How build an octagon whose left to right distance is 4'?

Question

I want to build an octagon with sides 4' across from each parallel side. What will be the length of each equal side? Answer in inches, please. Thank you.

Answer 1

each side will be the sq rt of 2 feet.Draw a square then a line cutting off each corner one foot in from the corner.sq. rt.( 1^2+1^2)=1.4142 ft or 16.97 inches (17 inches will probably be close enough.If it is to be all sides equal then each side needs to be 20 inches long.(19.9")(sqrt of(14^2+14^2))

Answer 2

How thick are the panels?

Answer 3

first and foremost.the sum of interior angles of an octagon is 1080 degrees which gives us 135 degrees per vertex. from this we can derive the angle of the triangle that could produce the edge of the octagon.

Using pythagorean theorem we could obtain the length of the side of the octagon.

y= distance from the vertex to the center of the octagon

x= half of the length of the side of the octagon

thus;

y= square root (x^2+2^2)

sin 22.5=x/y

(sin 22.5)^2= x^2/(x^2+4)

0.146x^2+0.5857=x^2

X^2=0.686

x=0.828

therefore 2x= 1.66

the length of the inner edge of your octagon shall be 1.66

if youre going to join it, ensure that youll give 22.5 degrees bend for its joint to be assured of snug fit....