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Question Number 48338 by ajfour last updated on 22/Nov/18

Commented by mr W last updated on 23/Nov/18

$$niceworking!$$

Commented by ajfour last updated on 22/Nov/18

$$Forexamplep=2,q=4,r=3.$$$$Findx=?.$$

Commented by ajfour last updated on 22/Nov/18

$$WillthisdoSir?$$

Commented by ajfour last updated on 22/Nov/18

Commented by ajfour last updated on 22/Nov/18

$$R^2{\cos }^{-1}\frac{h}{R}-h\sqrt{R^2-h^2}=p+q$$$$R^2{\cos }^{-1}\frac{h}{R}-k\sqrt{R^2-k^2}=r+p$$$$x=\pi R^2-p-q-r$$$$And\frac{\pi R^2}{4}=\frac{q+r}{2}+hk$$$$\Rightarrow \frac{\pi R^2}{4}=\frac{q+r}{2}+h^{\prime }{k}^{\prime }{R}^2$$$$R^2=\frac{(q+r)/2}{\frac{\pi }{4}-h^{\prime }{k}^{\prime }}$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$let\frac{h}{R}=h^{\prime }\mathrm{\&}\frac{p}{R^2}=p^{\prime }andsoon$$$${\cos }^{-1}{h}^{\prime }-h^{\prime }\sqrt{1-h^{\prime 2}}=\left(\frac{p+q}{q+r}\right)(\frac{\pi }{2}-2h^{\prime }{k}^{\prime })$$$${\cos }^{-1}{k}^{\prime }-k^{\prime }\sqrt{1-k^{\prime 2}}=\left(\frac{r+p}{q+r}\right)(\frac{\pi }{2}-2h^{\prime }{k}^{\prime })$$$$\mathit{x}=\frac{\mathit{\pi }(\mathit{q}+\mathit{r})}{(\frac{\mathit{\pi }}{2}-2{\mathit{h}}^{\prime }{\mathit{k}}^{\prime })}-(\mathit{p}+\mathit{q}+\mathit{r}).$$

Commented by MJS last updated on 22/Nov/18

$$\mathrm{tried}\mathrm{with}\mathrm{integrals}$$$$\mathrm{for}\mathrm{your}\mathrm{example}\mathrm{I}\mathrm{get}$$$$x=5.68110\wedge R=2.16174\vee x=144.410\wedge R=6.98799;$$$$\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{case}\mathrm{the}\mathrm{center}\mathrm{is}\mathrm{shifted}\mathrm{by}$$$$\left(\begin{array}{c}.143930\\ .253156\end{array}\right),\mathrm{in}\mathrm{the}\mathrm{second}\mathrm{case}\mathrm{by}\left(\begin{array}{c}.835304\\ .854448\end{array}\right)$$

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