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

Commented by Joel575 last updated on 09/Feb/17

The side length of the square is 1 cm.  5 circles have same radius.  What is the value of the circle′s radius?

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

Answered by sandy_suhendra last updated on 09/Feb/17

Commented by sandy_suhendra last updated on 09/Feb/17

let AC=GB=x and CD=CF=FG=r    so AD=AE=x−r        FB=EB=x+r  AE+EB=1  x−r+x+r=1 ⇒ x=(1/2)  AC=x=(1/2) and CB=2r+x=2r+(1/2)  Phytagoras theorem :  AC^2 +CB^2 =AB^2   ((1/2))^2 +(2r+(1/2))^2 =1^2   (1/4)+4r^2 +2r+(1/4)=1  4r^2 +2r−(1/2)=0  8r^2 +4r−1=0  r = ((−4±(√(16+32)))/(16))  r = ((−4+4(√3))/(16)) = ((−1+(√3))/4)  (r must be positive)

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

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