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With-a-center-on-a-given-circle-of-radius-r-an-arc-has-been-drawn-in-order-to-divide-the-circle-in-two-equal-in-area-parts-What-is-the-radius-of-the-arc-in-terms-of-r-radius-of-given-circle-




Question Number 26772 by Rasheed.Sindhi last updated on 29/Dec/17
With a center on a given circle of  radius r ,an arc has been drawn in order  to divide the circle in two equal  (in area) parts.  What is the radius of the arc in terms  of  r (radius of given circle)?
Withacenteronagivencircleofradiusr,anarchasbeendrawninordertodividethecircleintwoequal(inarea)parts.Whatistheradiusofthearcintermsofr(radiusofgivencircle)?
Answered by mrW1 last updated on 29/Dec/17
Let′s say the radius of the circle is  r and the radius of the arc is R.  The angle of the arc is θ.  r cos (θ/2)=(R/2)  ⇒R=2r cos (θ/2)  (ϕ/2)=π−2×(θ/2)  ⇒ϕ=2π−2θ    area of segment of arc BC:  A_1 =(R^2 /2)(θ−sin θ)=2r^2 cos^2  (θ/2)(θ−sin θ)  area of segment of circle BAC:  A_2 =(r^2 /2)[2π−2θ−sin (2π−2θ)]=r^2 (π−θ+sin θ cos θ)  A_1 +A_2 =((πr^2 )/2)  2r^2 cos^2  (θ/2)(θ−sin θ)+r^2 (π−θ+sin θ cos θ)=((πr^2 )/2)  (cos θ+1)(θ−sin θ)+(π/2)−θ+sin θ cos θ=0  sin θ−θ cos θ=(π/2)  ⇒θ≈1.9057 (109.2°)  ⇒R=2rcos (θ/2)≈1.16r    If  the arc divides the area of the  circle in ratio λ:1, then the  equation is  sin θ−θ cos θ=(1−λ)π  ⇒λ=1−((sin θ−θ cos θ)/π)  cos (θ/2)=(R/(2r))=γ  ⇒θ=2 cos^(−1) γ  sin (θ/2)=(√(1−γ^2 ))  ⇒sin θ=2γ(√(1−γ^2 ))  ⇒cos θ=2γ^2 −1  ⇒λ=f(γ)=1−((2γ(√(1−γ^2 ))−2(2γ^2 −1) cos^(−1) γ)/π)
LetssaytheradiusofthecircleisrandtheradiusofthearcisR.Theangleofthearcisθ.rcosθ2=R2R=2rcosθ2φ2=π2×θ2φ=2π2θareaofsegmentofarcBC:A1=R22(θsinθ)=2r2cos2θ2(θsinθ)areaofsegmentofcircleBAC:A2=r22[2π2θsin(2π2θ)]=r2(πθ+sinθcosθ)A1+A2=πr222r2cos2θ2(θsinθ)+r2(πθ+sinθcosθ)=πr22(cosθ+1)(θsinθ)+π2θ+sinθcosθ=0sinθθcosθ=π2θ1.9057(109.2°)R=2rcosθ21.16rIfthearcdividestheareaofthecircleinratioλ:1,thentheequationissinθθcosθ=(1λ)πλ=1sinθθcosθπcosθ2=R2r=γθ=2cos1γsinθ2=1γ2sinθ=2γ1γ2cosθ=2γ21λ=f(γ)=12γ1γ22(2γ21)cos1γπ
Commented by mrW1 last updated on 29/Dec/17
Commented by Rasheed.Sindhi last updated on 29/Dec/17
Great!   ThanX  a Lot Sir!
Great!ThanXaLotSir!
Commented by mrW1 last updated on 29/Dec/17
Thank you too Sir!  But I′m not able to find an analytic  solution.
ThankyoutooSir!ButImnotabletofindananalyticsolution.
Commented by mrW1 last updated on 30/Dec/17
Sir, do you have any idea if such an  arc can be drawn only with ruler and  compass? I can hardly believe that it  is possible.
Sir,doyouhaveanyideaifsuchanarccanbedrawnonlywithrulerandcompass?Icanhardlybelievethatitispossible.
Commented by Rasheed.Sindhi last updated on 30/Dec/17
To determine area of sector when  coordinates of three specifying  points are given. What′s the   formula?
Todetermineareaofsectorwhencoordinatesofthreespecifyingpointsaregiven.Whatstheformula?
Commented by Rasheed.Sindhi last updated on 30/Dec/17
Sorry it′s not only matter of   sector area!
Sorryitsnotonlymatterofsectorarea!
Commented by mrW1 last updated on 30/Dec/17
I don′t know such a formula.
Idontknowsuchaformula.
Commented by Rasheed.Sindhi last updated on 30/Dec/17
No sir, I myself am not sure!  Mr prakash jain can certainly  say something.
Nosir,Imyselfamnotsure!Mrprakashjaincancertainlysaysomething.
Commented by prakash jain last updated on 30/Dec/17
I am still thinking about that  problem.
Iamstillthinkingaboutthatproblem.

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