by **Guest** » Thu Feb 28, 2019 9:39 am

Let "x" be the length of a side of the octagon, centered on a side of the paper. That leaves 29-x cm on that side of the paper so (29- x)/2 on each side of the octagon. The other four sides of the octagon are the hypotenuses of the right triangles cut off. By the Pythagorean theorem that is [tex]\sqrt{\frac{(29- x)^2}{2}}[/tex] and that must be the same as the length of the other sides, x. So we have [tex]\sqrt{\frac{(29- x)^2}{2}}= x[/tex].

Squaring both sides, [tex]\frac{(29- x)^2}{2]= x^2[/tex]. [tex](29- x)^2= 841- 58x+ x^2= 2x^2[/tex], [tex]x^2+58x- 841= 0[/tex]. Solve that quadratic equation, using, perhaps, the quadratic formula. The length of a side of the octagon is the positive root. The circumference is 8 times the length of each side.

To determine the area, imagine drawing lines from the center of the octagon to each vertex. That divides the octagon into 8 isosceles triangles, with vertex angle 360/8= 45 degrees and base the length of a side, x, found above. Drawing the altitude of each triangle (from the center of the octagon to the center of each further divides the those 8 triangles into 16 right triangles with angle 22.5 degrees and the "opposite side" of length x/2. Taking "y" to be the length of that altitude, [tex]\frac{y}{x/2}= tan(22.5)[/tex] so [tex]y= tan(22.5)x/2[/tex]. The area of each of those 16 right triangles is [tex](1/2)xy= x^2 tan(22.5)/4[/tex]. The area of the octagon is 16 times that, [tex]4 x^2 tan(22.5)[/tex].

Let "a" be the length of the congruent sides of one of the triangles. By the cosine law, [tex]x^2= 2a^2- 2acos(45)= 2a^2- \sqrt{2}a= x[/tex], where x is the length of the