Question-191854

Question Number 191854 by Rupesh123 last updated on 01/May/23

Answered by a.lgnaoui last updated on 02/May/23

l image comporte 3 principale S s surfaces

∙ S 1 : S urface du quart ceecle C 1 (rayon r_{1} = 2)

∙ S 2 : S urface petit cercleC 2 (ryon : r_{2})

∙ S 3 : Surface portion cercle (rayon r_{3})

a vec r_{3} = AB = BC

S_{1} = \frac{π \times 4}{4} = π;

OF = OE + EF = OE + r_{2} \Rightarrow 2 = OE + r_{2}

{OE}^{2} = 2 r_{2}^{2} \Rightarrow OE = r_{2} \sqrt{2}

\Rightarrow r_{2} = 2 \sqrt{2} - 2

S_{2} = π r_{2}^{2} = 4 π {(\sqrt{2} - 1)}^{2}

\frac{S_{1} - S_{2}}{2} = A_{1} + \frac{A_{2}}{2}

Calcul A_{2}

A_{2} = {OH}^{2} - \frac{S_{2}}{4} = r_{2}^{2} - \frac{π r_{2}^{2}}{4} = \frac{3}{4} π r_{2}^{2}

donc A_{1} + A_{2} = \frac{S_{1} - S_{2}}{2} + \frac{3}{8} π r_{2}^{2}

= \frac{π - 4 π {(\sqrt{2} - 1)}^{2}}{2} + \frac{3}{8} π \times 4 {(\sqrt{2} - 1)}^{2}

A_{1} + A_{2} = (\sqrt{2} - 1) π

∙ calcul A_{3}

S urface Porti o n arc de cercle OAC

S urface (Arc de cercle BAC) - (Surface triangle OAB)

\Rightarrow Surface arc = π {AB}^{2} (\frac{θ}{2 π}) (θ en radian)

r_{3} = AB = 2 \sqrt{2} (B centre cercle C 3)

\sin θ = \cos θ = \frac{\sqrt{2}}{2} \Rightarrow θ = \frac{π}{4}

S_{3} = \frac{π \times 8}{8} = π

surface triangle = \frac{{AB}^{2}}{2} = 2

\Rightarrow A_{3} = (S_{3} - 4) A 3 = π - 2

Conclusion :

S = A_{1} + A_{2} + A_{3}

(\sqrt{2} - 1) π + π - 2

S = π \sqrt{2} - 2

Commented by a.lgnaoui last updated on 03/May/23