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cloe3112
@cloe3112
January 2021
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s’il vous plaît pouvez vous m’aider
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taalbabachir
1) AH² = AC² - HC² = 6² - 3² = 36 - 9 = 27
⇒ AH = √27 = √3 x 9 = 3√3
2) les triangles BSC et BAC sont égaux donc leurs hauteurs sont forcément égales
⇒ SH = AH = 3√3 ⇒ donc le triangle SAH est isocèle en H
3) sachant que OH = 1/3) x AH = 1/3) x 3√3 = √3
démontrer que SO = 2√6 cm
SO² = SH² - OH² = (3√3)² - (√3)² = 9 x 3 - 3 = 27 - 3 = 24
donc SO = √24 = √6 x 4 = 2√6 cm
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⇒ AH = √27 = √3 x 9 = 3√3
2) les triangles BSC et BAC sont égaux donc leurs hauteurs sont forcément égales
⇒ SH = AH = 3√3 ⇒ donc le triangle SAH est isocèle en H
3) sachant que OH = 1/3) x AH = 1/3) x 3√3 = √3
démontrer que SO = 2√6 cm
SO² = SH² - OH² = (3√3)² - (√3)² = 9 x 3 - 3 = 27 - 3 = 24
donc SO = √24 = √6 x 4 = 2√6 cm