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georgino06
@georgino06
June 2021
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Scolaire pour l'exercice 4 s'il vous plaît
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mohayati2006
1) ACE triangle rectangle en C
Pythagore AE²=AC²+EC²
EC²=AE²-AC² avec AE=6√5 et AC=2√5
donc EC²=(6√5)²-(2√5)²
EC²=36x5-4x5
=5(36-4)
=5x2(18-2)
EC²=10x16
donc EC=√(10x16)=4√10
2) ECB rectangle Pythagore EB²=EC²+CB²
EB²=160+100x5
EB²=660
EB=√660
3) P(AEB)=6√5+√660+2√5+10√5
4) AIR(AEB)=ECxAB/2
= 4√10 x(2√5+10√5)/2
=2√10 x12√5
= 24√50
4) AEB est rectangle AB²=AE²+EB²
(12√5)²=(6√5)²+660
144x5=36x5+660
720=180+660
720=840
donc AB²≠AE²+EB² donc AEB n'est pas rectangle
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Pythagore AE²=AC²+EC²
EC²=AE²-AC² avec AE=6√5 et AC=2√5
donc EC²=(6√5)²-(2√5)²
EC²=36x5-4x5
=5(36-4)
=5x2(18-2)
EC²=10x16
donc EC=√(10x16)=4√10
2) ECB rectangle Pythagore EB²=EC²+CB²
EB²=160+100x5
EB²=660
EB=√660
3) P(AEB)=6√5+√660+2√5+10√5
4) AIR(AEB)=ECxAB/2
= 4√10 x(2√5+10√5)/2
=2√10 x12√5
= 24√50
4) AEB est rectangle AB²=AE²+EB²
(12√5)²=(6√5)²+660
144x5=36x5+660
720=180+660
720=840
donc AB²≠AE²+EB² donc AEB n'est pas rectangle