Five circles

On the line segment CD = 6, there are five circles with one radius at regular intervals. Find the lengths of the lines AD, AF, AG, BD, and CE.

Correct answer:

d =  4.5826
f =  2.1794
g =  3.1225
b =  3.6056
c =  2.6458

Step-by-step explanation:

$$\begin{gathered} r=1 \\ h=v(A,CD) \\ A{S}_1{S}_2=\Delta 1-1-1,\alpha =\beta =\gamma =60°, \\ h=\sqrt{3}/2\cdot r=\sqrt{3}/2\cdot 1\doteq 0.866 \\ {x}_1=(4+1/2)\cdot r=(4+1/2)\cdot 1=\frac{9}{2}=4\frac{1}{2}=4.5 \\ d=|AD| \\ d=\sqrt{x_1^2+h^2}=\sqrt{4.5^2+0.866^2}=\sqrt{21}=4.5826 \end{gathered}$$

$$\begin{gathered} f=|AE| \\ {x}_2=4/2\cdot r=4/2\cdot 1=2 \\ f=\sqrt{x_2^2+h^2}=\sqrt{2^2+0.866^2}=2.1794 \end{gathered}$$

$$\begin{gathered} g=|AG| \\ {x}_3=6/2\cdot r=6/2\cdot 1=3 \\ g=\sqrt{x_3^2+h^2}=\sqrt{3^2+0.866^2}=3.1225 \end{gathered}$$

$$\begin{gathered} b=|BD| \\ {x}_4=7/2\cdot r=7/2\cdot 1=\frac{7}{2}=3\frac{1}{2}=3.5 \\ b=\sqrt{x_4^2+h^2}=\sqrt{3.5^2+0.866^2}=\sqrt{13}=3.6056 \end{gathered}$$

$$\begin{gathered} c=|CE| \\ {x}_5=5/2\cdot r=5/2\cdot 1=\frac{5}{2}=2\frac{1}{2}=2.5 \\ c=\sqrt{x_5^2+h^2}=\sqrt{2.5^2+0.866^2}=\sqrt{7}=2.6458 \end{gathered}$$

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