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licorne38
@licorne38
January 2021
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Bonjour, c'est l'exercice 53 qui est aussi dans le livre Hyperbole de seconde, je ni arrive pas non plus. merci
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MonsieurFirdown
Verified answer
Bonsoir
53
♧ 1/ À toi de faire ...
♧ 2/
♤ a.
V = 1 /3 × 2^2 × h
V= 4/3 × h
V = 8 cm^3
D'où h = 6 cm
♤ b.
♤ On sait que AEI est un triangle rectangle on a donc :
♤ D'après le théorème de Pythagore :
AE^2 = AI^2 + EI^2
D'où
11 = AI^2 + h^2
● (" AI = 1 /2 AC d'où AC^2 = AB^2 + BC^2 = 8 ")
11 = 8/ 4 + h^2
11 = 2 + h^2
D'où
h^2 = 11 - 2 = 9 d'où h = √9 = 3 cm
Voilà ^^
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licorne38
January 2021 | 0 Respostas
Bonjour, j'ai cette exercice a faire, c'est le 65 page 268 du livre Hyperbole de 2de. Je ni arrive pas du tout... Merci
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Bonsoir53
♧ 1/ À toi de faire ...
♧ 2/
♤ a.
V = 1 /3 × 2^2 × h
V= 4/3 × h
V = 8 cm^3
D'où h = 6 cm
♤ b.
♤ On sait que AEI est un triangle rectangle on a donc :
♤ D'après le théorème de Pythagore :
AE^2 = AI^2 + EI^2
D'où
11 = AI^2 + h^2
● (" AI = 1 /2 AC d'où AC^2 = AB^2 + BC^2 = 8 ")
11 = 8/ 4 + h^2
11 = 2 + h^2
D'où
h^2 = 11 - 2 = 9 d'où h = √9 = 3 cm
Voilà ^^