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Driehoek Meetkunde

Proporsie

Twee lynstukke is verdeel in die dieselfde proporsie as die verhoudings tussen die dele gelyk is.

A B B C = x y = k x k y = D E E F A B B C = x y = k x k y = D E E F
(1)
die lynstukke is in dieselfde proporsie die lynstukke is in dieselfde proporsie
(2)

Figuur 1
Figuur 1 (MG11C16_009.png)

As die lynstukke eweredig is, is die volgende ook waar

  1. C B A C = F E D F C B A C = F E D F
  2. A C · F E = C B · D F A C · F E = C B · D F
  3. ABBC=DEFEABBC=DEFE en BCAB=FEDEBCAB=FEDE
  4. ABAC=DEDFABAC=DEDF en ACAB=DFDEACAB=DFDE

Evenredigheid van Driehoeke

Driehoeke met gelyke hoogtes het gebiede wat in dieselfde proporsie is tot mekaar as die basisse van die driehoeke.

h 1 = h 2 gebied A B C gebied D E F = 1 2 B C × h 1 1 2 E F × h 2 = B C E F h 1 = h 2 gebied A B C gebied D E F = 1 2 B C × h 1 1 2 E F × h 2 = B C E F
(3)

Figuur 2
Figuur 2 (MG11C16_010.png)

  • 'n Spesiale geval van hierdie gebeur wanneer die basisse van die driehoeke gelyk is: Die Driehoeke met gelyke basisse tussen dieselfde ewewydige lyne het dieselfde gebied.
    gebiedABC=12·h·BC=gebiedDBCgebiedABC=12·h·BC=gebiedDBC
    (4)
    Figuur 3
    Figuur 3 (MG11C16_011.png)
  • Driehoeke op die dieselfde kant van dieselfde basis, met gelyke gebiede, lê tussen parallelle lyne.
    AsgebiedABC=gebiedBDC,AsgebiedABC=gebiedBDC,
    (5)
    DanADBC.DanADBC.
    (6)
    Figuur 4
    Figuur 4 (MG11C16_012.png)

Stelling 1 Proporsie Stelling: 'n streep parallel aan die een kant van' n driehoek verdeel die ander twee sye eweredig.

Figuur 5
Figuur 5 (MG11C16_013.png)

Gegee:ABC met lyn DE BC

nodig om te bewys:

A D D B = A E E C A D D B = A E E C
(7)

Bewys: Tekenh1h1 van E loodreg op AD, enh2h2 van D loodreg op AE.

Teken BE en CD.

gebied ADE gebied BDE = 1 2 A D · h 1 1 2 D B · h 1 = A D D B gebied ADE gebied CED = 1 2 A E · h 2 1 2 E C · h 2 = A E E C maar gebied BDE = gebied CED (gelyk aan basis en hoogte) gebied ADE gebied BDE = gebied ADE gebied CED A D D B = A E E C DE deel AB en AC proporsioneel. gebied ADE gebied BDE = 1 2 A D · h 1 1 2 D B · h 1 = A D D B gebied ADE gebied CED = 1 2 A E · h 2 1 2 E C · h 2 = A E E C maar gebied BDE = gebied CED (gelyk aan basis en hoogte) gebied ADE gebied BDE = gebied ADE gebied CED A D D B = A E E C DE deel AB en AC proporsioneel.
(8)

Sortgelyk,

A D A B = A E A C A B B D = A C C E A D A B = A E A C A B B D = A C C E
(9)

Na aanleiding van die stelling "Proporsie", kan ons die middelpunt stelling bewys.

Stelling 2 Middelpuntstelling:'n lyn wat die middelpunte van die twee kante van 'n driehoek aansluit is parallel aan die derde sy en gelyk aan die helfte van die lengte van die derde kant.

Bewys: Dit is 'n spesiale geval van die Proporsie Stelling (Stelling "Proporsie"). As AB = BD en AC = AE, en AD = AB + BD = 2AB AE = AC + CB = 2AC dan DE BC en BC = 2DE.

Figuur 6
Figuur 6 (MG11C16_014.png)

Stelling 3 Gelykvormigheid Stelling 1: Gelykhoekig driehoeke het hul sye in verhouding en is dus soortgelyk.

Figuur 7
Figuur 7 (MG11C16_015.png)

Gegee:ABC en DEF met A^=D^A^=D^; B^=E^B^=E^; C^=F^C^=F^

nodig om te bewys:

A B D E = A C D F A B D E = A C D F
(10)

Konstrueer: G op AB, so dat AG = DE, H op AC, so dat AH = DF

Bewys: In 's AGH en DEF

AG = DE (Konst.) AH = D ( Konst. ) A ^ = D ^ ( gegee ) AGH DEF ( SHS ) A G ^ H = E ^ = B ^ G H BC ( ooreenstemmende 's equal ) AG AB = A H A C ( proporsie stelling ) DE AB = D F A C ( AG = DE ; AH = DF ) ABC | | | DEF AG = DE (Konst.) AH = D ( Konst. ) A ^ = D ^ ( gegee ) AGH DEF ( SHS ) A G ^ H = E ^ = B ^ G H BC ( ooreenstemmende 's equal ) AG AB = A H A C ( proporsie stelling ) DE AB = D F A C ( AG = DE ; AH = DF ) ABC | | | DEF
(11)
leidraad:
|||||| bedoel “is soortgelyk aan"

Stelling 4 Gelykvormigheid Stelling 2: Driehoeke met sye in verhouding is gelykhoekig en daarom soortgelyke.

Figuur 8
Figuur 8 (MG11C16_016.png)

Gegee:ABC met lyn DE sodanig dat

A D D B = A E E C A D D B = A E E C
(12)

nodig om te bewys:DEBCDEBC; ADE ||||||ABC

Bewys: Teken h1h1 vanaf E loodreg op AD, en h2h2 vanaf D loodreg op AE.

Teken BE en CD.

gebied ADE gebied BDE = 1 2 A D · h 1 1 2 D B · h 1 = A D D B gebied ADE gebied CED = 1 2 A E · h 2 1 2 E C · h 2 = A E E C maar A D D B = A E E C (gegee) gebied ADE gebied BDE = gebied ADE gebied CED gebied BDE = gebied CED D E B C (dieselfde kant met gelyke basis DE, dieselfde gebied) A D ^ E = A B ^ C (ooreenstemmende 's) en A E ^ D = A C ^ B gebied ADE gebied BDE = 1 2 A D · h 1 1 2 D B · h 1 = A D D B gebied ADE gebied CED = 1 2 A E · h 2 1 2 E C · h 2 = A E E C maar A D D B = A E E C (gegee) gebied ADE gebied BDE = gebied ADE gebied CED gebied BDE = gebied CED D E B C (dieselfde kant met gelyke basis DE, dieselfde gebied) A D ^ E = A B ^ C (ooreenstemmende 's) en A E ^ D = A C ^ B
(13)
ADE en ABC is gelykhoekig ADE en ABC is gelykhoekig
(14)
A D E | | | A B C (AAA) A D E | | | A B C (AAA)
(15)

Stelling 5 Pythagoras se Stelling: Tdie vierkant op die skuinssy van 'n reghoekige driehoek is gelyk aan die som van die vierkante van die ander twee kante

Gegee: ABC metA^=90A^=90

Figuur 9
Figuur 9 (MG11C16_017.png)

Nodig om te bewys:BC2=AB2+AC2BC2=AB2+AC2

Proef:

Laat C ^ = x D A ^ C = 90 - x ( 'e van 'n ) D A ^ B = x A B ^ D = 90 - x ( 'e van 'n ) B D ^ A = C D ^ A = A ^ = 90 Laat C ^ = x D A ^ C = 90 - x ( 'e van 'n ) D A ^ B = x A B ^ D = 90 - x ( 'e van 'n ) B D ^ A = C D ^ A = A ^ = 90
(16)
ABD | | | CBA en CAD | | | CBA ( AAA ) ABD | | | CBA en CAD | | | CBA ( AAA )
(17)
A B C B = B D B A = A D C A en C A C B = C D C A = A D B A A B C B = B D B A = A D C A en C A C B = C D C A = A D B A
(18)
A B 2 = C B × B D en A C 2 = C B × C D A B 2 = C B × B D en A C 2 = C B × C D
(19)
A B 2 + A C 2 = C B ( B D + C D ) = C B ( C B ) = C B 2 i . e . B C 2 = A B 2 + A C 2 A B 2 + A C 2 = C B ( B D + C D ) = C B ( C B ) = C B 2 i . e . B C 2 = A B 2 + A C 2
(20)
Exercise 1: Driehoek Meetkunde 1

In GHI, GH LJ; GJ LK en JKKIJKKI = 5353. Bepaal HJKIHJKI.

Figuur 10
Figuur 10 (MG11C16_018.png)

Solution
  1. Stap 1. Identifiseer gelykvormige driehoeke :
    L I ^ J = G I ^ H J L ^ I = H G ^ I ( ooreenstemmende . e ) L I J | | | G I H ( gelykhoekige e ) L I ^ J = G I ^ H J L ^ I = H G ^ I ( ooreenstemmende . e ) L I J | | | G I H ( gelykhoekige e )
    (21)
    L I ^ K = G I ^ J K L ^ I = J G ^ I ( Ooreenstemmende e ) L I K | | | G I J ( Gelykhoekige e ) L I ^ K = G I ^ J K L ^ I = J G ^ I ( Ooreenstemmende e ) L I K | | | G I J ( Gelykhoekige e )
    (22)
  2. Stap 2. Use proportional sides :
    H J J I = G L L I ( L I J | | | G I H ) en G L L I = J K K I ( L I K | | | G I J ) = 5 3 H J J I = 5 3 H J J I = G L L I ( L I J | | | G I H ) en G L L I = J K K I ( L I K | | | G I J ) = 5 3 H J J I = 5 3
    (23)
  3. Stap 3. Rearrange to find the required ratio :
    H J K I = H J J I × J I K I H J K I = H J J I × J I K I
    (24)

    Ons moet die volgende bereken JIKIJIKI: Ons is gegee JKKI=53JKKI=53 Na herrangskiking, het ons JK=53KIJK=53KI En:

    J I = J K + K I = 5 3 K I + K I = 8 3 K I J I K I = 8 3 J I = J K + K I = 5 3 K I + K I = 8 3 K I J I K I = 8 3
    (25)

    Met die gebruik van hierdie verhouding:

    = 5 3 × 8 3 = 40 9 = 5 3 × 8 3 = 40 9
    (26)
Exercise 2: Triangle Geometry 2

PQRS is 'n trapesium, met PQ RS. Bewys dat PT ·· TR = ST ·· TQ.

Figuur 11
Figuur 11 (MG11C16_019.png)

Solution
  1. Stap 1. Identify similar triangles :
    P 1 ^ = S 1 ^ ( Alt . s ) Q 1 ^ = R 1 ^ ( Alt . s ) P T Q | | | S T R ( gelykhoekige e ) P 1 ^ = S 1 ^ ( Alt . s ) Q 1 ^ = R 1 ^ ( Alt . s ) P T Q | | | S T R ( gelykhoekige e )
    (27)
  2. Stap 2. Use proportional sides :
    P T T Q = S T T R ( P T Q | | | S T R ) P T · T R = S T · T Q P T T Q = S T T R ( P T Q | | | S T R ) P T · T R = S T · T Q
    (28)
Driehoekige Meetkunde
  1. Bereken SV
    Figuur 12
    Figuur 12 (MG11C16_020.png)
  2. CBYB=32CBYB=32. Vind DSSBDSSB.
    Figuur 13
    Figuur 13 (MG11C16_021.png)
  3. Gegee die volgende figuur met die volgende lengtes, vind AE, EC en BE. BC = 15 cm, AB = 4 cm, CD = 18 cm, en ED = 9 cm.
    Figuur 14
    Figuur 14 (MG11C16_022.png)
  4. Met behulp van die volgende figuur en lengtes, vind IJ en KJ. HI = 26 m, KL = 13 m, JL = 9 m en HJ = 32 m.
    Figuur 15
    Figuur 15 (MG11C16_023.png)
  5. Vind FH in die volgende figuur.
    Figuur 16
    Figuur 16 (MG11C16_024.png)
  6. BF = 25 m, AB = 13 m, AD = 9 m, DF = 18m. Bereken die lengtes van BC, CF, CD, CE en EF, en vind die verhouding DEACDEAC.
    Figuur 17
    Figuur 17 (MG11C16_025.png)
  7. As LM JK, bereken yy.
    Figuur 18
    Figuur 18 (MG11C16_026.png)

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