u00e0 m\u1eabu s\u1ed1 b\u1eb1ng 14+1=15 b\u1eaft \u0111\u1ea7u l\u00e0: 14\/1 ; 13\/2 ; 12\/3 ; 11\/4 ; \u2026.. ; 6\/9 ; 5\/10 ; 4\/11 ; 3\/12 ; 2\/13 ; 1\/14 Do d\u01b0 5 ph\u00e2n s\u1ed1 n\u00ean ta b\u1ecf \u0111i 5 ph\u00e2n s\u1ed1 \u1edf cu\u1ed1i s\u1ebd \u0111\u01b0\u1ee3c ph\u00e2n s\u1ed1 th\u1ee9 100. Ph\u00e2n s\u1ed1 th\u1ee9 100 l\u00e0 6\/9 B\u1ed4 SUNG B\u00e0i 6: T\u00ecm s\u1ed1 h\u1ea1ng th\u1ee9 99 c\u1ee7a d\u00e3y s\u1ed1: 1;6;15;28;...;120;153. Nh\u1eadn x\u00e9t: Th\u1ee9 1: 1 =1 Th\u1ee9 2: 6 = 1+5 Th\u1ee9 3: 15 = 1+5+9 Th\u1ee9 4: 28 = 1+5+9+13 Th\u1ee9 5: 45 = 1+5+9+13+17 \u2026\u2026\u2026\u2026.. Th\u1ee9 99: \u2026.. = 1+5+9+13+\u2026.. 99 s\u1ed1 h\u1ea1ng. Ta th\u1ea5y s\u1ed1 mang th\u1ee9 t\u1ef1 \u1edf m\u1ed7i s\u1ed1 h\u1ea1ng \u1ee9ng v\u1edbi T\u1ed4NG c\u1ee7a d\u00e3y s\u1ed1 c\u00e1ch \u0111\u1ec1u nhau 4 \u0111\u01a1n v\u1ecb \u0111\u01b0\u1ee3c b\u1eaft \u0111\u1ea7u t\u1eeb 1 c\u00f3 s\u1ed1 s\u1ed1 h\u1ea1ng b\u1eb1ng v\u1edbi s\u1ed1 th\u1ee9 t\u1ef1 c\u1ee7a d\u00e3y s\u1ed1 \u0111\u00f3. S\u1ed1 h\u1ea1ng th\u1ee9 99 c\u1ee7a d\u00e3y s\u1ed1: 1;5;9;13;\u2026. l\u00e0: 1 + (99 \u2013 1) x 4 = 393 S\u1ed1 h\u1ea1ng th\u1ee9 99 c\u1ee7a d\u00e3y s\u1ed1: 1;6;15;28;\u2026.. l\u00e0: (t\u1ed5ng 99 s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y s\u1ed1 c\u00e1ch \u0111\u1ec1u). (1 + 393) x 99 : 2 = 19503 7.D\u00c3Y S\u1ed0 B\u00e0i 1: S = 1,2 + 2,3 + 3,4 +.......+ 28,29 + 29,30. V\u1eady S =? A= 1,2 + 2,3\u2026\u2026... 7,8 +8,9 (c\u00f3 8 s\u1ed1 h\u1ea1ng, h\u01a1n k\u00e9m nhau 1,1) B= 9,10+10,11+\u2026\u2026.. 28,29 + 29,30 (c\u00f3 29-9+1=21 s\u1ed1 h\u1ea1ng, h\u01a1n k\u00e9m nhau 1,01) A= (1,2+8,9) x 8 : 2 =40,4 B= (9,10+29,30) x21 : 2 =403,2","S = A + B = 40,4 + 403,2 =443,6 B\u00e0i 2: M\u1ed7i chi\u1ebfc xe \u0111\u01b0\u1ee3c g\u1eafn m\u1ed9t bi\u1ec3n s\u1ed1 g\u1ed3m 4 ch\u1eef s\u1ed1. H\u1ecfi t\u1eeb s\u1ed1 0009 \u0111\u1ebfn s\u1ed1 9999 c\u00f3 bao nhi\u00eau s\u1ed1 m\u00e0 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 chia h\u1ebft cho 9. S\u1ed1 chia h\u1ebft cho 9 khi t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 chia h\u1ebft cho 9. C\u00e1c s\u1ed1 chia h\u1ebft cho 9 g\u1ed3m: 0009; 0018; 0027; \u2026\u2026; 9990; 9999 \u0110\u00e2y l\u00e0 d\u00e3y s\u1ed1 c\u00e1ch \u0111\u1ec1u nhau 9 \u0111\u01a1n v\u1ecb. S\u1ed1 c\u00f3 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 chia h\u1ebft cho 9 c\u00f3: (9999 \u2013 0009) : 9 +1 = 1111 (s\u1ed1) \u0110\u00e1p s\u1ed1: 1111 s\u1ed1. B\u00e0i 3: T\u1ed5ng c\u1ee7a m\u1ed9t d\u00e3y s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp b\u1eb1ng 2012. T\u00ecm s\u1ed1 b\u00e9 nh\u1ea5t trong d\u00e3y s\u1ed1 \u0111\u00f3. T\u1eeb c\u00f4ng th\u1ee9 t\u00ednh t\u1ed5ng d\u00e3y s\u1ed1 c\u00e1ch \u0111\u1ec1u (s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp c\u0169ng l\u00e0 d\u00e3y s\u1ed1 c\u00e1ch \u0111\u1ec1u nhau 1 \u0111\u01a1n v\u1ecb)T\u1ed5ng=(s\u1ed1 \u0111\u1ea7u+s\u1ed1 cu\u1ed1i)x s\u1ed1 s\u1ed1 h\u1ea1ng : 2 2 t\u1ed5ng = (s\u1ed1 \u0111\u1ea7u + s\u1ed1 cu\u1ed1i) x s\u1ed1 s\u1ed1 h\u1ea1ng Hai t\u1ed5ng c\u1ee7a d\u00e3y s\u1ed1: 2012 x 2 = 4024 4024 chia h\u1ebft cho: 2; 4; 8; 503; 1006; 2012 Ta c\u00f3 c\u00e1c c\u1eb7p: N\u1ebfu t\u1ed5ng s\u1ed1 \u0111\u1ea7u v\u00e0 s\u1ed1 cu\u1ed1i l\u00e0 2012 th\u00ec c\u00f3 2 s\u1ed1 h\u1ea1ng; l\u00e0 1006 th\u00ec c\u00f3 4 s\u1ed1 h\u1ea1ng; l\u00e0 503 th\u00ec c\u00f3 8 s\u1ed1 h\u1ea1ng. Trong d\u00e3y s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp c\u00f3 s\u1ed1 s\u1ed1 h\u1ea1ng l\u00e0 m\u1ed9t s\u1ed1 ch\u1eb5n th\u00ec t\u1ed5ng s\u1ed1 \u0111\u1ea7u v\u00e0 s\u1ed1 cu\u1ed1i l\u00e0 m\u1ed9t s\u1ed1 l\u1ebb. *.T\u1ed5ng s\u1ed1 \u0111\u1ea7u s\u1ed1 cu\u1ed1i c\u0169ng l\u00e0 t\u1ed5ng 2 s\u1ed1 \u1edf gi\u1eefa b\u1eb1ng 503. Ta \u0111\u01b0\u1ee3c 2 s\u1ed1 \u1edf gi\u1eefa l\u00e0 (503-1) : 2 = 251 v\u00e0 251+1=252 D\u00e3y s\u1ed1: 248; 249; 250; 251; 252; 253; 254; 255 S\u1ed1 b\u00e9 nh\u1ea5t l\u00e0: 248 B\u00e0i 4: Cho d\u00e3y s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp 1 ; 2 ; 3 ; 4 ... n. T\u00ecm n bi\u1ebft s\u1ed1 ch\u1eef s\u1ed1 c\u1ee7a d\u00e3y s\u1ed1 \u0111\u00f3 b\u1eb1ng 3.n Trong m\u1ed7i s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 th\u00ec s\u1ed1 ch\u1eef s\u1ed1 g\u1ea5p 3 l\u1ea7n m\u1ed7i s\u1ed1 \u1ea5y. M\u1ed7i s\u1ed1 s\u1ed1 c\u00f3 2 ch\u1eef s\u1ed1 thi\u1ebfu 1 ch\u1eef s\u1ed1 n\u1eefa m\u1edbi c\u00f3 s\u1ed1 ch\u1eef s\u1ed1 g\u1ea5p 3 l\u1ea7n m\u1ed7i s\u1ed1 \u0111\u00f3. M\u1ed7i s\u1ed1 s\u1ed1 c\u00f3 1 ch\u1eef s\u1ed1 thi\u1ebfu 2 ch\u1eef s\u1ed1 n\u1eefa m\u1edbi c\u00f3 s\u1ed1 ch\u1eef s\u1ed1 g\u1ea5p 3 l\u1ea7n m\u1ed7i s\u1ed1 \u0111\u00f3. Ta xem c\u00f3 bao nhi\u00eau s\u1ed1 c\u00f3 2 ch\u1eef s\u1ed1: T\u1eeb 10 \u0111\u1ebfn 99 c\u00f3 99-10+1= 90 (s\u1ed1). Nh\u01b0 v\u1eady c\u00f2n thi\u1ebfu 1x90=90 (ch\u1eef s\u1ed1) S\u1ed1 c\u00f3 1 ch\u1eef s\u1ed1: T\u1eeb 1 \u0111\u1ebfn 9 c\u00f3 9 s\u1ed1. Nh\u01b0 v\u1eady c\u00f2n thi\u1ebfu 2x9=18 (ch\u1eef s\u1ed1) M\u1ed7i s\u1ed1 c\u00f3 4 ch\u1eef s\u1ed1 th\u00ec d\u01b0 1 ch\u1eef s\u1ed1 khi s\u1ed1 ch\u1eef s\u1ed1 g\u1ea5p 3 l\u1ea7n m\u1ed7i s\u1ed1 \u0111\u00f3. V\u1eady s\u1ed1 4 ch\u1eef s\u1ed1 c\u1ea7n c\u00f3 l\u00e0: 90+18=108 (s\u1ed1) S\u1ed1 n l\u00e0: 1000+(108-1)=1107 Gi\u1ed1ng d\u1ea1ng b\u00e0i t\u00ecm s\u1ed1 trang s\u00e1ch, bi\u1ebft t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 \u0111\u1ec3 d\u00f9ng \u0111\u00e1nh s\u1ed1 trang g\u1ea5p 2 l\u1ea7n (3 l\u1ea7n) s\u1ed1 trang. B\u00e0i 5: T\u00ednh t\u1ed5ng t\u1ea5t c\u1ea3 c\u00e1c s\u1ed1 c\u00f3 hai ch\u1eef s\u1ed1 m\u00e0 m\u1ed7i s\u1ed1 \u0111\u00f3 chia cho5 d\u01b0 2?","S\u1ed1 c\u00f3 2 ch\u1ee9c \u1ed1 chia cho 5 d\u01b0 2 \u0111\u00f3 l\u00e0: 12; 17; 22; \u2026\u2026\u2026\u2026\u2026\u2026\u2026; 92; 97. \u0110\u00e2y l\u00e0 d\u00e3y s\u1ed1 c\u00e1ch \u0111\u1ec1u nhau 5 \u0111\u01a1n v\u1ecb, c\u00f3: (97-12):5+1=18 (s\u1ed1 h\u1ea1ng) T\u1ed5ng c\u00e1c s\u1ed1 n\u00e0y l\u00e0: (12+97)x18:2 = 981 B\u00e0i 6: S\u1ed1 ch\u1eef s\u1ed1 d\u00f9ng \u0111\u1ec3 \u0111\u00e1nh s\u1ed1 trang c\u1ee7a m\u1ed9t quy\u1ec3n s\u00e1ch l\u00e0 m\u1ed9t s\u1ed1 chia h\u1ebft cho s\u1ed1 trang c\u1ee7a cu\u1ed1n s\u00e1ch \u0111\u00f3. Bi\u1ebft r\u1eb1ng cu\u1ed1n s\u00e1ch \u0111\u00f3 tr\u00ean 100 trang v\u00e0 \u00edt h\u01a1n 500 trang. H\u1ecfi cu\u1ed1n s\u00e1ch \u0111\u00f3 c\u00f3 bao nhi\u00eau trang ? S\u1ed1 ch\u1eef s\u1ed1 chia h\u1ebft cho s\u1ed1 trang t\u1eeb 100 \u0111\u1ebfn 500 th\u01b0\u01a1ng ch\u1ec9 c\u00f3 th\u1ec3 l\u00e0 2 (v\u00ec th\u01b0\u01a1ng l\u00e0 3 cu\u1ed1n s\u00e1ch s\u1ebd c\u00f3 s\u1ed1 trang ph\u1ea3i h\u01a1n 1000_ s\u1ed1 c\u00f3 4 ch\u1eef s\u1ed1_). T\u1eeb trang 1 \u0111\u1ebfn trang 9 m\u1ed7i trang c\u00f3 1 ch\u1eef s\u1ed1. M\u1ed7i trang c\u00f2n thi\u1ebfu 1 ch\u1eef s\u1ed1 \u0111\u1ec3 g\u1ea5p 2 l\u1ea7n s\u1ed1 trang. T\u1eeb trang 10 \u0111\u1ebfn 99, m\u1ed7i trang c\u00f3 s\u1ed1 ch\u1eef s\u1ed1 g\u1ea5p 2 l\u1ea7n s\u1ed1 trang. \u0110\u1ec3 s\u1ed1 ch\u1eef s\u1ed1 g\u1ea5p 2 l\u1ea7n s\u1ed1 trang th\u00ec t\u1eeb trang 100 tr\u1edf l\u00ean ph\u1ea3i c\u00f3 9 trang \u0111\u1ec3 b\u00f9 v\u00e0o s\u1ed1 trang c\u00f3 1 ch\u1eef s\u1ed1. Quy\u1ec3n s\u00e1ch c\u00f3 100+(9-1) = 108 (trang) B\u00e0i 7: H\u00e3y cho bi\u1ebft c\u00f3 t\u1ea5t c\u1ea3 bao nhi\u00eau s\u1ed1 l\u1ebb c\u00f3 4 ch\u1eef s\u1ed1 chia h\u1ebft cho 9. S\u1ed1 c\u00f3 4 ch\u1eef s\u1ed1 t\u1eeb 1000 \u0111\u1ebfn 9999 C\u00e1c s\u1ed1 chia h\u1ebft cho 9 g\u1ed3m: 1008; 1017; 1026; \u2026\u2026\u2026; 9990; 9999 D\u00e3y s\u1ed1 tr\u00ean c\u00e1ch \u0111\u1ec1u nhau 9 \u0111\u01a1n v\u1ecb c\u00f3 s\u1ed1 \u0111\u1ea7u l\u00e0 1008 v\u00e0 s\u1ed1 cu\u1ed1i l\u00e0 9999 C\u00e1c s\u1ed1 chia h\u1ebft cho 9 c\u00f3: (999-1008):9+1 = 1000 (s\u1ed1) Trong \u0111\u00f3 xen k\u1ebb 1 s\u1ed1 ch\u1eb5n th\u00ec 1 s\u1ed1 l\u1ebb, n\u00ean s\u1ed1 l\u1ebb c\u00f3 4 ch\u1eef s\u1ed1 chia h\u1ebft cho 9 c\u00f3: 1000 : 2 = 500 (s\u1ed1) (ho\u1eb7c ta t\u00ednh s\u1ed1 s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y s\u1ed1: 1017; 1035; \u2026\u2026; 9981; 9999 c\u00f3: (9999-1017):18+1= 500 (s\u1ed1)) B\u00e0i 8: T\u00ednh trung b\u00ecnh c\u1ed9ng c\u1ee7a c\u00e1c s\u1ed1 l\u1ebb nh\u1ecf h\u01a1n 2012. C\u00e1c s\u1ed1 l\u1ebb nh\u1ecf h\u01a1n 2012, g\u1ed3m: 1; 3; 5; \u2026\u2026\u2026.; 2009; 2011 Th\u01b0\u1eddng th\u00ec ng\u01b0\u1eddi ta t\u00ednh TBC b\u1eb1ng c\u00e1ch l\u1ea5y t\u1ed5ng c\u00e1c s\u1ed1 \u0111\u00f3 chia cho s\u1ed1 c\u00e1c s\u1ed1 h\u1ea1ng. Nh\u01b0ng v\u1edbi d\u00e3y s\u1ed1 c\u00e1ch \u0111\u1ec1u th\u00ec TBC c\u1ee7a ch\u00fang th\u00ec b\u1eb1ng trung b\u00ecnh c\u1ed9ng c\u1ee7a s\u1ed1 \u0111\u1ea7u v\u00e0 s\u1ed1 cu\u1ed1i. TBC c\u1ee7a d\u00e3y s\u1ed1 \u0111\u00f3 l\u00e0: (1+2011) : 2 = 1006 B\u00e0i 9: T\u00ednh t\u1ed5ng c\u00e1c s\u1ed1 l\u1ebb nh\u1ecf h\u01a1n 200. C\u00e1c s\u1ed1 l\u1ebb nh\u1ecf h\u01a1n 120 l\u00e0: 1; 3; 5; \u2026\u2026\u2026.; 197; 199. \u0110\u00e2y l\u00e0 d\u00e3y s\u1ed1 c\u00e1ch \u0111\u1ec1u nhau 2 \u0111\u01a1n v\u1ecb, c\u00f3 s\u1ed1 \u0111\u1ea7u l\u00e0 1 v\u00e0 s\u1ed1 cu\u1ed1i l\u00e0 199 S\u1ed1 s\u1ed1 h\u1ea1ng: (199 \u2013 1) : 2 + 1 = 100 (s\u1ed1) T\u1ed5ng ch\u00fang l\u00e0: (1+199) x 100 : 2 = 10 000"]
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