Question 1111020
Here is the circle, centered at O, and  the octagon,
with sides AB, CD, EF, and GH extended in both directions
up to the point where extension meets extension:
{{{drawing(320,320,-16,16,-16,16,
red(circle(0,0,14.485)),
green(triangle(0,0,0,-14.485,6,-14.485)),
green(rectangle(0,-14.485,0.6,-13.885)),
green(triangle(-6,-14.485,-14.485,-6,0,0)),
green(triangle(-6,14.485,-14.485,6,0,0)),
green(triangle(6,14.485,14.485,6,0,0)),
green(triangle(6,-14.485,14.485,-6,0,0)),
green(rectangle(-14.485,-14.485,14.485,14.485)),
triangle(-6,-14.485,-14.485,-14.485,-14.485,-6),
triangle(-6,14.485,-14.485,14.485,-14.485,6),
line(-6,14.485,6,14.485),line(-6,-14.485,6,-14.485),
line(-14.485,-6,-14.485,6),line(14.485,-6,14.485,6),
line(-14.485,-6,-6,-14.485),line(-14.485,6,-6,14.485),
line(14.485,-6,6,-14.485),line(14.485,6,6,14.485),
locate(-6.3,-14.5,A),locate(5.7,-14.5,B),
locate(-6.3,16,F),locate(5.7,16,E),
locate(-15.7,-5.3,H),locate(-15.7,6.7,G),
locate(14.7,-5.3,C),locate(14.7,6.7,D),
locate(0.6,0.7,red(O)),locate(-0.3,-14.5,red(T)),
locate(-15.7,-13.6,P),locate(-15.7,15.5,Q),
locate(14.7,15.5,green(R)),locate(14.7,-13.6,green(S))
)}}}
a) Angle BAH is one of the internal angles of the octagon,
and it is the supplement of angle PAH,
which is one of the external angles of the octagon.
There is a formula about the internal angles of convex polygons,
but I find external angle measures easier to calculate.
The external angle of a convex polygon,
is the change of direction as you "turn the corner"
around one of the vertices of the polygon.
For example, as you you goo from B to A,
and turn around A to head to H,
your change of direction is angle PAH.
As you go a whole turn around any convex polygon,
your changes of direction add up to a whole turn, {{{360^o}}} ,
so its stands to reason that for a convex polygon
the sum of the measures of the external angles is {{{360^o}}} .
In the case of a regular octagon, like ABCDEFGH,
all {{{8}}} angles external have the same measure,
so angle {{{PAH=AHP=360^o/8=45^o}}} .
Angle BAH is supplementary to PAH, so
{{{BAH=180^o-PAH=180^o-35^o=highlight(135^o)}}} .
Also, in triangle PAH, as {{{PAH=AHP=45^o}}} ,
the measure of the other angle, APH is
{{{180^o-(PAH+AHP)=180^o-(45^o+45^o)=180^o-90^o=highlight(90^o)}}}
 
b) Triangle PAH is a right triangle with two congruent {{{45^o}}} anglem
and that means that the opposite sides are also congruent,
so {{{AP=PH}}} .
That triangle is an isosceles right triangle,
so, according to the Pythagorean theorem
{{{AP^2+PH^2=AH^2}}} .
Substituting {{{AP=PH}}} and {{{AH=12cm}}} ,
{{{2PH^2=(12cm)^2}}}
{{{2PH^2=144cm^2}}}
{{{PH^2=144cm^2/2}}} 
{{{PH^2=72cm^2}}} 
{{{PH=sqrt(72cm^2)=highlight(6sqrt(2))}}}{{{cm}}}
As ABCDEFGH is a regular octagon,
everything already prove/calculated about triangle PAH
is also true about triangles GQF, ERD , and CSB.
So, {{{GQ=PH=6sqrt(2)cm}}} , and as {{{HG=12cm}}} ,
{{{PQ=PH+HG+GQ+(6sqrt(2)+12+6sqrt(2))cm=highlight(12+12sqrt(2))}}}{{{cm}}} .
Of course, the other sides of quadrilateral PQRS also measure
{{{(12+12sqrt(2))cm}}} , and PQRS is a square.
{{{area(APH)=AP*PH/2=(6sqrt(2)cm)*(6sqrt(2)cm)/2=highlight(36cm^2)}}} ,
and each of the areas of GQF, ERD , and CSB are also {{{36cm^2}}}
{{{area(ABCDEFGH)=area(PQRS)-(area(PAH)+area(GQF)+area(ERD)+area(CSB))=area(PQRS)-4*area(PAH)}}}
{{{area(PQRS)=((12+12sqrt(2))cm)^2=(144+288+288sqrt(2))cm^2}}} , so
{{{area(ABCDEFGH)=(144+288+288sqrt(2))cm^2-4*36cm^2=(144+288+288sqrt(2))cm^2-144cm^2=highlight(288+288sqrt(2))}}}{{{cm^2}}} .
 
c) For a regular polygon, {{{area=perimeter*apothem/2}}}
The apothem of ABCDEFGH is {{{OT}}} , and is the radius of the circle.
{{{area(ABCDEFGH)=(288+288sqrt(2))cm^2}}} and
{{{perimeter(ABCDEFGH)=8*(12cm)=96cm}}} , so
{{{(288+288sqrt(2))cm^2=(96cm)(OT)/2}}} .
Solving for OT, we find the radius of the circle as
{{{OT=2*(288+288sqrt(2))/96}}}{{{cm=highlight(6+6sqrt(2))}}}{{{cm}}}
The area of the circle is
{{{pi*radius^2=pi(6+6sqrt(2))^2}}}{{{cm^2=pi(36+72+72sqrt(2))cm^2=pi(108+72sqrt(2))cm^2}}}
The ratio of that area to the area of the octagon is
{{{pi(108+72sqrt(2))/(288+288sqrt(2))=about0.948}}} .
As {{{0.948=94.8/100="94.8%"}}} ,
The area of the circle is approximately {{{highlight("94.8%")}}} of the area of the octagon.