Octagonal: calculate side length, circumference and area

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Octagonal: calculate side length, circumference and area

Postby Guest » Sun Jun 03, 2018 5:24 am

Square-shaped piece of paper is intended to make a regular octagon through the cutting of the vertices of a square.
The length of the piece of paper is 29 cm.
How long triangle cathetus have to be cut off from the vertices of the square?
Calculate the octagonal side length, circumference, and area.
Sorry for my english, if it's not understandable I'll try to explain it better.. but how do I do this?
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Re: Octagonal: calculate side length, circumference and area

Postby 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
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