Paper Folding Fractions
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This example demonstrates the concept of fractions through paper folding. It involves mathematical concepts such as fractions, inequalities, and rotation. Using GeoGebra commands like Take, Reverse, Element, etc., a series of geometric shapes can be constructed. Suitable for students at the junior high school level and above to learn about fractions and geometric construction.
GeoGebra Commands & Parameters
Take(Text[num], 3, Length[Text[num]]) → text1Reverse(list1) → list2Element(list2, A2) → B2Polygon((0, 0), (1, 0), (1, 0.2), (0, 0.2)) → poly1, f, g, h, iTextfield(num) → InputBox1Length(Text[num]) → aElement(list2, A1) → B1Element(list2, A3) → B3Element(list2, A4) → B4Element(list2, A5) → B5Element(list2, A6) → B6Element(list2, A7) → B7Element(list2, A8) → B8Element(list2, A9) → B9Element(list2, A10) → B10Element(list2, A11) → B11Element(list2, A12) → B12Element(list2, A13) → B13Element(list2, A14) → B14If(Defined[B2], If[B2 ≟ 1, Dilate[C1, dfactor, (1, 0)], Dilate[C1, dfactor, (0, 0)]], (NaN, NaN)) → C2Element(list2, A15) → B15If(Defined[B3], If[B3 ≟ 1, Dilate[C2, dfactor, (1, 0)], Dilate[C2, dfactor, (0, 0)]], (NaN, NaN)) → C3If(Defined[B4], If[B4 ≟ 1, Dilate[C3, dfactor, (1, 0)], Dilate[C3, dfactor, (0, 0)]], (NaN, NaN)) → C4If(Defined[B5], If[B5 ≟ 1, Dilate[C4, dfactor, (1, 0)], Dilate[C4, dfactor, (0, 0)]], (NaN, NaN)) → C5If(Defined[B6], If[B6 ≟ 1, Dilate[C5, dfactor, (1, 0)], Dilate[C5, dfactor, (0, 0)]], (NaN, NaN)) → C6If(Defined[B7], If[B7 ≟ 1, Dilate[C6, dfactor, (1, 0)], Dilate[C6, dfactor, (0, 0)]], (NaN, NaN)) → C7If(Defined[B8], If[B8 ≟ 1, Dilate[C7, dfactor, (1, 0)], Dilate[C7, dfactor, (0, 0)]], (NaN, NaN)) → C8If(Defined[B9], If[B9 ≟ 1, Dilate[C8, dfactor, (1, 0)], Dilate[C8, dfactor, (0, 0)]], (NaN, NaN)) → C9If(Defined[B10], If[B10 ≟ 1, Dilate[C9, dfactor, (1, 0)], Dilate[C9, dfactor, (0, 0)]], (NaN, NaN)) → C10If(Defined[B11], If[B11 ≟ 1, Dilate[C10, dfactor, (1, 0)], Dilate[C10, dfactor, (0, 0)]], (NaN, NaN)) → C11If(Defined[B12], If[B12 ≟ 1, Dilate[C11, dfactor, (1, 0)], Dilate[C11, dfactor, (0, 0)]], (NaN, NaN)) → C12If(Defined[B13], If[B13 ≟ 1, Dilate[C12, dfactor, (1, 0)], Dilate[C12, dfactor, (0, 0)]], (NaN, NaN)) → C13If(Defined[B14], If[B14 ≟ 1, Dilate[C13, dfactor, (1, 0)], Dilate[C13, dfactor, (0, 0)]], (NaN, NaN)) → C14If(Defined[B15], If[B15 ≟ 1, Dilate[C14, dfactor, (1, 0)], Dilate[C14, dfactor, (0, 0)]], (NaN, NaN)) → C15Segment(C1, C1 + (0, 0.2)) → D1Segment(C2, C2 + (0, 0.2)) → D2Segment(C3, C3 + (0, 0.2)) → D3Segment(C4, C4 + (0, 0.2)) → D4Segment(C5, C5 + (0, 0.2)) → D5Segment(C6, C6 + (0, 0.2)) → D6Segment(C7, C7 + (0, 0.2)) → D7Segment(C8, C8 + (0, 0.2)) → D8Segment(C9, C9 + (0, 0.2)) → D9Segment(C10, C10 + (0, 0.2)) → D10Segment(C11, C11 + (0, 0.2)) → D11Segment(C12, C12 + (0, 0.2)) → D12Segment(C13, C13 + (0, 0.2)) → D13Segment(C14, C14 + (0, 0.2)) → D14Segment(C15, C15 + (0, 0.2)) → D15RemoveUndefined(list3) → list4RemoveUndefined(list5) → list6Sequence(FractionText[x(Element[list6, t])], t, 1, Length[list6]) → list7Segment((0, 0), (0, 0.2)) → jSegment((1, 0), (1, 0.2)) → kVector((1, 0.25), (0, 0.25)) → E1If(B2 ≟ 1, Vector[B, C1 + (0, 0.25)], Vector[A, C1 + (0, 0.25)]) → E2If(B4 ≟ 1, Vector[B, C3 + (0, 0.25)], Vector[A, C3 + (0, 0.25)]) → E4If(B5 ≟ 1, Vector[B, C4 + (0, 0.25)], Vector[A, C4 + (0, 0.25)]) → E5If(B6 ≟ 1, Vector[B, C5 + (0, 0.25)], Vector[A, C5 + (0, 0.25)]) → E6If(B3 ≟ 1, Vector[B, C2 + (0, 0.25)], Vector[A, C2 + (0, 0.25)]) → E3If(B7 ≟ 1, Vector[B, C6 + (0, 0.25)], Vector[A, C6 + (0, 0.25)]) → E7If(B8 ≟ 1, Vector[B, C7 + (0, 0.25)], Vector[A, C7 + (0, 0.25)]) → E8If(B9 ≟ 1, Vector[B, C8 + (0, 0.25)], Vector[A, C8 + (0, 0.25)]) → E9If(B10 ≟ 1, Vector[B, C9 + (0, 0.25)], Vector[A, C9 + (0, 0.25)]) → E10If(B11 ≟ 1, Vector[B, C10 + (0, 0.25)], Vector[A, C10 + (0, 0.25)]) → E11If(B12 ≟ 1, Vector[B, C11 + (0, 0.25)], Vector[A, C11 + (0, 0.25)]) → E12If(B13 ≟ 1, Vector[B, C12 + (0, 0.25)], Vector[A, C12 + (0, 0.25)]) → E13If(B14 ≟ 1, Vector[B, C13 + (0, 0.25)], Vector[A, C13 + (0, 0.25)]) → E14If(B15 ≟ 1, Vector[B, C14 + (0, 0.25)], Vector[A, C14 + (0, 0.25)]) → E15