Donnes-AD-1-DB-6-BCD-45-BAC-90-Determiner-1-AC-2-AE-Solution-BAC-BAC-90-BC-2-AB-2-AC-2-CD-cote-commun-aux-BAC-et-BDC-DB-2-CD-2-BC-2-2BC-CDcos-45-1-

Question Number 179360 by a.lgnaoui last updated on 28/Oct/22

D o n n e s : AD = 1; DB = 6; ∡ B C D = 45 °; ∡ B A C = 90 °

D e t e r m i n e r

1) A C ?

2) A E ?

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S o l u t i o n

△ B A C ∡ B A C = 90 °

{B C}^{2} = {A B}^{2} + {A C}^{2}

C D (c o t e c o m m u n a u x △ B A C e t B D C)

{D B}^{2} = {C D}^{2} + {B C}^{2} - 2 B C \times C D \cos 45 ° (1)

△ C A D {C D}^{2} = {A D}^{2} + {A C}^{2}

(1) {D B}^{2} = ({A D}^{2} + {A C}^{2}) + ({A B}^{2} + {A C}^{2}) - 2 \sqrt{({A B}^{2} + {A C}^{2}) ({A D}^{2} + {A C}^{2})} \cos 45 °

{D B}^{2} = ({A D}^{2} + {A B}^{2} + 2 {A C}^{2} - \sqrt{2 ({A B}^{2} + {A C}^{2}) ({A D}^{2} + {A C}^{2})}

2 ({A B}^{2} + {A C}^{2}) ({A D}^{2} + {A C}^{2}) = {({A D}^{2} + {A B}^{2} + 2 {A C}^{2} - {D B}^{2})}^{2}

p o s o n s : A C = x

2 (49 + x^{2}) (1 + x^{2}) = {(1 + 49 + 2 x^{2} - 36)}^{2}

(x^{4} + 50 x^{2} + 49) = 2 (x^{4} + 14 x^{2} + 49)

x^{4} - 22 x^{2} + 49 = 0

x^{2} = 11 \pm 6 \sqrt{2} x = \sqrt{11 + 6 \sqrt{2}}

A C = 4, 4142135

2) A B e t A C c o u p e n t l e c e r c l e e n (D, B) e t (E, C)

N o u s a v o n s A D \times A B = A E \times A C

A E = \frac{A D \times A B}{A C} = \frac{7}{\sqrt{11 + 6 \sqrt{2}}} = \frac{7 \sqrt{11 + 6 \sqrt{2}}}{11 + 6 \sqrt{2}}

A E = 1, 585786

Answered by a.lgnaoui last updated on 28/Oct/22