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Question Number 98516 by bemath last updated on 14/Jun/20

Commented by Aziztisffola last updated on 14/Jun/20

$$\mathrm{Hi}\mathrm{sir}\mathrm{best}\mathrm{draw}!\mathrm{which}\mathrm{app}\mathrm{do}\mathrm{you}\mathrm{use}?)=\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+...}}}\iff {\mathrm{f}}^2\left(\mathrm{x}\right)=\mathrm{x}+\mathrm{f}\left(\mathrm{x}\right)$$$$\mathrm{I}\mathrm{need}\mathrm{the}\mathrm{best}\mathrm{one}.$$$$\mathrm{x})-\mathrm{f}\left(\mathrm{x}\right)-\mathrm{x}=0$$$$={(-1)}^2-4\times 1\times (-\mathrm{x})=1+4\mathrm{x}$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{1+\sqrt{1+4\mathrm{x}}}{2}(\mathrm{x}\mathrm{and}\mathrm{f}\left(\mathrm{x}\right)>0)$$$$\mathrm{f}{}^{\prime }\mathrm{x})=\frac{1}{2}\frac{4}{2\sqrt{1+4\mathrm{x}}}=\frac{1}{\sqrt{1+4\mathrm{x}}}$$$${}^{\prime }\left(5\right)=\frac{1}{\sqrt{1+4\times 5}}=\frac{1}{\sqrt{21}}$$$$$$

Commented by Quvonchbek last updated on 14/Jun/20

Answered by mr W last updated on 14/Jun/20

$$saycenterofbigcircleis(k,h)$$$${(k-4)}^2+{(h-3)}^2={(R-5)}^2...\left(i\right)$$$${(k-0)}^2+{(h-3)}^2={(R-3)}^2...\left(ii\right)$$$${(k-4)}^2+{(h-0)}^2={(R-4)}^2...\left(iii\right)$$$$\left(iii\right)-\left(i\right):$$$$6h-9=2R-9$$$$\Rightarrow h=\frac{R}{3}$$$$\left(ii\right)-\left(i\right):$$$$4k-8=2R-8$$$$\Rightarrow k=\frac{R}{2}$$$$putthisinto\left(ii\right):$$$$\frac{R^2}{4}+{(\frac{R}{3}-3)}^2={(R-3)}^2$$$$\Rightarrow R=\frac{144}{23}$$

Answered by john santu last updated on 14/Jun/20

Commented by john santu last updated on 14/Jun/20

$${(\mathrm{R}-3)}^2=4^2+{(\mathrm{R}-5)}^2-2\times 4\times \cos \alpha$$$${(\mathrm{R}-4)}^2=3^2+{(\mathrm{R}-5)}^2-2\times 3\times \sin \alpha$$$$[\sin {}^2\alpha =1-\cos {}^2\alpha ]$$$${\left(\frac{\mathrm{R}-9}{5(\mathrm{R}-5)}\right)}^2+{\left(\frac{\mathrm{R}-8}{2(\mathrm{R}-5)}\right)}^2=1$$$$\mathrm{R}=\frac{144}{23}.◼$$

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