Square to round. I have a question regarding surface development or the net of a shape.
Suppose you had a truncated cone that transitioned from a circle at the top with for example a radius of $50$ mm to a square base with sides of $200$ mm and height of the piece is $150$ mm. My question is if you flattened this shape out will the curved part of the pattern be part of a perfect circle? My guess is that it would form a circular arc of length $100 \pi$ mm but obviously not create a full circle rather a part of a circle of greater radius.
What would be the relationship between the original radius and the flattened development?

 A: Like $AR$ already drawn, draw perpendiculars $DN$, $GO$, $KP$ to the midpoints of the remaining sides of the square base, and extend them upward to concurrence at $Q$, as in the figure below. Points $R$, $N$, $O$, $P$, lie on the circle with radius $100$ inscribed in the square and parallel to the original circle. Thus $QR=QN=QO=QP$, and $AR=DN=GO=KP$, and hence $Q$ is the center of the circle thru  points $A$, $D$, $G$, $K$.
Since the intermediate points also lie on the original circle, then if all lengths are preserved in the flattening, it appears the remaining points $B$, $C$, $E$, $F$, $H$, $J$, $L$, $M$  must also lie on the circle with center $Q$ and radius $QA$.

The following may help to confirm this. $ABCDS$ and its three equal counterparts is the flattened surface of a portion of the oblique cone whose base is the original circle, whose altitude, as shown later, is equal to that of the cone with vertex $Q$ and original circle as base, and whose vertex is a vertex of the square, offset from base center $W$ by distance $ZS=100\sqrt 2$. This appears more clearly in the unflattened figure below.

Cutting the surface of this oblique cone from base to vertex along the two straight lines $AS$, $DS$ which meet the base $45^o$ on either side of the shortest straight line (e.g. $SY$ in the first figure), and flattening that portion of the surface, yields a circular sector unlike that of a right cone, which is concave toward the sector's vertex, but rather one like $ABYCDS$ (first figure), with circular arc $ABYCD$ convex toward the vertex.
Again then, all points $ABCDEFGHJKLM$ in the flattened figure lie on a circle whose radius $QD$ is the hypotenuse of a right triangle (second figure) whose base is the radius $WD$ of the original circle and whose height $QW$ is given by the proportion$$\frac{QW}{WD}=\frac{QZ}{ZN}=\frac{QW}{50}=\frac{QW+WZ}{100}=\frac{QW+150}{100}$$
so that $QW=WZ=150$, and hence $$QD=\sqrt{50^2+150^2}=50\sqrt{10}=r\sqrt {10}$$where $r=WD$ is the radius of the original circle. This is an instance of the general rule that$$\sqrt{x^2+(nx)^2}=x\sqrt{n^2+1}$$where here $x=r$ and $n=3$.
Reconsideration:
However, the first figure is not a flattened version of the original structure, as appears from the following. Since $SN=100$ and $DN=QD=50\sqrt {10}$, then$$\angle DSN=\arctan 1.58\approx 57.69^o$$And since $QN=100\sqrt {10}$, then $$\angle QSN=\arctan 3.16\approx 72.45^o$$Therefore$$\angle ASD=2(72.45-57.69)\approx29.52^o$$and in the heptagon $QRSTUVR'$,$$\angle RQR'=900-8\times 57.69^o-4\times 29.52^o-2\times 90^o\approx140.4^o$$Thus arc $ADGKA'\ne100\pi$ after all, i.e. sector $QADGKA'$ is not the flattening of a right cone with base radius $50$, height $150$, and slant side $50\sqrt{10}$, since the sector angle of that cone flattened $\approx 113.84^o$.
On the other hand, if we re-construct the first figure with $\angle AQA'=113.84^o$, as in the figure below,

then since radius $QR=100\sqrt{10}$ is unchanged, but the angle at $Q$ is smaller, it is clear that the perimeter of the square base $RS+ST+TU+UV+VR'<800$. Specifically, since $\angle DQG = \angle \frac {AQA'}{4}=28.46^o$, then $\angle TQO=14.23^o$, and$$TO=\tan14.23^o\times QO=.254\times 100\sqrt{10}=80.32$$making the side of the square base $TU=160.64<200$.
Thus it now seems that a 3-dimensional structure whose base is a square with side $200$, whose top is a circle of diameter $100$, and whose height is $150$, cannot be constructed by bending and creasing a flat 2-dimensional shape. If we begin with the correct sector $AQA'$, the flat shape yields a circular top with diameter $100$ but a square base with side less than the expected $200$. If we begin with the structure having circular top of diameter $100$ and square base of side $200$, and flatten it out, we get, as in my first figure, too great an angle at $Q$, i.e. a circular arc $>100\pi$. Is it right to conclude, then, that the structure cannot be flattened without stretching/distortion, and conversely, that no flat shape can yield the desired structure?
