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Magie78
@Magie78
May 2019
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Bonjour j’aurais besoin d’aide pour l’exercice 1 de ce devoir maison svp c’est niveau 3ème.
Merci d’avance
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taalbabachir
(FG) ⊥ (DF) et (AB) ⊥ (D⊥F) ⇒ (FG) // (AB)
⇒ application du théorème de Thalès
EG/EA = FG/AB ⇒ AB = EA x FG/EG = 9 x 3/6 = 4.5 cm
⇒ tan 30° = DB/AB ⇒ DB = AB x tan 30° = 4.5 x 0.57735 = 2.598076... 588 cm
⇒ BE² = AE² - AB² = 9² - 4.5² = 81 - 20.25 = 60.75
BE = √60.75 = 7.7942286...764 cm
l'aire du quadrilatère croisé ADFG est :
A = 1/2(2.598076... x 4.5) + 1/2(4.5 x 7.7942286...) + 1/2( 3 x 3√3)
= 1/2(11.6913432... + 35.0740287... + 15.5884572...)
= 31.17691455... cm²
1 votes
Thanks 1
magie78
Je n’ai pas compris ce que signifie « 1/2 » sur le calcul de l’aire
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⇒ application du théorème de Thalès
EG/EA = FG/AB ⇒ AB = EA x FG/EG = 9 x 3/6 = 4.5 cm
⇒ tan 30° = DB/AB ⇒ DB = AB x tan 30° = 4.5 x 0.57735 = 2.598076... 588 cm
⇒ BE² = AE² - AB² = 9² - 4.5² = 81 - 20.25 = 60.75
BE = √60.75 = 7.7942286...764 cm
l'aire du quadrilatère croisé ADFG est :
A = 1/2(2.598076... x 4.5) + 1/2(4.5 x 7.7942286...) + 1/2( 3 x 3√3)
= 1/2(11.6913432... + 35.0740287... + 15.5884572...)
= 31.17691455... cm²