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MULO-A Meetkunde 1959 Algemeen

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Uitwerkingen MULO-A meetkunde 1959 Algemeen

Opgave 1

a. o o In geldt: 90 In geldt: 90 ABD BAD B BAD MAP C ABC C B                   1 2 1 2 AC AB AM AC AM AB      

Verder geldt AMP BAC90o

De driehoeken ABC en MPA hebben dus twee hoeken en een zijde gelijk en zijn dus volgen het congruentiekenmerk zhz congruent. b. 1 1 2 4 In is middenparallel ( ) ABC ME ME AC AB MP AB ABC MPA            3 4 EPAB, dus 1 3 1 3 4 4 4 4 : : : 1: 3 ME EPAB AB  . c. : 1: 3 3 : 2 : 3. : 1: 2 2 ME EP EP ME AC EP ME AC AC ME           o

,immers (90 ) en (Z-hoeken, verwisselbare binnenhoeken)

ACD PED   D DCAD EPD

 

dus CD DE: AC PE: 2 : 3.

Opgave 2

De constructie kan als volgt verlopen:

1. Teken lijn m en neem daarop het willekeurige punt E.

2. Richt in E een loodlijn op en pas DE af. 3. Cirkel vanuit D het lijnstuk BD om. 4. Teken AE DE .

5. Construeer EBC60o.

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Opgave 3

// MC r MC AB MP AB AP PB r AB         

 . Omdat ACPBCP (zhz) geldt CA CB . a. In CDA geldt 2 2 2 2 2

36 100 64 8

CDADACCD   CD  CD . b. BD BC CD  10 8 2. 

Volgens de machtsstelling geldt 16 2 6 3

6 8 2 2

AD DE CD BD    DE   DE  .

c. 2 o o

6

tanDAB  DAB18, 4349482  PCB18, 4349482 

FCD90o18, 4349482o 71,56505118o .

In CDF geldt cos FCD CD cos 71,56505118o 8 CF 25, 29822128 25

CF CF        . d. Stel 2 3 8 . EF  x AF  x

Volgens de machtsstelling geldt: 2 2 2 3 ( 8 ) 25, 29822128 EF FA FC  x x   2 2 2 3 8 640 0 3 26 1920 0 xx   xx   2 2 3 26 26 4 3 1920 26 154 21 6 6 FE         .

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