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-

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Question Number 26772 by Rasheed.Sindhi last updated on 29/Dec/17

$$\mathrm{With}\mathrm{a}\mathrm{center}\text{boldsymbol}\mathrm{on}\mathrm{a}\mathrm{given}\mathrm{circle}\mathrm{of}$$$$\mathrm{radius}\mathrm{r},\mathrm{an}\mathrm{arc}\mathrm{has}\mathrm{been}\mathrm{drawn}\mathrm{in}\mathrm{order}$$$$\mathrm{to}\mathrm{divide}\mathrm{the}\mathrm{circle}\mathrm{in}\mathrm{two}\mathrm{equal}$$$$\left(\mathrm{in}\mathrm{area}\right)\mathrm{parts}.$$$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{arc}\mathrm{in}\mathrm{terms}$$$$\mathrm{of}\mathrm{r}\left(\mathrm{radius}\mathrm{of}\mathrm{given}\mathrm{circle}\right)?$$

Answered by mrW1 last updated on 29/Dec/17

$${Let}^{\prime }ssaytheradiusofthecircleis$$$$randtheradiusofthearcisR.$$$$Theangleofthearcis\theta .$$$$r\cos \frac{\theta }{2}=\frac{R}{2}$$$$\Rightarrow R=2r\cos \frac{\theta }{2}$$$$\frac{\phi }{2}=\pi -2\times \frac{\theta }{2}$$$$\Rightarrow \phi =2\pi -2\theta$$$$$$$$areaofsegmentofarcBC:$$$$A_1=\frac{R^2}{2}(\theta -\sin \theta )=2r^2{\cos }^2\frac{\theta }{2}(\theta -\sin \theta )$$$$areaofsegmentofcircleBAC:$$$$A_2=\frac{r^2}{2}[2\pi -2\theta -\sin (2\pi -2\theta )]=r^2(\pi -\theta +\sin \theta \cos \theta )$$$$A_1+A_2=\frac{\pi r^2}{2}$$$$2r^2{\cos }^2\frac{\theta }{2}(\theta -\sin \theta )+r^2(\pi -\theta +\sin \theta \cos \theta )=\frac{\pi r^2}{2}$$$$(\cos \theta +1)(\theta -\sin \theta )+\frac{\pi }{2}-\theta +\sin \theta \cos \theta =0$$$$\sin \theta -\theta \cos \theta =\frac{\pi }{2}$$$$\Rightarrow \theta \approx 1.9057(109.2°)$$$$\Rightarrow R=2r\cos \frac{\theta }{2}\approx 1.16r$$$$$$$$Ifthearcdividestheareaofthe$$$$circleinratio\lambda :1,thenthe$$$$equationis$$$$\sin \theta -\theta \cos \theta =(1-\lambda )\pi$$$$\Rightarrow \lambda =1-\frac{\sin \theta -\theta \cos \theta }{\pi }$$$$\cos \frac{\theta }{2}=\frac{R}{2r}=\gamma$$$$\Rightarrow \theta =2{\cos }^{-1}\gamma$$$$\sin \frac{\theta }{2}=\sqrt{1-{\gamma }^2}$$$$\Rightarrow \sin \theta =2\gamma \sqrt{1-{\gamma }^2}$$$$\Rightarrow \cos \theta =2{\gamma }^2-1$$$$\Rightarrow \lambda =f\left(\gamma \right)=1-\frac{2\gamma \sqrt{1-{\gamma }^2}-2(2{\gamma }^2-1){\cos }^{-1}\gamma }{\pi }$$

Commented by mrW1 last updated on 29/Dec/17

Commented by Rasheed.Sindhi last updated on 29/Dec/17

$$\mathcal{G}reat!$$$$\mathcal{T}han\mathcal{X}a\mathcal{L}ot\mathcal{S}ir!$$

Commented by mrW1 last updated on 29/Dec/17

$$ThankyoutooSir!$$$$But{I}^{\prime }mnotabletofindananalytic$$$$solution.$$

Commented by mrW1 last updated on 30/Dec/17

$$Sir,doyouhaveanyideaifsuchan$$$$arccanbedrawnonlywithrulerand$$$$compass?Icanhardlybelievethatit$$$$ispossible.$$

Commented by Rasheed.Sindhi last updated on 30/Dec/17

$$\mathrm{To}\mathrm{determine}\mathrm{area}\mathrm{of}\mathrm{sector}\mathrm{when}$$$$\mathrm{coordinates}\mathrm{of}\mathrm{three}\mathrm{specifying}$$$$\mathrm{points}\mathrm{are}\mathrm{given}.{\mathrm{What}}^{\prime }\mathrm{s}\mathrm{the}$$$$\mathrm{formula}?$$

Commented by Rasheed.Sindhi last updated on 30/Dec/17

$$\mathrm{Sorry}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{only}\mathrm{matter}\mathrm{of}$$$$\mathrm{sector}\mathrm{area}!$$

Commented by mrW1 last updated on 30/Dec/17

$$I{don}^{\prime }tknowsuchaformula.$$

Commented by Rasheed.Sindhi last updated on 30/Dec/17

$$\mathrm{No}\mathrm{sir},\mathrm{I}\mathrm{myself}\mathrm{am}\mathrm{not}\mathrm{sure}!$$$$\mathrm{Mr}\mathrm{prakash}\mathrm{jain}\mathrm{can}\mathrm{certainly}$$$$\mathrm{say}\mathrm{something}.$$

Commented by prakash jain last updated on 30/Dec/17

$$\mathrm{I}\mathrm{am}\mathrm{still}\mathrm{thinking}\mathrm{about}\mathrm{that}$$$$\mathrm{problem}.$$

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