Bài tập toán 4 theo các chuyên đề

["a, 100 ; 93 ; 85 ; 76 ; ... b, 10 ; 13 ; 18 ; 26 ; ... c, 0 ; 1 ; 2 ; 4 ; 7 ; 12 ; ... d, 0 ; 1 ; 4 ; 9 ; 18 ; ... e, 5 ; 6 ; 8 ; 10 ; ... f, 1 ; 6 ; 54 ; 648 ; ... g, 1 ; 3 ; 3 ; 9 ; 27 ; ... h, 1 ; 1 ; 3 ; 5 ; 17 ; ... B\u00e0i 2 : \u0110i\u1ec1n th\u00eam 7 s\u1ed1 h\u1ea1ng v\u00e0o t\u1ed5ng sau sao cho m\u1ed7i s\u1ed1 h\u1ea1ng trong t\u1ed5ng \u0111\u1ec1u l\u1edbn h\u01a1n s\u1ed1 h\u1ea1ng \u0111\u1ee9ng tr\u01b0\u1edbc n\u00f3 : 49 + ... ... = 420. Gi\u1ea3i th\u00edch c\u00e1ch t\u00ecm. B\u00e0i 3 : T\u00ecm hai s\u1ed1 h\u1ea1ng \u0111\u1ea7u c\u1ee7a c\u00e1c d\u00e3y sau : a, . . . , 39, 42, 45 ; b, . . . , 4, 2, 0 ; c, . . . , 23, 25, 27, 29 ; Bi\u1ebft r\u1eb1ng m\u1ed7i d\u00e3y c\u00f3 15 s\u1ed1 h\u1ea1ng. B\u00e0i 4 : a, \u0110i\u1ec1n c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng, sao cho t\u00edch c\u00e1c s\u1ed1 c\u1ee7a 3 \u00f4 li\u00ean ti\u1ebfp \u0111\u1ec1u b\u1eb1ng 2000 50 2 b, Cho 9 s\u1ed1 : 1, 2, 3, 4, 5, 6, 7, 8 v\u00e0 9. H\u00e3y \u0111i\u1ec1n m\u1ed7i s\u1ed1 v\u00e0o 1 \u00f4 tr\u00f2n sao cho t\u1ed5ng c\u1ee7a 3 s\u1ed1 \u1edf 3 \u00f4 th\u1eb3ng h\u00e0ng nhau \u0111\u1ec1u chia h\u1ebft cho 5. H\u00e3y gi\u1ea3i th\u00edch c\u00e1ch l\u00e0m. O OO OO O OO","O O OO O OO c, H\u00e3y \u0111i\u1ec1n s\u1ed1 v\u00e0o c\u00e1c \u00f4 tr\u00f2n sao cho t\u1ed5ng c\u1ee7a 3 \u00f4 li\u00ean ti\u1ebfp \u0111\u1ec1u b\u1eb1ng nhau. Gi\u1ea3i th\u00edch c\u00e1ch l\u00e0m.? D\u1ea1ng 3 : T\u00ecm s\u1ed1 s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y s\u1ed1 . * L\u01b0u \u00fd : - \u1edf d\u1ea1ng n\u00e0y th\u01b0\u1eddng s\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i to\u00e1n kho\u1ea3ng c\u00e1ch (tr\u1ed3ng c\u00e2y).Ta c\u00f3 c\u00f4ng th\u1ee9c sau: S\u1ed1 s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y = S\u1ed1 kho\u1ea3ng c\u00e1ch + 1 - N\u1ebfu quy lu\u1eadt c\u1ee7a d\u00e3y l\u00e0 : s\u1ed1 \u0111\u1ee9ng sau b\u1eb1ng s\u1ed1 h\u1ea1ng li\u1ec1n tr\u01b0\u1edbc c\u1ed9ng v\u1edbi s\u1ed1 kh\u00f4ng \u0111\u1ed5i th\u00ec : S\u1ed1 c\u00e1c s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y = (S\u1ed1 cu\u1ed1i \u2013 s\u1ed1 \u0111\u1ea7u) : K\/c + 1 *B\u00e0i t\u1eadp v\u1eadn d\u1ee5ng : B\u00e0i 1: Vi\u1ebft c\u00e1c s\u1ed1 l\u1ebb li\u00ean ti\u1ebfp t\u1eeb 211. S\u1ed1 cu\u1ed1i c\u00f9ng l\u00e0 971. H\u1ecfi vi\u1ebft \u0111\u01b0\u1ee3c bao nhi\u00eau s\u1ed1 ? Gi\u1ea3i: Hai s\u1ed1 l\u1ebb li\u00ean ti\u1ebfp h\u01a1n k\u00e9m nhau 2 \u0111\u01a1n v\u1ecb S\u1ed1 cu\u1ed1i h\u01a1n s\u1ed1 \u0111\u1ea7u s\u1ed1 \u0111\u01a1n v\u1ecb l\u00e0 : 971 \u2013 211 = 760 (\u0111\u01a1n v\u1ecb) 760 \u0111\u01a1n v\u1ecb c\u00f3 s\u1ed1 kho\u1ea3ng c\u00e1ch l\u00e0 : 760 : 2 = 380 (K\/ c) D\u00e3y s\u1ed1 tr\u00ean c\u00f3 s\u1ed1 s\u1ed1 h\u1ea1ng l\u00e0 : 380 +1 = 381 (s\u1ed1) \u0110\u00e1p s\u1ed1 :381 s\u1ed1 h\u1ea1ng B\u00e0i 2: Cho d\u00e3y s\u1ed1 11, 14, 17, ... , 68. a, H\u00e3y x\u00e1c \u0111\u1ecbnh d\u00e3y tr\u00ean c\u00f3 bao nhi\u00eau s\u1ed1 h\u1ea1ng ? b, N\u1ebfu ta ti\u1ebfp t\u1ee5c k\u00e9o d\u00e0i c\u00e1c s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y s\u1ed1 th\u00ec s\u1ed1 h\u1ea1ng th\u1ee9 1 996 l\u00e0 s\u1ed1 m\u1ea5y ? Gi\u1ea3i : a,Ta c\u00f3 : 14 \u2013 11 = 3 17 \u2013 14 = 3 V\u1eady quy lu\u1eadt c\u1ee7a d\u00e3y l\u00e0 : m\u1ed7i s\u1ed1 h\u1ea1ng \u0111\u1ee9ng sau b\u1eb1ng s\u1ed1 h\u1ea1ng \u0111\u1ee9ng tr\u01b0\u1edbc c\u1ed9ng v\u1edbi 3 .","S\u1ed1 c\u00e1c s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y l\u00e0 : ( 68 \u2013 11 ) : 3 + 1 = 20 (s\u1ed1 h\u1ea1ng) b, Ta nh\u1eadn x\u00e9t : S\u1ed1 h\u1ea1ng th\u1ee9 hai : 14 = 11 + 3 = 11 + (2 \u2013 1) x 3 S\u1ed1 h\u1ea1ng th\u1ee9 ba : 17 = 11 + 6 = 11 + (3 \u2013 1) x 3 S\u1ed1 h\u1ea1ng th\u1ee9 t\u01b0 : 20 = 11 + 9 = 11 + (4 \u2013 1) x 3 V\u1eady s\u1ed1 h\u1ea1ng th\u1ee9 1 996 l\u00e0 : 11 + (1 996 \u2013 1) x 3 = 5 996 \u0110\u00e1p s\u1ed1 : 20 s\u1ed1 h\u1ea1ng ; 5 996 B\u00e0i 3: Trong c\u00e1c s\u1ed1 c\u00f3 ba ch\u1eef s\u1ed1, c\u00f3 bao nhi\u00eau s\u1ed1 chia h\u1ebft cho 4 ? Gi\u1ea3i : Ta c\u00f3 nh\u1eadn x\u00e9t :s\u1ed1 nh\u1ecf nh\u1ea5t c\u00f3 ba ch\u1eef s\u1ed1 chia h\u1ebft cho 4l\u00e0 100 v\u00e0 s\u1ed1 l\u1edbn nh\u1ea5t c\u00f3 ba ch\u1eef s\u1ed1 chia h\u1ebft cho 4 l\u00e0 996. Nh\u01b0 v\u1eady c\u00e1c s\u1ed1 c\u00f3 ba ch\u1eef s\u1ed1 chia h\u1ebft cho 4 l\u1eadp th\u00e0nh m\u1ed9t d\u00e3y s\u1ed1 c\u00f3 s\u1ed1 h\u1ea1ng \u0111\u1ea7u l\u00e0 100, s\u1ed1 h\u1ea1ng cu\u1ed1i l\u00e0 996 v\u00e0 m\u1ed7i s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y (K\u1ec3 t\u1eeb s\u1ed1 h\u1ea1ng th\u1ee9 hai) b\u1eb1ng s\u1ed1 h\u1ea1ng \u0111\u1ee9ng k\u1ec1 tr\u01b0\u1edbc c\u1ed9ng v\u1edbi 4. V\u1eady c\u00e1c s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 chia h\u1ebft cho 4 l\u00e0 : (996 \u2013 100) : 4 + 1 = 225 (s\u1ed1) \u0110\u00e1p s\u1ed1 : 225 s\u1ed1 D\u1ea1ng 4 : T\u00ecm t\u1ed5ng c\u00e1c s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y s\u1ed1 * C\u00e1ch gi\u1ea3i N\u1ebfu c\u00e1c s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y s\u1ed1 c\u00e1ch \u0111\u1ec1u nhau th\u00ec t\u1ed5ng c\u1ee7a 2 s\u1ed1 h\u1ea1ng c\u00e1ch \u0111\u1ec1u s\u1ed1 h\u1ea1ng \u0111\u1ea7u v\u00e0 s\u1ed1 h\u1ea1ng cu\u1ed1i trong d\u00e3y \u0111\u00f3 b\u1eb1ng nhau. V\u00ec v\u1eady : T\u1ed5ng c\u00e1c s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y = t\u1ed5ng c\u1ee7a 1 c\u1eb7p 2 s\u1ed1 h\u1ea1ng c\u00e1ch \u0111\u1ec1u s\u1ed1 h\u1ea1ng \u0111\u1ea7u v\u00e0 cu\u1ed1i x s\u1ed1 h\u1ea1ng c\u1ee7a d\u00e3y : 2 * B\u00e0i t\u1eadp v\u1eadn d\u1ee5ng : B\u00e0i 1 : T\u00ednh t\u1ed5ng c\u1ee7a 100 s\u1ed1 l\u1ebb \u0111\u1ea7u ti\u00ean. Gi\u1ea3i : D\u00e3y c\u1ee7a 100 s\u1ed1 l\u1ebb \u0111\u1ea7u ti\u00ean l\u00e0 : 1 + 3 + 5 + 7 + 9 + . . . + 197 + 199. Ta c\u00f3 : 1 + 199 = 200 3 + 197 = 200 5 + 195 = 200 ... V\u1eady t\u1ed5ng ph\u1ea3i t\u00ecm l\u00e0 : 200 x 100 : 2 = 10 000 \u0110\u00e1p s\u1ed1 10 000. B\u00e0i 2 : Cho 1 s\u1ed1 t\u1ef1 nhi\u00ean g\u1ed3m c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp t\u1eeb 1 \u0111\u1ebfn 1983 \u0111\u01b0\u1ee3c vi\u1ebft theo th\u1ee9 t\u1ef1 li\u1ec1n nhau nh\u01b0 sau :","1 2 3 4 5 6 7 8 9 10 11 12 13 . . . 1980 1981 1982 1983 H\u00e3y t\u00ednh t\u1ed5ng t\u1ea5t c\u1ea3 c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a s\u1ed1 \u0111\u00f3. (\u0110\u1ec1 thi h\u1ecdc sinh gi\u1ecfi to\u00e0n qu\u1ed1c n\u0103m 1983) Gi\u1ea3i : C\u00e1ch 1. Ta nh\u1eadn x\u00e9t : * c\u00e1c c\u1eb7p s\u1ed1 : - 0 v\u00e0 1999 c\u00f3 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 l\u00e0 : 0 + 1 + 9 + 9 + 9 = 28 - 1 v\u00e0 1998 c\u00f3 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 l\u00e0 : 1 + 1 + 9 + 9 + 8 = 28 - 2 v\u00e0 1997 c\u00f3 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 l\u00e0 : 2 + 1 + 9 + 9 + 7 = 28 - 998 v\u00e0 1001 c\u00f3 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 l\u00e0 : 9 + 9 + 8 + 1 + 1 = 28 - 999 v\u00e0 1000 c\u00f3 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 l\u00e0 : 9 + 9 + 9 + 1 = 28 Nh\u01b0 v\u1eady trong d\u00e3y s\u1ed1 0, 1, 2, 3, 4, 5, . . . , 1997, 1998, 1999 Hai s\u1ed1 h\u1ea1ng c\u00e1ch \u0111\u1ec1u s\u1ed1 h\u1ea1ng \u0111\u1ea7u v\u00e0 s\u1ed1 h\u1ea1ng cu\u1ed1i \u0111\u1ec1u c\u00f3 t\u1ed5ng b\u1eb1ng 28. C\u00f3 1000 c\u1eb7p nh\u01b0 v\u1eady, do \u0111\u00f3 t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 t\u1ea1o n\u00ean d\u00e3y s\u1ed1 tr\u00ean l\u00e0 : 28 x 1000 = 28 000 * S\u1ed1 t\u1ef1 nhi\u00ean \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh b\u1eb1ng c\u00e1ch vi\u1ebft li\u00ean ti\u1ebfp c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean t\u1eeb 1984 \u0111\u1ebfn 1999 l\u00e0 (1 + 9 + 8 + 4) + (1 + 9 + 8 + 5) +... +(1 + 9 + 8 + 9) + (1 + 9 + 9 + 0) + ... + 22 23 27 19 (1 + 9 + 9 + 8) + (1 + 9 + 9 + 9) = 382 27 28 * V\u1eady t\u1ed5ng c\u00e1c ch\u1eef s\u1ed1 c\u1ee7a s\u1ed1 t\u1ef1 nhi\u00ean \u0111\u00e3 cho l\u00e0 : 28 000 \u2013 382 = 27 618. B\u00e0i 3 : Vi\u1ebft c\u00e1c s\u1ed1 ch\u1eb5n li\u00ean ti\u1ebfp : 2, 4, 6, 8, . . . , 2000 T\u00ednh t\u1ed5ng c\u1ee7a d\u00e3y s\u1ed1 tr\u00ean Gi\u1ea3i : D\u00e3y s\u1ed1 tr\u00ean 2 s\u1ed1 ch\u1eb5n li\u00ean ti\u1ebfp h\u01a1n k\u00e9m nhau 2 \u0111\u01a1n v\u1ecb. D\u00e3y s\u1ed1 tr\u00ean c\u00f3 s\u1ed1 s\u1ed1 h\u1ea1ng l\u00e0 : (2000 \u2013 2) : 2 + 1 = 1000 (s\u1ed1)","1000 s\u1ed1 c\u00f3 s\u1ed1 c\u1eb7p s\u1ed1 l\u00e0 : 1000 : 2 = 500 (c\u1eb7p) T\u1ed5ng 1 c\u1eb7p l\u00e0 : 2 + 2000 = 2002 T\u1ed5ng c\u1ee7a d\u00e3y s\u1ed1 l\u00e0 : 2002 x 500 = 100100. * B\u00e0i t\u1eadp v\u1ec1 nh\u00e0 B\u00e0i 1 : T\u00ednh t\u1ed5ng : a, 6 + 8 + 10 + ... + 1999. b, 11 + 13 + 15 + ... + 147 + 150 c, 3 + 6 + 9 + ... + 147 + 150. B\u00e0i 2 : Vi\u1ebft 80 s\u1ed1 ch\u1eb5n li\u00ean ti\u1ebfp b\u1eaft \u0111\u1ea7u t\u1eeb 72. S\u1ed1 cu\u1ed1i c\u00f9ng l\u00e0 s\u1ed1 n\u00e0o? B\u00e0i 3 : C\u00f3 bao nhi\u00eau s\u1ed1 : a, C\u00f3 3 ch\u1eef s\u1ed1 khi chia cho 5 d\u01b0 1? d\u01b0 2? b, C\u00f3 4 ch\u1eef s\u1ed1 chia h\u1ebft cho 3? c, C\u00f3 3 ch\u1eef s\u1ed1 nh\u1ecf h\u01a1n 500 m\u00e0 chia h\u1ebft cho 4? B\u00e0i 4 : Khi \u0111\u00e1nh s\u1ed1 th\u1ee9 t\u1ef1 c\u00e1c d\u00e3y nh\u00e0 tr\u00ean m\u1ed9t \u0111\u01b0\u1eddng ph\u1ed1, ng\u01b0\u1eddi ta d\u00f9ng c\u00e1c s\u1ed1 l\u1ebb li\u00ean ti\u1ebfp 1, 3, 5, 7, ... \u0111\u1ec3 \u0111\u00e1nh s\u1ed1 d\u00e3y th\u1ee9 nh\u1ea5t v\u00e0 c\u00e1c s\u1ed1 ch\u1eb5n li\u00ean ti\u1ebfp 2, 4, 6, 8, ... \u0111\u1ec3 \u0111\u00e1nh s\u1ed1 d\u00e3y th\u1ee9 hai. H\u1ecfi nh\u00e0 cu\u1ed1i c\u00f9ng trong d\u00e3y ch\u1eb5n c\u1ee7a \u0111\u01b0\u1eddng ph\u1ed1 \u0111\u00f3 l\u00e0 s\u1ed1 m\u1ea5y, n\u1ebfu khi \u0111\u00e1nh s\u1ed1 d\u00e3y n\u00e0y ng\u01b0\u1eddi ta \u0111\u00e3 d\u00f9ng 769 ch\u1eef c\u1ea3 th\u1ea3y? B\u00e0i 5 : Cho d\u00e3y c\u00e1c s\u1ed1 ch\u1eb5n li\u00ean ti\u1ebfp 2, 4, 6, 8, ... H\u1ecfi s\u1ed1 1996 l\u00e0 s\u1ed1 h\u1ea1ng th\u1ee9 m\u1ea5y c\u1ee7a d\u00e3y n\u00e0y? Gi\u1ea3i th\u00edch c\u00e1ch t\u00ecm. B\u00e0i 6 : T\u00ecm t\u1ed5ng c\u1ee7a : a, C\u00e1c s\u1ed1 c\u00f3 hai ch\u1eef s\u1ed1 chia h\u1ebft cho 3 ; b, C\u00e1c s\u1ed1 c\u00f3 hai ch\u1eef s\u1ed1 chia cho 4 d\u01b0 1 ; c, 100 s\u1ed1 ch\u1eb5n \u0111\u1ea7u ti\u00ean ; d, 10 s\u1ed1 l\u1ebb kh\u00e1c nhau l\u1edbn h\u01a1n 20 v\u00e0 nh\u1ecf h\u01a1n 40. D\u1ea1ng 5 : T\u00ecm s\u1ed1 h\u1ea1ng th\u1ee9 n * B\u00e0i t\u1eadp v\u1eadn d\u1ee5ng B\u00e0i 1 : Cho d\u00e3y s\u1ed1 : 1, 3, 5, 7, ... H\u1ecfi s\u1ed1 h\u1ea1ng th\u1ee9 20 c\u1ee7a d\u00e3y l\u00e0 s\u1ed1 n\u00e0o? Gi\u1ea3i : D\u00e3y \u0111\u00e3 cho l\u00e0 d\u00e3y s\u1ed1 l\u1ebb n\u00ean c\u00e1c s\u1ed1 li\u00ean ti\u1ebfp trong d\u00e3y c\u00e1ch nhau 1 kho\u1ea3ng c\u00e1ch l\u00e0 2 \u0111\u01a1n v\u1ecb. 20 s\u1ed1 h\u1ea1ng th\u00ec c\u00f3 s\u1ed1 kho\u1ea3ng c\u00e1ch l\u00e0 : 20 \u2013 1 = 19 \u01a0kho\u1ea3ng c\u00e1ch) 19 s\u1ed1 c\u00f3 s\u1ed1 \u0111\u01a1n v\u1ecb l\u00e0 :","19 x 2 = 38 (\u0111\u01a1n v\u1ecb) S\u1ed1 cu\u1ed1i c\u00f9ng l\u00e0 : 1 + 38 = 39 \u0110\u00e1p s\u1ed1 : S\u1ed1 h\u1ea1ng th\u1ee9 20 c\u1ee7a d\u00e3y l\u00e0 39 B\u00e0i 2 : Vi\u1ebft 20 s\u1ed1 l\u1ebb, s\u1ed1 cu\u1ed1i c\u00f9ng l\u00e0 2001. S\u1ed1 \u0111\u1ea7u ti\u00ean l\u00e0 s\u1ed1 n\u00e0o? Gi\u1ea3i : 2 s\u1ed1 l\u1ebb li\u00ean ti\u1ebfp h\u01a1n k\u00e9m nhau 2 \u0111\u01a1n v\u1ecb 20 s\u1ed1 l\u1ebb c\u00f3 s\u1ed1 kho\u1ea3ng c\u00e1ch l\u00e0 : 20 \u2013 1 = 19 (kho\u1ea3ng c\u00e1ch) 19 kho\u1ea3ng c\u00e1ch c\u00f3 s\u1ed1 \u0111\u01a1n v\u1ecb l\u00e0 : 19 x 2 = 38 (\u0111\u01a1n v\u1ecb) S\u1ed1 \u0111\u1ea7u ti\u00ean l\u00e0 : 2001 \u2013 38 = 1963 \u0110\u00e1p s\u1ed1 : s\u1ed1 \u0111\u1ea7u ti\u00ean l\u00e0 1963. C\u00f4ng th\u1ee9c : a, Cu\u1ed1i d\u00e3y : n = S\u1ed1 \u0111\u1ea7u + kho\u1ea3ng c\u00e1ch x (n \u2013 1) b, \u0110\u1ea7u d\u00e3y : n = S\u1ed1 cu\u1ed1i \u2013 kho\u1ea3ng c\u00e1ch x (n \u2013 1) * B\u00e0i t\u1eadp v\u1ec1 nh\u00e0 : B\u00e0i 1 : Vi\u1ebft c\u00e1c s\u1ed1 ch\u1eb5n b\u1eaft \u0111\u1ea7u t\u1eeb 2. S\u1ed1 cu\u1ed1i c\u00f9ng l\u00e0 938. D\u00e3y s\u1ed1 c\u00f3 bao nhi\u00eau s\u1ed1? B\u00e0i 2 : T\u00ednh : 2 + 4 + 6 + ... + 2000. B\u00e0i 3 : Cho d\u00e3y s\u1ed1 : 4, 8, 12, ... T\u00ecm s\u1ed1 h\u1ea1ng 50 c\u1ee7a d\u00e3y s\u1ed1 . B\u00e0i 4 : Vi\u1ebft 25 s\u1ed1 l\u1ebb li\u00ean ti\u1ebfp s\u1ed1 cu\u1ed1i c\u00f9ng l\u00e0 2001. H\u1ecfi s\u1ed1 \u0111\u1ea7u ti\u00ean l\u00e0 s\u1ed1 n\u00e0o? B\u00e0i 5 : T\u00ednh t\u1ed5ng : a, 6 + 8 + 10 + ... + 2000 b, 11 + 13 + 15 + ... + 1999. c, 3 + 6 + 9 + ... + 147 + 150. B\u00e0i 6 : Vi\u1ebft 80 s\u1ed1 ch\u1eb5n li\u00ean ti\u1ebfp b\u1eaft \u0111\u1ea7u t\u1eeb 72. H\u1ecfi s\u1ed1 cu\u1ed1i c\u00f9ng l\u00e0 s\u1ed1 n\u00e0o? B\u00e0i 7 : Cho d\u00e3y s\u1ed1 g\u1ed3m 25 s\u1ed1 h\u1ea1ng : . . ., 146, 150, 154. H\u1ecfi s\u1ed1 \u0111\u1ea7u ti\u00ean l\u00e0 s\u1ed1 n\u00e0o? D\u1ea1ng 6 : T\u00ecm s\u1ed1 ch\u1eef s\u1ed1 bi\u1ebft s\u1ed1 s\u1ed1 h\u1ea1ng * B\u00e0i t\u1eadp v\u1eadn d\u1ee5ng B\u00e0i 1 : Cho d\u00e3y s\u1ed1 1, 2, 3, 4, ..., 150. D\u00e3y n\u00e0y c\u00f3 bao nhi\u00eau ch\u1eef s\u1ed1 Gi\u1ea3i : D\u00e3y s\u1ed1 1, 2, 3, ..., 150 c\u00f3 150 s\u1ed1.","Trong 150 s\u1ed1 c\u00f3 + 9 s\u1ed1 c\u00f3 1 ch\u1eef s\u1ed1 + 90 s\u1ed1 c\u00f3 2 ch\u1eef s\u1ed1 + C\u00e1c s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 l\u00e0 : 150 \u2013 9 \u2013 90 = 51 (ch\u1eef s\u1ed1) D\u00e3y n\u00e0y c\u00f3 s\u1ed1 ch\u1eef s\u1ed1 l\u00e0 : 1 x 9 + 2 x 90 + 3 x 51 = 342 (ch\u1eef s\u1ed1) \u0110\u00e1p s\u1ed1 342 ch\u1eef s\u1ed1 B\u00e0i 2 : Vi\u1ebft c\u00e1c s\u1ed1 ch\u1eb5n li\u00ean ti\u1ebfp t\u1eef 2 \u0111\u1ebfn 1998 th\u00ec ph\u1ea3i vi\u1ebft bao nhi\u00eau ch\u1eef s\u1ed1? Gi\u1ea3i : D\u00e3y s\u1ed1 : 2, 4, ..., 1998 c\u00f3 s\u1ed1 s\u1ed1 h\u1ea1ng l\u00e0 : (1998 \u2013 2) : 2 + 1 = 999 (s\u1ed1) Trong 999 s\u1ed1 c\u00f3 : 4 s\u1ed1 ch\u1eb5n c\u00f3 1 ch\u1eef s\u1ed1 45 s\u1ed1 ch\u1eb5n c\u00f3 2 ch\u1eef s\u1ed1 450 s\u1ed1 ch\u1eb5n c\u00f3 3 ch\u1eef s\u1ed1 C\u00e1c s\u1ed1 ch\u1eb5n c\u00f3 4 ch\u1eef s\u1ed1 l\u00e0 : 999 \u2013 4 \u2013 45 \u2013 450 = 500 (s\u1ed1) S\u1ed1 l\u01b0\u1ee3ng ch\u1eef s\u1ed1 ph\u1ea3i vi\u1ebft l\u00e0 : 1 x 4 + 2 x 45 + 3 x 450 + 4 x 500 = 3444 (ch\u1eef s\u1ed1) \u0111\u00e1p s\u1ed1 : 3444 ch\u1eef s\u1ed1 Ghi nh\u1edb : \u0110\u1ec3 t\u00ecm s\u1ed1 ch\u1eef s\u1ed1 ta : + T\u00ecm xem trong d\u00e3y s\u1ed1 c\u00f3 bao nhi\u00eau s\u1ed1 s\u1ed1 h\u1ea1ng + Trong s\u1ed1 c\u00e1c s\u1ed1 \u0111\u00f3 c\u00f3 bao nhi\u00eau s\u1ed1 c\u00f3 1, 2, 3, 4, ... ch\u1eef s\u1ed1 D\u1ea1ng 7 :T\u00ecm s\u1ed1 s\u1ed1 h\u1ea1ng bi\u1ebft s\u1ed1 ch\u1eef s\u1ed1 * B\u00e0i t\u1eadp v\u1eadn d\u1ee5ng B\u00e0i 1 : M\u1ed9t quy\u1ec3n s\u00e1ch coc 435 ch\u1eef s\u1ed1. H\u1ecfi quy\u1ec3n s\u00e1ch \u0111\u00f3 c\u00f3 bao nhi\u00eau trang? Gi\u1ea3i : \u0110\u1ec3 \u0111\u00e1nh s\u1ed1 trang s\u00e1ch ng\u01b0\u1eddi ta b\u1eaft \u0111\u1ea7u \u0111\u00e1nh t\u1eef trang s\u1ed1 1. Ta th\u1ea5y \u0111\u1ec3 \u0111\u00e1nh s\u1ed1 trang c\u00f3 1 ch\u1eef s\u1ed1 ng\u01b0\u1eddi ta \u0111\u00e1nh m\u1ea5t 9 s\u1ed1 v\u00e0 m\u1ea5t : 1 x 9 = 9 (ch\u1eef s\u1ed1) S\u1ed1 trang s\u00e1ch c\u00f3 2 ch\u1eef s\u1ed1 l\u00e0 90 n\u00ean \u0111\u1ec3 \u0111\u00e1nh 90 trang n\u00e0y m\u1ea5t : 2 x 90 = 180 (ch\u1eef s\u1ed1) \u0110\u00e1nh quy\u1ec3n s\u00e1ch c\u00f3 435 ch\u1eef s\u1ed1 nh\u01b0 v\u1eady ch\u1ec9 \u0111\u1ebfn s\u1ed1 trang c\u00f3 3 ch\u1eef s\u1ed1. S\u1ed1 ch\u1eef s\u1ed1 \u0111\u1ec3 \u0111\u00e1nh s\u1ed1 trang s\u00e1ch c\u00f3 3 ch\u1eef s\u1ed1 l\u00e0: 435 \u2013 9 \u2013 180 = 246 (ch\u1eef s\u1ed1) 246 ch\u1eef s\u1ed1 th\u00ec \u0111\u00e1nh \u0111\u01b0\u1ee3c s\u1ed1 trang c\u00f3 3 ch\u1eef s\u1ed1 l\u00e0 : 246 : 3 = 82 (trang)","Quy\u1ec3n s\u00e1ch \u0111\u00f3 c\u00f3 s\u1ed1 trang l\u00e0 : 9 + 90 + 82 = 181 (trang) \u0111\u00e1p s\u1ed1 181 trang. B\u00e0i 2 : Vi\u1ebft c\u00e1c s\u1ed1 l\u1ebb li\u00ean ti\u1ebfp b\u1eaft \u0111\u1ea7u t\u1eeb s\u1ed1 87. H\u1ecfi n\u1ebfu ph\u1ea3i vi\u1ebft t\u1ea5t c\u1ea3 3156 ch\u1eef s\u1ed1 th\u00ec vi\u1ebft \u0111\u1ebfn s\u1ed1 n\u00e0o? Gi\u1ea3i : T\u1eeb 87 \u0111\u1ebfn 99 c\u00f3 c\u00e1c s\u1ed1 l\u1ebb l\u00e0 : (99 \u2013 87) : 2 + 1 = 7 (s\u1ed1) \u0110\u1ec3 vi\u1ebft 7 s\u1ed1 l\u1ebb c\u1ea7n : 2 x 7 = 14 (ch\u1eef s\u1ed1) C\u00f3 450 s\u1ed1 l\u1ebb c\u00f3 3 ch\u1eef s\u1ed1 n\u00ean c\u1ea7n : 3 x 450 = 1350 (ch\u1eef s\u1ed1) S\u1ed1 ch\u1eef s\u1ed1 d\u00f9ng \u0111\u1ec3 vi\u1ebft c\u00e1c s\u1ed1 l\u1ebb c\u00f3 4 ch\u1eef s\u1ed1 l\u00e0 : 3156 \u2013 14 \u2013 1350 = 1792 (ch\u1eef s\u1ed1) Vi\u1ebft \u0111\u01b0\u1ee3c c\u00e1c s\u1ed1 c\u00f3 4 ch\u1eef s\u1ed1 l\u00e0 : 1792 : 4 = 448 (s\u1ed1) Vi\u1ebft \u0111\u1ebfn s\u1ed1 : 999 + (448 \u2013 1) x 2 = 1893 D\u1ea1ng 8 : vi\u1ebft li\u00ean ti\u1ebfp m\u1ed9t nh\u00f3m ch\u1eef s\u1ed1 ho\u1eb7c ch\u1eef c\u00e1i B\u00e0i 1 : Vi\u1ebft li\u00ean ti\u1ebfp c\u00e1c ch\u1eef c\u00e1i A, N, L, \u01af, U th\u00e0nh d\u00e3y AN L\u01afU, AN L\u01afU, ... Ch\u1eef c\u00e3i th\u1ee9 1998 l\u00e0 ch\u1eef c\u00e1i g\u00ec? Gi\u1ea3i : \u0110\u1ec3 vi\u1ebft 1 nh\u00f3m AN L\u01afU ng\u01b0\u1eddi ta ph\u1ea3i vi\u1ebft 5 ch\u1eef c\u00e1i A, N, L, \u01af, U. N\u1ebfu x\u1ebfp 5 ch\u1eef c\u00e1i \u1ea5y v\u00e0o 1 nh\u00f3m ta c\u00f3 : Chia cho 5 kh\u00f4ng d\u01b0 l\u00e0 ch\u1eef c\u00e1i U Chia cho 5 d\u01b0 1 l\u00e0 ch\u1eef c\u00e1i A Chia cho 5 d\u01b0 2 l\u00e0 ch\u1eef c\u00e1i N Chia cho 5 d\u01b0 3 l\u00e0 ch\u1eef c\u00e1i L Chia cho 5 d\u01b0 4 l\u00e0 ch\u1eef c\u00e1i \u01af M\u00e0 : 1998 : 5 = 339 (nh\u00f3m) d\u01b0 3 V\u1eady ch\u1eef c\u00e1i th\u1ee9 1998 l\u00e0 ch\u1eef c\u00e1i L c\u1ee7a nh\u00f3m th\u1ee9 400 B\u00e0i 2 : M\u1ed9t ng\u01b0\u1eddi vi\u1ebft li\u00ean ti\u1ebfp nh\u00f3m ch\u1eef T\u1ed5 qu\u1ed1c vi\u1ec7t nam th\u00e0nh d\u00e3y T\u1ed5 qu\u1ed1c vi\u1ec7t nam T\u1ed5 qu\u1ed1c vi\u1ec7t nam ... a, Ch\u1eef c\u00e1i th\u1ee9 1996 trong d\u00e3y l\u00e0 ch\u1eef g\u00ec? b, Ng\u01b0\u1eddi ta \u0111\u1ebfm \u0111\u01b0\u1ee3c trong d\u00e3y c\u00f3 50 ch\u1eef T th\u00ec d\u00e3y \u0111\u00f3 c\u00f3 bao nhi\u00eau ch\u1eef \u00d4? bao nhi\u00eau ch\u1eef I","c, B\u1ea1n An \u0111\u1ebfm \u0111\u01b0\u1ee3c trong d\u00e3y c\u00f3 1995 ch\u1eef \u00d4. H\u1ecfi b\u1ea1n \u1ea5y \u0111\u1ebfm \u0111\u00fang hay sai? Gi\u1ea3i th\u00edch t\u1ea1i sao? d, Ng\u01b0\u1eddi ta t\u00f4 m\u00e0u c\u00e1c ch\u1eef c\u00e1i trong d\u00e3y theo th\u1ee9 t\u1ef1 : Xanh, \u0111\u1ecf, t\u00edm, v\u00e0ng. xanh, \u0111\u1ecf, ... H\u1ecfi ch\u1eef c\u00e1i th\u1ee9 1995 trong d\u00e3y t\u00f4 m\u00e0u g\u00ec? Gi\u1ea3i : a, Nh\u00f3m ch\u1eef T\u1ed4 QU\u1ed0C VI\u1ec6T NAM c\u00f3 13 ch\u1eef c\u00e1i. M\u00e0 1996 : 13 = 153 (nh\u00f3m) d\u01b0 7. Nh\u01b0 v\u1eady k\u1ec3 t\u1eeb ch\u1eef c\u00e1i \u0111\u1ea7u ti\u00ean \u0111\u1ebfn ch\u1eef c\u00e1i th\u1ee9 1996 trong d\u00e3y ng\u01b0\u1eddi ta \u0111\u00e3 vi\u1ebft 153 l\u1ea7n nh\u00f3m ch\u1eef T\u1ed4 QU\u1ed0C VI\u1ec6T NAM v\u00e0 7 ch\u1eef c\u00e1i ti\u1ebfp theo l\u00e0 : T\u1ed4 QU\u1ed0C V. Ch\u1eef c\u00e1i th\u1ee9 1996 trong d\u00e3y l\u00e0 ch\u1eef V. b, M\u1ed7i nh\u00f3m ch\u1eef T\u1ed4 QU\u1ed0C VI\u1ec6T NAM c\u00f3 2 ch\u1eef T v\u00e0 c\u0169ng c\u00f3 2 ch\u1eef \u00d4 v\u00e0 1 ch\u1eef I. v\u00ec v\u1eady, n\u1ebfu ng\u01b0\u1eddi ta \u0111\u1ebfm \u0111\u01b0\u1ee3c trong d\u00e3y c\u00f3 50 ch\u1eef T th\u00ec d\u00e3y \u0111\u00f3 c\u0169ng ph\u1ea3i c\u00f3 50 ch\u1eef \u00d4 v\u00e0 c\u00f3 25 ch\u1eef I. c, B\u1ea1n \u0111\u00f3 \u0111\u00e3 \u0111\u1ebfm sai, v\u00ec s\u1ed1 ch\u1eef \u00d4 trong d\u00e3y ph\u1ea3i l\u00e0 s\u1ed1 ch\u1eb5n d, Ta nh\u1eadn x\u00e9t : c\u00e1c m\u00e0u Xanh, \u0111\u1ecf, t\u00edm, v\u00e0ng g\u1ed3m c\u00f3 4 m\u00e0u. M\u00e0 1995 : 4 = 498 (nh\u00f3m) d\u01b0 3. Nh\u1eefng ch\u1eef c\u00e1i trong d\u00e3y c\u00f3 s\u1ed1 th\u1ee9 t\u1ef1 l\u00e0 s\u1ed1 chia cho 4 d\u01b0 3 th\u00ec \u0111\u01b0\u1ee3c t\u00f4 m\u00e0u t\u00edm V\u1eady ch\u1eef c\u00e1i th\u1ee9 1995 trong d\u00e3y \u0111\u01b0\u1ee3c t\u00f4 m\u00e0u t\u00edm. * B\u00e0i t\u1eadp v\u1ec1 nh\u00e0 : B\u00e0i 1 : D\u00e3y s\u1ed1 l\u1ebb t\u1eeb 9 \u0111\u1ebfn 1999 c\u00f3 bao nhi\u00eau ch\u1eef s\u1ed1 B\u00e0i 2 : Vi\u1ebft c\u00e1c s\u1ed1 ch\u1eb5n li\u00ean ti\u1ebfp b\u1eaft \u0111\u1ea7u t\u1eeb 60. H\u1ecfi n\u1ebfu vi\u1ebft 2590 ch\u1eef s\u1ed1 th\u00ec vi\u1ebft \u0111\u1ebfn s\u1ed1 n\u00e0o? B\u00e0i 3 : Ng\u01b0\u1eddi ta vi\u1ebft TO\u00c1N TU\u1ed4I TH\u01a0 th\u00e0nh d\u00e3y m\u1ed7i ch\u1eef s\u1ed1 vi\u1ebft 1 m\u00e0u theo th\u1ee9 t\u1ef1 xanh, \u0111\u1ecf, v\u00e0ng. H\u1ecfi ch\u1eef th\u1ee9 2000 l\u00e0 ch\u1eef g\u00ec, m\u00e0u g\u00ec? B\u00e0i 4 : M\u1ed9t ng\u01b0\u1eddi vi\u1ebft li\u00ean ti\u1ebfp nh\u00f3m ch\u1eef CH\u0102M H\u1eccC CH\u0102M L\u00c0M th\u00e0nh d\u00e3y CH\u0102M H\u1eccC CH\u0102M L\u00c0M CH\u0102M H\u1eccC CH\u0102M L\u00c0M ... a, Ch\u1eef c\u00e1i th\u1ee9 1000 trong d\u00e3y l\u00e0 ch\u1eef g\u00ec? b, N\u1ebfu ng\u01b0\u1eddi ta \u0111\u1ebfm \u0111\u01b0\u1ee3c trong d\u00e3y c\u00f3 1200 ch\u1eef H th\u00ec \u0111\u1ebfm \u0111\u01b0\u1ee3c ch\u1eef A? c, M\u1ed9t ng\u01b0\u1eddi \u0111\u1ebfm \u0111\u01b0\u1ee3c trong d\u00e3y c\u00f3 1996 ch\u1eef C. H\u1ecfi ng\u01b0\u1eddi \u0111\u00f3 \u0111\u1ebfm \u0111\u00fang hay sai? Gi\u1ea3i th\u00edch t\u1ea1i sao? B\u00e0i 5 : a, C\u00f3 bao nhi\u00eau s\u1ed1 ch\u1eb5n c\u00f34 ch\u1eef s\u1ed1? b, C\u00f3 bao nhi\u00eau s\u1ed1 c\u00f3 3 ch\u1eef s\u1ed1 \u0111\u1ec1u l\u1ebb? c, C\u00f3 bao nhi\u00eau s\u1ed1 c\u00f3 5 ch\u1eef s\u1ed1 m\u00e0 trong \u0111\u00f3 c\u00f3 \u00edt nh\u1ea5t hai ch\u1eef s\u1ed1 gi\u1ed1ng nhau? B\u00e0i 6 : cho d\u00e3y s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp : 1, 2, 3, 4, 5, ..., 1999 H\u1ecfi d\u00e3y s\u1ed1 c\u00f3 bao nhi\u00eau ch\u1eef s\u1ed1? B\u00e0i 7 : Cho d\u00e3y s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp: 1, 2, 3, 4, 5, ..., x. T\u00ecm x bi\u1ebft d\u00e3y s\u1ed1 c\u00f3 1989 ch\u1eef s\u1ed1","B\u00e0i 8 : Cho d\u00e3y s\u1ed1 ch\u1eb5n li\u00ean ti\u1ebfp : 2, 4, 6, 8, 10, ..., 2468. a, H\u1ecfi d\u00e3y c\u00f3 bao nhi\u00eau ch\u1eef s\u1ed1? b, T\u00ecm ch\u1eef s\u1ed1 th\u1ee9 2000 c\u1ee7a d\u00e3y \u0111\u00f3. B\u00e0i 9 : Cho d\u00e3y s\u1ed1 1,1; 2,2; 3,3; ...; 108,9; 110,0 a, D\u00e3y s\u1ed1 n\u00e0y c\u00f3 bao nhi\u00eau s\u1ed1 h\u1ea1ng? b, S\u1ed1 h\u1ea1ng th\u1ee9 50 c\u1ee7a d\u00e3y l\u00e0 s\u1ed1 h\u1ea1ng n\u00e0o? B\u00e0i 10 : Cho d\u00e3y 3, 18, 48, 93, 153, ... a, T\u00ecm s\u1ed1 h\u1ea1ng th\u1ee9 100 c\u1ee7a d\u00e3y. b, S\u1ed1 11703 l\u00e0 s\u1ed1 h\u1ea1ng th\u1ee9 bao nhi\u00eau c\u1ee7a d\u00e3y B\u00c0I 4 C\u00d4NG VI\u1ec6C CHUNG I. M\u1ee4C TI\u00caU TI\u1ebeT D\u1ea0Y : - HS n\u1eafm \u0111\u01b0\u1ee3c c\u00e1ch gi\u1ea3i c\u00e1c b\u00e0i to\u00e1n trong d\u1ea1ng n\u00e0y - L\u00e0m \u0111\u01b0\u1ee3c m\u1ed9t s\u1ed1 b\u00e0i t\u1eadp n\u00e2ng cao. - R\u00e8n k\u1ef9 n\u0103ng gi\u1ea3i to\u00e1n cho h\u1ecdc sinh . II. CHU\u1ea8N B\u1eca - C\u00e2u h\u1ecfi v\u00e0 b\u00e0i t\u1eadp thu\u1ed9c d\u1ea1ng v\u1eeba h\u1ecdc. - C\u00e1c ki\u1ebfn th\u1ee9c c\u00f3 li\u00ean quan. III. C\u00c1C HO\u1ea0T \u0110\u1ed8NG D\u1ea0Y H\u1eccC 1\/ \u1ed4n \u0111\u1ecbnh t\u1ed5 ch\u1ee9c l\u1edbp. 2\/ Ki\u1ec3m tra b\u00e0i c\u0169. G\u1ecdi h\u1ecdc sinh l\u00e0m b\u00e0i t\u1eadp v\u1ec1 nh\u00e0 gi\u1edd tr\u01b0\u1edbc, GV s\u1eeda ch\u1eefa. 3\/ Gi\u1ea3ng b\u00e0i m\u1edbi. 3.1 Ki\u1ebfn th\u1ee9c c\u1ea7n nh\u1edb. a. Lo\u1ea1i to\u00e1n n\u00e0y c\u0169ng th\u1ec3 hi\u1ec7n r\u00f5 m\u1ed1i quan h\u1ec7 \u0111\u1ea1i l\u01b0\u1ee3ng t\u1ec9 l\u1ec7 thu\u1eadn v\u00e0 t\u1ec9 l\u1ec7 ngh\u1ecbch trong c\u00e1c t\u00ecnh hu\u1ed1ng ph\u1ee9c t\u1ea1p h\u01a1n b\u00e0i to\u00e1n v\u1ec1 quy t\u1eafc tam su\u1ea5t. b. ch\u00fa \u00fd : - Ta c\u00f3 th\u1ec3 hi\u1ec3u 1 c\u00f4ng vi\u1ec7c nh\u01b0 l\u00e0 1 \u0111\u01a1n v\u1ecb. Do \u0111\u00f3 c\u00f3 th\u1ec3 bi\u1ec3u th\u1ecb 1 c\u00f4ng vi\u1ec7c th\u00e0nh nhi\u1ec1u ph\u1ea7n b\u1eb1ng nhau (ph\u00f9 h\u1ee3p v\u1edbi c\u00e1c \u0111i\u1ec1u ki\u1ec7n c\u1ee7a b\u00e0i to\u00e1n) \u0111\u1ec3 thu\u1eadn ti\u1ec7n cho vi\u1ec7c t\u00ednh to\u00e1n. - S\u1eed d\u1ee5ng ph\u00e2n s\u1ed1 \u0111\u01b0\u1ee3c coi l\u00e0 th\u01b0\u01a1ng c\u1ee7a ph\u00e9p chia hai s\u1ed1 t\u1ef1 nhi\u00ean. - B\u00e0i to\u00e1n n\u00e0yth\u01b0\u1eddng c\u00f3 \u0111\u1ea1i l\u01b0\u1ee3ng th\u1eddi gian. C\u1ea7n ph\u1ea3i bi\u1ebft chuy\u1ec3n \u0111\u1ed5i v\u00e0 s\u1eed d\u1ee5ng c\u00e1c \u0111\u01a1n v\u1ecb \u0111o th\u1eddi gian th\u00edch h\u1ee3p cho vi\u1ec7c t\u00ednh to\u00e1n. 3.2 B\u00e0i t\u1eadp v\u1eadn d\u1ee5ng.","B\u00e0i 1 : An v\u00e0 B\u00ecnh nh\u1eadn l\u00e0m chung m\u1ed9t c\u00f4ng vi\u1ec7c. N\u1ebfu m\u1ed9t m\u00ecnh An l\u00e0m th\u00ec sau 3 gi\u1edd s\u1ebd xong vi\u1ec7c, c\u00f2n n\u1ebfu B\u00ecnh l\u00e0m m\u1ed9t m\u00ecnh th\u00ec sau 6 gi\u1edd s\u1ebd xong vi\u1ec7c \u0111\u00f3. H\u1ecfi c\u1ea3 2 ng\u01b0\u1eddi c\u00f9ng l\u00e0m th\u00ec sau m\u1ea5y gi\u1edd s\u1ebd xong vi\u1ec7c \u0111\u00f3? Gi\u1ea3i : C\u00e1ch 1 : Bi\u1ec3u th\u1ecb c\u00f4ng vi\u1ec7c th\u00e0nh 6 ph\u1ea7n b\u1eb1ng nhau th\u00ec sau 1 gi\u1edd An l\u00e0m \u0111\u01b0\u1ee3c 2 ph\u1ea7n v\u00e0 B\u00ecnh l\u00e0m \u0111\u01b0\u1ee3c 1 ph\u1ea7n \u0111\u00f3. Do \u0111\u00f3, sau 1 gi\u1edd c\u1ea3 2 ng\u01b0\u1eddi c\u00f9ng l\u00e0m \u0111\u01b0\u1ee3c 2 + 1 = 3 (ph\u1ea7n) 1 gi\u1edd ||||||| I II Th\u1eddi gian \u0111\u1ec3 2 ng\u01b0\u1eddi c\u00f9ng l\u00e0n xong vi\u1ec7c \u0111\u00f3 l\u00e0 : 6 ; 3 = 2 (gi\u1edd) \u0110\u00e1p s\u1ed1 2 gi\u1edd C\u00e1ch 2 : N\u1ebfu An l\u00e0m m\u1ed9t m\u00ecnh th\u00ec sau 1 gi\u1edd l\u00e0m \u0111\u01b0\u1ee3c 1 c\u00f4ng vi\u1ec7c, n\u1ebfu B\u00ecnh l\u00e0m 1 m\u00ecnh 3 th\u00ec sau 1 gi\u1edd l\u00e0m \u0111\u01b0\u1ee3c 1 c\u00f4ng vi\u1ec7c. Do \u0111\u00f3, N\u1ebfu c\u1ea3 2 ng\u01b0\u1eddi c\u00f9ng l\u00e0m th\u00ec sau 1 gi\u1edd s\u1ebd l\u00e0m 6 \u0111\u01b0\u1ee3c s\u1ed1 ph\u1ea7n c\u00f4ng vi\u1ec7c l\u00e0 : 1 + 1 = 1 (c\u00f4ng vi\u1ec7c) 362 Th\u1eddi gian \u0111\u1ec3 2 ng\u01b0\u1eddi c\u00f9ng l\u00e0m xong vi\u1ec7c \u0111\u00f3 l\u00e0 : 1 : 1 = 2 (gi\u1edd) 2 \u0110\u00e1p s\u1ed1 2 gi\u1edd. B\u00e0i 2 : Ba ng\u01b0\u1eddi c\u00f9ng l\u00e0m m\u1ed9t c\u00f4ng vi\u1ec7c. Ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t c\u00f3 th\u1ec3 ho\u00e0n th\u00e0nh trong 3 tu\u1ea7n; ng\u01b0\u1eddi th\u1ee9 hai c\u00f3 th\u1ec3 ho\u00e0n th\u00e0nh m\u1ed9t c\u00f4ng vi\u1ec7c nhi\u1ec1u g\u1ea5p ba l\u1ea7n c\u00f4ng vi\u1ec7c \u0111\u00f3 trong 8 tu\u1ea7n; ng\u01b0\u1eddi th\u1ee9 ba c\u00f3 th\u1ec3 ho\u00e0n th\u00e0nh m\u1ed9t c\u00f4ng vi\u1ec7c nhi\u1ec1u g\u1ea5p 5 c\u00f4ng vi\u1ec7c \u0111\u00f3 trong 12 tu\u1ea7n. H\u1ecfi n\u1ebfu c\u1ea3 ba ng\u01b0\u1eddi c\u00f9ng l\u00e0m c\u00f4ng vi\u1ec7c ban \u0111\u1ea7u th\u00ec s\u1ebd ho\u00e0n th\u00e0nh trong bao nhi\u00eau gi\u1edd? n\u1ebfu m\u1ed7i tu\u1ea7n l\u00e0m 45 gi\u1edd? Gi\u1ea3i: Theo b\u00e0i ra ta c\u00f3 : Ng\u01b0\u1eddi th\u1ee9 hai l\u00e0m xong c\u00f4ng vi\u1ec7c ban \u0111\u1ea7u trong: 8 8 : 3 = (tu\u1ea7n) 3","Ng\u01b0\u1eddi th\u1ee9 ba l\u00e0m xong c\u00f4ng vi\u1ec7c ban \u0111\u1ea7u trong : 12 : 5 = 12 (tu\u1ea7n) 5 Trong m\u1ed9t tu\u1ea7n ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t l\u00e0m \u0111\u01b0\u1ee3c 1 c\u00f4ng vi\u1ec7c, ng\u01b0\u1eddi th\u1ee9 hai l\u00e0m \u0111\u01b0\u1ee3c 3\/8 3 c\u00f4ng vi\u1ec7c, ng\u01b0\u1eddi th\u1ee9 ba l\u00e0m d\u01b0\u1ee3c 5 c\u00f4ng vi\u1ec7c . V\u1eady c\u1ea3 ba ng\u01b0\u1eddi trong m\u1ed9t tu\u1ea7n s\u1ebd l\u00e0m 12 1 3 5 = 9 (c\u00f4ng vi\u1ec7c) \u0111\u01b0\u1ee3c: ++ 3 8 12 8 Th\u1eddi gian \u0111\u1ec3 c\u1ea3 ba ng\u01b0\u1eddi l\u00e0m xong c\u00f4ng vi\u1ec7c l\u00e0: 9 8 (tu\u1ea7n) 1: = 89 S\u1ed1 gi\u1edd c\u1ea3 ba ng\u01b0\u1eddi l\u00e0m xong c\u00f4ng vi\u1ec7c l\u00e0: 45 x 8 = 40 (gi\u1edd) 9 \u0110\u00e1p s\u1ed1 : 40 gi\u1edd B\u00e0i 3 : Hai v\u00f2i n\u01b0\u1edbc c\u00f9ng ch\u1ea3y v\u00e0o b\u1ec3 th\u00ec sau 1 gi\u1edd 12 ph\u00fat s\u1ebd \u0111\u1ea7y b\u1ec3. N\u1ebfu m\u1ed9t m\u00ecnh v\u00f2i th\u1ee9 nh\u1ea5t ch\u1ea3y th\u00ec sau 2 gi\u1edd s\u1ebd \u0111\u1ea7y b\u1ec3. H\u1ecfi m\u1ed9t m\u00ecnh v\u00f2i th\u1ee9 hai ch\u1ea3y th\u00ec m\u1ea5y gi\u1edd s\u1ebd \u0111\u1ea7y b\u1ec3? Gi\u1ea3i : \u0110\u1ed5i : 1 gi\u1edd 12 ph\u00fat = 72 ph\u00fat 2 gi\u1edd = 120 ph\u00fat C\u00e1ch 1: Bi\u1ec3u th\u1ecb l\u01b0\u1ee3ng n\u01b0\u1edbc \u0111\u1ea7y b\u1ec3 l\u00e0 360 ph\u1ea7n b\u1eb1ng nhau th\u00ec sau m\u1ed9t ph\u00fat c\u1ea3 hai v\u00f2i c\u00f9ng ch\u1ea3y \u0111\u01b0\u1ee3c s\u1ed1 ph\u1ea7n l\u00e0 : 360 : 72 = 5 (ph\u1ea7n) M\u1ed7i ph\u00fat v\u00f2i th\u1ee9 nh\u1ea5t ch\u1ea3y \u0111\u01b0\u1ee3c s\u1ed1 ph\u1ea7n l\u00e0: 360 : 120 = 3 (ph\u1ea7n) Do \u0111\u00f3 m\u1ed7i ph\u00fat v\u00f2i th\u1ee9 hai ch\u1ea3y \u0111\u01b0\u1ee3c s\u1ed1 ph\u1ea7n l\u00e0: 5 \u2013 3 = 2 (ph\u1ea7n) Th\u1eddi gian \u0111\u1ec3 v\u00f2i th\u1ee9 hai ch\u1ea3y \u0111\u01b0\u1ee3c \u0111\u1ea7y b\u1ec3 l\u00e0 : 360 : 2 = 180 (ph\u00fat) = 3 gi\u1edd C\u00e1ch 2 : M\u1ed9t ph\u00fat c\u1ea3 hai v\u00f2i ch\u1ea3y \u0111\u01b0\u1ee3c 1 (b\u1ec3 n\u01b0\u1edbc) 72 M\u1ed9t ph\u00fat m\u1ed9t m\u00ecnh v\u00f2i th\u1ee9 nh\u1ea5t ch\u1ea3y \u0111\u01b0\u1ee3c 1 b\u1ec3 n\u01b0\u1edbc. 120 Do \u0111\u00f3 m\u1ed9t ph\u00fat v\u00f2i th\u1ee9 hai ch\u1ea3y m\u1ed9t m\u00ecnh \u0111\u01b0\u1ee3c :","1\u2013 1 1 (b\u1ec3 n\u01b0\u1edbc) = 72 120 180 Th\u1eddi gian \u0111\u1ec3 v\u00f2i th\u1ee9 hai ch\u1ea3y m\u1ed9t m\u00ecnh \u0111\u1ea7y b\u1ec3 l\u00e0: 1 1 : = 180 (ph\u00fat) 180 = 3 gi\u1edd \u0110\u00e1p s\u1ed1 : 3 gi\u1edd B\u00e0i 4 : Ki\u00ean v\u00e0 Hi\u1ec1n c\u00f9ng l\u00e0m m\u1ed9t c\u00f4ng vi\u1ec7c c\u00f3 th\u1ec3 ho\u00e0n th\u00e0nh trong 10 ng\u00e0y. Sau 7 ng\u00e0y c\u00f9ng l\u00e0m th\u00ec Ki\u00ean ngh\u1ec9 vi\u1ec7c. Hi\u1ec1n ph\u1ea3i l\u00e0m n\u1ed1t ph\u1ea7n vi\u1ec7c c\u00f2n l\u1ea1i trong 9 ng\u00e0y. H\u1ecfi n\u1ebfu l\u00e0m ri\u00eang th\u00ec m\u1ed7i ng\u01b0\u1eddi l\u00e0m trong bao l\u00e2u ? Gi\u1ea3i : C\u00e1ch 1: Ki\u00ean v\u00e0 Hi\u1ec1n c\u00f9ng l\u00e0m 1 ng\u00e0y \u0111\u01b0\u1ee3c 1 c\u00f4ng vi\u1ec7c 10 Sau 7 ng\u00e0y c\u00f9ng l\u00e0m hai ng\u01b0\u1eddi \u0111\u00e3 l\u00e0m \u0111\u01b0\u1ee3c s\u1ed1 ph\u1ea7n c\u00f4ng vi\u1ec7c l\u00e0 : 1 x 7 = 7 (c\u00f4ng vi\u1ec7c) 10 10 Ph\u1ea7n vi\u1ec7c c\u00f2n l\u1ea1i l\u00e0 : 1 \u2013 7 = 3 (c\u00f4ng vi\u1ec7c) 10 10 M\u1ed7i ng\u00e0y Hi\u1ec1n l\u00e0m \u0111\u01b0\u1ee3c : 3 : 9 = 1 (c\u00f4ng vi\u1ec7c) 10 30 S\u1ed1 ng\u00e0y Hi\u1ec1n l\u00e0m m\u1ed9t m\u00ecnh \u0111\u1ec3 xong c\u00f4ng vi\u1ec7c l\u00e0: 1 1 : = 30 (ng\u00e0y) 30 M\u1ed7i ng\u00e0y Ki\u00ean l\u00e0m \u0111\u01b0\u1ee3c : 1 \u2013 1 = 1 (c\u00f4ng vi\u1ec7c) 10 30 15 S\u1ed1 ng\u00e0y Ki\u00ean l\u00e0m m\u1ed9t m\u00ecnh \u0111\u1ec3 xong c\u00f4ng vi\u1ec7c l\u00e0: 1 1 : = 15 (ng\u00e0y) 15 \u0110\u00e1p s\u1ed1 : Ki\u00ean 15 ng\u00e0y Hi\u1ec1n 30 ng\u00e0y 4. B\u00e0i t\u1eadp v\u1ec1 nh\u00e0 : B\u00e0i 1 :Ba v\u00f2i n\u01b0\u1edbc c\u00f9ng ch\u1ea3y v\u00e0o b\u1ec3 th\u00ec sau 1 gi\u1edd 20 ph\u00fat s\u1ebd \u0111\u1ea7y b\u1ec3. N\u1ebfu ri\u00eang v\u00f2i th\u1ee9 nh\u1ea5t ch\u1ea3y th\u00ec sau 6 gi\u1edd s\u1ebd \u0111\u1ea7y b\u1ec3, ri\u00eang v\u00f2i th\u1ee9 hai ch\u1ea3y th\u00ec sau 4 gi\u1edd s\u1ebd \u0111\u1ea7y b\u1ec3. H\u1ecfi ri\u00eang v\u00f2i th\u1ee9 ba ch\u1ea3y th\u00ec sau m\u1ea5y gi\u1edd s\u1ebd \u0111\u1ea7y b\u1ec3? B\u00e0i 2: M\u00e1y c\u00e0y th\u1ee9 nh\u1ea5t c\u1ea7n 9 gi\u1edd \u0111\u1ec3 c\u00e0y xong di\u1ec7n t\u00edch c\u00e1nh \u0111\u1ed3ng, m\u00e1y c\u00e0y th\u1ee9 hai c\u1ea7n 15 gi\u1edd \u0111\u1ec3 c\u00e0y xong di\u1ec7n t\u00edch c\u00e1nh \u0111\u1ed3ng \u1ea5y . Ng\u01b0\u1eddi ta cho m\u00e1y c\u00e0y th\u1ee9 nh\u1ea5t l\u00e0m vi\u1ec7c","trong 6 gi\u1edd r\u1ed3i ngh\u1ec9 \u0111\u1ec3 m\u00e1y c\u00e0y th\u1ee9 hai l\u00e0m ti\u1ebfp cho \u0111\u1ebfn khi c\u00e0y xong di\u1ec7n t\u00edch c\u00e1nh \u0111\u1ed3ng n\u00e0y. H\u1ecfi m\u00e1y c\u00e0y th\u1ee9 2 \u0111\u00e3 l\u00e0m trong bao l\u00e2u? B\u00e0i 3 : Hai v\u00f2i n\u01b0\u1edbc c\u00f9ng ch\u1ea3y v\u00e0o b\u1ec3 b\u01a1i sau 48 ph\u00fat s\u1ebd \u0111\u1ea7y b\u1ec3. M\u1ed9t m\u00ecnh v\u00f2i th\u1ee9 nh\u1ea5t ch\u1ea3y 2 gi\u1edd s\u1ebd \u0111\u1ea7y b\u1ec3. H\u00e3y t\u00ednh xem b\u1ec3 b\u01a1i n\u00e0y ch\u1ee9a \u0111\u01b0\u1ee3c bao nhi\u00eau m\u00e9t kh\u1ed1i n\u01b0\u1edbc, bi\u1ebft r\u1eb1ng m\u1ed7i ph\u00fat v\u00f2i th\u1ee9 hai ch\u1ea3y nhi\u1ec1u h\u01a1n v\u00f2i th\u1ee9 nh\u1ea5t 50 m3 n\u01b0\u1edbc. B\u00e0i 4: Ba ng\u01b0\u1eddi th\u1ee3 c\u00f9ng l\u00e0m m\u1ed9t c\u00f4ng vi\u1ec7c . N\u1ebfu ng\u01b0\u1eddi th\u1ee9 nh\u1ea5t l\u00e0m m\u1ed9t m\u00ecnh th\u00ec sau 8 gi\u1edd s\u1ebd xong c\u00f4ng vi\u1ec7c ; n\u1ebfu ng\u01b0\u1eddi th\u1ee9 ba l\u00e0m m\u1ed9t m\u00ecnh th\u00ec sau 6 gi\u1edd s\u1ebd xong vi\u1ec7c \u0111\u00f3 ;n\u1ebfu ng\u01b0\u1eddi th\u1ee9 hai l\u00e0m m\u1ed9t m\u00ecnh th\u00ec sau 3 gi\u1edd s\u1ebd xong vi\u1ec7c . H\u1ecfi c\u1ea3 ba ng\u01b0\u1eddi c\u00f9ng l\u00e0m th\u00ec sau bao l\u00e2u s\u1ebd xong c\u00f4ng vi\u1ec7c n\u00e0y ? B\u00e0i 5: C\u00f3 m\u1ed9t c\u00f4ng vi\u1ec7c m\u00e0 Ho\u00e0ng l\u00e0m m\u1ed9t m\u00ecnh th\u00ec sau 10 ng\u00e0y s\u1ebd xong vi\u1ec7c, Minh l\u00e0m m\u1ed9t m\u00ecnh th\u00ec sau 15 gi\u1edd s\u1ebd xong vi\u1ec7c \u0111\u00f3 . Anh l\u00e0m m\u1ed9t m\u00ecnh ph\u1ea3i c\u1ea7n s\u1ed1 ng\u00e0y g\u1ea5p 5 l\u1ea7n s\u1ed1 ng\u00e0y c\u1ee7a Ho\u00e0ng v\u00e0 Minh c\u00f9ng l\u00e0m \u0111\u1ec3 xong vi\u1ec7c \u0111\u00f3 . H\u1ecfi n\u1ebfu c\u1ea3 ba ng\u01b0\u1eddi c\u00f9ng l\u00e0m th\u00ec sau bao l\u00e2u s\u1ebd xong vi\u1ec7c n\u00e0y ? B\u00e0i 6:C\u00f3 ba v\u00f2i n\u01b0\u1edbc ch\u1ea3y v\u00e0o m\u1ed9t c\u00e1i b\u1ec3 c\u1ea1n n\u01b0\u1edbc . N\u1ebfu m\u1ed9t v\u00f2i th\u1ee9 nh\u1ea5t v\u00e0 v\u00f2i th\u1ee9 hai c\u00f9ng ch\u1ea3y trong 9 gi\u1edd th\u00ec \u0111\u01b0\u1ee3c 3 b\u1ec3 .N\u1ebfu m\u1edf v\u00f2i th\u1ee9 hai v\u00e0 v\u00f2i th\u1ee9 ba c\u00f9ng ch\u1ea3y trong 5 4 7 b\u1ec3 .N\u1ebfu v\u00f2i th\u1ee9 nh\u1ea5t v\u00e0 v\u00f2i th\u1ee9 ba ch\u1ea3y trong 6 gi\u1edd th\u00ec \u0111\u01b0\u1ee3c 3 b\u1ec3. gi\u1edd th\u00ec \u0111\u01b0\u1ee3c 12 5 N\u1ebfu m\u1edf c\u1ea3 ba v\u00f2i c\u00f9ng ch\u1ea3y th\u00ec sau bao l\u00e2u b\u1ec3 s\u1ebd \u0111\u1ea7y ? B\u00c0I 5 T\u1ec8 S\u1ed0 V\u00c0 T\u1ec8 S\u1ed0 PH\u1ea6N TR\u0102M. I. M\u1ee4C TI\u00caU TI\u1ebeT D\u1ea0Y : - HS n\u1eafm \u0111\u01b0\u1ee3c c\u00e1ch gi\u1ea3i c\u00e1c b\u00e0i to\u00e1n v\u1ec1 t\u1ec9 s\u1ed1 ph\u1ea7n tr\u0103m. - L\u00e0m \u0111\u01b0\u1ee3c m\u1ed9t s\u1ed1 b\u00e0i t\u1eadp n\u00e2ng cao. - R\u00e8n k\u1ef9 n\u0103ng gi\u1ea3i to\u00e1n cho h\u1ecdc sinh . II. CHU\u1ea8N B\u1eca - C\u00e2u h\u1ecfi v\u00e0 b\u00e0i t\u1eadp thu\u1ed9c d\u1ea1ng v\u1eeba h\u1ecdc. - C\u00e1c ki\u1ebfn th\u1ee9c c\u00f3 li\u00ean quan. III. C\u00c1C HO\u1ea0T \u0110\u1ed8NG D\u1ea0Y H\u1eccC 1\/ \u1ed4n \u0111\u1ecbnh t\u1ed5 ch\u1ee9c l\u1edbp. 2\/ Ki\u1ec3m tra b\u00e0i c\u0169. G\u1ecdi h\u1ecdc sinh l\u00e0m b\u00e0i t\u1eadp v\u1ec1 nh\u00e0 gi\u1edd tr\u01b0\u1edbc, GV s\u1eeda ch\u1eefa. 3\/ Gi\u1ea3ng b\u00e0i m\u1edbi. * B\u00e0i t\u1eadp v\u1eadn d\u1ee5ng B\u00e0i 1 : M\u1ed9t l\u1edbp c\u00f3 22 n\u1eef sinh v\u00e0 18 nam sinh. H\u00e3y t\u00ednh t\u1ec9 s\u1ed1 ph\u1ea7n tr\u0103m c\u1ee7a n\u1eef sinh so v\u1edbi t\u1ed5ng s\u1ed1 h\u1ecdc sinh c\u1ea3 l\u1edbp, t\u1ec9 s\u1ed1 ph\u1ea7n tr\u0103m c\u1ee7a nam sinh so v\u1edbi t\u1ed5ng s\u1ed1 h\u1ecdc sinh c\u1ee7a c\u1ea3 l\u1edbp. Gi\u1ea3i :","T\u1ed5ng s\u1ed1 h\u1ecdc sinh c\u1ee7a l\u1edbp l\u00e0 : 22 + 18 = 40 (h\u1ecdc sinh) T\u1ec9 s\u1ed1 h\u1ecdc sinh n\u1eef so v\u1edbi h\u1ecdc sinh c\u1ee7a l\u1edbp l\u00e0 : 22 55 22 : 40 = 0,55 = 55% ( = = 55% ) 40 100 T\u1ec9 s\u1ed1 h\u1ecdc sinh nam so v\u1edbi h\u1ecdc sinh c\u1ee7a l\u1edbp l\u00e0 : 18 : 40 = 0,45 = 45% \u0110\u00e1p s\u1ed1 : 55% v\u00e0 45% B\u00e0i 2 : M\u1ed9t s\u1ed1 sau khi gi\u1ea3m \u0111i 20% th\u00ec ph\u1ea3i t\u0103ng th\u00eam bao nhi\u00eau ph\u1ea7n tr\u0103m s\u1ed1 m\u1edbi \u0111\u1ec3 l\u1ea1i \u0111\u01b0\u1ee3c s\u1ed1 c\u0169. Gi\u1ea3i : M\u1ed9t s\u1ed1 gi\u1ea3m \u0111i 20% t\u1ee9c l\u00e0 gi\u1ea3m \u0111i 1 gi\u00e1 tr\u1ecb c\u1ee7a s\u1ed1 \u0111\u00f3. 5 S\u1ed1 c\u0169 : | | | | | | S\u1ed1 m\u1edbi : | | | | | V\u1eady ph\u1ea3i t\u0103ng s\u1ed1 m\u1edbi th\u00eam 1 c\u1ee7a n\u00f3 t\u1ee9c l\u00e0 25% th\u00ec \u0111\u01b0\u1ee3c s\u1ed1 ban \u0111\u1ea7u. 4 B\u00e0i 3 : M\u1ed9t s\u1ed1 t\u0103ng th\u00eam 25% th\u00ec ph\u1ea3i gi\u1ea3m \u0111i bao nhi\u00eau ph\u1ea7n tr\u0103m \u0111\u1ec3 l\u1ea1i \u0111\u01b0\u1ee3c s\u1ed1 c\u0169. Gi\u1ea3i : M\u1ed9t s\u1ed1 t\u0103ng th\u00eam 25% t\u1ee9c l\u00e0 t\u0103ng th\u00eam 1 c\u1ee7a n\u00f3 4 S\u1ed1 c\u0169 : | | | | | S\u1ed1 m\u1edbi : | | | | | | V\u1eady s\u1ed1 m\u1edbi ph\u1ea3i gi\u1ea3m \u0111i 1 gi\u00e1 tr\u1ecb c\u1ee7a n\u00f3 t\u1ee9c l\u00e0 20% c\u1ee7a n\u00f3 th\u00ec lai \u0111\u01b0\u1ee3c s\u1ed1 ban 5 \u0111\u1ea7u. B\u00e0i 4 : L\u01b0\u1ee3ng n\u01b0\u1edbc trong c\u1ecf t\u01b0\u01a1i l\u00e0 55%, trong c\u1ecf kh\u00f4 l\u00e0 10%. H\u1ecfi ph\u01a1i 100 kg c\u1ecf t\u01b0\u01a1i ta \u0111\u01b0\u1ee3c bao nhi\u00eau ki l\u00f4 gam c\u1ecf kh\u00f4. Gi\u1ea3i : L\u01b0\u1ee3ng c\u1ecf c\u00f3 trong c\u1ecf t\u01b0\u01a1i l\u00e0 : 100 \u2013 55 = 45% Hay 100 kg c\u1ecf t\u01b0\u01a1i c\u00f3 45 kg c\u1ecf. Nh\u01b0ng trong c\u1ecf kh\u00f4 c\u00f2n c\u00f3 10% n\u01b0\u1edbc. N\u00ean 45 kg c\u1ecf l\u00e0 90% kh\u1ed1i l\u01b0\u1ee3ng trong c\u1ecf kh\u00f4. V\u1eady 100 kg c\u1ecf t\u01b0\u01a1i thu \u0111\u01b0\u1ee3c s\u1ed1 c\u1ecf kh\u00f4 l\u00e0 : 45x100 = 50 (kg) \u0110\u00e1p s\u1ed1 50 kg. 90","B\u00e0i 5 : N\u01b0\u1edbc bi\u1ec3n ch\u1ee9a 4% mu\u1ed1i. C\u1ea7n \u0111\u1ed5 th\u00eam bao nhi\u00eau gam n\u01b0\u1edbc l\u00e3 v\u00e0o 400 gam n\u01b0\u1edbc bi\u1ec3n \u0111\u1ec3 t\u1ec9 l\u1ec7 mu\u1ed1i trong dung d\u1ecbch l\u00e0 2%. Gi\u1ea3i : L\u01b0\u1ee3ng n\u01b0\u1edbc mu\u1ed1i c\u00f3 trong 400g n\u01b0\u1edbc bi\u1ec3n l\u00e0 : 400 x 4 : 100 = 16 (g) Dung d\u1ecbch ch\u1ee9a 2 % mu\u1ed1i l\u00e0 : C\u1ee9 c\u00f3 100 g n\u01b0\u1edbc th\u00ec c\u00f3 2 g mu\u1ed1i 16 g mu\u1ed1i c\u1ea7n s\u1ed1 l\u01b0\u1ee3ng n\u01b0\u1edbc l\u00e0 : 100 : 2 x 16 = 800 (g) L\u01b0\u1ee3ng n\u01b0\u1edbc ph\u1ea3i th\u00eam l\u00e0 : 800 \u2013 400 = 400 (g) \u0110\u00e1p s\u1ed1 400 g. B\u00e0i 6 : Di\u1ec7n t\u00edch c\u1ee7a 1 h\u00ecnh ch\u1eef nh\u1eadt s\u1ebd thay \u0111\u1ed5i th\u1ebf n\u00e0o n\u1ebfu t\u0103ng chi\u1ec1u d\u00e0i c\u1ee7a n\u00f3 l\u00ean 10 % v\u00e0 b\u1edbt chi\u1ec1u r\u1ed9ng c\u1ee7a n\u00f3 \u0111i 10 % Gi\u1ea3i : G\u1ecdi s\u1ed1 \u0111o chi\u1ec1u d\u00e0i l\u00e0 100 x a S\u1ed1 \u0111o chi\u1ec1u r\u1ed9ng l\u00e0 100 x b S\u1ed1 \u0111o di\u1ec7n t\u00edch l\u00e0 : 10 000 x a x b S\u1ed1 \u0111o chi\u1ec1u d\u00e0i m\u1edbi l\u00e0 : 110 x a s\u1ed1 \u0111o chi\u1ec1u r\u1ed9ng m\u1edbi l\u00e0 : 90 x b S\u1ed1 \u0111o di\u1ec7n t\u00edch m\u1edbi l\u00e0 : 9900 x a x b S\u1ed1 \u0111o di\u1ec7n t\u00edch m\u1edbi k\u00e9m s\u1ed1 \u0111o di\u1ec7n t\u00edch c\u0169 l\u00e0 : 10 000 x a x b \u2013 9 900 x a x b = 100 x a x b T\u1ee9c l\u00e0 k\u00e9m di\u1ec7n t\u00edch c\u0169 l\u00e0 : 100xaxb = 10% 10000xaxb B\u00e0i 7 : L\u01b0\u1ee3ng n\u01b0\u1edbc trong h\u1ea1t t\u01b0\u01a1i l\u00e0 20%. C\u00f3 200 kg h\u1ea1t t\u01b0\u01a1i sau khi ph\u01a1i kh\u00f4 nh\u1eb9 \u0111i 30 kg. T\u00ednh t\u1ec9 s\u1ed1 % n\u01b0\u1edbc trong h\u1ea1t \u0111\u00e3 ph\u01a1i kh\u00f4. Gi\u1ea3i : L\u01b0\u1ee3ng n\u01b0\u1edbc ban \u0111\u1ea7u ch\u1ee9a trong 200 g h\u1ea1t t\u01b0\u01a1i l\u00e0 : 200 : 100 x 20 = 40 (kg) S\u1ed1 l\u01b0\u1ee3ng h\u1ea1t ph\u01a1i kh\u00f4 c\u00f2n : 200 \u2013 30 = 170 (kg) L\u01b0\u1ee3ng n\u01b0\u1edbc c\u00f2n l\u1ea1i trong 170 kg h\u1ea1t \u0111\u00e3 ph\u01a1i kh\u00f4 l\u00e0 : 40 \u2013 30 = 10 (kg) T\u1ec9 s\u1ed1 % n\u01b0\u1edbc ch\u1ee9a trong h\u1ea1t \u0111\u00e3 ph\u01a1i kh\u00f4 l\u00e0 : 10 : 170 = 5,88%","\u0110\u00e1p s\u1ed1 5,88 % B\u00e0i 8 : Gi\u00e1 hoa ng\u00e0y t\u1ebft t\u0103ng 20% so v\u1edbi th\u00e1ng 11. Th\u00e1ng gi\u00eang gi\u00e1 hoa l\u1ea1i h\u1ea1 20%. H\u1ecfi Gi\u00e1 hoa th\u00e1ng gi\u00eang so v\u1edbi gi\u00e1 hoa th\u00e1ng 11 th\u00ec th\u00e1ng n\u00e0o \u0111\u1eaft h\u01a1n v\u00e0 \u0111\u1eaft h\u01a1n bao nhi\u00eau ph\u1ea7n tr\u0103m. Gi\u1ea3i : Gi\u00e1 hoa ng\u00e0y t\u1ebft so v\u1edbi th\u00e1ng 11 l\u00e0 : 100 + 20 = 120 (%) Gi\u00e1 hoa sau t\u1ebft c\u00f2n l\u00e0 : 100 \u2013 20 = 80 (% hoa sau t\u1ebft so v\u1edbi th\u00e1ng 11 l\u00e0 : 120 x 80 = 96 (%) 100 100 Gi\u00e1 hoa sau t\u1ebft so v\u1edbi th\u00e1ng 11 l\u00e0 : 100 \u2013 96 = 4 (%) \u0110\u00e1p s\u1ed1 4 % B\u00e0i 9 : M\u1ed9t ng\u01b0\u1eddi mua m\u1ed9t k\u1ef3 phi\u1ebfu lo\u1ea1i 3 th\u00e1ng v\u1edbi l\u00e3i xu\u1ea5t 1,9% 1 th\u00e1ng v\u00e0 gi\u00e1 tr\u1ecb k\u1ef3 phi\u1ebfu 6000 000 \u0111\u1ed3ng. H\u1ecfi sau 3 th\u00e1ng ng\u01b0\u1eddi \u0111\u00f3 l\u0129nh v\u1ec1 bao nhi\u00eau ti\u1ec1n c\u1ea3 v\u1ed1n l\u1eabn l\u00e3i. Bi\u1ebft r\u1eb1ng, ti\u1ec1n v\u1ed1n th\u00e1ng tr\u01b0\u1edbc nh\u1eadp th\u00e0nh v\u1ed1n c\u1ee7a th\u00e1ng sau. Gi\u1ea3i : V\u1ed1n c\u1ee7a th\u00e1ng sau so v\u1edbi th\u00e1ng li\u1ec1n tr\u01b0\u1edbc l\u00e0 : 100 + 1,9 = 101,9 (%) Ti\u1ec1n v\u1ed1n \u0111\u1ea7u th\u00e1ng th\u1ee9 hai l\u00e0 : 6000000x101,9 = 6 114 0000 (\u0110) 100 Ti\u1ec1n v\u1ed1n \u0111\u1ea7u th\u00e1ng th\u1ee9 3 l\u00e0 : 6114000x101,9 = 6230 166 (\u0110) 100 Ti\u1ec1n v\u1ed1n v\u00e0 l\u00e3i sau 3 th\u00e1ng l\u00e0 : 6230166x101,9 = 6348539,154 (\u0110) 100 \u0110\u00e1p s\u1ed1 6348539,154 \u0111\u1ed3ng B\u00e0i 10 : Gi\u00e1 c\u00e1c lo\u1ea1i rau th\u00e1ng 3 th\u01b0\u1eddng \u0111\u1eaft h\u01a1n th\u00e1ng hai l\u00e0 10%. Gi\u00e1 rau th\u00e1ng 4 l\u1ea1i r\u1ebb h\u01a1n th\u00e1ng 3 l\u00e0 10%. Gi\u00e1 rau th\u00e1ng 2 \u0111\u1eaft hay r\u1ebb h\u01a1n gi\u00e1 rau th\u00e1ng 4? Gi\u1ea3i : N\u1ebfu gi\u00e1 rau th\u00e1ng 2 l\u00e0 100% Nh\u01b0 v\u1eady gi\u00e1 rau th\u00e1ng 3 l\u00e0 : 100 + 10 = 110 (%) Gi\u00e1 rau th\u00e1ng 2 Gi\u00e1 rau th\u00e1ng 4 l\u00e0 : 100 \u2013 10 = 90 (%) gi\u00e1 rau th\u00e1ng 3 v\u00e0 b\u1eb1ng :","110 + 90 = 99% gi\u00e1 rau th\u00e1ng 2 100 100 Nh\u01b0 v\u1eady rau th\u00e1ng t\u01b0 r\u1ebb h\u01a1n rau th\u00e1ng hai. * B\u00e0i t\u1eadp v\u1ec1 nh\u00e0 : B\u00e0i 1 : M\u1ed9t c\u1eeda s\u00e1ch, h\u1ea1 gi\u00e1 10% gi\u00e1 s\u00e1ch nh\u00e2n ng\u00e0y 1\/6 tuy v\u1eady c\u1eeda h\u00e0ng v\u1eabn c\u00f2n l\u00e3i 8%. H\u1ecfi : Ng\u00e0y th\u01b0\u1eddng th\u00ec c\u1eeda h\u00e0ng \u0111\u01b0\u1ee3c l\u00e3i bao nhi\u00eau ph\u1ea7n tr\u0103m. B\u00e0i 2 : M\u1ed9t ng\u01b0\u1eddi b\u00e1n h\u00e0ng \u0111\u01b0\u1ee3c l\u1eddi 15% gi\u00e1 b\u00e1n. H\u1ecfi ng\u01b0\u1eddi \u1ea5y \u0111\u01b0\u1ee3c l\u1eddi bao nhi\u00eau ph\u1ea7n tr\u0103m gi\u00e1 mua? B\u00e0i 3 : M\u1ed9t c\u1eeda h\u00e0ng b\u00e1n g\u1ea1o \u0111\u01b0\u1ee3c l\u00e3i 25% gi\u00e1 mua. H\u1ecfi c\u1eeda h\u00e0ng \u0111\u01b0\u1ee3c l\u00e3i bao nhi\u00eau ph\u1ea7n tr\u0103m gi\u00e1 b\u00e1n. B\u00e0i 4 : Cu\u1ed1i n\u0103m h\u1ecdc, m\u1ed9t c\u1eeda h\u00e0ng h\u1ea1 gi\u00e1 b\u00e1n v\u1edf 20%. H\u1ecfi v\u1edbi c\u00f9ng m\u1ed9t s\u1ed1 ti\u1ec1n nh\u01b0 c\u0169, m\u1ed9t h\u1ecdc sinh s\u1ebd mua th\u00eam \u0111\u01b0\u1ee3c bao nhi\u00eau ph\u1ea7n tr\u0103m s\u1ed1 v\u1edf. B\u00e0i 5 : T\u00ecm di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt, bi\u1ebft r\u1eb1ng n\u1ebfu chi\u1ec1u d\u00e0i t\u0103ng 20% s\u1ed1 \u0111o v\u00e0 chi\u1ec1u r\u1ed9ng gi\u1ea3m 20% s\u1ed1 \u0111o th\u00ec di\u1ec7n t\u00edch b\u1ecb gi\u1ea3m \u0111i 30m2 B\u00e0i 6 : S\u1ea3n l\u01b0\u1ee3ng l\u00faa c\u1ee7a khu v\u1ef1c A h\u01a1n khu v\u1ef1c B l\u00e0 26% m\u1eb7c d\u00f9 di\u1ec7n t\u00edch c\u1ee7a khu v\u1ef1c A ch\u1ec9 l\u1edbn h\u01a1n khu v\u1ef1c B l\u00e0 5 %. H\u1ecfi n\u0103ng su\u1ea5t thu ho\u1ea1ch c\u1ee7a khu v\u1ef1c A nhi\u1ec1u h\u01a1n khu v\u1ef1c B l\u00e0 m\u1ea5y ph\u1ea7n tr\u0103m? B\u00e0i 7 : Kh\u1ed1i l\u01b0\u1ee3ng c\u00f4ng vi\u1ec7c t\u0103ng 80%. H\u1ecfi ph\u1ea3i t\u0103ng s\u1ed1 ng\u01b0\u1eddi lao \u0111\u1ed9ng th\u00eam bao nhi\u00eau ph\u1ea7n tr\u0103m \u0111\u1ec3 n\u0103ng su\u1ea5t lao \u0111\u1ed9ng t\u0103ng 20%? B\u00e0i 8 : M\u1ee9c l\u01b0\u01a1ng c\u1ee7a c\u00f4ng nh\u00e2n t\u0103ng 20%, gi\u00e1 h\u00e0ng gi\u1ea3m 20%. H\u1ecfi v\u1edbi m\u1ee9c l\u01b0\u01a1ng m\u1edbi n\u00e0y th\u00ec l\u01b0\u1ee3ng h\u00e0ng m\u1edbi s\u1ebd mua \u0111\u01b0\u1ee3c nhi\u1ec1u h\u01a1n h\u00e0ng c\u0169 bao nhi\u00eau ph\u1ea7n tr\u0103m? B\u00c0I 6 H\u00ccNH H\u1eccC A\/ C\u00c1C B\u00c0I TO\u00c1N V\u1ec0 NH\u1eacN D\u1ea0NG C\u00c1C H\u00ccNH I. M\u1ee4C TI\u00caU TI\u1ebeT D\u1ea0Y : - HS n\u1eafm \u0111\u01b0\u1ee3c m\u1ed9t s\u1ed1 t\u00ednh ch\u1ea5t c\u1ee7a c\u00e1c h\u00ecnh \u0111\u00e3 h\u1ecdc - Nh\u1eadn d\u1ea1ng \u0111\u01b0\u1ee3c c\u00e1c h\u00ecnh v\u00e0 gi\u1ea3i \u0111\u01b0\u1ee3c c\u00e1c b\u00e0i to\u00e1n c\u00f3 li\u00ean quan - R\u00e8n k\u1ef9 n\u0103ng gi\u1ea3i to\u00e1n, quan s\u00e1t, t\u00ednh to\u00e1n cho h\u1ecdc sinh . II. CHU\u1ea8N B\u1eca - C\u00e2u h\u1ecfi v\u00e0 b\u00e0i t\u1eadp thu\u1ed9c d\u1ea1ng v\u1eeba h\u1ecdc. - C\u00e1c ki\u1ebfn th\u1ee9c c\u00f3 li\u00ean quan. III. C\u00c1C HO\u1ea0T \u0110\u1ed8NG D\u1ea0Y H\u1eccC 1\/ \u1ed4n \u0111\u1ecbnh t\u1ed5 ch\u1ee9c l\u1edbp. 2\/ Ki\u1ec3m tra b\u00e0i c\u0169.","G\u1ecdi h\u1ecdc sinh l\u00e0m b\u00e0i t\u1eadp v\u1ec1 nh\u00e0 gi\u1edd tr\u01b0\u1edbc, GV s\u1eeda ch\u1eefa. 3\/ Gi\u1ea3ng b\u00e0i m\u1edbi. 3.1 C\u00e1c ki\u1ebfn th\u1ee9c c\u1ea7n nh\u1edb : A B - N\u1ed1i hai \u0111i\u1ec3m A, B ta \u0111\u01b0\u1ee3c \u0111o\u1ea1n th\u1eb3ng AB | | A C C - H\u00ecnh tam gi\u00e1c c\u00f3 3 \u0111\u1ec9nh, 3 c\u1ea1nh v\u00e0 3 g\u00f3c. . H\u00ecnh tam gi\u00e1c ABC c\u00f3 3 \u0111\u1ec9nh l\u00e0 A, B, C ; C C\u00f3 3 c\u1ea1nh l\u00e0 AB, BC v\u00e0 CA; C\u00f3 3 g\u00f3c l\u00e0 g\u00f3c A, g\u00f3c B v\u00e0 g\u00f3c C. - H\u00ecnh t\u1ee9 gi\u00e1c c\u00f3 4 \u0111\u1ec9nh, 4 c\u1ea1nh v\u00e0 4 g\u00f3c. B T\u1ee9 gi\u00e1c ABCD c\u00f3 4 \u0111\u1ec9nh l\u00e0 A, B, C v\u00e0 D ; B C\u00f3 4 c\u1ea1nh l\u00e0 AB, BC, CD v\u00e0 DA ; C\u00f3 4 g\u00f3c l\u00e0 g\u00f3c A, g\u00f3c B v\u00e0 g\u00f3c D - H\u00ecnh vu\u00f4ng c\u00f3 4 g\u00f3c vu\u00f4ng v\u00e0 c\u00f3 4 c\u1ea1nh b\u1eb1ng A nhau. - H\u00ecnh ch\u1eef nh\u1eadt ABCD c\u00f3 4 g\u00f3c D vu\u00f4ng ; Hai c\u1ea1nh AD v\u00e0 BC l\u00e0 B chi\u1ec1u d\u00e0i, hai c\u1ea1nh AB v\u00e0 CD l\u00e0 chi\u1ec1u r\u1ed9ng. AD 3.2) B\u00e0i t\u1eadp v\u1eadn d\u1ee5ng B\u00e0i 1 : Cho tam gi\u00e1c ABC. Tr\u00ean c\u1ea1nh BC ta l\u1ea5y 6 \u0111i\u1ec3m. N\u1ed1i \u0111\u1ec9nh A v\u1edbi m\u1ed7i \u0111i\u1ec3m v\u1eeba ch\u1ecdn. H\u1ecfi \u0111\u1ebfm \u0111\u01b0\u1ee3c bao nhi\u00eau h\u00ecnh tam gi\u00e1c. Gi\u1ea3i : A A 12 12 3 C B DE BC A","1 23 4 5 6 7 B DE P G H I C Ta nh\u1eadn x\u00e9t : - khi l\u1ea5y 1 \u0111i\u1ec3m th\u00ec t\u1ea1o th\u00e0nh 2 tam gi\u00e1c \u0111\u01a1n ABD v\u00e0 ADC. S\u1ed1 tam gi\u00e1c \u0111\u1ebfm \u0111\u01b0\u1ee3c l\u00e0 3 : ABC, ADB v\u00e0 ADC. Ta c\u00f3 : 1 + 2 = 3 (tam gi\u00e1c) - khi l\u1ea5y 2 \u0111i\u1ec3m th\u00ec t\u1ea1o th\u00e0nh 3 tam gi\u00e1c \u0111\u01a1n v\u00e0 s\u1ed1 tam gi\u00e1c \u0111\u1ebfm \u0111\u01b0\u1ee3c l\u00e0 6 : ABC, ABD, ADE, ABE, ADC v\u00e0 AEC. Ta c\u00f3 : 1+ 2 + 3 = 6 (tam gi\u00e1c) V\u1eady khi l\u1ea5y 6 \u0111i\u1ec3m ta s\u1ebd c\u00f3 7 tam gi\u00e1c \u0111\u01a1n \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh v\u00e0 s\u1ed1 tam gi\u00e1c \u0111\u1ebfm \u0111\u01b0\u1ee3c l\u00e0 : 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 (tam gi\u00e1c) C\u00e1ch 2 :- N\u1ed1i A v\u1edbi m\u1ed7i \u0111i\u1ec3m D, E, \u2026, C ta \u0111\u01b0\u1ee3c m\u1ed9t tam gi\u00e1c c\u00f3 c\u1ea1nh AD. C\u00f3 6 \u0111i\u1ec3m nh\u01b0 v\u1eady n\u00ean c\u00f3 6 tam gi\u00e1c chung c\u1ea1nh AD (kh\u00f4ng k\u1ec3 tam gi\u00e1c ADB v\u00ec \u0111\u00e3 t\u00ednh r\u1ed3i) L\u1eadp lu\u1eadn t\u01b0\u01a1ng t\u1ef1 nh\u01b0 tr\u00ean theo th\u1ee9 t\u1ef1 ta c\u00f3 5, 4, 3, 2, 1 tam gi\u00e1c chung c\u1ea1nh AE, AP, \u2026, AI. V\u1eady s\u1ed1 tam gi\u00e1c t\u1ea1o th\u00e0nh l\u00e0 : 7 + 6 + 5 + 4 +3 +2 + 1 = 28 (tam gi\u00e1c). B\u00e0i t\u1eadp 2 : Cho h\u00ecnh ch\u1eef nh\u1eadt ABCD. Chia m\u1ed7i c\u1ea1nh AD v\u00e0 BC th\u00e0nh 4 ph\u1ea7n b\u1eb1ng nhau, AB v\u00e0 CD th\u00e0nh 3 ph\u1ea7n b\u1eb1ng nhau, r\u1ed3i n\u1ed1i c\u00e1c \u0111i\u1ec3m chia nh\u01b0 h\u00ecnh v\u1ebd. Ta \u0111\u1ebfm \u0111\u01b0\u1ee3cbao nhi\u00eau h\u00ecnh ch\u1eef nh\u1eadt tr\u00ean h\u00ecnh v\u1ebd? BC MN EP AD Gi\u1ea3i : Tr\u01b0\u1edbc h\u1ebft Ta x\u00e9t c\u00e1c h\u00ecnh ch\u1eef nh\u1eadt t\u1ea1o b\u1edfi hai \u0111o\u1ea1n AD, EP v\u00e0 c\u00e1c \u0111o\u1ea1n n\u1ed1i c\u00e1c \u0111i\u1ec3m tr\u00ean hai c\u1ea1nh AD v\u00e0 BC. B\u1eb1ng c\u00e1ch t\u01b0\u01a1ng t\u1ef1 nh\u01b0 tronh v\u00ed d\u1ee5 1 ta t\u00ednh \u0111\u01b0\u1ee3c 10 h\u00ecnh. T\u01b0\u01a1ng t\u1ef1 ta t\u00ednh \u0111\u01b0\u1ee3c s\u1ed1 h\u00ecnh ch\u1eef nh\u1eadt t\u1ea1o th\u00e0nh do hai \u0111o\u1ea1n EP v\u00e0 MN, do MN v\u00e0 BC \u0111\u1ec1u b\u1eb1ng 10.","Ti\u1ebfp theo ta t\u00ednh s\u1ed1 h\u00ecnh ch\u1eef nh\u1eadt t\u1ea1o th\u00e0nh do hai \u0111o\u1ea1n AD v\u00e0 MN, EP v\u00e0 BC v\u1edbi c\u00e1c \u0111o\u1ea1n n\u1ed1i c\u00e1c \u0111i\u1ec3m tr\u00ean hai c\u1ea1nh AD v\u00e0 BC \u0111\u1ec1u b\u1eb1ng 10. V\u00ec v\u1eady : S\u1ed1 h\u00ecnh ch\u1eef nh\u1eadt \u0111\u1ebfm \u0111\u01b0\u1ee3c tr\u00ean h\u00ecnh v\u1ebd l\u00e0 : 10 + 10 + 10 + 10 + 10 + 10 = 60 (h\u00ecnh) \u0110\u00e1p s\u1ed1 60 h\u00ecnh. B\u00e0i t\u1eadp 3 :C\u1ea7n \u00edt nh\u1ea5t bao nhi\u00eau \u0111i\u1ec3m \u0111\u1ec3 khi n\u1ed1i l\u1ea1i ta \u0111\u01b0\u1ee3c 5 h\u00ecnh t\u1ee9 gi\u00e1c ? Gi\u1ea3i : E N\u1ebfu ta ch\u1ec9 c\u00f3 4 \u0111i\u1ec3m ( trong \u0111\u00f3 lh\u00f4ng c\u00f3 * 3 \u0111i\u1ec3m n\u00e0o c\u00f9ng n\u1eb1m tr\u00ean 1 \u0111o\u1ea1n th\u1eb3ng) A B th\u00ec n\u1ed1i l\u1ea1i ch\u1ec9 \u0111\u01b0\u1ee3c 1 h\u00ecnh t\u1ee9 gi\u00e1c. * * - N\u1ebfu ta ch\u1ecdn 5 \u0111i\u1ec3m, ch\u1eb3ng h\u1ea1n A, B, C, D, E (trong \u0111\u00f3 kh\u00f4ng c\u00f3 3 \u0111i\u1ec3m n\u00e0o n\u1eb1m tr\u00ean c\u00f9ng m\u1ed9t \u0111o\u1ea1n th\u1eb3ng) th\u00ec : - N\u1ebfu ta ch\u1ecdn A l\u00e0 1 \u0111\u1ec9nh th\u00ec khi * * ch\u1ecdn th\u00eam 3 trong s\u1ed1 4 \u0111i\u1ec3m c\u00f2n l\u1ea1i DC B, C, D, E v\u00e0 n\u1ed1i l\u1ea1i ta s\u1ebd \u0111\u01b0\u1ee3c m\u1ed9t t\u1ee9 gi\u00e1c c\u00f3 m\u1ed9t \u0111\u1ec9nh l\u00e0 A. C\u00f3 4 c\u00e1ch ch\u1ecdn 3 \u0111i\u1ec3m trong s\u1ed1 4 \u0111i\u1ec3m B, C, D, E \u0111\u1ec3 gh\u00e9p v\u1edbi A. V\u1eady c\u00f3 4 t\u1ee9 gi\u00e1c \u0111\u1ec9nh A. - C\u00f3 1 t\u1ee9 gi\u00e1c kh\u00f4ng nh\u1eadn A l\u00e0m \u0111\u1ec9nh, d\u00f3 l\u00e0 BCDE. T\u1eeb k\u1ebft qu\u1ea3 tr\u00ean \u0111\u00e2y ta suy ra Khi c\u00f3 5 \u0111i\u1ec3m ta \u0111\u01b0\u1ee3c 5 t\u1ee9 gi\u00e1c. V\u1eady \u0111\u1ec3 c\u00f3 5 h\u00ecnh t\u1ee9 gi\u00e1c ta c\u1ea7n \u00edt nh\u1ea5t 5 \u0111i\u1ec3m kh\u00e1c nhau (trong \u0111\u00f3 kh\u00f4ng c\u00f3 3 \u0111i\u1ec3m n\u00e0o n\u1eb1m tr\u00ean c\u00f9ng m\u1ed9t \u0111o\u1ea1n th\u1eb3ng) B\u00e0i 4 : Cho 5 \u0111i\u1ec3m A, B, C, D, E trong \u0111\u00f3 kh\u00f4ng c\u00f3 3 \u0111i\u1ec3m n\u00e0o n\u1eb1m tr\u00ean c\u00f9ng m\u1ed9t \u0111o\u1ea1n th\u1eb3ng. H\u1ecfi khi n\u1ed1i c\u00e1c \u0111i\u1ec3m tr\u00ean ta \u0111\u01b0\u1ee3c bao nhi\u00eau \u0111o\u1ea1n th\u1eb3ng? C\u0169ng h\u1ecfi nh\u01b0 th\u1ebf khi c\u00f3 6 \u0111i\u1ec3m, 10 \u0111i\u1ec3m. B\u00e0i 5 : \u0110\u1ec3 c\u00f3 10 \u0111o\u1ea1n th\u1eb3ng ta c\u1ea7n \u00edt nh\u1ea5t bao nhi\u00eau \u0111i\u1ec3m ? 4\/ B\u00e0i t\u1eadp v\u1ec1 nh\u00e0 B\u00e0i 1 : Cho tam gi\u00e1c ABC. Tr\u00ean c\u1ea1nh BC ta l\u1ea5y : a) 5 \u0111i\u1ec3m ; b) 10 \u0111i\u1ec3m ; c) 100 \u0111i\u1ec3m . H\u1ecfi c\u00f3 bao nhi\u00eau tam gi\u00e1c \u0111\u01b0\u1ee3c h\u00ecnh th\u00e0nh ? B\u00e0i 2 : C\u1ea7n \u00edt nmh\u1ea5t bao nhi\u00eau \u0111i\u1ec3m \u0111\u1ec3 n\u1ed1i lai ta \u0111\u01b0\u1ee3c : a) 4 h\u00ecnh tam gi\u00e1c ? b) 5 h\u00ecnh tam gi\u00e1c","B\u00e0i 3 : cho h\u00ecnh ch\u1eef nh\u1eadt ABCD. Tr\u00ean c\u1ea1nh AB l\u1ea5y 5 \u0111i\u1ec3m v\u00e0 tr\u00ean c\u1ea1nh CD l\u1ea5y 6 \u0111i\u1ec3m. N\u1ed1i \u0111\u1ec9nh C v\u00e0 \u0111\u1ec9nh D v\u1edbi m\u1ed7i \u0111i\u1ec3m thu\u1ed9c c\u1ea1nh AB. N\u1ed1i \u0111\u1ec9nh A v\u00e0 \u0111\u1ec9nh B v\u1edbi m\u1ed7i \u0111i\u1ec3m thu\u1ed9c c\u1ea1nh CD. H\u1ecfi c\u00f3 bao nhi\u00eau tam gi\u00e1c c\u00f3 c\u00e1c \u0111\u1ec9nh n\u1eb1m tr\u00ean c\u00e1c c\u1ea1nh c\u1ee7a h\u00ecnh ch\u1eef nh\u1eadt \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh ? B\u00e0i 4 : Cho h\u00ecnh thang ABCD. A C Chia c\u1ea1nh \u0111\u00e1y AB v\u00e0 CD th\u00e0nh 3 ph\u1ea7n b\u1eb1ng nhau v\u00e0 c\u00e1c c\u1ea1nh b\u00ean AB, CD th\u00e0nh 4 ph\u1ea7n b\u1eb1ng nhau nh\u01b0 h\u00ecnh v\u1ebd. Ta \u0111\u1ebfm \u0111\u01b0\u1ee3c bao nhi\u00eau h\u00ecnh thang tr\u00ean h\u00ecnh v\u1ebd ? A D B\u00e0i 5 : Cho tam gi\u00e1c ABC. Tr\u00ean m\u1ed7i c\u1ea1nh c\u1ee7a tam gi\u00e1c ta l\u1ea5y m\u1ed9t \u0111i\u1ec3m r\u1ed3i n\u1ed1i 3 \u0111i\u1ec3m \u0111\u00f3 v\u1edbi nhau. Tr\u00ean c\u00e1c c\u1ea1nh c\u1ee7a m\u1ed7i tam gi\u00e1c v\u1eeba t\u1ea1o th\u00e0nh ta l\u1ea1i l\u1ea5y m\u1ed9t \u0111i\u1ec3m r\u1ed3i n\u1ed1i 3 \u0111i\u1ec3m \u0111\u00f3 v\u1edbi nhau. Ti\u1ebfp t\u1ee5c nh\u01b0 th\u1ebf 3 l\u1ea7n th\u00ec d\u1eebng l\u1ea1i. H\u1ecfi khi \u0111\u00f3 ta \u0111\u1ebfm \u0111\u01b0\u1ee3c t\u1ea5t c\u1ea3 bao nhi\u00eau tam gi\u00e1c ? B\u00c0I 7 C\u00c1C B\u00c0I TO\u00c1N V\u1ec0 DI\u1ec6N T\u00cdCH C\u00c1C H\u00ccNH I - H\u00ccNH TAM GI\u00c1C I. M\u1ee4C TI\u00caU TI\u1ebeT D\u1ea0Y : - HS n\u1eafm \u0111\u01b0\u1ee3c m\u1ed9t s\u1ed1 t\u00ednh ch\u1ea5t c\u1ee7a h\u00ecnh tam gi\u00e1c - Gi\u1ea3i \u0111\u01b0\u1ee3c c\u00e1c b\u00e0i to\u00e1n v\u1ec1 di\u1ec7n t\u00edch h\u00ecnh tam gi\u00e1c - R\u00e8n k\u1ef9 n\u0103ng gi\u1ea3i to\u00e1n, quan s\u00e1t, t\u00ednh to\u00e1n cho h\u1ecdc sinh . II. CHU\u1ea8N B\u1eca - C\u00e2u h\u1ecfi v\u00e0 b\u00e0i t\u1eadp thu\u1ed9c d\u1ea1ng v\u1eeba h\u1ecdc. - C\u00e1c ki\u1ebfn th\u1ee9c c\u00f3 li\u00ean quan. III. C\u00c1C HO\u1ea0T \u0110\u1ed8NG D\u1ea0Y H\u1eccC 1\/ \u1ed4n \u0111\u1ecbnh t\u1ed5 ch\u1ee9c l\u1edbp. 2\/ Ki\u1ec3m tra b\u00e0i c\u0169. G\u1ecdi h\u1ecdc sinh l\u00e0m b\u00e0i t\u1eadp v\u1ec1 nh\u00e0 gi\u1edd tr\u01b0\u1edbc, GV s\u1eeda ch\u1eefa. 3\/ Gi\u1ea3ng b\u00e0i m\u1edbi. 3.1 Ki\u1ebfn th\u1ee9c c\u1ea7n nh\u1edb. - H\u00ecnh tam gi\u00e1c c\u00f3 3 c\u1ea1nh, 3 \u0111\u1ec9nh. \u0110\u1ec9nh l\u00e0 \u0111i\u1ec3m 2 c\u1ea1nh ti\u1ebfp gi\u00e1p nhau. C\u1ea3 3 c\u1ea1nh \u0111\u1ec1u c\u00f3 th\u1ec3 l\u1ea5y l\u00e0m \u0111\u00e1y. - Chi\u1ec1u cao c\u1ee7a h\u00ecnh tam gi\u00e1c l\u00e0 \u0111o\u1ea1n th\u1eb3ng h\u1ea1 t\u1eeb \u0111\u1ec9nh xu\u1ed1ng \u0111\u1eafy v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi \u0111\u1eafy. Nh\u01b0 v\u1eady m\u1ed7i tam gi\u00e1c c\u00f3 3 chi\u1ec1u cao. C\u00f4ng th\u1ee9c t\u00ednh : S = (a x h) : 2 h=sx2:a a=sx2:h","- Hai tam gi\u00e1c c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau khi ch\u00fang c\u00f3 \u0111\u00e1y b\u1eb1ng nhau (ho\u1eb7c \u0111\u00e1y chung), chi\u1ec1u cao b\u1eb1ng nhau (ho\u1eb7c chung chi\u1ec1u cao). - Hai tam gi\u00e1c c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau th\u00ec chi\u1ec1u cao c\u1ee7a 2 tam gi\u00e1c \u1ee9ng v\u1edbi 2 c\u1ea1nh \u0111\u1eafy b\u1eb1ng nhau \u0111\u00f3 c\u0169ng b\u1eb1ng nhau. Hai tam gi\u00e1c c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau khi \u0111\u00e1y tam gi\u00e1c P g\u1ea5p \u0111\u00e1y tam gi\u00e1c Q g\u1ea5p chi\u1ec1u cao tam gi\u00e1c P b\u1ea5y nhi\u00eau l\u1ea7n. B\u00e0i t\u1eadp \u1ee9ng d\u1ee5ng B\u00e0i 1 : Cho tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch l\u00e0 150 cm2. N\u1ebfu k\u00e9o d\u00e0i \u0111\u00e1y BC (v\u1ec1 ph\u00eda B) 5 cm th\u00ec di\u1ec7n t\u00edch s\u1ebd t\u0103ng th\u00eam 37,5 cm2 . T\u00ednh \u0111\u00e1y BC c\u1ee7a tam gi\u00e1c. Gi\u1ea3i : A B H C 5 cm D C\u00e1ch 1 : T\u1eeb A k\u1ebb \u0111\u01b0\u1eddng cao AH c\u1ee7a \u2206 ABC th\u00ec AH c\u0169ng l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a \u2206 ABD \u0110\u01b0\u1eddng cao AH l\u00e0 : 37,5 x 2 : 5 = 15 (cm) \u0110\u00e1y BC l\u00e0 : 150 x 2 : 15 = 20 (cm) \u0110\u00e1p s\u1ed1 20 cm. C\u00e1ch 2 : T\u1eeb A h\u1ea1 \u0111\u01b0\u1eddng cao AH vu\u00f4ng g\u00f3c v\u1edbi BC . \u0110\u01b0\u1eddng cao AH l\u00e0 \u0111\u01b0\u1eddng cao chung c\u1ee7a hai tam gi\u00e1c ABC v\u00e0 ABD . M\u00e0 : T\u1ec9 s\u1ed1 2 di\u1ec7n t\u00edch tam gi\u00e1c l\u00e0 : S \u2206 ABC 150 = =4 S \u2206 ABD 37,5","Hai tam gi\u00e1c c\u00f3 t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch l\u00e0 4 m\u00e0 ch\u00fang c\u00f3 chung \u0111\u01b0\u1eddng cao,n\u00ean t\u1ec9 s\u1ed1 2 \u0111\u00e1y c\u0169ng l\u00e0 4. V\u1edfy \u0111\u00e1y BC l\u00e0 : 5 x 4 = 20 (cm) \u0110\u00e1p s\u1ed1 20 cm. B\u00e0i 2 : Cho tam gi\u00e1c ABC vu\u00f4ng \u1edf A c\u00f3 c\u1ea1nh AB d\u00e0i 24 cm, c\u1ea1nh AC d\u00e0i 32 cm. \u0110i\u1ec3m M n\u1eb1m tr\u00ean c\u1ea1nh AC. T\u1eeb M k\u1ebb \u0111\u01b0\u1eddng song song v\u1edbi c\u1ea1nh AB c\u1eaft BC t\u1ea1i N. \u0110o\u1ea1n MN d\u00e0i 16 cm. T\u00ednh \u0111o\u1ea1n MA. Gi\u1ea3i : N\u1ed1i AN. Ta c\u00f3 tam gi\u00e1c NCA c\u00f3 NM l\u00e0 \u0111\u01b0\u1eddng cao v\u00ec MN AB n\u00ean MN c\u0169ng CA C Di\u1ec7n t\u00edch tam gi\u00e1c NCA l\u00e0 32 x 16 : 2 = 256 (cm2) Di\u1ec7n t\u00edch tam gi\u00e1c ABC l\u00e0 : 24 x 32 : 2 = 348 (cm2) Di\u1ec7n t\u00edch tam gi\u00e1c NAB l\u00e0 M N 384 \u2013 256 = 128 (cm2) Chi\u1ec1u cao NK h\u1ea1 t\u1eeb N xu\u1ed1ng AB l\u00e0 : 2 A B 128 x 2 : 24 = 10 (cm) 3 V\u00ec MN || AB n\u00ean t\u1ee9 gi\u00e1c MNBA l\u00e0 h\u00ecnh thang vu\u00f4ng. Do v\u1eady MA c\u0169ng b\u1eb1ng 10 2 cm 3 \u0110\u00e1p s\u1ed1 10 2 cm 3 B\u00e0i 3 : Cho tam gi\u00e1c ABC vu\u00f4ng \u1edf A. C\u1ea1nh AB d\u00e0i 28 cm, c\u1ea1nh AC d\u00e0i 36 cm M l\u00e0 m\u1ed9t \u0111i\u1ec3m tr\u00ean AC v\u00e0 c\u00e1ch A l\u00e0 9 cm. T\u1eeb M k\u1ebb \u0111\u01b0\u1eddng song song v\u1edbi AB v\u00e0 \u0111\u01b0\u1eddng n\u00e0y c\u1eaft c\u1ea1nh BC t\u1ea1i N. T\u00ednh \u0111o\u1ea1n MN. Gi\u1ea3i : C V\u00ec MN || AB n\u00ean MN AC N t\u1ea1i M. T\u1ee9 gi\u00e1cMNAB l\u00e0 h\u00ecnh thang vu\u00f4ng. N\u1ed1i NA. T\u1eeb N h\u1ea1 NH AB th\u00ec NH l\u00e0 chi\u1ec1u cao c\u1ee7a tam gi\u00e1c NBA M v\u00e0 c\u1ee7a h\u00ecnh thang MNBA n\u00ean NH = MA v\u00e0 l\u00e0 9 cm.","A HB Di\u1ec7n t\u00edch tam gi\u00e1c NBA l\u00e0 : 28 x 9 : 2 = 126 (cm2) Di\u1ec7n t\u00edch tam gi\u00e1c ABC l\u00e0 : 36 x 28 : 2 = 504 (cm2) Di\u1ec7n t\u00edch tam gi\u00e1c NAC l\u00e0 : 504 \u2013 126 = 378 (cm2) \u0110o\u1ea1n MN d\u00e0i l\u00e0 : 378 x 2 : 36 = 21 (cm) B\u00e0i 4 : Tam gi\u00e1c ABC c\u00f3 di\u1ec7n t\u00edch l\u00e0 90 cm2, D l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa AB. Tr\u00ean AC l\u1ea5y \u0111i\u1ec3m E sao cho AE g\u1ea5p \u0111\u00f4i EC. T\u00ednh di\u1ec7n t\u00edch AED. Gi\u1ea3i : A + N\u1ed1i DC ta c\u00f3 - SCAD = 12 SCAB D (v\u00ec c\u00f9ng chi\u1ec1u cao h\u1ea1 t\u1eeb C xu\u1ed1ng E AB v\u00e0 \u0111\u00e1y DB = DA C = 90 : 2 = 45 cm2) B 2 SDAE = 3 SADC (V\u00ec c\u00f9ng chi\u1ec1u cao h\u1ea1 t\u1eeb D xu\u1ed1ng AC v\u00e0 \u0111\u00e1y 2 45x2 = 30 (cm2) E = AC) = 33 \u0110\u00e1p s\u1ed1 SAED = 30 cm2 B\u00e0i 5 : Cho tam gi\u00e1c ABC, tr\u00ean AB l\u1ea5y \u0111i\u1ec3m D, E sao cho AD = DE = EB. Tr\u00ean AC l\u1ea5y \u0111i\u1ec3m H, K sao cho AK = HK = KC. Tr\u00ean BC l\u1ea5y \u0111i\u1ec3m M, N sao cho BM = MC = NC. T\u00ednh di\u1ec7n t\u00edch DEMNKH? Bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c ABC l\u00e0 270 cm2. Gi\u1ea3i : A D3 H E K 2 1 B","MN C + SABC \u2013 (S1 + S2 + S3) = SDEMNHK - N\u1ed1i C v\u1edbi E, ta t\u00ednh \u0111\u01b0\u1ee3c : 1 (V\u00ec c\u00f9ng chi\u1ec1u cao h\u1ea1 t\u1eeb C xu\u1ed1ng AB, \u0111\u00e1y BE = 1 BC). S =S CEB 3 CAB 3 1 Hay S1 = 9 SABC . + T\u01b0\u01a1ng t\u1ef1 ta t\u00ednh : 12 S1 = S2 = S3 = 9 SABC v\u00e0 b\u1eb1ng 270 : 9 = 30 (cm ) + T\u1eeb \u0111\u00f3 ta t\u00ednh \u0111\u01b0\u1ee3c : SDEMNKH = 180 (cm2) \u0110\u00e1p s\u1ed1 180 cm2 B\u00e0i 6 : Cho tam gi\u00e1c ABC, c\u00f3 BC = 60 cm, \u0111\u01b0\u1eddng cao AH = 30 cm. Tr\u00ean AB l\u1ea5y \u0111i\u1ec3m E v\u00e0 D sao cho AE = ED = DB. Tr\u00ean AC l\u1ea5y \u0111i\u1ec3m G v\u00e0 K sao cho AG = GK = KC. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh DEGK? Gi\u1ea3i : A N\u1ed1i BK ta c\u00f3 : EG - SABC = 60 x 30 : 2 = 900 (cm2) D K - SBKA = 23 SBAC (V\u00ec c\u00f9ng chi\u1ec1u cao h\u1ea1 2 B C t\u1eeb B xu\u1ed1ng AC v\u00e0 \u0111\u00e1y KA = AC) 3 SBKA = 900 : 3 x 2 = 600 (cm2) N\u1ed1i EK ta c\u00f3 : - SEAG = SKDB (v\u00ec c\u00f9ng chi\u1ec1u cao h\u1ea1 t\u1eeb E xu\u1ed1ng AH. \u0110\u00e1y GA- GK) -V\u00e0SKED = SKDB (V\u00ec c\u00f9ng chi\u1ec1u cao h\u1ea1 t\u1eeb K xu\u1ed1ng EB v\u00e0 \u0111\u00e1y DE=DB). 1 - Do \u0111\u00f3 S +S =S +S =S EGK KED EAG KDB 2 BAK - V\u1eady SEGK + SKED = 600 : 2 = 300 (cm2) Hay SEGKD = 300cm2 \u0110\u00e1p s\u1ed1 SEGKA = 300 cm2 B\u00e0i 7 : Cho tam gi\u00e1c MNP, F l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ea1nh NP. E l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ea1nh MN. Hai \u0111o\u1ea1n MF v\u00e0 PE c\u1eaft nhau t\u1ea1i I. H\u00e3y t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c IMN? Bi\u1ebft SMNP = 180 cm2 . Gi\u1ea3i : M","N\u1ed1i NI, ta c\u00f3 : 1. - SPME = SPNE (V\u00ec c\u00f3 c\u00f9ng chi\u1ec1u cao h\u1ea1 t\u1eeb P xu\u1ed1ng MN, \u0111\u00e1y EM = EN) - SIME = SINE (v\u00ec c\u00f3 c\u00f9ng chi\u1ec1u cao h\u1ea1 t\u1eeb I xu\u1ed1ng MN, \u0111\u00e1y EM = EN) E - Do \u0111\u00f3 SIMP = SINP I (Hi\u1ec7u hai di\u1ec7n t\u00edch b\u1eb1ng nhau) 2. SMNE = SPMF (V\u00ec c\u00f3 c\u00f9ng chi\u1ec1u cao h\u1ea1 t\u1eeb M xu\u1ed1ng NP, N P \u0111\u00e1y FN = FP F m\u00e0 SINF = SIFP (v\u00ec c\u00f3 c\u00f9ng chi\u1ec1u cao h\u1ea1 t\u1eeb I xu\u1ed1ng NP, \u0111\u00e1y FN = FP) Do \u0111\u00f3 SIMN = SIMP (Gi\u1ea3i th\u00edch nh\u01b0 tr\u00ean). K\u1ebft h\u1ee3p (1) v\u00e0 (2) ta c\u00f3 : 1 S = S = S = S : 3 = S = 180 : 3 = 60 (cm2) IMP INP IMN ABC 3 ABC B\u00e0i 8 : Cho tam gi\u00e1c ABC. \u0110i\u1ec3m M l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ea1nh AB. Tr\u00ean c\u1ea1nh AC l\u1ea5y AN b\u1eb1ng 1\/2 NC. Hai \u0111o\u1ea1n th\u1eb3ng BN v\u00e0 CM c\u1eaft nhau t\u1ea1i K. H\u00e3y t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c AKC? Bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c KAB b\u1eb1ng 42 dm2. Gi\u1ea3i : A H N\u1ed1i AK, ta c\u00f3 N + SCAM = SCMB (v\u00ec c\u00f3 c\u00f9ng chi\u1ec1u cao h\u1ea1 t\u1eeb C xu\u1ed1ng AB, \u0111\u00e1y MA = MB) MI - M\u00e0 SKAM = SKBM (v\u00ec c\u00f3 c\u00f9ng K chi\u1ec1u cao h\u1ea1 t\u1eeb K xu\u1ed1ng AB, \u0111\u00e1y MA = MB) B C - V\u1eady SAKC = SBKC1(v\u00ec c\u00f9ng l\u00e0 hi\u1ec7u c\u1ee7a hai tam gi\u00e1c c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau1) +S = S KAN 2 KCN (v\u00ec c\u00f9ng chi\u1ec1u cao h\u1ea1 t\u1eeb K xu\u1ed1ng AC, \u0111\u00e1y AN = 2 NC) N\u1ebfu coi A, C l\u00e0 \u0111\u1ec9nh th\u00ec 2 tam gi\u00e1c c\u00f3 di\u1ec7n t\u00edch g\u1ea5p \u0111\u00f4i m\u00e0 chung \u0111\u00e1y (AK) v\u1eady chi\u1ec1u cao c\u0169ng ph\u1ea3i g\u1ea5p \u0111\u00f4i nhau. Do \u0111\u00f3 : 1 AI = CH. 2 1 - SAKB = SCKB (chung \u0111\u00e1y BK, chi\u1ec1u cao AI = CH) 2 V\u1eady SAKC = SBKC = SABK x 2 = 42 x 2 = 84 (dm2)","* B\u00e0i t\u1eadp v\u1ec1 nh\u00e0 B\u00e0i 1 : M\u1ed9t th\u1eeda \u0111\u1ea5t h\u00ecnh tam gi\u00e1c c\u00f3 chi\u1ec1u cao l\u00e0 10 m. H\u1ecfi n\u1ebfu k\u00e9o d\u00e0i \u0111\u00e1y th\u00eam 4 m th\u00ec di\u1ec7n t\u00edch s\u1ebd t\u0103ng th\u00eam bao nhi\u00eau m2? B\u00e0i 2 : M\u1ed9t th\u1eeda \u0111\u1ea5t h\u00ecnh tam gi\u00e1c c\u00f3 \u0111\u00e1y l\u00e0 25 m. N\u1ebfu k\u00e9o d\u00e0i \u0111\u00e1y th\u00eam 5 m th\u00ec di\u1ec7n t\u00edch s\u1ebd t\u0103ng th\u00eam l\u00e0 50 m2. T\u00ednh di\u1ec7n t\u00edch m\u1ea3nh \u0111\u1ea5t khi ch\u01b0a m\u1edf r\u1ed9ng. B\u00e0i 3 : Cho tam gi\u00e1c ABC vu\u00f4ng \u1edf A, c\u1ea1nh AB d\u00e0i 54 cm, c\u1ea1nh AC d\u00e0i 60 m. \u0110i\u1ec3m M tr\u00ean AB c\u00e1ch A l\u00e0 10 m. T\u1eeb M k\u1ebb \u0111\u01b0\u1eddng song song v\u1edbi AC c\u1eaft c\u1ea1nh BC t\u1ea1i N. T\u00ednh \u0111o\u1ea1n MN. B\u00e0i 4 : Cho tam gi\u00e1c ABC c\u00f3 BC = 6 cm. L\u1ea5y D l\u00e0 \u0111i\u1ec3m \u1edf ch\u00ednh gi\u1eefa c\u1ee7a AC, k\u00e9o d\u00e0i AB m\u1ed9t \u0111o\u1ea1n BE = AB. N\u1ed1i D v\u1edbi E, DE c\u1eaft BC \u1edf M. T\u00ednh BM? B\u00e0i 5 : Cho tam gi\u00e1c ABC, c\u00f3 AB = 6 cm. Tr\u00ean AC l\u1ea5y \u0111i\u1ec3m D sao cho AD g\u1ea5p \u0111\u00f4i DC. Tr\u00ean BC l\u1ea5y \u0111i\u1ec3m E sao cho BE = 1\/2 EC, K\u00e9o d\u00e0i DE v\u00e0 AB c\u1eaft nhau \u1edf G. T\u00ednh BG? B\u00e0i 6 : Cho tam gi\u00e1c ABC, \u0111i\u1ec3m D n\u1eb1m tr\u00ean c\u1ea1nh AC, \u0111i\u1ec3m E n\u1eb1m tr\u00ean c\u1ea1nh BC sao cho : AD = DC, BE = 3\/2 EC. C\u00e1c \u0111o\u1ea1n th\u1eb3ng AE v\u00e0 BD c\u1eaft nhau \u1edf K. a) BK g\u1ea5p m\u1ea5y l\u1ea7n KD? b) Bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c ABC b\u1eb1ng 80 m2. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh DKEC? II - H\u00ccNH THANG I. M\u1ee4C TI\u00caU TI\u1ebeT D\u1ea0Y : - HS n\u1eafm \u0111\u01b0\u1ee3c m\u1ed9t s\u1ed1 t\u00ednh ch\u1ea5t c\u1ee7a h\u00ecnh thang - Gi\u1ea3i \u0111\u01b0\u1ee3c c\u00e1c b\u00e0i to\u00e1n v\u1ec1 di\u1ec7n t\u00edch h\u00ecnh thang - R\u00e8n k\u1ef9 n\u0103ng gi\u1ea3i to\u00e1n, quan s\u00e1t, t\u00ednh to\u00e1n cho h\u1ecdc sinh . II. CHU\u1ea8N B\u1eca - C\u00e2u h\u1ecfi v\u00e0 b\u00e0i t\u1eadp thu\u1ed9c d\u1ea1ng v\u1eeba h\u1ecdc. - C\u00e1c ki\u1ebfn th\u1ee9c c\u00f3 li\u00ean quan. III. C\u00c1C HO\u1ea0T \u0110\u1ed8NG D\u1ea0Y H\u1eccC 1\/ \u1ed4n \u0111\u1ecbnh t\u1ed5 ch\u1ee9c l\u1edbp. 2\/ Ki\u1ec3m tra b\u00e0i c\u0169. G\u1ecdi h\u1ecdc sinh l\u00e0m b\u00e0i t\u1eadp v\u1ec1 nh\u00e0 gi\u1edd tr\u01b0\u1edbc, GV s\u1eeda ch\u1eefa. 3\/ Gi\u1ea3ng b\u00e0i m\u1edbi. 3.1 Ki\u1ebfn th\u1ee9c c\u1ea7n nh\u1edb. - M\u1ed9t t\u1ee9 gi\u00e1c c\u00f3 hai c\u1ea1nh \u0111\u00e1y l\u1edbn, \u0111\u00e1y b\u00e9 song song v\u1edbi nhau g\u1ecdi l\u00e0 h\u00ecnh thang (H\u00ecnh vu\u00f4ng, h\u00ecnh ch\u1eef nh\u1eadt c\u0169ng coi l\u00e0 d\u1ea1ng h\u00ecnh thang \u0111\u1eb7c bi\u1ec7t) - \u0110o\u1ea1n th\u1eb3ng gi\u1eefa hai \u0111\u00e1y c\u1ee7a h\u00ecnh thang v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi hai \u0111\u00e1y l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a h\u00ecnh thang. M\u1ecdi chi\u1ec1u cao c\u1ee7a h\u00ecnh thang \u0111\u1ec1u b\u1eb1ng nhau. + C\u00e1c lo\u1ea1i h\u00ecnh thang - H\u00ecnh thang vu\u00f4ng c\u00f3 m\u1ed9t c\u1ea1nh b\u00ean vu\u00f4ng g\u00f3c v\u1edbi hai \u0111\u00e1y c\u1ee7a h\u00ecnh thang. H\u00ecnh thang vu\u00f4ng c\u00f3 hai g\u00f3c vu\u00f4ng.","- H\u00ecnh thang c\u00e2n c\u00f3 2 c\u1ea1nh b\u00ean b\u1eb1ng nhau. - C\u00e1c h\u00ecnh thang kh\u00f4ng c\u00f3 \u0111i\u1ec1u \u0111\u1eb7c bi\u1ec7t tr\u00ean g\u1ecdi l\u00e0 h\u00ecnh thang th\u01b0\u1eddng C\u00d4NG TH\u1ee8C S = (a + b) x h : 2 h = S x 2 : (a + b) a+b=Sx2:h 3.2 B\u00e0i t\u1eadp v\u1eadn d\u1ee5ng B\u00e0i 1 :Cho h\u00ecnh thang ABCD. Hai \u0111\u01b0\u1eddng ch\u00e9o AC v\u00e0 BD c\u1eaft nhau t\u1ea1i I. T\u00ecm c\u00e1c c\u1eb7p tam gi\u00e1c c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau. Ta c\u00f3 3 cap tam gi\u00e1c c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau l\u00e0 A B C S ADB = SABC I (v\u00ec c\u00f9ng \u0111\u00e1y AB x chi\u1ec1u cao chia 2) SACD = SBCD D SAID = SIBC V\u00ec ch\u00fang \u0111\u1ec1u l\u00e0 ph\u1ea7n di\u1ec7n t\u00edch c\u00f2n l\u1ea1i c\u1ee7a 2 tam gi\u00e1c c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau v\u00e0 c\u00f3 chung 1 ph\u1ea7n di\u1ec7n t\u00edch. (Tam gi\u00e1c ICD ho\u1eb7c AIB) B\u00e0i 2 : Cho h\u00ecnh thang ABCD c\u00f3 \u0111\u00e1y nh\u1ecf AB l\u00e0 27 cm, \u0111\u00e1y l\u1edbn CD l\u00e0 48 cm. N\u1ebfu k\u00e9o d\u00e0i \u0111\u00e1y nh\u1ecf th\u00eam 5 cm th\u00ec di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh t\u0103ng 40 cm2. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang \u0111\u00e3 cho. Gi\u1ea3i : c\u00e1ch1 A 27 B 5E \u2206 CBE c\u00f3 : \u0110\u00e1y BE = 5 cm, chi\u1ec1u cao l\u00e0 chi\u1ec1u cao c\u1ee7a h\u00ecnh thang ABCD . 40 cm2 V\u1eady chi\u1ec1u cao c\u1ee7a h\u00ecnh thang ABCD l\u00e0 : 40 x 2 : 5 = 16 (cm) Di\u1ec7n t\u00edch h\u00ecnh thang ABCD l\u00e0 : D 48 C","(27 + 48) x 16 : 2 = 600 (cm2) C\u00e1ch 2 : T\u1ed5ng hai \u0111\u00e1y h\u00ecnh thang g\u1ea5p \u0111\u00e1y BE l\u00e0 : (27 + 48) : 5 = 15 (l\u1ea7n) Hai h\u00ecnh (thang v\u00e0 tam gi\u00e1c) c\u00f3 chi\u1ec1u cao chung n\u00ean di\u1ec7n t\u00edch h\u00ecnh thang g\u1ea5p 15 l\u1ea7n di\u1ec7n t\u00edch \u2206 BCE Di\u1ec7n t\u00edch tam gi\u00e1c BCE l\u00e0 : 40 x 15 = 600 (cm2) B\u00e0i 3 : Cho h\u00ecnh thang ABCD c\u00f3 \u0111\u00e1y l\u1edbn CD l\u00e0 20 cm, \u0111\u00e1y nh\u1ecf AB l\u00e0 15 cm. M l\u00e0 m\u1ed9t \u0111i\u1ec3m tr\u00ean AB c\u00e1ch B l\u00e0 5 cm. N\u1ed1i M v\u1edbi C. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang m\u1edbi AMCD. Bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c MBC l\u00e0 280 cm2. Gi\u1ea3i : A MB \u0110\u00e1y m\u1edbi AM l\u00e0 : 15 \u2013 5 = 10 (cm) T\u1ed5ng hai \u0111\u00e1y AM v\u00e0 CD l\u00e0 : 10 + 20 = 30 (cm) A MB Chi\u1ec1u cao h\u00ecnh thang ABCD l\u00e0 : 280 x 2 : 5 = 112 (cm) D C Di\u1ec7n t\u00edch h\u00ecnh thang ABCD l\u00e0 : 30 x 112 : 2 = 1680 (cm2) C\u00e1ch 2 N\u1ed1i A v\u1edbi C Ta c\u00f3 \u0111o\u1ea1n AM l\u00e0 : 15 \u2013 5 = 10 (cm) Di\u1ec7n t\u00edch tam gi\u00e1c ACM g\u1ea5p 2 l\u1ea7n \u0111i\u1ec7n t\u00edch tam gi\u00e1c MCB \uf0de Di\u1ec7n t\u00edch tam gi\u00e1c ACM = 280 x 2 = 560 (cm2) (v\u00ec AM g\u1ea5p BM hai l\u1ea7n v\u00e0 \u0111\u01b0\u1eddng cao hai tam gi\u00e1c b\u1eb1ng nhau) \u2206 DAC v\u00e0 \u2206 MCB c\u00f3 : DC g\u1ea5p MB l\u00e0 20 : 5 = 4 ( l\u1ea7n) \u0110\u01b0\u1eddng cao chung n\u00ean di\u1ec7n t\u00edch tam gi\u00e1c DAC g\u1ea5p di\u1ec7n t\u00edch tam gi\u00e1c MCB 4 l\u1ea7n. Di\u1ec7n t\u00edch tam gi\u00e1c ADC l\u00e0 : 280 x 4 = 1120 (cm2) B\u00e0i 4 : M\u1ed9t th\u1eeda ru\u1ed9ng h\u00ecnh thang c\u00f3 di\u1ec7n t\u00edch l\u00e0 361,8 m2. \u0110\u00e1y l\u1edbn h\u01a1n \u0111\u00e1y nh\u1ecf l\u00e0 13,5 m. H\u00e3y t\u00ednh \u0111\u1ed9 d\u00e0i c\u1ee7a m\u1ed7i \u0111\u00e1y, bi\u1ebft r\u1eb1ng n\u1ebfu t\u0103ng \u0111\u00e1y l\u1edbn th\u00eam 5,6 m th\u00ec di\u1ec7n t\u00edch th\u1eeda ru\u1ed9ng s\u1ebd t\u0103ng th\u00eam 3,6 m2. Gi\u1ea3i :","Chi\u1ec1u cao c\u1ee7a h\u00ecnh thang l\u00e0 : A B 33,6 x 2 : 5,6 = 12 (m) T\u1ed5ng hai \u0111\u00e1y h\u00ecnh thang l\u00e0 : 361,8 x2 : 12 = 60,3 (m) 33,6 m2 \u0111\u00e1y nh\u1ecf c\u1ee7a h\u00ecnh thang l\u00e0 : (60,3 \u2013 13,5) : 2 = 23,4 (m) \u0110\u00e1y l\u1edbn c\u1ee7a h\u00ecnh thang l\u00e0 : 23,4 + 13,5 = 36,9 (m). E DH C B\u00e0i 5 : M\u1ed9t h\u00ecnh thang c\u00f3 chi\u1ec1u cao l\u00e0 10 m, hi\u1ec7u 2 \u0111\u00e1y l\u00e0 22 m. K\u00e9o d\u00e0i \u0111\u00e1y nh\u1ecf b\u1eb1ng \u0111\u00e1y l\u1edbn \u0111\u1ec3 h\u00ecnh \u0111\u00e3 cho th\u00e0nh h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chi\u1ec1u daid b\u1eb1ng \u0111\u00e1y l\u1edbn, chi\u1ec1u r\u1ed9ng b\u1eb1ng chi\u1ec1u cao h\u00ecnh thang. Di\u1ec7n t\u00edch \u0111\u01b0\u1ee3c m\u1edf r\u1ed9ng th\u00eam b\u1eb1ng 1\/7 di\u1ec7n t\u00edch h\u00ecnh thang c\u0169. Ph\u1ea7n m\u1edf r\u1ed9ng v\u1ec1 ph\u00eda tay ph\u1ea3i c\u00f3 di\u1ec7n t\u00edch l\u00e0 90 m2. T\u00ednh \u0111\u00e1y l\u1edbn c\u1ee7a h\u00ecnh thang ban \u0111\u1ea7u. Gi\u1ea3i : EA BG \u0110\u00e1y BG c\u1ee7a \u2206 CBG l\u00e0 : 90 x 2 : 10 = 18 (m) 90 cm2 \u0110\u00e1y EA c\u1ee7a \u2206 DAE l\u00e0 : 22 \u2013 18 = 4 (m) Di\u1ec7n t\u00edch 2 ph\u1ea7n m\u1edf r\u1ed9ng l\u00e0 : 20 + 90 = 110 (m2) Di\u1ec7n t\u00edch h\u00ecnh thang ABCD l\u00e0 : 110 x 7 = 770 (m2) D C T\u1ed5ng hai \u0111\u00e1y AB v\u00e0 CD l\u00e0 : 770 x 2 : 10 = 154 (m) \u0110\u00e1y CD l\u00e0 : (154 + 22) : 2 = 88 (m) B\u00e0i 6 : Cho h\u00ecnh thang vu\u00f4ng ABCD, c\u00f3 \u0111\u00e1y nh\u1ecf AB l\u00e0 40 m. L\u1ea5y E tr\u00ean AD, G tr\u00ean BC sao cho EG chia h\u00ecnh thang ABCD l\u00e0m hai h\u00ecnh thang c\u00f3 \u0111\u01b0\u1eddng cao AE l\u00e0 30 m v\u00e0 ED l\u00e0 10 m. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thangABGE v\u00e0 EGCD. Gi\u1ea3i : N\u1ed1i G v\u1edbi A, G v\u1edbi D A 40 m B Di\u1ec7n t\u00edch ABCD l\u00e0 : (40 \uf02b 60)x40 = 2000 (m2) 40 m 2 Di\u1ec7n t\u00edch \u2206 GBA l\u00e0 : (40 x 30) : 2 = 600 (m2)","Di\u1ec7n tich \u2206 GDC l\u00e0 : G 60 x 10 : 2 = 300 (m2) 10 m Di\u1ec7n t\u00edch \u2206 AGD l\u00e0 : D C 2000 \u2013 (600+300) = 1100 (m2) 60 m V\u1eady EG l\u00e0: 1100 x 2 : 40 = 55 (m ) Di\u1ec7n t\u00edch ABGE l\u00e0 : (55 + 40 ) x 30 : 2 = 1425 (m2) Di\u1ec7n t\u00edch EGCD l\u00e0: ( 60 + 55) x 10 : 2 = 575 (m2) B\u00e0i 6: Cho h\u00ecnh thang ABCD c\u00f3 di\u1ec7n t\u00edch l\u00e0 60m2 , \u0111i\u1ec3m M, N, P, Q l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa c\u1ee7a c\u00e1c c\u1ea1nh AB, BC, CD, DA T\u00ednh di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c MNPQ. Gi\u1ea3i : MQ k\u00e9o d\u00e0i c\u1eaft DC t\u1ea1i F; MN k\u00e9o d\u00e0i c\u1eaft DC t\u1ea1i E Ta c\u00f3 di\u1ec7n t\u00edch h\u00ecnh thang ABCD b\u1eb1ng di\u1ec7n t\u00edch tam gi\u00e1c FME Di\u1ec7n t\u00edch \u2206 MPF =di\u1ec7n t\u00edch \u2206 MPE (\u0111\u00e1y b\u1eb1ng nhau, \u0111\u01b0\u1eddng cao chung) Di\u1ec7n t\u00edch \u2206 MNP = di\u1ec7n t\u00edch \u2206NPE A MB (\u0111\u00e1y MN = NE, \u0111\u01b0\u1eddng cao chung) Di\u1ec7n t\u00edch \u2206PMQ = di\u1ec7n t\u00edch \u2206PQF (\u0111\u00e1y QM= QF, \u0111\u01b0\u1eddng cao chung) Q N N\u00ean di\u1ec7n t\u00edch MNPQ = 1\/2 di\u1ec7n t\u00edch \u2206FME . Hay di\u1ec7n t\u00edch MNPQ =1\/2 di\u1ec7n t\u00edch h\u00ecnh thangABCD v\u00e0 b\u1eb1ng F E 60 : 2 = 30 (cm2) D PC \u0110\u00e1p s\u1ed1: 30 cm2 B\u00e0i 7: T\u00ecm di\u1ec7n t\u00edch c\u1ee7a m\u1ed9t h\u00ecnh thangbi\u1ebft r\u1eb1ng n\u1ebfu k\u00e9o d\u00e0i \u0111\u00e1y b\u00e9 2m v\u1ec1 m\u1ed9t ph\u00eda th\u00ec ta \u0111\u01b0\u1ee3c h\u00ecnh vu\u00f4ng c\u00f3 chu vi 24m. Gi\u1ea3i: Theo b\u00e0i ra h\u00ecnh thang vu\u00f4ng. \u0110\u00e1y A B 2m M l\u1edbn b\u1eb1ng c\u1ea1nh h\u00ecnh vu\u00f4ng AMCD v\u00e0 chi\u1ec1u cao h\u00ecnh thang c\u0169ng b\u1eb1ng c\u1ea1nh h\u00ecnh vu\u00f4ng. C\u1ea1nh h\u00ecnh vu\u00f4ng AMCD l\u00e0: 24 : 4 =6 (m) D C \u0110\u00e1y b\u00e9 h\u00ecnh thang ABCDl\u00e0: 6 \u2013 2 = 4(m) Di\u1ec7n t\u00edch h\u00ecnh thang ABCD l\u00e0: (6 \uf02b 4)x6 = 30 (m2) \u0110\u00e1p s\u1ed1 :30m2 2","B\u00e0i 8 : Cho h\u00ecnh thang ABCD c\u00f3 \u0111\u00e1y b\u00e9 AB b\u1eb1ng 18 cm, \u0111\u00e1y l\u1edbn CD b\u1eb1ng 3\/2 \u0111\u00e1y b\u00e9 AB. Tr\u00ean AB l\u1ea5y \u0111i\u1ec3m M sao cho AM = 12 cm. N\u1ed1i M v\u1edbi C. T\u00ecm di\u1ec7n t\u00edch h\u00ecnh thang AMCD, bi\u1ebft di\u1ec7n t\u00edch h\u00ecnh thang ABCD h\u01a1n di\u1ec7n t\u00edch h\u00ecnh thang AMCD l\u00e0 42 cm2. Gi\u1ea3i : \u0110\u00e1y l\u1edbn h\u00ecnh thang ABCD l\u00e0 : 3 A MB 18 x = 27 (cm) 2 \u0110\u1ed9 d\u00e0i \u0111o\u1ea1n MB l\u00e0 : 18 \u2013 12 = 6 (cm) MB ch\u00ednh l\u00e0 \u0111\u00e1y c\u1ee7a \u2206 MBC, chi\u1ec1u cao c\u1ee7a \u2206 MBC ( c\u0169ng l\u00e0 chi\u1ec1u cao c\u1ee7a h\u00ecnh thang AMCD) 42x2 D C = 14 (cm) 6 Di\u1ec7n t\u00edch h\u00ecnh thang AMCD l\u00e0 : (12 \uf02b 27)x14 = 273 (cm2) 2 \u0110\u00e1p s\u1ed1 273 cm2 4.B\u00e0i t\u1eadp v\u1ec1 nh\u00e0 B\u00e0i 1 : M\u1ed9t th\u1eeda ru\u1ed9ng h\u00ecnh thang c\u00f3 trung b\u00ecnh c\u1ed9ng 2 \u0111\u00e1y l\u00e0 32 m. N\u1ebfu \u0111\u00e1y l\u1edbn t\u0103ng 16 m, \u0111\u00e1y nh\u1ecf t\u0103ng 10 m th\u00ec di\u1ec7n t\u00edch th\u1eeda ru\u1ed9ng s\u1ebd t\u0103ng th\u00eam 130 m2. T\u00ednh di\u1ec7n t\u00edch th\u1eeda ru\u1ed9ng \u0111\u00f3. B\u00e0i 2 : Cho h\u00ecnh thang ABCD c\u00f3 \u0111\u00e1y nh\u1ecf AB. Hai \u0111\u01b0\u1eddng ch\u00e9o AC, BD c\u1eaft nhau t\u1ea1i 0. T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang \u0111\u00f3 bi\u1ebft di\u1eb9n t\u00edch h\u00ecnh tam gi\u00e1cAOB l\u00e0 15 cm2, di\u1ec7n t\u00edch tam gi\u00e1c BOC l\u00e0 30 cm2. B\u00e0i 3 : M\u1ed9t mi\u1ebfng \u0111\u1ea5t h\u00ecnh thang c\u00f3 di\u1ec7n t\u00edch 705,5 m2, \u0111\u00e1y l\u1edbn h\u01a1n \u0111\u00e1y b\u00e9 8 m, n\u1ebfu \u0111\u00e1y l\u1edbn \u0111\u01b0\u1ee3c t\u0103ng th\u00eam 6 m th\u00ec mi\u1ebfng \u0111\u1ea5t c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng 756,5 m2. T\u00ednh \u0111\u1ed9 d\u00e0i m\u1ed7i \u0111\u00e1y h\u00ecnh thang. B\u00e0i 4 : Trung b\u00ecnh c\u1ed9ng hai \u0111\u00e1y c\u1ee7a m\u1ed9t th\u1eeda ru\u1ed9ng h\u00ecnh thang b\u1eb1ng 34 m. N\u1ebfu t\u0103ng \u0111\u00e1y b\u00e9 th\u00eam 12 m th\u00ec di\u1ec7n t\u00edch th\u1eeda ru\u1ed9ng t\u0103ng th\u00eam 114 m2. H\u00e3y t\u00ecm di\u1ec7n t\u00edch th\u1eeda ru\u1ed9ng. B\u00e0i 5 : Cho h\u00ecnh thang ABCD \u0111\u00e1y AB = 30 cm v\u00e0 CD = 45 cm. AC v\u00e0 BD c\u1eaft nhau t\u1ea1i O. Cho bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c OAB l\u00e0 180 cm2. H\u00e3y t\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang. B\u00e0i 6 : Cho h\u00ecnh thang ABCD, hai \u0111\u00e1y AB v\u00e0 CD. C\u00e1c c\u1ea1nh b\u00ean AD v\u00e0 BC k\u00e9o d\u00e0i c\u1eaft nhau \u1edf K. Cho bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c KCD g\u1ea5p 1,5 l\u1ea7n di\u1ec7n t\u00edch tam gi\u00e1c KAC. T\u00ednh c\u00e1c c\u1ea1nh \u0111\u00e1y c\u1ee7a h\u00ecnh thang \u0111\u00f3 n\u1ebfu bi\u1ebft di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh thang l\u00e0 375 cm2 v\u00e0 chi\u1ec1u cao c\u1ee7a n\u00f3 l\u00e0 10 cm. III - C\u00c1C B\u00c0I TO\u00c1N V\u1ec0 C\u1eaeT GH\u00c9P H\u00ccNH","I. M\u1ee4C TI\u00caU TI\u1ebeT D\u1ea0Y : - HS n\u1eafm \u0111\u01b0\u1ee3c m\u1ed9t s\u1ed1 t\u00ednh ch\u1ea5t c\u1ee7a h\u00ecnh thang - Gi\u1ea3i \u0111\u01b0\u1ee3c c\u00e1c b\u00e0i to\u00e1n v\u1ec1 di\u1ec7n t\u00edch h\u00ecnh thang - R\u00e8n k\u1ef9 n\u0103ng gi\u1ea3i to\u00e1n, quan s\u00e1t, t\u00ednh to\u00e1n cho h\u1ecdc sinh . II. CHU\u1ea8N B\u1eca - C\u00e2u h\u1ecfi v\u00e0 b\u00e0i t\u1eadp thu\u1ed9c d\u1ea1ng v\u1eeba h\u1ecdc. - C\u00e1c ki\u1ebfn th\u1ee9c c\u00f3 li\u00ean quan. III. C\u00c1C HO\u1ea0T \u0110\u1ed8NG D\u1ea0Y H\u1eccC 1\/ \u1ed4n \u0111\u1ecbnh t\u1ed5 ch\u1ee9c l\u1edbp. 2\/ Ki\u1ec3m tra b\u00e0i c\u0169. G\u1ecdi h\u1ecdc sinh l\u00e0m b\u00e0i t\u1eadp v\u1ec1 nh\u00e0 gi\u1edd tr\u01b0\u1edbc, GV s\u1eeda ch\u1eefa. 3\/ Gi\u1ea3ng b\u00e0i m\u1edbi. 3.1. L\u01b0u \u00fd C\u00e1c b\u00e0i to\u00e1n v\u1ec1 c\u1eaft gh\u00e9p h\u00ecnh th\u01b0\u1eddng g\u1eb7p d\u01b0\u1edbi hai d\u1ea1ng : 1) B\u1eb1ng m\u1ed9t s\u1ed1 n\u00e9t k\u1ebb h\u00e3y chia m\u1ed9t h\u00ecnh cho tr\u01b0\u1edbc ra th\u00e0nh nh\u1eefng ph\u1ea7n c\u00f3 di\u1ec7n t\u00edch t\u1ec9 l\u1ec7 v\u1edbi c\u00e1c s\u1ed1 cho tr\u01b0\u1edbc. 2) B\u1eb1ng m\u1ed9t s\u1ed1 nh\u1ea5t c\u1eaft h\u00e3y chia m\u1ed9t h\u00ecnh cho tr\u01b0\u1edbc th\u00e0nh h\u1eefng m\u1ea3nh nh\u1ecf \u0111\u1ec3 gh\u00e9p l\u1ea1i ta \u0111\u01b0\u1ee3c m\u1ed9t h\u00ecnh c\u00f3 h\u00ecnh d\u1ea1ng cho tr\u01b0\u1edbc. Ph\u01b0\u01a1ng ph\u00e1p chung \u0111\u1ec3 gi\u1ea3i c\u00e1c b\u00e0i to\u00e1n n\u00e0y, ta s\u1ebd minh ho\u1ea1 b\u1eb1ng c\u00e1c v\u00ed d\u1ee5 c\u1ee5 th\u1ec3 d\u01b0\u1edbi \u0111\u00e2y. 3.2. B\u00e0i t\u1eadp v\u1eadn d\u1ee5ng B\u00e0i 1 : H\u00e3y chia m\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt th\u00e0nh 4 h\u00ecnh tam gi\u00e1c c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau ? Gi\u1ea3i : Xu\u1ea5t ph\u00e1t t\u1eeb nh\u1eadn x\u00e9t : - Hai tam gi\u00e1c c\u00f3 c\u00f9ng chi\u1ec1u cao v\u00e0 s\u1ed1 \u0111o c\u1ee7a \u0111\u00e1y b\u1eb1ng nhau th\u00ec b\u1eb1ng nhau. - Hai tam gi\u00e1c c\u00f3 chung \u0111\u00e1y v\u00e0 s\u1ed1 \u0111o c\u1ee7a \u0111\u01b0\u1eddng cao b\u1eb1ng nhau th\u00ec di\u1ec7n t\u00edch b\u1eb1ng nhau. A B Ta gi\u1ea3i b\u00e0i to\u00e1n tr\u00ean . Tr\u01b0\u1edbc h\u1ebft ta k\u1ebb \u0111\u01b0\u1eddng ch\u00e9o AC \u0111\u1ec3 h\u00ecnh ch\u1eef nh\u1eadt th\u00e0nh hai tam gi\u00e1c c\u00f3di\u1ec7n t\u00edch b\u1eb1ng nhau. CD B\u00e2y gi\u1edd ta chia m\u1ed7i tam gi\u00e1c ABC v\u00e0 ADC th\u00e0nh hai tam gi\u00e1c c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau. Nh\u01b0 v\u1eady ta \u0111\u01b0\u1ee3c m\u1ed9t l\u1eddi gi\u1ea3i c\u1ee7a b\u00e0i to\u00e1n. C\u00e1ch 1 A B Ch\u1ecdn AC l\u00e0m \u0111\u00e1y chung c\u1ee7a 2 tam gi\u00e1c s\u1ebd chia ra. Nh\u01b0 v\u1eady \u0111\u1ec3 \u0111\u01b0\u1ee3c 2 tam gi\u00e1c b\u1eb1ng nhau c\u00f3 c\u00f9ng \u0111\u01b0\u1eddng cao h\u1ea1","t\u1eeb B (v\u00e0 t\u1eeb D) xu\u1ed1ng AC th\u00ec ph\u1ea3i chia \u0111\u00e1y AC th\u00e0nh 2 ph\u1ea7n b\u1eb1ng nhau b\u1edfi O D \u0111i\u1ec3m O. N\u1ed1i BO v\u00e0 DO ta \u0111\u01b0\u1ee3c c\u00e1c tam M gi\u00e1c ABO, BOC, COD v\u00e0 DOA tho\u1ea3 C C m\u00e3n c\u00e1c \u0111i\u1ec1u ki\u1ec7n c\u1ee7a \u0111\u1ec1 b\u00e0i. ND C\u00e1ch 2 Ch\u1ecdn 2 c\u1ea1nh BC v\u00e0 AD l\u00e0m \u0111\u00e1y c\u1ee7a 2 tam gi\u00e1c s\u1ebd chia ra. Nh\u01b0 v\u1eady c\u00e1c tam gi\u00e1c \u0111\u01b0\u1ee3c chia ra t\u1eeb B tam gi\u00e1c ABC c\u00f3 chung \u0111\u01b0\u1eddng cao AB cho n\u00ean ta ph\u1ea3i chia \u0111\u00e1y BC th\u00e0nh 2 ph\u1ea7n c\u00f3 s\u1ed1 \u0111o b\u1eb1ng nhau b\u1edfi \u0111i\u1ec3m M.T\u01b0\u01a1ng t\u1ef1 chia AD b\u1edfi \u0111i\u1ec3m N. N\u1ed1i AM, CN ta \u0111\u01b0\u1ee3c 4 tam gi\u00e1c ABM, AMC, CAN v\u00e0 CND tho\u1ea3 A m\u00e3n \u0111i\u1ec1u ki\u1ec7n c\u1ee7a \u0111\u1ec1 b\u00e0i B C P H C\u00e1ch 3 Ch\u1ecdn hai c\u1ea1nh AB v\u00e0 CD l\u00e0m \u0111\u00e1y c\u1ee7a tam gi\u00e1c A D s\u1ebd chia ra. Nh\u01b0 v\u1eady c\u00e1c tam gi\u00e1c \u0111\u01b0\u1ee3c chia t\u1eeb tam gi\u00e1c ABC c\u00f3 chung \u0111\u01b0\u1eddng cao CB th\u00e0nh 2 ph\u1ea7n c\u00f3 s\u1ed1 \u0111o b\u1eb1ng nhau b\u1edfi \u0111i\u1ec3m P. T\u01b0\u01a1ng t\u1ef1 ta chia CD th\u00e0nh 2 ph\u1ea7n b\u1edfi \u0111i\u1ec3m H. N\u1ed1i CP v\u00e0 AH ta \u0111\u01b0\u1ee3c 4 tam gi\u00e1c ACP, CPB, ADH, v\u00e0 AHC tho\u1ea3 m\u00e3n \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec1 b\u00e0i. C\u00e1ch 4 Ph\u1ed1i h\u1ee3p c\u00e1ch 1 v\u00e0 c\u00e1ch 2 nh\u01b0 h\u00ecnh v\u1ebd Ngo\u00e0i ra c\u00f2n c\u00f3 th\u1ec3 chia theo c\u00e1c c\u00e1ch kh\u00e1c. B\u00e0i 2 : Cho m\u1ea3nh b\u00eca h\u00ecnh t\u1ee9 gi\u00e1c ABCD. B\u1eb1ng m\u1ed9t l\u1ea7n c\u1eaft (kh\u00f4ng nh\u1ea5c k\u00e9o) h\u00e3y chia m\u1ea3nh b\u00eca \u0111\u00f3 th\u00e0nh hai ph\u1ea7n c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau. Gi\u1ea3i : K\u1ebb \u0111\u01b0\u1eddng ch\u00e9o BD. B\u1eb1ng l\u1eadp lu\u1eadn nh\u01b0 C trong v\u00ed d\u1ee5 8, ch\u1ecdn \u0111i\u1ec3m gi\u1eefa O c\u1ee7a BD. N\u1ed1i AO, CO. Ta c\u1eaft m\u1ea3nh b\u00eca theo n\u00e9t v\u1ebd chi\u1ec1u B m\u0169i t\u00ean s\u1ebd \u0111\u01b0\u1ee3c 2 m\u1ea3nh b\u00eca ABCO v\u00e0 ADCO tho\u1ea3 m\u00e3n \u0111i\u1ec1u ki\u1ec7n c\u1ee7a \u0111\u1ec1 b\u00e0i. O AD","4. B\u00e0i t\u1eadp v\u1ec1 nh\u00e0 B\u00e0i 1 : Cho 1 m\u1ea3nh b\u00eca h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chi\u1ec1u d\u00e0i 9 cm v\u00e0 chi\u1ec1u r\u1ed9ng 4 cm. b\u1eb1ng 1 nh\u00e1t c\u1eaft (kh\u00f4ng nh\u1ea5c k\u00e9o) h\u00e3y chia m\u1ea3nh b\u00eca th\u00e0nh 2 m\u1ea3nh \u0111\u1ec3 gh\u00e9p l\u1ea1i \u0111\u01b0\u1ee3c m\u1ed9t h\u00ecnh vu\u00f4ng c\u00f3 c\u00f9ng di\u1ec7n t\u00edch. B\u00e0i 2 : H\u00e3y c\u1eaft m\u1ed9t m\u1ea3nh b\u00eca h\u00ecnh ch\u1eef nh\u1eadt th\u00e0nh hai m\u1ea3nh \u0111\u1ec3 gh\u00e9p l\u1ea1i ta \u0111\u01b0\u1ee3c m\u1ed9t h\u00ecnh thang c\u00f3 : a) \u0111\u00e1y l\u1edbn g\u1ea5p 3 l\u1ea7n \u0111\u00e1y nh\u1ecf ; b) \u0110\u00e1y l\u1edbn g\u1ea5p 5 l\u1ea7n \u0111\u00e1y nh\u1ecf. B\u00e0i 3 : H\u00e3y c\u1eaft m\u1ed9t m\u1ea3nh b\u00eca h\u00ecnh thang th\u00e0nh c\u00e1c m\u1ea3nh nh\u1ecf \u0111\u1ec3 gh\u00e9p l\u1ea1i ta \u0111\u01b0\u1ee3c : a) M\u1ed9t tam gi\u00e1c b) M\u1ed9t h\u00ecnh thang c) M\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt B\u00e0i 4 : Cho hai m\u1ea3nh b\u00eca h\u00ecnh vu\u00f4ng. H\u00e3y c\u1eaft hai m\u1ea3nh b\u00eca \u0111\u00f3 th\u00e0nh c\u00e1c m\u1ea3nh nh\u1ecf \u0111\u1ec3 gh\u00e9p l\u1ea1i ta \u0111\u01b0\u1ee3c m\u1ed9t h\u00ecnh vu\u00f4ng. B\u00e0i 5 : Cho m\u1ed9t mi\u1ebfng t\u00f4n h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chi\u1ec1u d\u00e0i g\u1ea5p hai l\u1ea7n chi\u1ec1u r\u1ed9ng. h\u00e3y c\u1eaft mi\u1ebfng t\u00f4n \u0111\u00f3 \u0111\u1ec3 gh\u00e9p l\u1ea1i \u0111\u01b0\u1ee3c m\u1ed9t mi\u1ebfng t\u00f4n h\u00ecnh vu\u00f4ng. IV - H\u00ccNH TR\u00d2N I. M\u1ee4C TI\u00caU TI\u1ebeT D\u1ea0Y : - HS n\u1eafm \u0111\u01b0\u1ee3c c\u00e1ch t\u00ednh di\u1ec7n t\u00edch h\u00ecnh tr\u00f2n v\u00e0 c\u00e1c y\u1ebfu t\u1ed1 c\u00f3 li\u00ean quan - Gi\u1ea3i \u0111\u01b0\u1ee3c c\u00e1c b\u00e0i to\u00e1n v\u1ec1 h\u00ecnh tr\u00f2n - R\u00e8n k\u1ef9 n\u0103ng gi\u1ea3i to\u00e1n, quan s\u00e1t, t\u00ednh to\u00e1n cho h\u1ecdc sinh . II. CHU\u1ea8N B\u1eca - C\u00e2u h\u1ecfi v\u00e0 b\u00e0i t\u1eadp thu\u1ed9c d\u1ea1ng v\u1eeba h\u1ecdc. - C\u00e1c ki\u1ebfn th\u1ee9c c\u00f3 li\u00ean quan. III. C\u00c1C HO\u1ea0T \u0110\u1ed8NG D\u1ea0Y H\u1eccC 1\/ \u1ed4n \u0111\u1ecbnh t\u1ed5 ch\u1ee9c l\u1edbp. 2\/ Ki\u1ec3m tra b\u00e0i c\u0169. G\u1ecdi h\u1ecdc sinh l\u00e0m b\u00e0i t\u1eadp v\u1ec1 nh\u00e0 gi\u1edd tr\u01b0\u1edbc, GV s\u1eeda ch\u1eefa. 3\/ Gi\u1ea3ng b\u00e0i m\u1edbi. 3.1. Ki\u1ebfn th\u1ee9c c\u1ea7n nh\u1edb : - C\u00e1c c\u00f4ng th\u1ee9c : C = d x 3,14 C = r x 2 x 3,14 S = r x r x 3,14 r = C : 3,14 : 2","- Hai h\u00ecnh tr\u00f2n c\u00f3 b\u00e1n k\u00ednh (ho\u1eb7c \u0111\u01b0\u1eddng k\u00ednh) g\u1ea5p nhau bao nhi\u00eau l\u1ea7n th\u00ec chu vi c\u1ee7a ch\u00fang c\u0169ng g\u1ea5p nhau bao nhi\u00eau l\u1ea7n. - Hai h\u00ecnh tr\u00f2n c\u00f3 t\u1ec9 s\u1ed1 chu vi l\u00e0 k th\u00ec t\u1ec9 s\u1ed1 b\u00e1n k\u00ednh (ho\u1eb7c \u0111\u01b0\u1eddng k\u00ednh) b\u1eb1ng k th\u00ec t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch c\u1ee7a ch\u00fang l\u00e0 k x k 3.2 B\u00e0i t\u1eadp v\u1eadn d\u1ee5ng B\u00e0i 1 : T\u00ecm di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng bi\u1ebft di\u1ec7n t\u00edch h\u00ecnh tr\u00f2n l\u00e0 50,24 cm2. G\u1ecdi r l\u00e0 b\u00e1n k\u00ednh c\u1ee7a h\u00ecnh tr\u00f2n Di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh tr\u00f2n l\u00e0 : AB r x r x 3,14 Theo b\u00e0i ra ta c\u00f3 : r x r x 3,14 = 50,24 r x r = 16 rxr=4x4 \uf0der=4 DC S\u1ed1 \u0111o \u0111o\u1ea1n th\u1eb3ng BD l\u00e0 : 4 x 2 = 8 (cm) Di\u1ec7n t\u00edch tam gi\u00e1c ABD l\u00e0 : 8x4 = 16 (cm2) 2 Di\u1ec7n t\u00edch h\u00ecnh vu\u00f4ng ABCD l\u00e0 : 16 x 2 = 32 (cm2) B\u00e0i 2 : M\u1ed9t mi\u1ebfng b\u00eca h\u00ecnh tr\u00f2n c\u00f3 chu vi 37,68 cm. t\u00ednh di\u1ec7n t\u00edch mi\u1ebfng b\u00eca \u0111\u00f3 : Gi\u1ea3i : B\u00e1n k\u00ednh mi\u1ebfng b\u00eca l\u00e0 : 37,68 : 3,14 : 2 = 6 (cm) Di\u1ec7n t\u00edch mi\u1ebfng b\u00eca l\u00e0 : 6 x 6 x 3,14 = 113,04 (cm2) \u0110\u00e1p s\u1ed1 113,04 cm2 B\u00e0i 3 : H\u00ecnh tr\u00f2n A c\u00f3 chu vi 219,8 cm, h\u00ecnh tr\u00f2n B c\u00f3 di\u1ec7n t\u00edch 113,04 cm2. H\u00ecnh tr\u00f2n n\u00e0o c\u00f3 b\u00e1n k\u00ednh l\u1edbn h\u01a1n? Gi\u1ea3i : B\u00e1n k\u00ednh h\u00ecnh tr\u00f2n A l\u00e0 : 219,8 : 3,14 : 2 = 35 (cm) = 3,5 dm. G\u1ecdi r l\u00e0 b\u00e1n k\u00ednh h\u00ecnh tr\u00f2n B ta c\u00f3 : r x r = 113,04 : 3,14 = 36 (dm) \uf0de r = 6 dm V\u00ec 6 > 3,5 n\u00ean b\u00e1n k\u00ednh h\u00ecnh tr\u00f2n B l\u1edbn h\u01a1n b\u00e1n k\u00ednh h\u00ecnh tr\u00f2n A","B\u00e0i 4 : Bi\u1ebft t\u1ec9 s\u1ed1 b\u00e1n k\u00ednh c\u1ee7a 2 h\u00ecnh tr\u00f2n l\u00e0 3\/4.H\u00e3y t\u00ednh t\u1ec9 s\u1ed1 2 chu vi, 2 di\u1ec7n t\u00edch c\u1ee7a 2 h\u00ecnh tr\u00f2n \u0111\u00f3. Gi\u1ea3i : G\u1ecdi r1 l\u00e0 b\u00e1n k\u00ednh c\u1ee7a h\u00ecnh tr\u00f2n th\u1ee9 nh\u1ea5t, r2 l\u00e0 b\u00e1n k\u00ednh c\u1ee7a h\u00ecnh tr\u00f2n th\u1ee9 hai G\u1ecdi C1 v\u00e0 S1 l\u00e0 chu vi v\u00e0 di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh tr\u00f2n th\u1ee9 nh\u1ea5t G\u1ecdi C2 v\u00e0 S2 l\u00e0 chu vi v\u00e0 di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh tr\u00f2n th\u1ee9 hai th\u00ec : C1 3,14xr1x2 r1 3 = 3,14x2xr2xr2 = r2 = 4 C2 T\u1ec9 s\u1ed1 chu vi hai \u0111\u01b0\u1eddng tr\u00f2n b\u1eb1ng 3\/4 S1 = 3,14xr1xr2 = r1 x r1 = 3 x 3 = 9 S 2 3,14xr2xr2 r2 r2 4 4 16 4. B\u00e0i t\u1eadp v\u1ec1 nh\u00e0 B\u00e0i 1 : Cho hai h\u00ecnh tr\u00f2n \u0111\u1ed3ng t\u00e2m, h\u00ecnh tr\u00f2n th\u1ee9 nh\u1ea5t c\u00f3p chu vi 18,84 cm ; H\u00ecnh tr\u00f2n th\u1ee9 hai c\u00f3 chu vi 31,2 cm. H\u00e3y t\u00ednh di\u1ec7n t\u00edch h\u00ecnh v\u00e0nh khuy\u00ean do hai h\u00ecnh tr\u00f2n t\u1ea1o th\u00e0nh. B\u00e0i 2 : Di\u1ec7n t\u00edch c\u1ee7a 1 h\u00ecnh tr\u00f2n s\u1ebd thay \u0111\u1ed5i nh\u01b0 th\u1ebf n\u00e0o n\u1ebfu ta t\u0103ng b\u00e1n k\u00ednh c\u1ee7a n\u00f3 l\u00ean 3 l\u1ea7n. B\u00e0i 3 : Hai h\u00ecnh tr\u00f2n c\u00f3 hi\u1ec7u hai chu vi b\u1eb1ng 6,908 dm. T\u00ecm hi\u1ec7u 2 b\u00e1n k\u00ednh c\u1ee7a hai h\u00ecnh tr\u00f2n \u0111\u00f3. V -DI\u1ec6N T\u00cdCH XUNG QUANH, DI\u1ec6N T\u00cdCH TO\u00c0N PH\u1ea6N, TH\u1ec2 T\u00cdCH H\u00ccNH H\u1ed8P CH\u1eee NH\u1eacT, H\u00ccNH L\u1eacP PH\u01af\u01a0NG, H\u00ccNH TR\u1ee4 I. M\u1ee4C TI\u00caU TI\u1ebeT D\u1ea0Y : - HS n\u1eafm \u0111\u01b0\u1ee3c c\u00e1ch t\u00ednh di\u1ec7n t\u00edch xung quanh, di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n v\u00e0 th\u1ec3 t\u00edch c\u1ee7a c\u00e1c h\u00ecnh : h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt, l\u1eadp ph\u01b0\u01a1ng, h\u00ecnh tr\u1ee5. - V\u1eadn d\u1ee5ng l\u00e0m \u0111\u01b0\u1ee3c c\u00e1c b\u00e0i t\u1eadp. - R\u00e8n k\u1ef9 n\u0103ng gi\u1ea3i to\u00e1n, quan s\u00e1t, t\u00ednh to\u00e1n cho h\u1ecdc sinh . II. CHU\u1ea8N B\u1eca - C\u00e2u h\u1ecfi v\u00e0 b\u00e0i t\u1eadp thu\u1ed9c d\u1ea1ng v\u1eeba h\u1ecdc. - C\u00e1c ki\u1ebfn th\u1ee9c c\u00f3 li\u00ean quan. III. C\u00c1C HO\u1ea0T \u0110\u1ed8NG D\u1ea0Y H\u1eccC 1\/ \u1ed4n \u0111\u1ecbnh t\u1ed5 ch\u1ee9c l\u1edbp. 2\/ Ki\u1ec3m tra b\u00e0i c\u0169. G\u1ecdi h\u1ecdc sinh l\u00e0m b\u00e0i t\u1eadp v\u1ec1 nh\u00e0 gi\u1edd tr\u01b0\u1edbc, GV s\u1eeda ch\u1eefa. 3\/ Gi\u1ea3ng b\u00e0i m\u1edbi. 3.1. Ki\u1ebfn th\u1ee9c c\u1ea7n nh\u1edb : A \u2013 H\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt :","H\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt c\u00f3 6 m\u1eb7t l\u00e0 c\u00e1c h\u00ecnh ch\u1eef nh\u1eadt, c\u00f3 3 k\u00edch th\u01b0\u1edbc l\u00e0 chi\u1ec1u d\u00e0i a, chi\u1ec1u r\u1ed9ng b, chi\u1ec1u cao c. Sxq = Pm\u0111 x h = (a + b) x 2 x c STP = Sxq + S2\u0111 = Sxq + a + b x 2 V=axbxc B \u2013 H\u00ecnh l\u1eadp ph\u01b0\u01a1ng H\u00ecnh l\u1eadp ph\u01b0\u01a1ng c\u00f3 6 m\u1eb7t l\u00e0 c\u00e1c h\u00ecnh vu\u00f4ng b\u1eb1ng nhau. T\u1ea5t c\u1ea3 c\u00e1c c\u1ea1nh c\u1ee7a h\u00ecnh l\u1eadp ph\u01b0\u01a1ng \u0111\u1ec1u b\u1eb1ng nhau. Sxq = a x a x 4 STP = a x a x 6 V=axaxa C \u2013 H\u00ecnh tr\u1ee5 h\u00ecnh tr\u1ee5 c\u00f3 hai \u0111\u00e1y l\u00e0 hai h\u00ecnh tr\u00f2n b\u1eb1ng nhau Sxq = r x 2 x 3,14 x h STP = Sxq + r x r x 3,14 x 2 V = r x r x 3,14 x h 3.2. B\u00e0i t\u1eadp v\u1eadn d\u1ee5ng B\u00e0i 1 : C\u00f3 8 h\u00ecnh l\u1eadp ph\u01b0\u01a1ng, m\u1ed7i h\u00ecnh c\u00f3 c\u1ea1nh b\u1eb1ng 2 cm. X\u1ebfp 8 h\u00ecnh \u0111\u00f3 th\u00e0nh 1 h\u00ecnh l\u1eadp ph\u01b0\u01a1ng l\u1edbn. T\u00ecm di\u1ec7n t\u00edch xung quanh, dio\u1ec7n t\u00edch to\u00e0n ph\u1ea7n v\u00e0 th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh l\u1eadp ph\u01b0\u01a1ng l\u1edbn. Gi\u1ea3i : 8 h\u00ecnh l\u1eadp ph\u01b0\u01a1ng ta x\u1ebfp th\u00e0nh h\u00ecnh l\u1eadp ph\u01b0\u01a1ng l\u1edbn bao g\u1ed3m c\u00f3 2 t\u1ea7ng m\u1ed7i t\u1ea7ng c\u00f3 4 h\u00ecnh l\u1eadp ph\u01b0\u01a1ng nh\u1ecf C\u1ea1nh c\u1ee7a h\u00ecnh l\u1eadp ph\u01b0\u01a1ng nh\u1ecf l\u00e0 2 n\u00ean c\u1ea1nh c\u1ee7a h\u00ecnh l\u1eadp ph\u01b0\u01a1ng l\u1edbn l\u00e0 : 2 x 2 = 4 (cm) Di\u1ec7n t\u00edch xung quanh l\u00e0 : 4 x 4 x 4 = 64 (cm2) Di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n l\u00e0 : 4 x 4 x 6 = 96 (cm2) Th\u1ec3 t\u00edch l\u00e0 : 4 x 4 x 4 = 64 (cm2) B\u00e0i 2 : C\u00f3 27 h\u00ecnh l\u1eadp ph\u01b0\u01a1ng, m\u1ed7i h\u00ecnh c\u00f3 th\u1ec3 t\u00edch 8 cm3. X\u1ebfp 27 h\u00ecnh \u0111\u00f3 th\u00e0nh m\u1ed9t h\u00ecnh l\u1eadp ph\u01b0\u01a1ng l\u1edbn. h\u1ecfi h\u00ecnh l\u1eadp ph\u01b0\u01a1ng l\u1edbn c\u00f3 c\u1ea1nh l\u00e0 bao nhi\u00eau? Gi\u1ea3i : Ta c\u00f3 : 8=2x2x2 V\u1eady m\u1ed7i h\u00ecnh l\u1eadp ph\u01b0\u01a1ng nh\u1ecf c\u00f3 \u0111\u00e1y b\u1eb1ng 2 cm.","X\u1ebfp 27 h\u00ecnh l\u1eadp ph\u01b0\u01a1ng nh\u1ecf th\u00e0nh m\u1ed9t h\u00ecnh l\u1eadp ph\u01b0\u01a1ng l\u1edbn c\u00f3 3 t\u1ea7ng m\u1ed7i t\u1ea7ng c\u00f3 3 h\u00e0ng, m\u1ed7i h\u00e0ng c\u00f3 3 h\u00ecnh l\u1eadp ph\u01b0\u01a1ng nh\u1ecf. N\u00ean c\u1ea1nh c\u1ee7a h\u00ecnh l\u1eadp ph\u01b0\u01a1ng l\u1edbn l\u00e0 : 2 x 3 = 6 (cm) \u0110\u00e1p s\u1ed1 6 cm B\u00e0i 3 : M\u1ed9t h\u00ecnh l\u1eadp ph\u01b0\u01a1ng c\u00f3 di\u1ec7n t\u00edch xung quanh b\u1eb1ng 64 cm2. T\u00ednh th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh l\u1eadp ph\u01b0\u01a1ng \u0111\u00f3. Gi\u1ea3i : Di\u1ec7n t\u00edch m\u1ed9t m\u1eb7t c\u1ee7a h\u00ecnh l\u1eadp ph\u01b0\u01a1ng l\u00e0 : 64 : 4 = 16 (cm2) Ta th\u1ea5y 16 = 4 x 4 \uf0de c\u1ea1nh c\u1ee7a h\u00ecnh l\u1eadp ph\u01b0\u01a1ng l\u00e0 4 Th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh l\u1eadp ph\u01b0\u01a1ng l\u00e0 : 4 x 4 x 4 = 64 (cm3) \u0110\u00e1p s\u1ed1 64 cm3 B\u00e0i 4 : M\u1ed9t b\u1ec3 ch\u1ee9a n\u01b0\u1edbc h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt, \u0111o \u1edf trong l\u00f2ng b\u1ec3 th\u1ea5y chi\u1ec1u d\u00e0i b\u1eb1ng 2,5 m ; chi\u1ec1u r\u1ed9ng b\u1eb1ng 1,4 m ; chi\u1ec1u cao g\u1ea5p 1,5 l\u1ea7n chi\u1ec1u r\u1ed9ng. H\u1ecfi b\u1ec3 ch\u1ee9a \u0111\u1ea7y n\u01b0\u1edbc th\u00ec \u0111\u01b0\u1ee3c bao nhi\u00eau l\u00edt. Gi\u1ea3i : Chi\u1ec1u cao c\u1ee7a b\u1ec3 n\u01b0\u1edbc l\u00e0 : 1,4 x 1,5 = 2,1 (m) Th\u1ec3 t\u00edch b\u1ec3 n\u01b0\u1edbc l\u00e0 : 2,5 x 1,4 x 2,1 = 7,35 (m3) ta c\u00f3 : 7,35 m3 = 7350 dm3 = 7350 l\u00edt \u0110\u00e1p s\u1ed1 7350 l\u00edt B\u00e0i 5 : M\u1ed9t c\u00e1i th\u00f9ng h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt c\u00f3 \u0111\u00e1y l\u00e0 h\u00ecnh vu\u00f4ng c\u00f3 chu vi l\u00e0 20 dm. Ng\u01b0\u1eddi ta \u0111\u1ed5 v\u00e0o th\u00f9ng 150 l\u00edt d\u1ea7u. H\u1ecfi chi\u1ec1u cao c\u1ee7a d\u1ea7u trong th\u00f9ng l\u00e0 bao nhi\u00eau? Gi\u1ea3i : C\u1ea1nh c\u1ee7a \u0111\u00e1y th\u00f9ng l\u00e0 : 20 : 4 = 5 (dm) Di\u1ec7n t\u00edch \u0111\u00e1y th\u00f9ng l\u00e0 : 5 x 5 = 25 (dm2) Ta c\u00f3 : 150 l\u00edt = 150 dm3 Chi\u1ec1u cao c\u1ee7a d\u1ea7u trong th\u00f9ng l\u00e0 : 150 : 25 = 6 (dm) \u0110\u00e1p s\u1ed1 6 dm.","B\u00e0i 6 : M\u1ed9t phi\u1ebfn \u0111\u00e1 h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt c\u00f3 chu vi \u0111\u00e1y b\u1eb1ng 60 dm, chi\u1ec1u d\u00e0i b\u1eb1ng 3\/2 chi\u1ec1u r\u1ed9ng v\u00e0 chi\u1ec1u cao b\u1eb1ng 1\/2 chi\u1ec1u d\u00e0i. Phi\u1ebfn \u0111\u00e1 c\u00e2n n\u1eb7ng4471,2 kg. H\u1ecfi 1 dm3 \u0111\u00e1 n\u1eb7ng bao nhi\u00eau ki l\u00f4 gam? Gi\u1ea3i : N\u1eeda chu vi phi\u1ebfn \u0111\u00e1 l\u00e0 : 60 : 2 = 30 (dm) Chi\u1ec1u d\u00e0i c\u1ee7a phi\u1ebfn \u0111\u00e1 l\u00e0 : 30 : (3 + 2) x 3 = 18 (dm) Chi\u1ec1u r\u1ed9ng c\u1ee7a phi\u1ebfn \u0111\u00e1 l\u00e0 : 30 \u2013 18 = 12 (dm) Chi\u1ec1u cao c\u1ee7a phi\u1ebfn \u0111\u00e1 l\u00e0 : 18 : 2 = 9 (dm) Th\u1ec3 t\u00edch c\u1ee7a phi\u1ebfn \u0111\u00e1 l\u00e0 : 18 x 12 x 9 = 1944 (dm3) 1 dm3 \u0111\u00e1 n\u1eb7ng l\u00e0 : 4471,2 : 1944 = 2,3 (kg) \u0111\u00e1p s\u1ed1 2,3 kg B\u00e0i 7: M\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chi\u1ec1u cao 6 dm. N\u1ebfu t\u0103ng chi\u1ec1u cao th\u00eam 2 dm th\u00ec th\u1ec3 t\u00edch h\u1ed9p t\u0103ng th\u00eam 96 dm3. T\u00ednh th\u1ec3 t\u00edch h\u1ed9p. Gi\u1ea3i : Di\u1ec7n t\u00edch \u0111\u00e1y c\u1ee7a h\u1ed9p ch\u1eef nh\u1eadt l\u00e0 : 96 : 2 = 48 (dm2) Th\u1ec3 t\u00edch h\u1ed9p ch\u1eef nh\u1eadt l\u00e0 : 48 x 6 = 228 (dm3) C\u00e1ch 2 6 dm so v\u1edbi 2 dm th\u00ec g\u1ea5p : 6 : 2 = 3 (l\u1ea7n) Ph\u1ea7n t\u0103ng th\u00eam v\u00e0 h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt c\u00f3 chung di\u1ec7n t\u00edch \u0111\u00e1y v\u00e0 chi\u1ec1u cao h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt g\u1ea5p 3 l\u00e0an ph\u1ea7n t\u0103ng th\u00eam n\u00ean th\u1ec3 t\u00edch h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt c\u0169ng ph\u1ea3i g\u1ea5p 3 l\u1ea7n th\u1ec3 t\u00edch t\u0103ng th\u00eam. v\u1eady th\u1ec3 t\u00edch h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt l\u00e0 : 96 x 3 = 288 (dm3) \u0110\u00e1p s\u1ed1 : 288 dm3 B\u00e0i 8 : M\u1ed9t c\u0103n ph\u00f2ng d\u00e0i 8 m, r\u1ed9ng 6 m cao 5 m. Ng\u01b0\u1eddi ta mu\u1ed1n qu\u00e9t v\u00f4i tr\u1ea7n nh\u00e0 v\u00e0 4 m\u1eb7t t\u01b0\u1eddng trong ph\u00f2ng. Tr\u00ean 4 m\u1ef1t t\u01b0\u1eddng c\u00f3 2 c\u1eeda ra v\u00e0o m\u1ed7i c\u1eeda r\u1ed9ng 1,6 m cao 2,2 m v\u00e0 4 c\u1eeda s\u1ed5, m\u1ed7i c\u1eeda s\u1ed5 r\u1ed9ng 1,2 m cao 1,5 m. Ti\u1ec1n thu\u00ea qu\u00e9t v\u00f4i 1 m\u00e9t vu\u1ed3ng h\u1ebft 1500 \u0111\u1ed3ng. H\u1ecfi ti\u1ec1n c\u00f4ng qu\u00e9t v\u00f4i c\u0103n ph\u00f2ng \u0111\u00f3 h\u1ebft bao nhi\u00eau ?","Gi\u1ea3i : Di\u1ec7n t\u00edch 4 m\u1eb7t t\u01b0\u1eddng c\u1ee7a c\u0103n ph\u00f2ng l\u00e0 : (9 + 6) x 2 x 5 = 150 (m2) Di\u1ec7n t\u00edch tr\u1ea7n nh\u00e0 l\u00e0 : 9 x 6m = 54 (m2) Di\u1ec7n t\u00edch 4 c\u1eeda s\u1ed5 l\u00e0 : 1,2 x 1,5 x 4 = 7,2 (m2) Di\u1ec7n t\u00edch 2 c\u1eeda ra v\u00e0o l\u00e0 : 2,2 x 1,6 x 2 = 7,04 (m2) Di\u1ec7n t\u00edch c\u1ea7n qu\u00e9t v\u00f4i l\u00e0 : (150 + 54) \u2013 (7,2 + 7,04) = 189,76 (m2) Ti\u1ec1n c\u00f4ng m\u01b0\u1edbn qu\u00e9t v\u00f4i l\u00e0 : 1500 x 189,76 = 284640 (\u0111\u1ed3ng) \u0110\u00e1p s\u1ed1 284640 \u0111\u1ed3ng B\u00e0i 9 : M\u1ed9t ph\u00f2ng h\u1ecdp d\u00e0i 8 m, r\u1ed9ng 5 m, cao 4 m. H\u1ecfi ph\u1ea3i m\u1edf r\u1ed9ng chi\u1ec1u d\u00e0i ra th\u00eam bao nhi\u00eau \u0111\u1ec3 phg\u00f2ng h\u1ecdp c\u00f3 th\u1ec3 ch\u1ee9a \u0111\u01b0\u1ee3c 60 ng\u01b0\u1eddi v\u00e0 m\u1ed7i ng\u01b0\u1eddi c\u00f3 \u0111\u1ee7 4,5 m2 kh\u00f4ng kh\u00ed \u0111\u1ec3 \u0111\u1ea3m b\u1ea3o s\u1ee9c kho\u1ebb ? Gi\u1ea3i : Th\u1ec3 t\u00edch c\u1ee7a h\u1ed9i tr\u01b0\u1eddng sau khi m\u1edf r\u1ed9ng l\u00e0 : 4,5 x 60 = 270 (m3) Di\u1ec7n t\u00edch m\u1eb7t b\u00ean c\u1ee7a h\u1ed9i tr\u01b0\u1eddng l\u00e0 : 5 x 4 = 20 (m2) Chi\u1ec1u d\u00e0i c\u1ee7a h\u1ed9i tr\u01b0\u1eddng sau khi m\u1edf r\u1ed9ng l\u00e0 : 270 : 20 = 13,5 (m) Chi\u1ec1u d\u00e0i ph\u1ea3i m\u1edf r\u1ed9ng th\u00eam l\u00e0 : 13,5 \u2013 8 = 5,5(m) \u0110\u00e1p s\u1ed1 5,5 m B\u00e0i 10 : C\u00e1i b\u1ec3 ch\u1ee9a n\u01b0\u1edbc nh\u00e0 em c\u00f3 h\u00ecnh ch\u1eef nh\u1eadt, \u0111o trong l\u00f2ng b\u1ec3 \u0111\u01b0\u1ee3c chi\u1ec1u d\u00e0i 1,5 m, chi\u1ec1u r\u1ed9ng l\u00e0 1,2 m v\u00e0 chi\u1ec1u cao l\u00e0 0,9 m. B\u1ec3 \u0111\u00e3 h\u1ebft n\u01b0\u1edbc, ch\u1ecb em v\u1eeba \u0111\u1ed5 v\u00e0o b\u1ec3 30 g\u00e1nh n\u01b0\u1edbc m\u1ed7i g\u00e1nh 45 l\u00edt. H\u1ecfi m\u1eb7t n\u01b0\u1edbc c\u00f2n c\u00e1ch mi\u1ec7ng b\u1ec3 bao nhi\u00eau v\u00e0 c\u1ea7n \u0111\u1ed5 th\u00eam bao nhi\u00eau g\u00e1nh n\u01b0\u1edbc n\u1eefa \u0111\u1ec3 \u0111\u1ea7y b\u1ec3 ? Gi\u1ea3i : S\u1ed1 l\u00edt n\u01b0\u1edbc \u0111\u00e3 \u0111\u1ed5 v\u00e0o b\u1ec3 l\u00e0 : 45 x 30 = 1350 (l\u00edt) = 1350 dm3 = m1,35 m3 Di\u1ec7n t\u00edch \u0111\u00e1y b\u1ec3 l\u00e0 : 1,5 x 1,2 = 1,8 (m2)","M\u1eb7t n\u01b0\u1edbc c\u00e1ch \u0111\u00e1y b\u1ec3 l\u00e0 : 1,35 : 1,8 = 0,75 (m) M\u1eb7t n\u01b0\u1edbc trong b\u1ec3 c\u00e1ch mi\u1ec7ng b\u1ec3 l\u00e0 : 0,9 \u2013 0,75 = 0,15 (m) Th\u1ec3 t\u00edch b\u1ec3 l\u00e0 : 1,8 x 0,9 = 1,62 (m3) = 1620 l\u00edt S\u1ed1 g\u00e1nh n\u01b0\u1edbc c\u1ea7n \u0111\u1ed5 \u0111\u1ea7y b\u1ec3 l\u00e0 : 1620 : 45 = 36 (g\u00e1nh) \u0110\u1ec3 \u0111\u1ea7y b\u1ec3 c\u1ea7n \u0111\u1ed5 th\u00eam l\u00e0 : 36 \u2013 30 = 6 (g\u00e1nh) \u0110\u00e1p s\u1ed1 0,15 m v\u00e0 6 g\u00e1nh. B\u00e0i 11 : X\u1ebfp 8 h\u00ecnh l\u1eadp ph\u01b0\u01a1ng nh\u1ecf c\u00f3 c\u1ea1nh 4 cm th\u00e0nh m\u1ed9t h\u00ecnh l\u1eadp ph\u01b0\u01a1ng l\u1edbn r\u1ed3i s\u01a1n t\u1ea5t c\u1ea3 c\u00e1c c\u1ea1nh c\u1ee7a h\u00ecnh l\u1eadp ph\u01b0\u01a1ng l\u1edbn. H\u1ecfi m\u1ed7i h\u00ecnh l\u1eadp ph\u01b0\u01a1ng nh\u1ecf c\u00f3 m\u1ea5y m\u1eb7t \u0111\u01b0\u1ee3c s\u01a1n v\u00e0 di\u1ec7n t\u00edch \u0111\u01b0\u1ee3c s\u01a1n c\u1ee7a m\u1ed7i HLP nh\u1ecf l\u00e0 bao nhi\u00eau? Gi\u1ea3i : X\u1ebfp 8 HLP nh\u1ecf th\u00e0nh 1 HLP l\u1edbn g\u1ed3m 2 t\u1ea7ng, m\u1ed7i t\u1ea7ng g\u1ed3m 4 h\u00ecnh l\u1eadp ph\u01b0\u01a1ng nh\u1ecf, v\u00ec th\u1ebf m\u1ed7i HLP nh\u1ecf \u0111\u1ec1u c\u00f3 3 m\u1eb7t \u0111\u01b0\u1ee3c gh\u00e9p v\u1edbi c\u00e1c h\u00ecnh l\u1eadp ph\u01b0\u01a1ng kh\u00e1c. C\u00e1c m\u1eb7t \u0111\u01b0\u1ee3c gh\u00e9p kh\u00f4ng \u0111\u01b0\u1ee3c s\u01a1n. V\u00ec HLP c\u00f3 6 m\u1eb7t n\u00ean s\u1ed1 m\u1eb7t \u0111\u01b0\u1ee3c s\u01a1n l\u00e0 : 6 \u2013 3 = 3 (m\u1eb7t) Di\u1ec7n t\u00edch m\u1ed9t m\u1eb7t c\u1ee7a HLP nh\u1ecf l\u00e0 : 4 x 4 = 16 (cm2) Di\u1ec7n t\u00edch m\u1ed7i HLP nh\u1ecf \u0111\u01b0\u1ee3c s\u01a1n l\u00e0 : 16 x 3 = 48 (cm2) \u0110\u00e1p s\u1ed1 48 cm2 B\u00e0i 12 : Ng\u01b0\u1eddi ta x\u1ebb 1 kh\u00fac g\u1ed7 h\u00ecnh tr\u1ee5 d\u00e0i 5 m c\u00f3 \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y 0,6 m th\u00e0nh 1 kh\u1ed1i h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt c\u00f3 \u0111\u00e1y l\u00e0 h\u00ecnh vu\u00f4ng v\u00e0 \u0111\u01b0\u1eddng ch\u00e9o c\u1ee7a \u0111\u00e1y b\u1eb1ng \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a kh\u00fac g\u1ed7. T\u00ednh th\u1ec3 t\u00edch c\u1ee7a 4 t\u1ea5m b\u00eca g\u1ed7 \u0111\u01b0\u1ee3c x\u1ebb ra? Gi\u1ea3i : Ta chia \u0111\u00e1y c\u1ee7a kh\u00fac g\u1ed7 HHCN th\u00e0nh 2 tam gi\u00e1c c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng nhau. M\u1ed7i tam gi\u00e1c c\u00f3 m\u1ed9t c\u1ea1nh \u0111\u00e1y b\u1eb1ng \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a kh\u00fac g\u1ed7 v\u00e0 chi\u1ec1u cao c\u1ee7a tam gi\u00e1c \u1ee9ng v\u1edbi c\u1ea1nh \u0111\u00e1y \u0111\u00f3 b\u1eb1ng 0,6 : 2 = o,3 (m) Di\u1ec7n t\u00edch tam gi\u00e1c l\u00e0 : 0,6x0,3 = 0,09 (m2) 2 Di\u1ec7n t\u00edch c\u1ee7a kh\u00fac g\u1ed7 HHCN l\u00e0 : 0,09 x 2 = 0,18 (m2)","Th\u1ec3 t\u00edch kh\u1ed1i g\u1ed7 HHCN l\u00e0 : 0,18 x 5 = 0,9 (m3) Th\u1ec3 t\u00edch kh\u00fac g\u1ed7 h\u00ecnh tr\u1ee5 l\u00e0 : 0,3 x 0,3 x 3,14 x 5 = 1,413 (m3) Th\u1ec3 t\u00edch 4 t\u1ea5m \u0111\u01b0\u1ee3c x\u1ebb ra l\u00e0 : 1,413 \u2013 0,9 = 0,513 (m3) \u0110\u00e1p s\u1ed1 0,513 m3 B\u00e0i 13 : Di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n 1 c\u00e1i h\u1ed9p kh\u00f4ng c\u00f3 n\u1eafp h\u00ecnh l\u1eadp ph\u01b0\u01a1ng l\u00e0 500 cm2. T\u00ednh c\u1ea1nh c\u00e1i h\u1ed9p \u0111\u00f3. N\u1ebfu t\u0103ng c\u1ea1nh h\u1ed9p n\u00e0y l\u00ean 2 l\u1ea7n th\u00ec di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n t\u0103ng l\u00ean m\u1ea5y l\u1ea7n ? Gi\u1ea3i : Di\u1ec7n t\u00edch 1 m\u1eb7t l\u00e0 : 500 : 5 = 100 (cm2) V\u00ec 100 = 10 x 10 n\u00ean c\u1ea1nh HLP l\u00e0 10 cm : C\u1ea1nh h\u1ed9p khi t\u0103ng l\u00ean 2 l\u1ea7n l\u00e0 : 10 x 2 = 20 (cm) Di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n c\u1ee7a h\u1ed9p m\u1edbi l\u00e0 : (20 x 20) x 5 = 2000 (cm2) So v\u1edbi tr\u01b0\u1edbc di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n t\u0103ng s\u1ed1 l\u1ea7n l\u00e0 : 2000 : 500 = 4 (l\u1ea7n) \u0110\u00e1p s\u1ed1 4 l\u1ea7n. B\u00e0i 14 : T\u00ednh th\u1ec3 t\u00edch h\u00ecnh l\u1eadp ph\u01b0\u01a1ng bi\u1ebft di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n v\u00e0 di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh \u0111\u00f3 l\u00e0 128 cm2. Gi\u1ea3i : Hi\u1ec7u di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n v\u00e0 di\u1ec7n t\u00fach xung quanh b\u1eb1ng 2 l\u1ea7n di\u1ec7n t\u00edch \u0111\u00e1y. V\u1eady di\u1ec7n t\u00edch \u0111\u00e1y l\u00e0: 128 : 2 = 64 (cm2) V\u00ec 64 = 8 x 8 \uf0de c\u1ea1nh HLP l\u00e0 8 cm : Th\u1ec3 t\u00edch h\u00ecnh l\u1eadp ph\u01b0\u01a1ng l\u00e0 : 8 x 8 x 8 = 512 (cm3) \u0110\u00e1p s\u1ed1 512 cm3 4\/ B\u00e0i t\u1eadp v\u1ec1 nh\u00e0 : B\u00e0i 1 : M\u1ed9t HLP c\u00f3 di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n b\u1eb1ng 384 cm2. T\u00ednh di\u1ec7n t\u00edch xung quanh v\u00e0 th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh l\u1eadp ph\u01b0\u01a1ng \u0111\u00f3 . B\u00e0i 2 : M\u1ed9t c\u00e1i b\u1ec3 HHCN ch\u1ee9a 1500 l\u00edt n\u01b0\u1edbc th\u00ec \u0111\u1ea7y b\u1ec3, bi\u1ebft \u0111\u00e1y b\u1ec3 c\u00f3 chu vi 8 m, chi\u1ec1u d\u00e0i b\u1eb1ng 5\/3 chi\u1ec1u r\u1ed9ng. T\u00ednh chi\u1ec1u cao c\u1ee7a b\u1ec3? B\u00e0i 3 : Ng\u01b0\u1eddi ta \u0111\u00e0o m\u1ed9t c\u00e1i gi\u1ebfng h\u00ecnh tr\u1ee5 s\u00e2u 6 m c\u00f3 chu vi \u0111\u00e1y b\u1eb1ng 6,28 m, ph\u1ea7n \u0111\u1ea5t l\u1ea5y l\u00ean t\u1eeb gi\u1ebfng ng\u01b0\u1eddi ta \u0111em \u0111\u1eafp v\u00e0o m\u1ed9t c\u00e1i s\u00e2n h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chi\u1ec1u d\u00e0i 8 m, r\u1ed9ng 5 m. H\u1ecfi s\u00e2n \u0111\u01b0\u1ee3c \u0111\u1eafp th\u00eam 1 l\u1edbp \u0111\u1ea5t d\u00e0y bao nhi\u00eau? B\u00e0i 4 : Ph\u1ea3i x\u1ebfp bao nhi\u00eau h\u00ecnh l\u1eadp ph\u01b0\u01a1ng c\u1ea1nh 1 cm \u0111\u1ec3 \u0111\u01b0\u1ee3c 1 h\u00ecnh l\u1eadp ph\u01b0\u01a1ng c\u00f3 di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n l\u00e0 150 m2","B\u00e0i 5 : M\u1ed9t kh\u00fac g\u1ed7 h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt c\u00f3 k\u00edch th\u01b0\u1edbc : d\u00e0i 3 dm, r\u1ed9ng 2,5 dm, cao 2 dm \u0111\u01b0\u1ee3c s\u01a1n c\u1ea3 6 m\u1eb7t v\u00e0 \u0111em c\u1eaft th\u00e0nh c\u00e1c kh\u1ed1i h\u1ed9p nh\u1ecf c\u00f3 k\u00edch th\u01b0\u1edbc b\u1eb1ng d\u00e0i 3 cm, r\u1ed9ng 2,5 cm, cao 2 cm l\u00e0m \u0111\u1ed3 ch\u01a1i cho tr\u1ebb em. H\u1ecfi : C\u1eaft \u0111\u01b0\u1ee3c bao nhi\u00eau kh\u1ed1i h\u1ed9p nh\u1ecf (m\u1ea1ch c\u1eaft kh\u00f4ng \u0111\u00e1ng k\u1ec3). B\u00e0i 6 : Hai v\u1eadt th\u1ec3 c\u00f3 h\u00ecnh l\u1eadp ph\u01b0\u01a1ng v\u00e0 c\u00f9ng ch\u1ea5t li\u1ec7u nh\u01b0ng k\u00edch th\u01b0\u1edbc g\u1ea5p nhau 3 l\u1ea7n. T\u1ed5ng kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a 2 v\u1eadt th\u1ec3 l\u00e0 21 kg. T\u00ednh kh\u1ed1i l\u01b0\u1ee3ng m\u1ed7i v\u1eadt th\u1ec3 ."]