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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
die lynstukke is in dieselfde proporsie

As die lynstukke eweredig is, is die volgende ook waar

  1. C B A C = F E D F
  2. A C · F E = C B · D F
  3. A B B C = D E F E en B C A B = F E D E
  4. A B A C = D E D F en A C A B = D F D E

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

  • '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.
    gebied A B C = 1 2 · h · B C = gebied D B C
  • Driehoeke op die dieselfde kant van dieselfde basis, met gelyke gebiede, lê tussen parallelle lyne.
    As gebied ABC = gebied BDC,
    Dan AD BC.

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

Gegee : ABC met lyn DE BC

nodig om te bewys :

A D D B = A E E C

Bewys : Teken h 1 van E loodreg op AD, en h 2 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.

Sortgelyk,

A D A B = A E A C A B B D = A C C E

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 = 2ABAE = AC + CB = 2AC dan DE BC en BC = 2DE.

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

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

nodig om te bewys :

A B D E = A C D F

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
| | | bedoel “is soortgelyk aan"

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

Gegee : ABC met lyn DE sodanig dat

A D D B = A E E C

nodig om te bewys : D E B C ; ADE | | | ABC

Bewys : Teken h 1 vanaf E loodreg op AD, en h 2 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
ADE en ABC is gelykhoekig
A D E | | | A B C (AAA)

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 met A ^ = 90

Nodig om te bewys : B C 2 = A B 2 + A C 2

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
ABD | | | CBA en CAD | | | CBA ( AAA )
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 2 = C B × B D en A C 2 = C B × C D
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

In GHI, GH LJ; GJ LK en J K K I = 5 3 . Bepaal H J K I .

  1. L I ^ J = G I ^ H J L ^ I = H G ^ I ( ooreenstemmende . e ) L I J | | | G I H ( gelykhoekige e )
    L I ^ K = G I ^ J K L ^ I = J G ^ I ( Ooreenstemmende e ) L I K | | | G I J ( Gelykhoekige e )
  2. 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
  3. H J K I = H J J I × J I K I

    Ons moet die volgende bereken J I K I : Ons is gegee J K K I = 5 3 Na herrangskiking, het ons J K = 5 3 K I En:

    J I = J K + K I = 5 3 K I + K I = 8 3 K I J I K I = 8 3

    Met die gebruik van hierdie verhouding:

    = 5 3 × 8 3 = 40 9

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

  1. P 1 ^ = S 1 ^ ( Alt . s ) Q 1 ^ = R 1 ^ ( Alt . s ) P T Q | | | S T R ( gelykhoekige e )
  2. P T T Q = S T T R ( P T Q | | | S T R ) P T · T R = S T · T Q

Driehoekige meetkunde

  1. Bereken SV
  2. C B Y B = 3 2 . Vind D S S B .
  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.
  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.
  5. Vind FH in die volgende figuur.
  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 D E A C .
  7. As LM JK, bereken y .

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Source:  OpenStax, Siyavula textbooks: wiskunde (graad 11). OpenStax CNX. Sep 20, 2011 Download for free at http://cnx.org/content/col11339/1.4
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