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Question-188078




Question Number 188078 by Rupesh123 last updated on 25/Feb/23
Answered by a.lgnaoui last updated on 25/Feb/23
from Graphe:△ABC  Equilaterale  big circle(radius OF=(√(4+((EF^2 )/4) )) )  sin (π/6)=(1/2)=(2/(OF))⇒(OF=4;EF=4(√(3.))  )  Area(big circleC_B =16π)  Length of △ABC=8(√3)  ⇒A=96(√3)  Area(A_1 +A_2 +A_3 )=96(√3) −16π  Area(C_B −C_S );C_S =Smal circle=  16π−4π=12π  A′_1 +A_2 +A′_3 =[Area(smal triagle)−4π]  =4(√3) ×(4(√3)×((√3)/2))−4π=24(√3) −4π  Total Area=(A_1 +A_2 +A_3 )+(A_1 ′+A_2 ′+A_3 ′)  (96(√3) −16π)+(24(√3) −4π)=120(√3) −20π       ShadedArea=145
$${from}\:{Graphe}:\bigtriangleup{ABC}\:\:{Equilaterale} \\ $$$${big}\:{circle}\left({radius}\:{OF}=\sqrt{\mathrm{4}+\frac{{EF}^{\mathrm{2}} }{\mathrm{4}}\:}\:\right) \\ $$$$\mathrm{sin}\:\frac{\pi}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{2}}{{OF}}\Rightarrow\left({OF}=\mathrm{4};{EF}=\mathrm{4}\sqrt{\mathrm{3}.}\:\:\right) \\ $$$${Area}\left({big}\:{circleC}_{{B}} =\mathrm{16}\pi\right) \\ $$$${Length}\:{of}\:\bigtriangleup{ABC}=\mathrm{8}\sqrt{\mathrm{3}}\:\:\Rightarrow{A}=\mathrm{96}\sqrt{\mathrm{3}} \\ $$$${Area}\left({A}_{\mathrm{1}} +{A}_{\mathrm{2}} +{A}_{\mathrm{3}} \right)=\mathrm{96}\sqrt{\mathrm{3}}\:−\mathrm{16}\pi \\ $$$${Area}\left({C}_{{B}} −{C}_{{S}} \right);{C}_{{S}} ={Smal}\:{circle}= \\ $$$$\mathrm{16}\pi−\mathrm{4}\pi=\mathrm{12}\pi \\ $$$${A}'_{\mathrm{1}} +{A}_{\mathrm{2}} +{A}'_{\mathrm{3}} =\left[{Area}\left({smal}\:{triagle}\right)−\mathrm{4}\pi\right] \\ $$$$=\mathrm{4}\sqrt{\mathrm{3}}\:×\left(\mathrm{4}\sqrt{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)−\mathrm{4}\pi=\mathrm{24}\sqrt{\mathrm{3}}\:−\mathrm{4}\pi \\ $$$${Total}\:{Area}=\left({A}_{\mathrm{1}} +{A}_{\mathrm{2}} +{A}_{\mathrm{3}} \right)+\left({A}_{\mathrm{1}} '+{A}_{\mathrm{2}} '+{A}_{\mathrm{3}} '\right) \\ $$$$\left(\mathrm{96}\sqrt{\mathrm{3}}\:−\mathrm{16}\pi\right)+\left(\mathrm{24}\sqrt{\mathrm{3}}\:−\mathrm{4}\pi\right)=\mathrm{120}\sqrt{\mathrm{3}}\:−\mathrm{20}\pi \\ $$$$ \\ $$$$\:\:\:\boldsymbol{{ShadedArea}}=\mathrm{145} \\ $$$$ \\ $$
Commented by Rupesh123 last updated on 25/Feb/23
Please check your solution well!
Commented by a.lgnaoui last updated on 25/Feb/23
thank you.
$${thank}\:{you}. \\ $$
Commented by mr W last updated on 25/Feb/23
clearly wrong!  the shaded area can not be more than  the big equilateral triangle!
$${clearly}\:{wrong}! \\ $$$${the}\:{shaded}\:{area}\:{can}\:{not}\:{be}\:{more}\:{than} \\ $$$${the}\:{big}\:{equilateral}\:{triangle}! \\ $$
Commented by a.lgnaoui last updated on 25/Feb/23
Area of  big triangle   =(8(√3))×((8(√3) ×(√3))/2)=32×3(√3) =166,267  166,267>145 (logic)  the shaded Area =(Bleu)
$${Area}\:{of}\:\:{big}\:{triangle}\: \\ $$$$=\left(\mathrm{8}\sqrt{\mathrm{3}}\right)×\frac{\mathrm{8}\sqrt{\mathrm{3}}\:×\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{32}×\mathrm{3}\sqrt{\mathrm{3}}\:=\mathrm{166},\mathrm{267} \\ $$$$\mathrm{166},\mathrm{267}>\mathrm{145}\:\left({logic}\right) \\ $$$${the}\:{shaded}\:{Area}\:=\left({Bleu}\right) \\ $$
Commented by mr W last updated on 25/Feb/23
you have paper and ruler, why not  draw such an equilateral and measure  its base and height and then calculate  its area? you′ll see its area is about  83. when it is logic for you that the   shaded area is 145, then i must say  you have no logic.
$${you}\:{have}\:{paper}\:{and}\:{ruler},\:{why}\:{not} \\ $$$${draw}\:{such}\:{an}\:{equilateral}\:{and}\:{measure} \\ $$$${its}\:{base}\:{and}\:{height}\:{and}\:{then}\:{calculate} \\ $$$${its}\:{area}?\:{you}'{ll}\:{see}\:{its}\:{area}\:{is}\:{about} \\ $$$$\mathrm{83}.\:{when}\:{it}\:{is}\:{logic}\:{for}\:{you}\:{that}\:{the}\: \\ $$$${shaded}\:{area}\:{is}\:\mathrm{145},\:{then}\:{i}\:{must}\:{say} \\ $$$${you}\:{have}\:{no}\:{logic}. \\ $$
Answered by mr W last updated on 25/Feb/23
radius of small circle r=2  radius of big circle R=4  side length small equilateral s=4(√3)  side length big equilateral S=8(√3)  shaded area=(big equilateral−big circle)+(small equilateral−small circle)     =5×(small equilateral−small circle)     =5×((((√3) s^2 )/4)−πr^2 )     =5×((((√3)×16×3)/4)−π×4)     =60(√3)−20π≈41.092
$${radius}\:{of}\:{small}\:{circle}\:{r}=\mathrm{2} \\ $$$${radius}\:{of}\:{big}\:{circle}\:{R}=\mathrm{4} \\ $$$${side}\:{length}\:{small}\:{equilateral}\:{s}=\mathrm{4}\sqrt{\mathrm{3}} \\ $$$${side}\:{length}\:{big}\:{equilateral}\:{S}=\mathrm{8}\sqrt{\mathrm{3}} \\ $$$${shaded}\:{area}=\left({big}\:{equilateral}−{big}\:{circle}\right)+\left({small}\:{equilateral}−{small}\:{circle}\right) \\ $$$$\:\:\:=\mathrm{5}×\left({small}\:{equilateral}−{small}\:{circle}\right) \\ $$$$\:\:\:=\mathrm{5}×\left(\frac{\sqrt{\mathrm{3}}\:{s}^{\mathrm{2}} }{\mathrm{4}}−\pi{r}^{\mathrm{2}} \right) \\ $$$$\:\:\:=\mathrm{5}×\left(\frac{\sqrt{\mathrm{3}}×\mathrm{16}×\mathrm{3}}{\mathrm{4}}−\pi×\mathrm{4}\right) \\ $$$$\:\:\:=\mathrm{60}\sqrt{\mathrm{3}}−\mathrm{20}\pi\approx\mathrm{41}.\mathrm{092} \\ $$
Commented by Rupesh123 last updated on 25/Feb/23
Excellent, sir!

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