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Question-90594




Question Number 90594 by byaw last updated on 24/Apr/20
Commented by mr W last updated on 24/Apr/20
Commented by mr W last updated on 24/Apr/20
when posting an image, you can   rotate it at first as shown above.
$${when}\:{posting}\:{an}\:{image},\:{you}\:{can}\: \\ $$$${rotate}\:{it}\:{at}\:{first}\:{as}\:{shown}\:{above}. \\ $$
Commented by mahdi last updated on 25/Apr/20
AB=2r=x+1⇒r=((x+1)/2)  CD=(√(OD^2 −OC^2 ))=(√(r^2 −(x−r)^2 ))  =(√(2rx−x^2 ))=(√(2(((x+1)/2))x−x^2 ))=(√x)
$$\mathrm{AB}=\mathrm{2r}=\mathrm{x}+\mathrm{1}\Rightarrow\mathrm{r}=\frac{\mathrm{x}+\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{CD}=\sqrt{\mathrm{OD}^{\mathrm{2}} −\mathrm{OC}^{\mathrm{2}} }=\sqrt{\mathrm{r}^{\mathrm{2}} −\left(\mathrm{x}−\mathrm{r}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\mathrm{2rx}−\mathrm{x}^{\mathrm{2}} }=\sqrt{\mathrm{2}\left(\frac{\mathrm{x}+\mathrm{1}}{\mathrm{2}}\right)\mathrm{x}−\mathrm{x}^{\mathrm{2}} }=\sqrt{\mathrm{x}} \\ $$

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