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](Q197675.png)
$$\boldsymbol{\mathrm{Calcul}}\:\boldsymbol{\mathrm{de}}\:\:\:\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{BCDE}}\right) \\ $$$$ \\ $$$$\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{BCDE}}\right)=\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{AMN}}\right)−\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{DEMN}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{ABC}}\right) \\ $$$$ \\ $$$$\bullet\boldsymbol{\mathrm{Calcul}}\:\:\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{DEMN}}\right) \\ $$$$\measuredangle\mathrm{MON}=\frac{\mathrm{2}\pi}{\mathrm{9}}\:\:\:\left(\mathrm{O}\:\mathrm{Centre}\:\mathrm{cercle}\right) \\ $$$$\boldsymbol{\alpha}=\measuredangle\mathrm{ION}\:=\frac{\pi}{\mathrm{9}}=\mathrm{2}\measuredangle\mathrm{OAN}=\mathrm{2}\measuredangle\mathrm{ONA} \\ $$$$\Rightarrow\measuredangle\mathrm{OAN}=\frac{\pi}{\mathrm{18}}=\measuredangle\boldsymbol{\mathrm{HAE}};\:\:\:\boldsymbol{\mathrm{MN}}=\mathrm{2}\boldsymbol{\mathrm{IN}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\frac{\pi}{\mathrm{9}}=\frac{\mathrm{IN}}{\mathrm{R}}=\frac{\mathrm{1}}{\mathrm{2R}}\:\:\:\: \\ $$$$\mathrm{sin}\:\frac{\pi}{\mathrm{18}}==\frac{\mathrm{HE}}{\mathrm{AE}}=\frac{\mathrm{FN}}{\mathrm{EN}}=\frac{\mathrm{IN}−\mathrm{HE}}{\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{HE} \\ $$$$\Rightarrow\boldsymbol{\mathrm{HE}}=\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{sin}\:\frac{\pi}{\mathrm{18}}=\left(\mathrm{0},\mathrm{321}\right) \\ $$$$\mathrm{cos}\:\frac{\pi}{\mathrm{18}}=\frac{\mathrm{EF}}{\mathrm{EN}}=\frac{\mathrm{HI}}{\mathrm{1}}\Rightarrow\:\:\boldsymbol{\mathrm{HI}}=\mathrm{cos}\:\frac{\pi}{\mathrm{18}} \\ $$$$\:\:\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{DEMN}}\right)=\left(\frac{\boldsymbol{\mathrm{DE}}+\boldsymbol{\mathrm{MN}}}{\mathrm{2}}\right)\boldsymbol{\mathrm{HI}}= \\ $$$$\:\left(\boldsymbol{\mathrm{HE}}+\boldsymbol{\mathrm{IN}}\right)×\boldsymbol{\mathrm{HI}}=\left[\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{sin}\:\frac{\boldsymbol{\pi}}{\mathrm{18}}\right)\right]\mathrm{cos}\:\frac{\pi}{\mathrm{18}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{DEMN}}\right)\:\:\:\:\:=\bar {\mathrm{0}},\mathrm{80} \\ $$$$\:\:\: \\ $$$$ \\ $$$$\bullet\boldsymbol{\mathrm{Cakcul}}\:\boldsymbol{\mathrm{de}}\:\:\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{ABC}}\right) \\ $$$$\:\:\:\mathrm{1}−\bigtriangleup\:\:\mathrm{ASL}\:\:\mathrm{et}\:\:\mathrm{AB}\:\mathrm{L}\:\: \\ $$$$\measuredangle\mathrm{AOS}=\frac{\mathrm{2}\pi}{\mathrm{9}}\Rightarrow\:\measuredangle\mathrm{ASO}=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{9}}=\frac{\mathrm{7}\pi}{\mathrm{18}} \\ $$$$\Rightarrow\measuredangle\mathrm{SAL}=\frac{\pi}{\mathrm{2}}−\frac{\mathrm{7}\pi}{\mathrm{18}}=\frac{\pi}{\mathrm{9}}\:; \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{AL}}=\mathrm{2sin}\:\frac{\mathrm{7}\pi}{\mathrm{18}} \\ $$$$\mathrm{d}\:\mathrm{apres}\:\mathrm{cercle}\:\:\:\boldsymbol{\mathrm{SA}}\mid\mid\:\boldsymbol{\mathrm{LT}}\:\Rightarrow\measuredangle\mathrm{ALB}=\frac{\pi}{\mathrm{9}} \\ $$$$\measuredangle\mathrm{BAC}=\frac{\pi}{\mathrm{9}}\:\:;\measuredangle\mathrm{LAB}=\measuredangle\mathrm{SAO}−\left(\frac{\pi}{\mathrm{9}}+\frac{\pi}{\mathrm{18}}\right) \\ $$$$\:\:=\frac{\mathrm{7}\pi}{\mathrm{18}}−\frac{\mathrm{3}\pi}{\mathrm{18}}=\frac{\mathrm{2}\pi}{\mathrm{9}}\:\:\:\:\measuredangle\mathrm{ABL}=\pi−\frac{\mathrm{3}\pi}{\mathrm{9}}=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\bigtriangleup\mathrm{ABL}\:\:\:\frac{\mathrm{AB}}{\mathrm{sin}\:\frac{\pi}{\mathrm{9}}}=\frac{\mathrm{BL}}{\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{9}}}=\frac{\mathrm{AL}}{\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$$$\begin{cases}{\mathrm{AB}=\frac{\mathrm{ALsin}\:\frac{\pi}{\mathrm{9}}}{\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}}}=\frac{\mathrm{2sin}\:\frac{\mathrm{7}\boldsymbol{\pi}}{\mathrm{18}}×\mathrm{sin}\:\frac{\pi}{\mathrm{9}}}{\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}}}}\\{\mathrm{BL}=\frac{\mathrm{ALsin}\:\mathrm{2}\frac{\pi}{\mathrm{9}}}{\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}}}=\frac{\mathrm{2sin}\:\frac{\mathrm{7}\pi}{\mathrm{18}}\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{9}}}{\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}}}}\end{cases} \\ $$$$ \\ $$$$\measuredangle\mathrm{ABL}=\frac{\mathrm{2}\pi}{\mathrm{3}}\Rightarrow\:\:\:\:\measuredangle\mathrm{ABC}=\frac{\pi}{\mathrm{3}} \\ $$$$\mathrm{dinc}\:\:\:\measuredangle\mathrm{ACB}=\pi−\left(\frac{\pi}{\mathrm{3}}+\frac{\pi}{\mathrm{9}}\right)=\frac{\mathrm{5}\pi}{\mathrm{9}} \\ $$$$\bigtriangleup\boldsymbol{\mathrm{ACL}}\:\:\:\:\frac{\mathrm{AC}}{\mathrm{sin}\:\frac{\pi}{\mathrm{9}}}=\frac{\mathrm{AL}}{\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{9}}} \\ $$$$\Rightarrow\boldsymbol{\mathrm{AC}}=\frac{\mathrm{2sin}\:\frac{\mathrm{7}\boldsymbol{\pi}}{\mathrm{18}}×\mathrm{sin}\:\frac{\pi}{\mathrm{9}}}{\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{9}}} \\ $$$$\bullet\boldsymbol{\mathrm{Calcul}}\:\boldsymbol{\mathrm{de}}\:\:\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{ABC}}\right) \\ $$$$\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{ABC}}\right)=\boldsymbol{\mathrm{AB}}×\boldsymbol{\mathrm{AC}}\mathrm{sin}\:\frac{\boldsymbol{\pi}}{\mathrm{9}};\:\:\:\left(\frac{\pi}{\mathrm{9}}=\measuredangle\mathrm{BAC}\right) \\ $$$$=\frac{\mathrm{2sin}\:\frac{\mathrm{7}\pi}{\mathrm{18}}×\mathrm{sin}\:\frac{\pi}{\mathrm{9}}}{\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}}}×\frac{\mathrm{2sin}\:\frac{\mathrm{7}\pi}{\mathrm{18}}×\mathrm{sin}^{\mathrm{2}} \:\frac{\pi}{\mathrm{9}}}{\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{9}}} \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{ABC}}\right)\:\:\:\:\:\:=\frac{\mathrm{sin}\:\:^{\mathrm{2}} \frac{\mathrm{2}\pi}{\mathrm{9}}\mathrm{sin}\frac{\pi}{\mathrm{9}}}{\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}}\mathrm{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\frac{\pi}{\mathrm{9}}\right)\:} \\ $$$$ \\ $$$$\bullet\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{AMN}}\right)=\left(\boldsymbol{\mathrm{AI}}×\boldsymbol{\mathrm{IN}}\right) \\ $$$$\:\:\boldsymbol{\mathrm{AI}}=\boldsymbol{\mathrm{R}}+\boldsymbol{\mathrm{OI}}=\boldsymbol{\mathrm{R}}+\boldsymbol{\mathrm{R}}\mathrm{cos}\:\:\:\frac{\boldsymbol{\pi}}{\mathrm{18}}\:\:\:\:\:\boldsymbol{\mathrm{IN}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\boldsymbol{\mathrm{S}}=\frac{\boldsymbol{\mathrm{R}}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{cos}\:\frac{\boldsymbol{\pi}}{\mathrm{9}}\right) \\ $$$$\:\:\mathrm{sin}\:\frac{\boldsymbol{\pi}}{\mathrm{9}}=\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{\mathrm{R}}}\Rightarrow\:\:\:\boldsymbol{\mathrm{R}}=\frac{\mathrm{1}}{\mathrm{2sin}\:\frac{\boldsymbol{\pi}}{\mathrm{9}}} \\ $$$$\Rightarrow\:\:\:\:\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{AMN}}\right)\:\:\:=\frac{\left(\mathrm{1}+\mathrm{cos}\:\frac{\boldsymbol{\pi}}{\mathrm{9}}\right)}{\mathrm{4sin}\:\frac{\boldsymbol{\pi}}{\mathrm{9}}}=\mathrm{1},\mathrm{44} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{Alors}}\:\:\:\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{BCDE}}\right)= \\ $$$$\:\:\:\:\:\frac{\mathrm{1}+\mathrm{cos}\:\frac{\boldsymbol{\pi}}{\mathrm{18}}}{\mathrm{4sin}\:\frac{\boldsymbol{\pi}}{\mathrm{9}}}−\left[\left(\mathrm{1}−\mathrm{sin}\frac{\pi}{\mathrm{18}}\:\right)\mathrm{cos}\:\frac{\pi}{\mathrm{18}}+\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{sin}\:^{\mathrm{2}} \frac{\mathrm{2}\pi}{\mathrm{9}}\mathrm{sin}\frac{\pi}{\mathrm{9}}\:}{\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}}\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\frac{\pi}{\mathrm{8}}\right)}\:\:\right] \\ $$$$=\mathrm{1},\mathrm{44}−\left(\mathrm{0},\mathrm{80}+\mathrm{0},\mathrm{16}\right)= \\ $$$$ \\ $$$$\Rightarrow\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{BCDE}}\right)=\mathrm{0},\mathrm{48} \\ $$
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
$${wrong}! \\ $$$${even}\:{without}\:{any}\:{calculation}\:{we}\:{can} \\ $$$${see}\:{that}\:{the}\:{shaded}\:{area}\:{must}\:{be}\: \\ $$$${less}\:{than}\:\mathrm{1}^{\mathrm{2}} ,\:{see}\:{below}. \\ $$
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
![AF=AG=(1/(2 cos 80°))=(1/(2 sin 10°)) AD=AE=(1/(2 sin 10°))−1 ((AC)/(sin 40°))=((AH)/(sin 100°)) ⇒AC=((sin 40°)/(cos 10°)) ((AB)/(sin 40°))=((AH)/(sin 120°)) ⇒AB=((2 sin 40°)/( (√3))) [BCED]=ΔADE−ΔABC =((sin 20°)/2)(AD×AE−AB×AC) =((sin 20°)/2)[((1/(2 sin 10°))−1)^2 −((2 sin 40°)/( (√3)))×((sin 40°)/(cos 10°))] ≈0.52117619](Q197688.png)
$${AF}={AG}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{cos}\:\mathrm{80}°}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{10}°} \\ $$$${AD}={AE}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{10}°}−\mathrm{1} \\ $$$$\frac{{AC}}{\mathrm{sin}\:\mathrm{40}°}=\frac{{AH}}{\mathrm{sin}\:\mathrm{100}°} \\ $$$$\Rightarrow{AC}=\frac{\mathrm{sin}\:\mathrm{40}°}{\mathrm{cos}\:\mathrm{10}°} \\ $$$$\frac{{AB}}{\mathrm{sin}\:\mathrm{40}°}=\frac{{AH}}{\mathrm{sin}\:\mathrm{120}°} \\ $$$$\Rightarrow{AB}=\frac{\mathrm{2}\:\mathrm{sin}\:\mathrm{40}°}{\:\sqrt{\mathrm{3}}} \\ $$$$\left[{BCED}\right]=\Delta{ADE}−\Delta{ABC} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{sin}\:\mathrm{20}°}{\mathrm{2}}\left({AD}×{AE}−{AB}×{AC}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{sin}\:\mathrm{20}°}{\mathrm{2}}\left[\left(\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{10}°}−\mathrm{1}\right)^{\mathrm{2}} −\frac{\mathrm{2}\:\mathrm{sin}\:\mathrm{40}°}{\:\sqrt{\mathrm{3}}}×\frac{\mathrm{sin}\:\mathrm{40}°}{\mathrm{cos}\:\mathrm{10}°}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\approx\mathrm{0}.\mathrm{52117619} \\ $$