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Lucasvdd
@Lucasvdd
May 2019
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Bonjour je n’arrive pas à l’ex 29 en maths je suis en 4 eme et c’est pour un dm merci de me répondre
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taalbabachir
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Le volume de la pyramide est : surface de sa base x hauteur
la surface de la base est triangulaire DJC
il faut calculer JC, on applique le théorème de Pythagore
DC² = DJ² + JC² ⇒ JC² = DC² - DJ² = 6² - 3² = 36 + 9 = 45 ⇒ JC = √45 = 3√5
v = aire de la base x h = 1/2) x 3 x 3√5 x 6 = 27√5 = 60.4 cm³
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Verified answer
Le volume de la pyramide est : surface de sa base x hauteurla surface de la base est triangulaire DJC
il faut calculer JC, on applique le théorème de Pythagore
DC² = DJ² + JC² ⇒ JC² = DC² - DJ² = 6² - 3² = 36 + 9 = 45 ⇒ JC = √45 = 3√5
v = aire de la base x h = 1/2) x 3 x 3√5 x 6 = 27√5 = 60.4 cm³