Question 739054
NOTE:
I do not know what is expected of you because I do not know if this is homework for fifth grade, for college, or somewhere in between. I assume that using trigonometric functions is acceptable. I suspect that the intention is showing (by examples) how the perimeter approaches the circumference of a circle as the number of sides increases. (Note how {{{2pi=about6.28}}} is always in between the perimeter for the inscribed and circumscribed polygons, an how both approach {{{2pi=about6.28}}} as the number of sides increases)
 
THE SOLUTION:
We can only solve the problem if those are regular (symmetrical) polygons, because otherwise the perimeter would change with the shape.
Connecting the vertices to the  center, we can split a polygon with {{{n}}} sides into {{{n}}} congruent isosceles triangles. If we draw the altitudes, we split each isosceles triangle into two congruent right triangles, for a total of {{{2n}}} right triangles.
{{{drawing(300,200,-9,9,-1,11,
triangle(-7.265,0,7.265,0,0,10),
green(rectangle(0,0,0.4,0.4)),green(line(0,0,0,10)),
locate(-7.5,-0.1,A),locate(7.3,-0.1,B),
locate(-0.2,-0.1,P),locate(-0.2,10.9,O)
)}}} {{{O}}} is the center of the circle and the polygon; {{{AB}}} is the side of the polygon
The angles AOP and POB measure {{{pi/n}}} (or {{{180^o/n}}} if you prefer degrees rather than radians)
 
FOR AN INSCRIBED POLYGON:
A and B are points on the circle and OA and OB are radii with length 
{{{OA=OB=1}}}
The length of the side of the polygon is
{{{AB=AP+PB=2*PB=2*(OB*sin(pi/n))=2*1*sin(pi/n)=2sin(pi/n)}}}
and the perimeter is
{{{highlight(P(n)=2n*sin(pi/n))}}}
{{{P(3)=6sin(pi/3)=6sin(60^o)=6(sqrt(3)/2)=3sqrt(3)=about5.20}}}
{{{P(4)=8sin(pi/4)=8sin(45^o)=8(sqrt(2)/2)=4sqrt(2)=about5.66}}}
{{{P(5)=10sin(pi/5)=10sin(36^o)=about5.88}}}
{{{P(6)=12sin(pi/6)=12sin(30^o)=12(1/2)=6}}}
{{{P(8)=16sin(pi/8)=16sin(22.5^o)=about6.12}}}
 
FOR A CIRCUMSCRIBED POLYGON:
P is on the circle, OP is a radius with lenght {{{OP=1}}}
The length of the side of the polygon is
{{{AB=AP+PB=2*PB=2*(OP*tan(pi/n))=2*1*tan(pi/n)=2tan(pi/n)}}}
and the perimeter is
{{{highlight(P(n)=2n*tan(pi/n))}}}
{{{P(3)=6tan(pi/3)=6tan(60^o)=6sqrt(3)=6sqrt(3)=about10.39}}}
{{{P(4)=8tan(pi/4)=8tan(45^o)=8*1=8}}}
{{{P(5)=10tan(pi/5)=10tan(36^o)=about7.27}}}
{{{P(6)=12tan(pi/6)=12tan(30^o)=12(sqrt(3)/3)=4sqrt(3)=about6.93}}}
{{{P(8)=16tan(pi/8)=16tan(22.5^o)=about6.63}}}