Question 997069
It may look like this:
{{{drawing(300,200,-3,3,-1,3,
arc(0,0,4,4,180,360),
arc(0,0,5.6,5.6,180,360),
line(2,0,2.8,0),line(1.92,0.56,2.69,0.79),
line(1.68,1.08,2.36,1.51),line(1.31,1.51,1.83,2.12),
line(0.83,1.82,1.16,2.55),line(0.28,1.97,0.4,2.77),
line(-2,0,-2.8,0),line(-1.92,0.56,-2.69,0.79),
line(-1.68,1.08,-2.36,1.51),line(-1.31,1.51,-1.83,2.12),
line(-0.83,1.82,-1.16,2.55),line(-0.28,1.97,-0.4,2.77),
locate(2.1,0,0.8),locate(0.95,0,red(2)),
red(arrow(0,0,2,0)),red(arrow(2,0,0,0))
)}}} or maybe the 12 straight segments are placed between the two semicircles
in a different arrangement.
There are {{{12}}} straight pieces measuring {{{0.8meters}}} each, for a total of
{{{12(0.8meters)=9.6meters}}} .
There are also {{{2}}} half-circle arcs.
The inside arc has a radius of {{{4meters/2=2 meters}}} .
The outside arc has a radius of {{{2meters+0.8meters=2.8meters}}} .
Since the circumference of a circle of radius {{{R}}} can be calculated as {{{2pi*R}}} ,
the length of a semicircle of radius {{{R}}} can be calculated as
{{{pi*R}}} .
So, the length needed for the inside semicircle is
{{{pi*(2meters)=about6.28meters}}} ,
and the length needed for the outside semicircle is
{{{pi*(2.8meters)=about8.80meters}}} .
The total length of the bent iron used to make the arch is about
{{{9.6meters+6.28meters+8.80meters=highlight(24.68meters)}}} .