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Number of Diagonals in Regular Polygons and Representations

Number of Diagonals in Regular Polygons and Representations. How about symbolically finding a functional relationship?. Recursive relationship?. 7 sides give 14 diagonals ↑. 6 sides give 9 diagonals ↑. 5 sides give 5 diagonals ↑. 4 sides give 2 diagonals ↑.

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Number of Diagonals in Regular Polygons and Representations

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  1. Number of Diagonals in Regular Polygons and Representations How about symbolically finding a functional relationship? Recursive relationship? 7 sides give 14 diagonals ↑ 6 sides give 9 diagonals ↑ 5 sides give 5 diagonals ↑ 4 sides give 2 diagonals ↑ 3 sides give no diagonals ↑ What about a polygon with 100 sides? ↑______________________________↑ 7 sides give 14 diagonals ↑_____________________↑ 6 sides gives 9 diagonals ↑______________↑ 5 sides gives 5 diagonals ↑________↑ 4 side gives 2 diagonals ↑ 3 sides gives no diagonals

  2. Deriving a functional relationship... Diagonal numbers

  3. How does this set compare to other known sets? Prime numbers? Even numbers? Odd numbers? Triangular numbers? Square numbers? Diagonal numbers Triangular numbers

  4. Comparing the two models… Triangularnumbers Diagonal Numbers Whatarethedifferencesbetweenthetwomodels? What are the similarities between the two models? But… So…. The setnumbers do not match The setvalues do not match In bothmodelseachsuccessivestepisone larger thanthepreviousstep Can we say this symbollically? Or in symbollic language… Letsintroducesomesymbolsfortheterms…. And Then, nd≠ ntandNd≠Nt Letting the step number be n and the step value associated withstep n be S(n), S(n) = S(n-1) + 1 Letthenumberofsides in anygivenpolygonbend LettheTriangularsetnumberbent LettheTriangularsetvalueassociatedwithntbeNt Letthenumberofdiagonals in a polygonofndsidesbeNd

  5. Triangularnumbers Diagonal Numbers More specifically, howcanwe express theseinequalities? Summarisingwhatweknow…. Rearranging (1), nt = nd- 2 (4) nd= nt+ 2 (1) Rearranging (2), Nt = Nd + 1 (5) Foreach, relatedstep, thenumberofsides in a polygonistwomorethanthetriangulartermnumber. From equation one it follows that the triangular term number is equal to the number of sides in a polygon minus 2. Nd= Nt– 1 (2) Substituting (4) and (5) into (3), Nd +1 = (nt- 2)[(nt– 2) +1] (6) 2 Toget a formulainvolvingonlythetermsrelatingtothepolygon/numberofdiagonalsset, wecansubstituteformulae (4) and (5) intoformula (3) Fromequationtwoitfollowsthatthetriangularsetvalueisequaltothenumberofdiagonals plus one. Anythingelse? nd= nt+ 2 Formulaforthetriangularset Nt = n(n+1) (3) 2 Simplifying (6), Nd = (nt- 2)(nt– 1) -1 (7) 2 and Simplifying (7), Nd = nt2- 3nt + 2 -2 (8) 2 Foreachrelatedstep, thenumberofdiagonalsis 1 lessthanthetriangularsetvalue Nd= Nt- 1 Simplifying (8), Nd = nt2- 3ntQED 2

  6. Nd = nt2 - 3nt 2 Now check! Substitute somevalues: In a polygonof 3 sides, nd= 3, so accordingtoourformulathenumberofdiagonalsis: Nd(3) = (3)2 - 3x(3) = 9 -9 = 0 2 2 Diagonal Numbers In a polygonof 4 sides, nd= 4, so accordingtoourformulathenumberofdiagonalsis: Nd(4) = (4)2 - 3x(4) = 16 -12 = 4 = 2 2 2 2 What about if nt is 2? In a polygonof 7 sides, nd= 4, so accordingtoourformulathenumberofdiagonalsis: So, Nd= nt2 - 3nt 2 Ifnt≥ 3 Nd(7) = (7)2 - 3x(7) = 49 -21 = 28 = 14 2 2 2

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