Circles!!!

$\text{You can compare the formulas for the area of heptadecagons. This way one doesn't assume the diameters of the little circles are straight.}$ $\text{Hence, minimizing the error from the assumption that the equal legs of the isosceles triangle are (R-r)}$ $\text{when this equal legs are actually (R-r+w) where w is a relatively small value in this problem. }$

$\text{Area of heptadecagon = }\dfrac {17 · s^2}{4 · \tan \dfrac {\pi}{17}} = \dfrac {17·A·s}{2}$

$\text{, where A = apothem and s = side length}$

$\text{Since the circle and the imaginary heptadecagon are concentric, we can say that,}$

$\text{A = R-r}$

$\text{And by connecting the little diameters, we can say}$

$\text{s = 2r}$

$\text{Subtituing the expressions, we get :}$

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$\text{Area of heptadecagon = }\dfrac {17 · 4r^2}{4 · \tan \dfrac {\pi}{17}} = \dfrac {17·(R-r)·2r}{2}$

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$\text{With some algebra we can find "r"... at this point, we can input R = 4}$

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$\text{Area of heptadecagon = }\dfrac {2r}{4 · \tan \dfrac {\pi}{17}} = \dfrac {4-r}{2}$

$\text{r =} \dfrac {4}{1+\cot \dfrac {\pi}{17}}$

$\text{, then r is around } \text{0.62997}$ $\text{which rounded to 3 decimals is } \text{0.630}$

This solution is just way longer and based on the fact or assumption (maybe?) that the apothem of the heptadecagon is (R-r), so maybe I can optimize a little bit the solution using the isosceles triangle exposed in your answer, Thanks.