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Calculate Side of Octagon, Trigonometry

by Bryan
(UK)











































if the diameter of a swimming pool is 1.8m and it needs enclosing in an octagonal surround made up of eight individual panels. How long should each panel be?

Comments for Calculate Side of Octagon, Trigonometry

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Jul 27, 2011
Calculate Side of Octagon
by: Staff


The question:

by Bryan
(UK)

if the diameter of a swimming pool is 1.8m and it needs enclosing in an octagonal surround made up of eight individual panels. How long should each panel be?


The answer:

First, draw a diagram of the problem.


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http://www.solving-math-problems.com/images/trig-01-stmt.png


Second, compute the interior and exterior angles for the octagon


Interior angle = (n-2) × 180° / n

Interior angle = (8-2) × 180° / 8

Interior angle = (6) × 180° / 8

Interior angle = 135°

So the Exterior Angle is just 180° - Interior Angle

So the Exterior Angle is just 180° - 135°

So the Exterior Angle is just 45°



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http://www.solving-math-problems.com/images/trig-02-ext.png


Third, draw right triangles with the following angles on the upper left and upper right segments of the octagon: 90°, 45°, 45°

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http://www.solving-math-problems.com/images/trig-03-angles.png



x = the length of each section of the octagon


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http://www.solving-math-problems.com/images/trig-04-sides.png



If the length of each panel = x, the lengths of the other two sides of the right triangles is x/sqrt(2).


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http://www.solving-math-problems.com/images/trig-05-tri-sides.png



Solving for x (the length of each panel in the regular octagon)

x/sqrt(2) + x + x/sqrt(2) = 1.8

2x/sqrt(2) + x = 1.8

x*[2/sqrt(2) + 1] = 1.8

x*[2/sqrt(2) + 1]/[2/sqrt(2) + 1] = 1.8/[2/sqrt(2) + 1]

x*1 = 1.8/[2/sqrt(2) + 1]

x = 1.8/[2/sqrt(2) + 1]

x = 0.745584 meters

the final answer is:

the length of each panel should be 0.745584 meters



check the answer

0.745584*cos(45°) + 0.745584 + 0.745584*cos(45°) = 1.8 meters

0.745584*0.7071067812 + 0.745584 + 0.745584*0.7071067812
= 1.8 meters, OK




Thanks for writing.

Staff
www.solving-math-problems.com


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