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Question Number 10435 by Joel575 last updated on 09/Feb/17

Commented by Joel575 last updated on 09/Feb/17

$$\mathrm{The}\mathrm{side}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{square}\mathrm{is}1\mathrm{cm}.$$$$5\mathrm{circles}\mathrm{have}\mathrm{same}\mathrm{radius}.$$$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}{\mathrm{circle}}^{\prime }\mathrm{s}\mathrm{radius}?$$

Answered by sandy_suhendra last updated on 09/Feb/17

Commented by sandy_suhendra last updated on 09/Feb/17

$$\mathrm{let}\mathrm{AC}=\mathrm{GB}=\mathrm{x}\mathrm{and}\mathrm{CD}=\mathrm{CF}=\mathrm{FG}=\mathrm{r}$$$$\mathrm{so}\mathrm{AD}=\mathrm{AE}=\mathrm{x}-\mathrm{r}$$$$\mathrm{FB}=\mathrm{EB}=\mathrm{x}+\mathrm{r}$$$$\mathrm{AE}+\mathrm{EB}=1$$$$\mathrm{x}-\mathrm{r}+\mathrm{x}+\mathrm{r}=1\Rightarrow \mathrm{x}=\frac{1}{2}$$$$\mathrm{AC}=\mathrm{x}=\frac{1}{2}\mathrm{and}\mathrm{CB}=2\mathrm{r}+\mathrm{x}=2\mathrm{r}+\frac{1}{2}$$$$\mathrm{Phytagoras}\mathrm{theorem}:$$$${\mathrm{AC}}^2+{\mathrm{CB}}^2={\mathrm{AB}}^2$$$${\left(\frac{1}{2}\right)}^2+{(2\mathrm{r}+\frac{1}{2})}^2=1^2$$$$\frac{1}{4}+{4\mathrm{r}}^2+2\mathrm{r}+\frac{1}{4}=1$$$${4\mathrm{r}}^2+2\mathrm{r}-\frac{1}{2}=0$$$${8\mathrm{r}}^2+4\mathrm{r}-1=0$$$$\mathrm{r}=\frac{-4\pm \sqrt{16+32}}{16}$$$$\mathrm{r}=\frac{-4+4\sqrt{3}}{16}=\frac{-1+\sqrt{3}}{4}$$$$\left(\mathrm{r}\mathrm{must}\mathrm{be}\mathrm{positive}\right)$$

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