Why will arc not accept \n1 as a radius?

Question

This MWE does work which tells me the let operation is correctly defined.

    \documentclass{amsart}
    \usepackage{tikz}
    \usetikzlibrary{calc}
    \begin{document}
    \begin{tikzpicture}
    \path
    let \n1 ={2}
    in
    coordinate [label= $N$] (B) at (\n1,\n1); %to show that \n1 does work as a value
     \draw (0,0)  arc [start angle=0, end angle=180, radius=2];
    \end{tikzpicture}
    \end{document}
\end{document}

The one below differs from that only by using \n1 to specify the radius. But it returns an error message of

! Undefined control sequence. \pgfk@/tikz/x radius ->\n

\documentclass{amsart}
    \usepackage{tikz}
    \usetikzlibrary{calc}
    \begin{document}
    \begin{tikzpicture}
    \path
    let \n1 ={2}
    in
    coordinate [label= $N$] (B) at (\n1,\n1); %to show that \n1 does work as a value
     \draw (0,0)  arc [start angle=0, end angle=180, radius=\n1];
    \end{tikzpicture}
    \end{document}
\end{document}

As @Zarko says, the syntax of the first picture is not exactly what the manual recommends. But this syntax is working for me in a specific tikzpicture with a much more complicated let operation. Here it is. Notice that commands after the let operation succeed at use thing base on \p1 through \p4 (so I think savenumber might work for me, if I figure out how) but the arc command will not let me use a value \n1:

\documentclass{amsart}
\usepackage{tikz}
\usetikzlibrary{calc,intersections}
\begin{document}
    \[
    \begin{tikzpicture}
    \path % Given endpoints \p1 and \p2 and the placement of the divide       everything but the semicircle radius is calculated
    let
    \p1 = (-2.2,0),
    \p2 = (2.2,0),
    \p{divide} = ($ (\p1) !.7! (\p2) $),
    \p{center} = ($ (\p1) !.5! (\p2) $), % center of the semicircle
    \p3 = ($ (\x1,{\x{divide}-\x2}) $),
    \p4 =  ($ (\x{divide},\y3) $)
    in
    coordinate [label= left:$B$] (B) at (\p1)
    coordinate  [label= right:$F$] (F) at (\p2)
    coordinate  [label= left:$C$] (C) at (\p3)
    coordinate  [label= right:$D$] (D) at (\p4)
    coordinate  [label= below right:$E$] (E) at (\p{divide})
    coordinate (M) at (\p{center});  %the midpoint M is not labelled in the drawing

    \path [name path=R] (E) -- ($(E) + 1*(0,2.5)$);%upper bound here is set by hand

    \draw (F) [name path=P] arc [start angle=0, end angle=180, radius=2.2];%radius here is set by hand

    \draw [name intersections={of=P and R, by={[label=above:$H$]H}}];

    \draw[dashed]  (0,0) -- (H);
    \draw (B) -- (F) -- (H) --(B) -- (C) -- (D) -- (H);  %all the solid straight lines
    \end{tikzpicture}
    \]
    \end{document}

I need this tikzpicture for use in a publication. I could just use it the way I have currently written it and publish. But this version is inelegant because when I want to alter it a little to see what proportions look best I have to recalculate one radius by hand. If there is a way to get arc to accept \n1 as a radius value the same way as ``coordinate'' does does, with this syntax for let, then I will use that.

Answer 1

You have wrong syntax. Try the following:

\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{calc}

\begin{document}
\begin{tikzpicture}
\draw let \n1 = {2} in
        coordinate[label=above:$N$] (B) at (\n1,\n1)
        (0,0)  arc [start angle=0, end angle=180, radius=\n1];
\end{tikzpicture}
\end{document}

Edit: from TikZ manual:

Note that the effect of a let operation is local to the body of the let operation.

Addendum: I really have the problem to understand what cause your problems. Based on my guessing, i made the following MWE, which should show, that you can use my first MWE in your image on the following way (it is just dummy case, which should be simply extend to your real image):

\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{calc,intersections}

\begin{document}
\begin{tikzpicture}
\draw[name path=A]
        let \n1 = {2} in
        coordinate[label=above:$N$] (B) at ( \n1,\n1)
        coordinate[label=below:$O$] (O) at (-\n1,0)
        (0,0)  arc [start angle=0, end angle=180, radius=\n1];
\draw[name path=C] (O) -- (B);
\fill[red,
      name intersections={of=A and C, by={D}}] (D) circle (2pt)
                                                   node[right=2mm] {intersection D};
\fill[gray] (O) circle (2pt)    (B) circle (2pt);
\end{tikzpicture}
\end{document}

Answer 2

You can save whatever value you like into macros with various ways. You don't need to hack any let in machinary. But your picture is also pretty algorithmic so you don't need to save intermediate values that much, here is an alternative take

\documentclass[tikz]{standalone}
\usetikzlibrary{calc,intersections}
\begin{document}
\begin{tikzpicture}[]
\def\myradius{2.2}
\draw[name path=P] (0,0) coordinate(o)+(\myradius,0) coordinate[label=0:$F$] (f) 
    arc(0:180:\myradius) coordinate[label=180:$B$](b) --($(b)!0.7!(f)$) 
    coordinate[label=-45:$E$] (e) --cycle;
\draw(e)--($(e)!1!-90:(f)$) coordinate[label=0:$C$] -| (b) coordinate[label=180:$D$,pos=0.5];
\path [name path=R,overlay] (e) -- ++(0,1.5*\myradius);
\draw [dashed,name intersections={of=P and R, by=h}] (h)coordinate[label=60:$H$] -- (o);
\draw (b) -- (h) -- (f) (e) -- (h);
\end{tikzpicture}
\end{document}

Now two variables, \myradius and 0.7 defines the whole picture.