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Designing an Octagon Hay Feeder

Designing an 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. Definitions. Polygon

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Designing an Octagon Hay Feeder

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  1. Designing an Octagon Hay Feeder Dickson Octagon Hay Feeder

  2. 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 Dickson Octagon Hay Feeder

  3. 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 Dickson Octagon Hay Feeder

  4. 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) Dickson Octagon Hay Feeder

  5. 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 Dickson Octagon Hay Feeder

  6. 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° Dickson Octagon Hay Feeder

  7. 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 Dickson Octagon Hay Feeder

  8. 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” Dickson Octagon Hay Feeder

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