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Question Number 197622 by sonukgindia last updated on 24/Sep/23

	Answered by a.lgnaoui last updated on 26/Sep/23
	$Calcul de S(BCDE) S(BCDE)=S(AMN)−S(DEMN) −S(ABC) •Calcul S(DEMN) ∡MON=((2π)/9) (O Centre cercle) 𝛂=∡ION =(π/9)=2∡OAN=2∡ONA ⇒∡OAN=(π/(18))=∡HAE; MN=2IN=(1/2) sin (π/9)=((IN)/R)=(1/(2R)) sin (π/(18))==((HE)/(AE))=((FN)/(EN))=((IN−HE)/1)=(1/2)−HE ⇒HE=(1/2)−sin (π/(18))=(0,321) cos (π/(18))=((EF)/(EN))=((HI)/1)⇒ HI=cos (π/(18)) S(DEMN)=(((DE+MN)/2))HI= (HE+IN)×HI=[((1/2)−sin (𝛑/(18)))]cos (π/(18)) S(DEMN) =0^� ,80 •Cakcul de S(ABC) 1−△ ASL et AB L ∡AOS=((2π)/9)⇒ ∡ASO=(π/2)−(π/9)=((7π)/(18)) ⇒∡SAL=(π/2)−((7π)/(18))=(π/9) ; AL=2sin ((7π)/(18)) d apres cercle SA∣∣ LT ⇒∡ALB=(π/9) ∡BAC=(π/9) ;∡LAB=∡SAO−((π/9)+(π/(18))) =((7π)/(18))−((3π)/(18))=((2π)/9) ∡ABL=π−((3π)/9)=((2π)/3) △ABL ((AB)/(sin (π/9)))=((BL)/(sin ((2π)/9)))=((AL)/(sin ((2π)/3))) { ((AB=((ALsin (π/9))/(sin ((2π)/3)))=((2sin ((7𝛑)/(18))×sin (π/9))/(sin ((2π)/3))))),((BL=((ALsin 2(π/9))/(sin ((2π)/3)))=((2sin ((7π)/(18))sin ((2π)/9))/(sin ((2π)/3))))) :} ∡ABL=((2π)/3)⇒ ∡ABC=(π/3) dinc ∡ACB=π−((π/3)+(π/9))=((5π)/9) △ACL ((AC)/(sin (π/9)))=((AL)/(sin ((5π)/9))) ⇒AC=((2sin ((7𝛑)/(18))×sin (π/9))/(sin ((5π)/9))) •Calcul de S(ABC) S(ABC)=AB×ACsin (𝛑/9); ((π/9)=∡BAC) =((2sin ((7π)/(18))×sin (π/9))/(sin ((2π)/3)))×((2sin ((7π)/(18))×sin^2 (π/9))/(sin ((5π)/9))) S(ABC) =((sin ^2 ((2π)/9)sin(π/9))/(sin ((2π)/3)sin(((2π)/3)−(π/9)) )) •S(AMN)=(AI×IN) AI=R+OI=R+Rcos (𝛑/(18)) IN=(1/2) S=(R/2)(1+cos (𝛑/9)) sin (𝛑/9)=(1/(2R))⇒ R=(1/(2sin (𝛑/9))) ⇒ S(AMN) =(((1+cos (𝛑/9)))/(4sin (𝛑/9)))=1,44 Alors S(BCDE)= ((1+cos (𝛑/(18)))/(4sin (𝛑/9)))−[(1−sin(π/(18)) )cos (π/(18))+ ((sin^2 ((2π)/9)sin(π/9) )/(sin ((2π)/3)sin (((2π)/3)−(π/8)))) ] =1,44−(0,80+0,16)= ⇒ S(BCDE)=0,48$
	$Calcul de S (BCDE)$ $S (BCDE) = S (AMN) - S (DEMN)$ $- S (ABC)$ $∙ Calcul S (DEMN)$ $∡ MON = \frac{2 π}{9} (O Centre cercle)$ $α = ∡ ION = \frac{π}{9} = 2 ∡ OAN = 2 ∡ ONA$ $\Rightarrow ∡ OAN = \frac{π}{18} = ∡ HAE; MN = 2 IN = \frac{1}{2}$ $\sin \frac{π}{9} = \frac{IN}{R} = \frac{1}{2 R}$ $\sin \frac{π}{18} == \frac{HE}{AE} = \frac{FN}{EN} = \frac{IN - HE}{1} = \frac{1}{2} - HE$ $\Rightarrow HE = \frac{1}{2} - \sin \frac{π}{18} = (0, 321)$ $\cos \frac{π}{18} = \frac{EF}{EN} = \frac{HI}{1} \Rightarrow HI = \cos \frac{π}{18}$ $S (DEMN) = (\frac{DE + MN}{2}) HI =$ $(HE + IN) \times HI = [(\frac{1}{2} - \sin \frac{π}{18})] \cos \frac{π}{18}$ $S (DEMN) = \bar{0}, 80$ $∙ Cakcul de S (ABC)$ $1 - △ ASL et AB L$ $∡ AOS = \frac{2 π}{9} \Rightarrow ∡ ASO = \frac{π}{2} - \frac{π}{9} = \frac{7 π}{18}$ $\Rightarrow ∡ SAL = \frac{π}{2} - \frac{7 π}{18} = \frac{π}{9};$ $AL = 2 \sin \frac{7 π}{18}$ $d apres cercle SA ∣∣ LT \Rightarrow ∡ ALB = \frac{π}{9}$ $∡ BAC = \frac{π}{9}; ∡ LAB = ∡ SAO - (\frac{π}{9} + \frac{π}{18})$ $= \frac{7 π}{18} - \frac{3 π}{18} = \frac{2 π}{9} ∡ ABL = π - \frac{3 π}{9} = \frac{2 π}{3}$ $△ ABL \frac{AB}{\sin \frac{π}{9}} = \frac{BL}{\sin \frac{2 π}{9}} = \frac{AL}{\sin \frac{2 π}{3}}$ ${\begin{cases} AB = \frac{ALsin \frac{π}{9}}{\sin \frac{2 π}{3}} = \frac{2 \sin \frac{7 π}{18} \times \sin \frac{π}{9}}{\sin \frac{2 π}{3}} \\ BL = \frac{ALsin 2 \frac{π}{9}}{\sin \frac{2 π}{3}} = \frac{2 \sin \frac{7 π}{18} \sin \frac{2 π}{9}}{\sin \frac{2 π}{3}} \end{cases}$ $∡ ABL = \frac{2 π}{3} \Rightarrow ∡ ABC = \frac{π}{3}$ $dinc ∡ ACB = π - (\frac{π}{3} + \frac{π}{9}) = \frac{5 π}{9}$ $△ ACL \frac{AC}{\sin \frac{π}{9}} = \frac{AL}{\sin \frac{5 π}{9}}$ $\Rightarrow AC = \frac{2 \sin \frac{7 π}{18} \times \sin \frac{π}{9}}{\sin \frac{5 π}{9}}$ $∙ Calcul de S (ABC)$ $S (ABC) = AB \times AC \sin \frac{π}{9}; (\frac{π}{9} = ∡ BAC)$ $= \frac{2 \sin \frac{7 π}{18} \times \sin \frac{π}{9}}{\sin \frac{2 π}{3}} \times \frac{2 \sin \frac{7 π}{18} \times \sin^{2} \frac{π}{9}}{\sin \frac{5 π}{9}}$ $S (ABC) = \frac{\sin^{2} \frac{2 π}{9} \sin \frac{π}{9}}{\sin \frac{2 π}{3} \sin (\frac{2 π}{3} - \frac{π}{9})}$ $∙ S (AMN) = (AI \times IN)$ $AI = R + OI = R + R \cos \frac{π}{18} IN = \frac{1}{2}$ $S = \frac{R}{2} (1 + \cos \frac{π}{9})$ $\sin \frac{π}{9} = \frac{1}{2 R} \Rightarrow R = \frac{1}{2 \sin \frac{π}{9}}$ $\Rightarrow S (AMN) = \frac{(1 + \cos \frac{π}{9})}{4 \sin \frac{π}{9}} = 1, 44$ $Alors S (BCDE) =$ $\frac{1 + \cos \frac{π}{18}}{4 \sin \frac{π}{9}} - [(1 - \sin \frac{π}{18}) \cos \frac{π}{18} +$ $\frac{\sin^{2} \frac{2 π}{9} \sin \frac{π}{9}}{\sin \frac{2 π}{3} \sin (\frac{2 π}{3} - \frac{π}{8})}]$ $= 1, 44 - (0, 80 + 0, 16) =$ $\Rightarrow S (BCDE) = 0, 48$
	Commented by a.lgnaoui last updated on 26/Sep/23

	Commented by a.lgnaoui last updated on 26/Sep/23

	Commented by mr W last updated on 26/Sep/23

	$w r o n g!$ $e v e n w i t h o u t a n y c a l c u l a t i o n w e c a n$ $s e e t h a t t h e s h a d e d a r e a m u s t b e$ $l e s s t h a n 1^{2}, s e e b e l o w .$
	Commented by mr W last updated on 26/Sep/23

	Answered by mr W last updated on 26/Sep/23

	Commented by mr W last updated on 26/Sep/23

	$A F = A G = \frac{1}{2 \cos 80 °} = \frac{1}{2 \sin 10 °}$ $A D = A E = \frac{1}{2 \sin 10 °} - 1$ $\frac{A C}{\sin 40 °} = \frac{A H}{\sin 100 °}$ $\Rightarrow A C = \frac{\sin 40 °}{\cos 10 °}$ $\frac{A B}{\sin 40 °} = \frac{A H}{\sin 120 °}$ $\Rightarrow A B = \frac{2 \sin 40 °}{\sqrt{3}}$ $[B C E D] = Δ A D E - Δ A B C$ $= \frac{\sin 20 °}{2} (A D \times A E - A B \times A C)$ $= \frac{\sin 20 °}{2} [{(\frac{1}{2 \sin 10 °} - 1)}^{2} - \frac{2 \sin 40 °}{\sqrt{3}} \times \frac{\sin 40 °}{\cos 10 °}]$ $\approx 0.52117619$
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