IkfUvaHpC
JNGAYNAY
s o 3-2012
'iri'^ii.i.ii'.in.ii
17
Sir dung phifdng phap dien tfch
de ren luyen, phat trien nang lut suy luan va chuhg minh cho hpc sinh tieu hpc
L3(
u> luan va chung minh la nang luc ca ban
*cua nguoi hoc va lam toan, Trong d^y hpc 1 can, CO the hieu nang lire suy luan va chung minh la kha nang van dung cac phep suy luan de tien hanh chung minh mot menh de toan hoc hoac giai quyet mot van de nao do cua toan hoc, Ren luyen, phat trien nang lye suy luan va chung minh la mot nhiem vu quan trong trong cong tae day va hoe Toan a truong pho thong. Viec ren luyen, phat trien nang lire suy luan va chung minh se giup hgc sinh (HS) nhan thuc the gioi ,\ung quanh mot each khach quan, bien ehung. Co rat nhieu bien phap ren luyen, phat trien nang luc suy luan va ehung minh eho HS, tuy nhien, su dung phuong phap dien tich (PPDT) de giai cac bai toan Hinh hoc eo rat nhieu thuan Igi trong viec ren luyen, phat trien nang luc suy luan va chung minh eho HS Tieu hgc,
1. Khai niem vc phuong phap dien tich PPDT la mgt phuang phap giai toan, diing de giai cac bai toan ve dien tich ma khong su dung tr\rc tiep eong thuc ve tinh dien tich cae hinh. PPDT la ea sa cua viec giai bai tap toan ve eat ghep hinh.
2. Ren luyen, phat then nang lure suy luan va chirng minh cho HS qua vice sir dung PPDT de giai cac bai toan Hinh hoc & Tieu hoc
De ren luyen, phat trien nang luc suy luan va ehung minh eho HS Tieu hge thong qua vice giai cae bai toan Hinh hge bang PPDT, giao vien (GV) can chu trgng viec xay dung he thong bai tap nham ren luyen, phat trien nang luc suy luan va ehung minh cho HS tren ca sa eae tinh chat hinh hge sau: 1) Neu mgt hinh duge chia ra thanh cac hinh nho thi tong dien tich cua eae hinh nho bang dien tich ciia hinh Ion ban dau; 2) Neu ghep cae hinh nho lai de duge mgt hinh lan thi dien tich hinh Ion thu duge bdng tdng dien tich eua eae hinh nho dem ghep lai; 3) Neu hai hinh co dien tich bang nhau cung bot di mgt phan
TS. NGUYEN MANH CHUNG LE THI HOA
L A MTHI NGA
TrUdng Dgi hpc Hong DQc
chung thi hai phan con lai cung eo dipn tich bang nhau; 4) Neu ta them vao hai hinh co dien ti'eh bang nhau mgt phan chung thi ta nhan duge hai hinh co dien tich bang nhau; 5) Khi so do canh day khong doi thi dien tich va chieu cao ciia mgt tam giae la hai dai lugng ti le thuan; 6) Khi chieu cao khong doi thi so do dien tich va so do canh day la hai dai lugng ti le thuan; 7) Khi so do dien tich khong doi thi so do canh day va chieu cao eiia mgt tam giae la hai dai lugng ti le nghjch; 8) Neu hai tam giae co chieu dai canh day va duong eao bang nhau thi dien tich cua chung bang nhau; 9) Neu hai tam giae eo dien bang nhau va eo so do canh day bang nhau thi eo chieu eao bang nhau; 10) Neu hai tam giae co dien tich bang nhau va eo chieu eao cung bang nhau thi eo so do canh day bang nhau..,
Tren ea sdf nhung tinh chat tren, GV can ehu trgng viec xay dung he thong bai tap nham ren luyen, phat trien nang lire suy luan dien dich, suy luan tucmg tu, cdc phep chung minh trye tiep, gidn tiep cho HS.
3. Mot so vi du ve vice ren luyen, phat trien nang luc suy luan va chung minh cho HS qua vice sir dung PPDT de giai cac bai toan Hinh hoc if Tieu hoc
Bai viet nay chi tap trung vao van de khai thdc mot bdi todn vd^tim nhieu lai gidi cho mot bdi todn, de thay duge y nghTa quan trgng eua van de ren luyen, phat trien nang lire suy luan va chung minh cho HS.
3.1. Khai thdc mgt bdi todn Vi du 1: Xet bdi todn sau:
Bdi todn 1: Cho tam giae A13C, tren BC lay M sao cho BM = MC, N la di^m tren canh AC sao eho NC = 2xNA. Keo dai MN cdt BA tai P Chiing to rdng AP - AB.
18i
iijfe«a-i SO 3-2012 - _ — JNGAY NtDmvoHc
Bai giai:
Noi BN. i:\\ ta c6: S|.UM =" SMPC (Vi BM = MC v;\ chung cliicii cao ha tu P).
SBMN = SMNC (Vi BM = MC vii chun^ chieu cao ha tu N).
Do do SpuM • SHIMM '^ SMI'C • S MNC,
t u c la S PUN = S PNC ( I )
Lai cd S PNC " 2 X S AI'N (Vi NC = 2 x NA va cluing chieu cao ha tir P) (2)
Tu (1) va (2) co: 2 x S AI'N = S PUN = S AUN
Su\ ra AP = PB (Do 2 tam giae co chung chieu cao h^i tu N c6 dien tich bang nhau nen 2 da> bdng nhau).
Thay Joi \ i tri cua M. N la cd bai toan sau:
Bdi todn 2: Cho tam giae ABC co AB = 2cm, tren BC lay M sao cho BM =^ 3 x MC. N la diem tren AC sao cho AN = 2xNC. Keo dai MN cdt BA tai P
a) Tinh AP.
b) So sanh PN vdi NM, Huang dan giai:
Tuang tu" bai i ta chung minh duge 3x S PNC = S
b) Tinh ti so dg dai PN va PB Hudng dan giai:
Tuang ti,r bai I ta chung minh duge S BCM =
2XSMCA va S BPM ~ 2XSMPA-
SfPN ~ 3XSNPA ; SDPC = 3XSAPU
Tu dd, so sanh duge S MI, va Spoc va tinh ti so dp dai PN va PB,
Bai loan 5: Cho tam giae ABC, tren BC la> M sao cho PBN va dua vao cac tinh chat, dac diem cua 2 tam giae cd MC = 2 x BM, N la dicrn tren canh AC sao cho AN = 3xNC.
dien tich bang nhau de giai bai toan.
Thay doi vj tri cua M, N ta cd bai toan sau:
Keo dai MN cdt BA tai P. biet AB = 6cm,Tinh PB.
Bdi todn 6: Cho tam giae ABC. tren AB lay M Bai todn 3: Cho tam giae ABC, tren BC ISy M sao sao cho BM = 3xMA. N la diciii tren canh AC sao cho eho MC= 2xMB, N la diem tren AC sao cho AN
4xNC.Keo dai MN cdt BA tai P
a) So sanh S APM va SMPC
b) So sanh AB vdi PB.
Hudng dan giai:
Tuang tu" bai I ta chung minh duge S APM 4xSMPcvaAB = 8xPB.
Thay doi vj tri cua M, N,P ta cd bai toan sau:
Bai tudn -/: Cho tam giae ABC, tren AB lay M sao cho MB = 2 x AM, N la diem tren canh AC sao cho NC = 3 x AN. Keo dai BN cdt CM tai P
a) So sanh S ABC va Spuc- Giai:
AB = 3 X AE; AC = 3 X AD nen ta co:
^AHI) - ^ACE ~ T ^ /!«(• ( V ^ ^i:iiO ~ ^IXO ( ^ } '
Tu (1), (2), (3) va (4) suy ra S^^^^ =S^m-, tir dd suy ra duang cao xuat phat tu B den OA bang duong cao xuat phat tQ C
= NC = 2 x AN. K e o d a i N B c a t M C t a i P a) So sanh SAPB va S MV
b) Tinh ti s6 dp dai PM va PC.
Nhu vay, tu mgt bai loan chi ean iha> doi vi tri cac diem ta duge cac bai toan khac nhau nhung each _ giai lai lien quan den nhau, dieu na> giup HS cd kha ndng vgn dgng de khai thac mgi bai toan nhdm hieu sau sdc bai toan cung nhu ren luyen tu duy logic cho HS.
/'/ du 2: XL'I bdi loan dud duy.
Bai todn I: Cho tam giae ABC, tren cac canh AB, AC lan lugt ISy cac 6\km E, D sao cho AB = 3 x AE;
AC = 3 X AD. Cac doan thdng BD va CE cdt nhau tai O, AO cat BC tgi M. Chirng minh rdng BM = MC
dSn OA, do do S, MOH 'MOC MB = MC
DmjvaHoc
^NGAYNAY
SO 3-2012
Tu ket qua bai todn 1 ta cd:
<\ - <r ~ . ^^Ai:c -
4 3 ABC
Tu (5) va (6) ta cd
S,oo=S,,,,^OA = OM(7).
Tir ket qua nay, giu nguyen gia thiet bdi todn I vd thay k^t luan ciia bai toan 1 bdng kit qua (7) ta cd bdi todn 2 sau day:
Bdi todn 2: Cho tam gidc ABC, tren cdc cgnh AB, AC ldn lupt Idy cac dilm E, D sao cho AB = 3 x AE; AC = 3 X AD. Cac doan thdng BD va CE cat nhau t^i O, AO cdt BC tai M. Chirng minh rdng AO = DM.
Bd sung ket qua ciia bai toan 1 vao gia thiet cua nd vd thay ddi ket luan ta cd bai todn 3 nhu sau:
Bai loan 3: Cho tam giae ABC, tren cdc canh AB, AC, BC lan lupt Idy cac di6m E, D, M sao cho AB = 3 x AE; AC = 3 X AD, MB = MC. Chung minh rdng ba doan thang BD, CE va AM ciing cdt nhau tai 1 dilm.
Bai toan nay dupe chuyin ve bai toan 1 bdng each cho BD va CE cdt nhau tai O, AO cdt BC tai N, ta dl dang
Ldi giai cua bai toan nay nhu sau:
chung minh dupe M triing N do dd ba doan thdng AM, . BD vd CE ciing cdt nhau t^i O.
'7^^ABC \^) ^^T^je^c=Rlrq6a?!}(^t4vg them hai kit qua sau:
Bdi todn 4: Cho tam gi^c ABC, tren cdc canh AB, AC l^n lupt Idy cdc dilm E, D sao cho AB = 3 x AE; AC = 3 X AD. Cdc dogn thdng BD vd CE cdt nhau tai O, Tren AO k^o ddi l^y dilm M sao cho OA = OM. Chirng minh rdng ba dilm B, M, C thdng hdng,
Bdi todn 5: Cho tam giae ABC, tren cdc canh AB, AC ldn lupt ISy cdc dilm E, D sao cho AB = 3 x AE; AC = 3 X AD, Cac doan thdng BD vd CE cdt nhau^tai O, Gpi M Id dilm tren BC sao cho MB = MC, Chirng minh rdng ba dilm A, O, M thdng hdng,
Vi?c gidi bdi todn 4 va 5 hodn todn tuang tu nhu each giai bdi todn 3,
Coi kit qud bai todn 1 la gia thilt vd ddo gia thilt ciia nd thdnh kit luan ta cd bdi todn ddo cua bai toan I nhu sau:
Bdi todn 6: Cho tam giae ABC, gpi M la trung dilm cua BC, O la trung diem cua AM. BO keo dai cdt AC tai D, CO k^o dai cdt AB tai E.
Chung minh rdng: AB = 3 x AE; AC = 3 x AD,
Vi OA = OM nen S^^^ = S^,^^„; 5^^^. = S^^^
0)
Vi BM = MC nen 5^„,,, = S,,^^ (2)
Tir (1) vd (2) suy ra iS^oa ~ ~^'^BOC • ^^ ^d suy ra dudng ^ cao xuat phat tir C den BD bdng 2 ldn dudng cao xudt phdt tir A din BD, do dd: S^.y -2xS BIX' 'BDA CD^lxAD.
Tuong tir: S(.,,^=2xS^i.y,^>BE = 2xAE-
AC^ADx3
>AB = AEx3.
Tu nhung ket qua tren, neu thay doi ti so d gid thiet eua bai toan 1 ta se ed nhifng ket qua tuang tu kha thu vi.
Nhu vay, tir mgt bai toan chi can thay ddi mgt so diem trong gia thiet hoac ket luan cua bai toan goc ta duge cac bai toan khde nhau nhung each giai lai lien quan den nhau, dieu nay giup HS ed kha nang van dung cac phep suy luan dien dich, suy luan tuang tu de khai thac mgt bai toan nhdm hieu sau sac bai toan cung nhu ren luyen tu duy logic, tu duy sang tao cho HS.
3.2. Tim cdc cdch gidi khac nhau ciia mgt bdi todn Xet bdi todn: Cho hinh thang vuong A BCD vuong gdc d A va D, day be AB = 12cm, day lan CD = 16cm, canh dai AD = 8cm, M la mot diem tren AD each D la
2cin. Tu M ke dudng thdng song song vdi day hinh thang cat BC tai N. Tinh dien tich tu giae MNCD, biet MN vuong gdc vdi AD.
* Phan tich: Do MN song song vai AB va CD nen tur giae MNCD la hinh thang do do de tinh di?n tich MNCD ta cd 2 hudng:
Huang thic nhdt: tinh dp dai MN.
Hudng thu 2: dien tich MNCD bang tong dien tich 2 tam giae.
* Hudng thir nhat ta cd cae each giai sau:
Cdchl:
VI MNCD la hinh thang nen dien tich tam giae NCD la: 16x2:2 = 16(cm ) Vi ABMN la hinh thang nen dien tfch tam giae ABN la: 12x (8-2):2 = 36(em^)
Dien tich hinh thang ABCD la: (12+16) x8:2 = 112 (cm^) Dien tich tam giae AND la: 12 - (16+36) = 60 (cm^)
20
l».', so 3-2012Chieu cao MN ciia tam gidc ADN la: 60x2:8 = 15(cm) DivMi tich hinh thang MNCD la: (15+16) x2:2 = 31 (cm^) Cdch 2:
Tir B ke dudng thdng song song vdi AD cdt MN tsii 11 va eat DC tai E. tu N ke Nl vuong goc vdi DC. Noi N vai i;,
Tacd:DE = Mll = A B - 12em.
BE = AD = 8cm. E C - 16-12 = 4cm; NI = MD = 2cm.
SBI:C = ECx BE:2 = 4x8:2 = 16 (cm^)
SNI;C- - IX^- 1N:2 = 4x2:2 = 4(cm^) SuNii=16-4=12(cm')
Chieu cao NH cua tam giae BNI: Id: 12x2:8 = 3cm.
Do dai MNIa: 12 + 3= 15 cm
Dien tich hinh thang MNCD la: (15+16) x 2:2 = 31 (cm^) Cdch 3:
Dien tich hinh thang ABCD la: (12 + 16) x 8:2 = 112 (cm^) Gpi dp dai MN la a, ta cd:
* Hudng thu hai ta cd cae each giai sau:
Cdch 4:
Tinh nhu each 1 ta cd: SADN = 60cm^
1 y sd eua MD va AD la: 2:8 = 1/4 nen:
SMND= 1/4 SADN => SMND= 1/4,60 = 15 cm^
Dien tich hinh thang MNCD la: 15+16 - 31 (cm^) Cdch 5:
Dien tich tam giae MDC la: 16x2:2 = 16 (em*)
Dien tich hinh thang ABCD la: (12 +16) x 8:2 = 112 (em^) Dien tich tam gidc BMC la: 112 - (16 + 36) = 60 (cm^) AM = BH = 8 - 2 - 6 c m ;
MD = CE = 2em.
A B
D
H Kf
E ]
y\
[ ( ; •SMNCD = (a+16) X 2:2 = a+16 (cm')
SAnNM = (12+a) X (8-2):2 = 3x (12+a) (cm^)
Md SABCD ~ SMNCD + SABNM
nena+I6+3x(I2+a)= 112=>a= 15(cm)
Dipn tich hinh thang MNCD Id: 15+16 = 31 (cm")
BH gap CE so lan la:
6:2= 3 (lan) nen SBMN =3XSNMC do do: SBMC=4XSNMC
Dien tich tam gidc MDC la: 60:4 = 15(cm^) Dien tich hinh thang MNCD la: 15+16 = 31 (cm^) Vdri 2 hudng suy nghT ma cd tai 5 each giai kliac nhau, mdi each giai deu van dung cac kien thue da hgc mgt each linh hoat. Viec giup HS tim ra cae hudng giai mgt bai toan cung nhu tim ra nhieu each giai cho 1 bai toan la rat quan trgng, giup HS ren luyen tu duy logic, sir sang tao.
Trong day hgc Toan d Tieu hge, viec ren luyen, phat triln nang luc suy luan va chung minh cho HS la mgt muc tieu quan trgng, de dat duge dieu dd, GV cdn c6 phuang phap hgp ly va su dau tu ky ludng, nhdt la trong viec bdi duong HS gidi.
Tren day, bai viet dd gioi thi^u mpt so vi du ve ren liiy?n, phat trien ndng \\xc suy lugn va chirng minh cho HS Tieu hgc thong qua giai cac bai toan Hinh hge bdng PPDT. Qua dd, cho thdy GV can quan tam tdi viec boi dudng ndng lyre suy luan va ehumg minh cho HS Tieu hge thi PPDT la mgt cong cy kha hOru hieu dl dat duge myc tieu nay.
Tai iifu tham khao:
1. Tran Dien Hien, Bdi dudng hoc sinh gidi todn Tiiu hgc, NXB Dai hge Su pham, H. 2009.
2. Do Trung Hipu, D6 Dinh Hoan, Vu Duong Thyy, Vu Qudc Chung, Phuang phdp day hoc mon Todn a Tiiu hoc. NXB Dai hgc Su pham, H. 2009.
3. Sdeh gido khoa Todn 1, 2, 3, 4, 5, NXB Giao dye, H. 2009.
