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Question Number 215147 by AlagaIbile last updated on 30/Dec/24

Answered by A5T last updated on 30/Dec/24

Commented by A5T last updated on 30/Dec/24

$$\mathrm{OP}=\sqrt{{(4.5-\mathrm{r})}^2-{(2.5-\mathrm{r})}^2}=\sqrt{2(7-2\mathrm{r})}$$$$\Rightarrow \mathrm{GO}=\sqrt{{\mathrm{OP}}^2+{\mathrm{GP}}^2}=\sqrt{2(7-2\mathrm{r})+{(2+\mathrm{r})}^2}=\sqrt{{\mathrm{r}}^2+18}$$$$\Rightarrow \mathrm{GC}=\sqrt{{\mathrm{GO}}^2-{\mathrm{OC}}^2}=\sqrt{{\mathrm{r}}^2+18-{\mathrm{r}}^2}=\sqrt{18}=3\sqrt{2}$$

Answered by mr W last updated on 30/Dec/24

Commented by mr W last updated on 30/Dec/24

$${let}^{\prime }slookatthegeneralcase.$$$$R=\frac{a+b}{2}$$$$OB=R-a=\frac{b-a}{2}$$$${GA}^2={(\frac{a+b}{2}-r)}^2-{(\frac{b-a}{2}-r)}^2+{(a+r)}^2$$$$=a(a+b)+r^2$$$${GC}^2={GA}^2-r^2=a(a+b)$$$$\Rightarrow GC=\sqrt{a(a+b)}=\sqrt{2\times (2+7)}=3\sqrt{2}$$

Commented by mr W last updated on 30/Dec/24

$$GCisindependentfromthesizeof$$$$thesmallcircle.$$

Commented by mr W last updated on 30/Dec/24

Commented by ajfour last updated on 30/Dec/24

https://youtu.be/qdhFOxn2SzU?si=QwAhflQzh-f_rJ8L my first video lecture on Inductance

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