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Dickson Octagon Hay Feeder Designing an Octagon Hay Feeder.

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Presentation on theme: "Dickson Octagon Hay Feeder Designing an Octagon Hay Feeder."— Presentation transcript:

1 Dickson Octagon Hay Feeder Designing an Octagon Hay Feeder

2 Dickson Octagon Hay Feeder Objectives The student will Translate word phrases & sentences into expressions and equations Solve linear equations (determine angle and degree of cut) Draw and analyze dimensional figures Use tools to construct figures

3 Dickson Octagon Hay Feeder Definitions Polygon –A simple closed curve made up of segments (each called a “side”) Complimentary angles –Two angles whose sum is 90 degrees Perpendicular –Two lines intersecting in a 90 degree angle

4 Dickson Octagon Hay Feeder Definitions (cont.) Proportion –A statement that ratios are equal Unit of measure –Linear feet or inches Octagon –Eight sided polygon Pentagon (5 sides) and Hexagon (6 sides)

5 Dickson Octagon Hay Feeder Formula for Interior Angle of a Polygon Angle = (N-2) 180° N (number of sides minus two, multiplied by 180 degrees, then divided by the number of sides) N = number of sides

6 Dickson Octagon Hay Feeder Formula for “cut of angle” Divide the degree of angle by 2 Subtract the degree from 90 Ex. Degree of cut = 135°÷ 2 67.5° = 90° - 67.5° = 22.5°

7 Dickson Octagon Hay Feeder Optional Formula for “cut of angle” Shortcut for finding the “cut of angle” is to use 180º divided by the number of sides  (180º  N) Example: “cut of angle” for a pentagon 180º  5 = 36º

8 Dickson Octagon Hay Feeder Finding the Length of Side To find the length of the sides of an octagon feeder is to use the formula 2rTan(180º/N) or dTan(180º/N), where r is radius and d is diameter. Ex.: 7’ diameter feeder 7Tan(180º/8) = 7Tan(22.5º) = 7(.4142) = 2.9” or 2’ 11”


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