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Lunard
@Lunard
May 2019
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Bonsoir,
Pourriez vous m'aider,pour ce problème ci dessous:
Il me faut répondre à la question c)
Merci d'avance
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Guitout
Verified answer
En utilisant le théorème de Thalès, on obtient la relation suivante :
OR/OR'=OE/OE'=RE/R'E'
Ensuite tu regardes les valeurs que tu as :
-RE
-OE'
-OE=EE'+E'O
Ce qui donne (EE'+E'O)/OE'=RE/R'E' <=> (EE'+E'O)/(OE'xRE)=1/R'E' <=> R'E'=RExOE'/(EE'+E'O)
R'E'=8x9/(15+9)=72/24=3cm
R'E'=3cm
3 votes
Thanks 0
taalbabachir
A. calculer la longueur de la flèche réduite R'E'
application du théorème de Thalès
OE'/OE = R'E'/RE ⇔ OE x R'E' = OE' x RE ⇒ R'E' =OE' x RE/OE
OE = OE' + E'E = 9 + 15 = 24 cm
R'E' =OE' x RE/OE = 9 x 8/24 = 3 cm
b. quel est le coefficient de réduction
OE'/OE = R'E'/RE = k
9/24 = 3/8 = k
k = 0.375
. OE''/OE = E''R''/ER ⇔ OE'' x RE = OE x E''R'' ⇒ OE'' = OE x E''R'' /RE
OE'' = 24 x 4/8 = 12 cm
4 votes
Thanks 3
lunard
merci mais il me faut la réponse du petit c
Guitout
Taalbabachir à déjà, il a juste oublier de mettre c. ^^ c'est la partie après k=0.375
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Lista de comentários
Verified answer
En utilisant le théorème de Thalès, on obtient la relation suivante :OR/OR'=OE/OE'=RE/R'E'
Ensuite tu regardes les valeurs que tu as :
-RE
-OE'
-OE=EE'+E'O
Ce qui donne (EE'+E'O)/OE'=RE/R'E' <=> (EE'+E'O)/(OE'xRE)=1/R'E' <=> R'E'=RExOE'/(EE'+E'O)
R'E'=8x9/(15+9)=72/24=3cm
R'E'=3cm
application du théorème de Thalès
OE'/OE = R'E'/RE ⇔ OE x R'E' = OE' x RE ⇒ R'E' =OE' x RE/OE
OE = OE' + E'E = 9 + 15 = 24 cm
R'E' =OE' x RE/OE = 9 x 8/24 = 3 cm
b. quel est le coefficient de réduction
OE'/OE = R'E'/RE = k
9/24 = 3/8 = k
k = 0.375
. OE''/OE = E''R''/ER ⇔ OE'' x RE = OE x E''R'' ⇒ OE'' = OE x E''R'' /RE
OE'' = 24 x 4/8 = 12 cm