Drawing an isosceles trapezoid with perpendicular diagonals












8














I need to draw an isosceles trapezoid with perpendicular diagonals, but I am not sure where to start. Here is my code for a square with perpendicular diagonals (if it is even helpful).



begin{tikzpicture}[scale=5.5]
coordinate[label=left:$W$] (W) at (0,0);
coordinate[label=right:$X$] (X) at (1,0);
coordinate[label=:$Y$] (Y) at (1, 1);
coordinate[label=:$Z$] (Z) at (0, 1);
coordinate[label=:$O$] (O) at (0.5, 0.5);
draw (W)--(X)--(Y)--(Z)--(W)--(Y)--(Z)--(X);
draw (O) -- node[sloped] {$|$} (Y);
draw (O) -- node[sloped] {$|$} (Z);
draw (O) -- node[sloped] {$|$} (X);
draw (O) -- node[sloped] {$|$} (W);
end{tikzpicture}


Would I have to mathematically find the coordinates of one such trapezoid or is there another way to do it? All help is appreciated!










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  • How would I implement it? I just learned Latex, so I am a complete beginner.
    – M. C.
    17 hours ago












  • Remember that not any such trapezium has such diagonals. You need to identify under such conditions it has.
    – Sigur
    16 hours ago










  • Observe that the diagonals determine an isosceles right triangle.
    – Sigur
    16 hours ago










  • Yes, I checked the code you wrote and it works out well. I guess my problem statement was a bit unclear. Here is what I meant to say: I need a trapezoid with equal base angles and hence equal left and right side lengths. I also want the diagonals to be drawn and to be perpendicular.
    – M. C.
    16 hours ago












  • @M.C., so that is the case. If trapezium has congruent base angles so ABO should be like that. But, if you want to specify the base angles, so that is another history.
    – Sigur
    16 hours ago
















8














I need to draw an isosceles trapezoid with perpendicular diagonals, but I am not sure where to start. Here is my code for a square with perpendicular diagonals (if it is even helpful).



begin{tikzpicture}[scale=5.5]
coordinate[label=left:$W$] (W) at (0,0);
coordinate[label=right:$X$] (X) at (1,0);
coordinate[label=:$Y$] (Y) at (1, 1);
coordinate[label=:$Z$] (Z) at (0, 1);
coordinate[label=:$O$] (O) at (0.5, 0.5);
draw (W)--(X)--(Y)--(Z)--(W)--(Y)--(Z)--(X);
draw (O) -- node[sloped] {$|$} (Y);
draw (O) -- node[sloped] {$|$} (Z);
draw (O) -- node[sloped] {$|$} (X);
draw (O) -- node[sloped] {$|$} (W);
end{tikzpicture}


Would I have to mathematically find the coordinates of one such trapezoid or is there another way to do it? All help is appreciated!










share|improve this question







New contributor




M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • How would I implement it? I just learned Latex, so I am a complete beginner.
    – M. C.
    17 hours ago












  • Remember that not any such trapezium has such diagonals. You need to identify under such conditions it has.
    – Sigur
    16 hours ago










  • Observe that the diagonals determine an isosceles right triangle.
    – Sigur
    16 hours ago










  • Yes, I checked the code you wrote and it works out well. I guess my problem statement was a bit unclear. Here is what I meant to say: I need a trapezoid with equal base angles and hence equal left and right side lengths. I also want the diagonals to be drawn and to be perpendicular.
    – M. C.
    16 hours ago












  • @M.C., so that is the case. If trapezium has congruent base angles so ABO should be like that. But, if you want to specify the base angles, so that is another history.
    – Sigur
    16 hours ago














8












8








8


2





I need to draw an isosceles trapezoid with perpendicular diagonals, but I am not sure where to start. Here is my code for a square with perpendicular diagonals (if it is even helpful).



begin{tikzpicture}[scale=5.5]
coordinate[label=left:$W$] (W) at (0,0);
coordinate[label=right:$X$] (X) at (1,0);
coordinate[label=:$Y$] (Y) at (1, 1);
coordinate[label=:$Z$] (Z) at (0, 1);
coordinate[label=:$O$] (O) at (0.5, 0.5);
draw (W)--(X)--(Y)--(Z)--(W)--(Y)--(Z)--(X);
draw (O) -- node[sloped] {$|$} (Y);
draw (O) -- node[sloped] {$|$} (Z);
draw (O) -- node[sloped] {$|$} (X);
draw (O) -- node[sloped] {$|$} (W);
end{tikzpicture}


Would I have to mathematically find the coordinates of one such trapezoid or is there another way to do it? All help is appreciated!










share|improve this question







New contributor




M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I need to draw an isosceles trapezoid with perpendicular diagonals, but I am not sure where to start. Here is my code for a square with perpendicular diagonals (if it is even helpful).



begin{tikzpicture}[scale=5.5]
coordinate[label=left:$W$] (W) at (0,0);
coordinate[label=right:$X$] (X) at (1,0);
coordinate[label=:$Y$] (Y) at (1, 1);
coordinate[label=:$Z$] (Z) at (0, 1);
coordinate[label=:$O$] (O) at (0.5, 0.5);
draw (W)--(X)--(Y)--(Z)--(W)--(Y)--(Z)--(X);
draw (O) -- node[sloped] {$|$} (Y);
draw (O) -- node[sloped] {$|$} (Z);
draw (O) -- node[sloped] {$|$} (X);
draw (O) -- node[sloped] {$|$} (W);
end{tikzpicture}


Would I have to mathematically find the coordinates of one such trapezoid or is there another way to do it? All help is appreciated!







tikz-graphdrawing






share|improve this question







New contributor




M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question







New contributor




M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question






New contributor




M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 17 hours ago









M. C.M. C.

514




514




New contributor




M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • How would I implement it? I just learned Latex, so I am a complete beginner.
    – M. C.
    17 hours ago












  • Remember that not any such trapezium has such diagonals. You need to identify under such conditions it has.
    – Sigur
    16 hours ago










  • Observe that the diagonals determine an isosceles right triangle.
    – Sigur
    16 hours ago










  • Yes, I checked the code you wrote and it works out well. I guess my problem statement was a bit unclear. Here is what I meant to say: I need a trapezoid with equal base angles and hence equal left and right side lengths. I also want the diagonals to be drawn and to be perpendicular.
    – M. C.
    16 hours ago












  • @M.C., so that is the case. If trapezium has congruent base angles so ABO should be like that. But, if you want to specify the base angles, so that is another history.
    – Sigur
    16 hours ago


















  • How would I implement it? I just learned Latex, so I am a complete beginner.
    – M. C.
    17 hours ago












  • Remember that not any such trapezium has such diagonals. You need to identify under such conditions it has.
    – Sigur
    16 hours ago










  • Observe that the diagonals determine an isosceles right triangle.
    – Sigur
    16 hours ago










  • Yes, I checked the code you wrote and it works out well. I guess my problem statement was a bit unclear. Here is what I meant to say: I need a trapezoid with equal base angles and hence equal left and right side lengths. I also want the diagonals to be drawn and to be perpendicular.
    – M. C.
    16 hours ago












  • @M.C., so that is the case. If trapezium has congruent base angles so ABO should be like that. But, if you want to specify the base angles, so that is another history.
    – Sigur
    16 hours ago
















How would I implement it? I just learned Latex, so I am a complete beginner.
– M. C.
17 hours ago






How would I implement it? I just learned Latex, so I am a complete beginner.
– M. C.
17 hours ago














Remember that not any such trapezium has such diagonals. You need to identify under such conditions it has.
– Sigur
16 hours ago




Remember that not any such trapezium has such diagonals. You need to identify under such conditions it has.
– Sigur
16 hours ago












Observe that the diagonals determine an isosceles right triangle.
– Sigur
16 hours ago




Observe that the diagonals determine an isosceles right triangle.
– Sigur
16 hours ago












Yes, I checked the code you wrote and it works out well. I guess my problem statement was a bit unclear. Here is what I meant to say: I need a trapezoid with equal base angles and hence equal left and right side lengths. I also want the diagonals to be drawn and to be perpendicular.
– M. C.
16 hours ago






Yes, I checked the code you wrote and it works out well. I guess my problem statement was a bit unclear. Here is what I meant to say: I need a trapezoid with equal base angles and hence equal left and right side lengths. I also want the diagonals to be drawn and to be perpendicular.
– M. C.
16 hours ago














@M.C., so that is the case. If trapezium has congruent base angles so ABO should be like that. But, if you want to specify the base angles, so that is another history.
– Sigur
16 hours ago




@M.C., so that is the case. If trapezium has congruent base angles so ABO should be like that. But, if you want to specify the base angles, so that is another history.
– Sigur
16 hours ago










3 Answers
3






active

oldest

votes


















5














Let ABXY be an isosceles trapezium with perpendicular diagonals AX and BY.



Then, the triangle AOB is isosceles and right at O (ie., the angle at O is right). So, the base angles should have 45 degrees.



Since AB and XY are parallel, to construct the trapezium it is enough to choose the lengths r_1 = OX and r_2 = OA.



If the origin of the coordinate system is O=(0,0) then the vertices can be given in polar coordinates by:



A=(-135:r_2)    B=(- 45:r_2)     X=( 45:r_1)     Y=(135:r_1)


enter image description here



Below, the MWE where the commands



newcommand{radioi}{1cm}
newcommand{radioii}{2cm}


determines the radius r_1 and r_2.



enter image description here



MWE



documentclass[margin=2mm]{standalone}
usepackage{tikz}

begin{document}
begin{tikzpicture}[scale=2]
newcommand{radioi}{1cm}
newcommand{radioii}{2cm}
coordinate[label=below:$O$] (O) at ( 0:0 );
coordinate[label=left:$A$] (A) at (-135:radioii);
coordinate[label=right:$B$] (B) at (- 45:radioii);
coordinate[label=right:$X$] (X) at ( 45:radioi );
coordinate[label=left:$Y$] (Y) at ( 135:radioi );

draw (A) -- (B) -- (X) -- (Y) -- cycle;
draw[dashed] (A) -- (X) (B) -- (Y);
end{tikzpicture}
end{document}





share|improve this answer































    5














    Conceptually the same as @Sigur's nice answer but with slightly different parametrization. There are two free parameters, which can be taken to be the length of the upper edge and the height. They go into commands of the sort



    draw[isosceleles trapezium=of width 2 and height 3 and name my trap];


    This is illustrated in



    documentclass[tikz,border=3.14mm]{standalone}
    begin{document}
    begin{tikzpicture}[isosceleles trapezium/.style args={of width #1 and height #2
    and name #3}{insert path={
    (45:{#1/sqrt(2)}) coordinate(#3-TR) -- (-45:{sqrt(#2*#2-#1*#1/2)}) coordinate(#3-BR)
    -- (-135:{sqrt(#2*#2-#1*#1/2)}) coordinate(#3-BL) -- (135:{#1/sqrt(2)}) coordinate(#3-TL) -- cycle}}]
    draw[isosceleles trapezium=of width 2 and height 3 and name my trap];
    draw[dashed] (my trap-TL) -- (my trap-BR) (my trap-TR) -- (my trap-BL);
    draw[latex-latex] ([yshift=2mm]my trap-TL) -- ([yshift=2mm]my trap-TR)
    node[midway,fill=white] {$w$};
    draw[latex-latex] ([xshift=-2mm]my trap-TL -| my trap-BL) --
    ([xshift=-2mm]my trap-BL) node[midway,fill=white] {$h$};
    begin{scope}[xshift=6cm,rotate=30]
    draw[isosceleles trapezium=of width 1 and height 2.5 and name another trap];
    draw[dashed] (another trap-TL) -- (another trap-BR) (another trap-TR) -- (another trap-BL);
    end{scope}
    end{tikzpicture}
    end{document}


    enter image description here



    So the first parameter is w and the second one h. In addition, there is the name which is used to name the corners such that you can draw the diagonals, say. And of course you can rotate the thing and so on.






    share|improve this answer























    • Very interesting.
      – Sigur
      16 hours ago










    • @Sigur I'll be happy to remove my answer.
      – marmot
      16 hours ago










    • Why? It is elegant and advanced for beginners. Good exercise to study the code.
      – Sigur
      15 hours ago










    • @Sigur Conceptually it is the same as your nice post. I slightly modified the parametrization to make my answer, which came later, not completely pointless.
      – marmot
      15 hours ago










    • You can define the polar command using as two arguments the length of semi-diagonals, as we did before. The computation is simpler and no tan and sqrt needed. Anyway, nice constructions.
      – Sigur
      14 hours ago



















    4














    A PSTricks solution just for fun purposes.



    documentclass[pstricks,12pt]{standalone}
    usepackage{pst-eucl}
    begin{document}
    foreach a in {0,10,...,350}{%
    pspicture(-7,-7)(7,7)
    pstGeonode(0,0){O}(2;a){A}([offset=6]{A}O){B}
    pstRotation[RotAngle=-90]{O}{A}[D]
    pstRotation[RotAngle=90]{O}{B}[C]
    psline(A)(B)(C)(D)(A)(C)(D)(B)
    endpspicture}
    end{document}


    enter image description here



    Algorithm




    • Define two points O and A.

    • Define point B such that OA is perpendicular to OB.

    • Define C as the image of rotating point B about O counter-clockwise.

    • Define D as the image of rotating point A about O clockwise.

    • Draw the lines.






    share|improve this answer























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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5














      Let ABXY be an isosceles trapezium with perpendicular diagonals AX and BY.



      Then, the triangle AOB is isosceles and right at O (ie., the angle at O is right). So, the base angles should have 45 degrees.



      Since AB and XY are parallel, to construct the trapezium it is enough to choose the lengths r_1 = OX and r_2 = OA.



      If the origin of the coordinate system is O=(0,0) then the vertices can be given in polar coordinates by:



      A=(-135:r_2)    B=(- 45:r_2)     X=( 45:r_1)     Y=(135:r_1)


      enter image description here



      Below, the MWE where the commands



      newcommand{radioi}{1cm}
      newcommand{radioii}{2cm}


      determines the radius r_1 and r_2.



      enter image description here



      MWE



      documentclass[margin=2mm]{standalone}
      usepackage{tikz}

      begin{document}
      begin{tikzpicture}[scale=2]
      newcommand{radioi}{1cm}
      newcommand{radioii}{2cm}
      coordinate[label=below:$O$] (O) at ( 0:0 );
      coordinate[label=left:$A$] (A) at (-135:radioii);
      coordinate[label=right:$B$] (B) at (- 45:radioii);
      coordinate[label=right:$X$] (X) at ( 45:radioi );
      coordinate[label=left:$Y$] (Y) at ( 135:radioi );

      draw (A) -- (B) -- (X) -- (Y) -- cycle;
      draw[dashed] (A) -- (X) (B) -- (Y);
      end{tikzpicture}
      end{document}





      share|improve this answer




























        5














        Let ABXY be an isosceles trapezium with perpendicular diagonals AX and BY.



        Then, the triangle AOB is isosceles and right at O (ie., the angle at O is right). So, the base angles should have 45 degrees.



        Since AB and XY are parallel, to construct the trapezium it is enough to choose the lengths r_1 = OX and r_2 = OA.



        If the origin of the coordinate system is O=(0,0) then the vertices can be given in polar coordinates by:



        A=(-135:r_2)    B=(- 45:r_2)     X=( 45:r_1)     Y=(135:r_1)


        enter image description here



        Below, the MWE where the commands



        newcommand{radioi}{1cm}
        newcommand{radioii}{2cm}


        determines the radius r_1 and r_2.



        enter image description here



        MWE



        documentclass[margin=2mm]{standalone}
        usepackage{tikz}

        begin{document}
        begin{tikzpicture}[scale=2]
        newcommand{radioi}{1cm}
        newcommand{radioii}{2cm}
        coordinate[label=below:$O$] (O) at ( 0:0 );
        coordinate[label=left:$A$] (A) at (-135:radioii);
        coordinate[label=right:$B$] (B) at (- 45:radioii);
        coordinate[label=right:$X$] (X) at ( 45:radioi );
        coordinate[label=left:$Y$] (Y) at ( 135:radioi );

        draw (A) -- (B) -- (X) -- (Y) -- cycle;
        draw[dashed] (A) -- (X) (B) -- (Y);
        end{tikzpicture}
        end{document}





        share|improve this answer


























          5












          5








          5






          Let ABXY be an isosceles trapezium with perpendicular diagonals AX and BY.



          Then, the triangle AOB is isosceles and right at O (ie., the angle at O is right). So, the base angles should have 45 degrees.



          Since AB and XY are parallel, to construct the trapezium it is enough to choose the lengths r_1 = OX and r_2 = OA.



          If the origin of the coordinate system is O=(0,0) then the vertices can be given in polar coordinates by:



          A=(-135:r_2)    B=(- 45:r_2)     X=( 45:r_1)     Y=(135:r_1)


          enter image description here



          Below, the MWE where the commands



          newcommand{radioi}{1cm}
          newcommand{radioii}{2cm}


          determines the radius r_1 and r_2.



          enter image description here



          MWE



          documentclass[margin=2mm]{standalone}
          usepackage{tikz}

          begin{document}
          begin{tikzpicture}[scale=2]
          newcommand{radioi}{1cm}
          newcommand{radioii}{2cm}
          coordinate[label=below:$O$] (O) at ( 0:0 );
          coordinate[label=left:$A$] (A) at (-135:radioii);
          coordinate[label=right:$B$] (B) at (- 45:radioii);
          coordinate[label=right:$X$] (X) at ( 45:radioi );
          coordinate[label=left:$Y$] (Y) at ( 135:radioi );

          draw (A) -- (B) -- (X) -- (Y) -- cycle;
          draw[dashed] (A) -- (X) (B) -- (Y);
          end{tikzpicture}
          end{document}





          share|improve this answer














          Let ABXY be an isosceles trapezium with perpendicular diagonals AX and BY.



          Then, the triangle AOB is isosceles and right at O (ie., the angle at O is right). So, the base angles should have 45 degrees.



          Since AB and XY are parallel, to construct the trapezium it is enough to choose the lengths r_1 = OX and r_2 = OA.



          If the origin of the coordinate system is O=(0,0) then the vertices can be given in polar coordinates by:



          A=(-135:r_2)    B=(- 45:r_2)     X=( 45:r_1)     Y=(135:r_1)


          enter image description here



          Below, the MWE where the commands



          newcommand{radioi}{1cm}
          newcommand{radioii}{2cm}


          determines the radius r_1 and r_2.



          enter image description here



          MWE



          documentclass[margin=2mm]{standalone}
          usepackage{tikz}

          begin{document}
          begin{tikzpicture}[scale=2]
          newcommand{radioi}{1cm}
          newcommand{radioii}{2cm}
          coordinate[label=below:$O$] (O) at ( 0:0 );
          coordinate[label=left:$A$] (A) at (-135:radioii);
          coordinate[label=right:$B$] (B) at (- 45:radioii);
          coordinate[label=right:$X$] (X) at ( 45:radioi );
          coordinate[label=left:$Y$] (Y) at ( 135:radioi );

          draw (A) -- (B) -- (X) -- (Y) -- cycle;
          draw[dashed] (A) -- (X) (B) -- (Y);
          end{tikzpicture}
          end{document}






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 13 hours ago

























          answered 16 hours ago









          SigurSigur

          24.4k355138




          24.4k355138























              5














              Conceptually the same as @Sigur's nice answer but with slightly different parametrization. There are two free parameters, which can be taken to be the length of the upper edge and the height. They go into commands of the sort



              draw[isosceleles trapezium=of width 2 and height 3 and name my trap];


              This is illustrated in



              documentclass[tikz,border=3.14mm]{standalone}
              begin{document}
              begin{tikzpicture}[isosceleles trapezium/.style args={of width #1 and height #2
              and name #3}{insert path={
              (45:{#1/sqrt(2)}) coordinate(#3-TR) -- (-45:{sqrt(#2*#2-#1*#1/2)}) coordinate(#3-BR)
              -- (-135:{sqrt(#2*#2-#1*#1/2)}) coordinate(#3-BL) -- (135:{#1/sqrt(2)}) coordinate(#3-TL) -- cycle}}]
              draw[isosceleles trapezium=of width 2 and height 3 and name my trap];
              draw[dashed] (my trap-TL) -- (my trap-BR) (my trap-TR) -- (my trap-BL);
              draw[latex-latex] ([yshift=2mm]my trap-TL) -- ([yshift=2mm]my trap-TR)
              node[midway,fill=white] {$w$};
              draw[latex-latex] ([xshift=-2mm]my trap-TL -| my trap-BL) --
              ([xshift=-2mm]my trap-BL) node[midway,fill=white] {$h$};
              begin{scope}[xshift=6cm,rotate=30]
              draw[isosceleles trapezium=of width 1 and height 2.5 and name another trap];
              draw[dashed] (another trap-TL) -- (another trap-BR) (another trap-TR) -- (another trap-BL);
              end{scope}
              end{tikzpicture}
              end{document}


              enter image description here



              So the first parameter is w and the second one h. In addition, there is the name which is used to name the corners such that you can draw the diagonals, say. And of course you can rotate the thing and so on.






              share|improve this answer























              • Very interesting.
                – Sigur
                16 hours ago










              • @Sigur I'll be happy to remove my answer.
                – marmot
                16 hours ago










              • Why? It is elegant and advanced for beginners. Good exercise to study the code.
                – Sigur
                15 hours ago










              • @Sigur Conceptually it is the same as your nice post. I slightly modified the parametrization to make my answer, which came later, not completely pointless.
                – marmot
                15 hours ago










              • You can define the polar command using as two arguments the length of semi-diagonals, as we did before. The computation is simpler and no tan and sqrt needed. Anyway, nice constructions.
                – Sigur
                14 hours ago
















              5














              Conceptually the same as @Sigur's nice answer but with slightly different parametrization. There are two free parameters, which can be taken to be the length of the upper edge and the height. They go into commands of the sort



              draw[isosceleles trapezium=of width 2 and height 3 and name my trap];


              This is illustrated in



              documentclass[tikz,border=3.14mm]{standalone}
              begin{document}
              begin{tikzpicture}[isosceleles trapezium/.style args={of width #1 and height #2
              and name #3}{insert path={
              (45:{#1/sqrt(2)}) coordinate(#3-TR) -- (-45:{sqrt(#2*#2-#1*#1/2)}) coordinate(#3-BR)
              -- (-135:{sqrt(#2*#2-#1*#1/2)}) coordinate(#3-BL) -- (135:{#1/sqrt(2)}) coordinate(#3-TL) -- cycle}}]
              draw[isosceleles trapezium=of width 2 and height 3 and name my trap];
              draw[dashed] (my trap-TL) -- (my trap-BR) (my trap-TR) -- (my trap-BL);
              draw[latex-latex] ([yshift=2mm]my trap-TL) -- ([yshift=2mm]my trap-TR)
              node[midway,fill=white] {$w$};
              draw[latex-latex] ([xshift=-2mm]my trap-TL -| my trap-BL) --
              ([xshift=-2mm]my trap-BL) node[midway,fill=white] {$h$};
              begin{scope}[xshift=6cm,rotate=30]
              draw[isosceleles trapezium=of width 1 and height 2.5 and name another trap];
              draw[dashed] (another trap-TL) -- (another trap-BR) (another trap-TR) -- (another trap-BL);
              end{scope}
              end{tikzpicture}
              end{document}


              enter image description here



              So the first parameter is w and the second one h. In addition, there is the name which is used to name the corners such that you can draw the diagonals, say. And of course you can rotate the thing and so on.






              share|improve this answer























              • Very interesting.
                – Sigur
                16 hours ago










              • @Sigur I'll be happy to remove my answer.
                – marmot
                16 hours ago










              • Why? It is elegant and advanced for beginners. Good exercise to study the code.
                – Sigur
                15 hours ago










              • @Sigur Conceptually it is the same as your nice post. I slightly modified the parametrization to make my answer, which came later, not completely pointless.
                – marmot
                15 hours ago










              • You can define the polar command using as two arguments the length of semi-diagonals, as we did before. The computation is simpler and no tan and sqrt needed. Anyway, nice constructions.
                – Sigur
                14 hours ago














              5












              5








              5






              Conceptually the same as @Sigur's nice answer but with slightly different parametrization. There are two free parameters, which can be taken to be the length of the upper edge and the height. They go into commands of the sort



              draw[isosceleles trapezium=of width 2 and height 3 and name my trap];


              This is illustrated in



              documentclass[tikz,border=3.14mm]{standalone}
              begin{document}
              begin{tikzpicture}[isosceleles trapezium/.style args={of width #1 and height #2
              and name #3}{insert path={
              (45:{#1/sqrt(2)}) coordinate(#3-TR) -- (-45:{sqrt(#2*#2-#1*#1/2)}) coordinate(#3-BR)
              -- (-135:{sqrt(#2*#2-#1*#1/2)}) coordinate(#3-BL) -- (135:{#1/sqrt(2)}) coordinate(#3-TL) -- cycle}}]
              draw[isosceleles trapezium=of width 2 and height 3 and name my trap];
              draw[dashed] (my trap-TL) -- (my trap-BR) (my trap-TR) -- (my trap-BL);
              draw[latex-latex] ([yshift=2mm]my trap-TL) -- ([yshift=2mm]my trap-TR)
              node[midway,fill=white] {$w$};
              draw[latex-latex] ([xshift=-2mm]my trap-TL -| my trap-BL) --
              ([xshift=-2mm]my trap-BL) node[midway,fill=white] {$h$};
              begin{scope}[xshift=6cm,rotate=30]
              draw[isosceleles trapezium=of width 1 and height 2.5 and name another trap];
              draw[dashed] (another trap-TL) -- (another trap-BR) (another trap-TR) -- (another trap-BL);
              end{scope}
              end{tikzpicture}
              end{document}


              enter image description here



              So the first parameter is w and the second one h. In addition, there is the name which is used to name the corners such that you can draw the diagonals, say. And of course you can rotate the thing and so on.






              share|improve this answer














              Conceptually the same as @Sigur's nice answer but with slightly different parametrization. There are two free parameters, which can be taken to be the length of the upper edge and the height. They go into commands of the sort



              draw[isosceleles trapezium=of width 2 and height 3 and name my trap];


              This is illustrated in



              documentclass[tikz,border=3.14mm]{standalone}
              begin{document}
              begin{tikzpicture}[isosceleles trapezium/.style args={of width #1 and height #2
              and name #3}{insert path={
              (45:{#1/sqrt(2)}) coordinate(#3-TR) -- (-45:{sqrt(#2*#2-#1*#1/2)}) coordinate(#3-BR)
              -- (-135:{sqrt(#2*#2-#1*#1/2)}) coordinate(#3-BL) -- (135:{#1/sqrt(2)}) coordinate(#3-TL) -- cycle}}]
              draw[isosceleles trapezium=of width 2 and height 3 and name my trap];
              draw[dashed] (my trap-TL) -- (my trap-BR) (my trap-TR) -- (my trap-BL);
              draw[latex-latex] ([yshift=2mm]my trap-TL) -- ([yshift=2mm]my trap-TR)
              node[midway,fill=white] {$w$};
              draw[latex-latex] ([xshift=-2mm]my trap-TL -| my trap-BL) --
              ([xshift=-2mm]my trap-BL) node[midway,fill=white] {$h$};
              begin{scope}[xshift=6cm,rotate=30]
              draw[isosceleles trapezium=of width 1 and height 2.5 and name another trap];
              draw[dashed] (another trap-TL) -- (another trap-BR) (another trap-TR) -- (another trap-BL);
              end{scope}
              end{tikzpicture}
              end{document}


              enter image description here



              So the first parameter is w and the second one h. In addition, there is the name which is used to name the corners such that you can draw the diagonals, say. And of course you can rotate the thing and so on.







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited 13 hours ago

























              answered 16 hours ago









              marmotmarmot

              90.4k4104195




              90.4k4104195












              • Very interesting.
                – Sigur
                16 hours ago










              • @Sigur I'll be happy to remove my answer.
                – marmot
                16 hours ago










              • Why? It is elegant and advanced for beginners. Good exercise to study the code.
                – Sigur
                15 hours ago










              • @Sigur Conceptually it is the same as your nice post. I slightly modified the parametrization to make my answer, which came later, not completely pointless.
                – marmot
                15 hours ago










              • You can define the polar command using as two arguments the length of semi-diagonals, as we did before. The computation is simpler and no tan and sqrt needed. Anyway, nice constructions.
                – Sigur
                14 hours ago


















              • Very interesting.
                – Sigur
                16 hours ago










              • @Sigur I'll be happy to remove my answer.
                – marmot
                16 hours ago










              • Why? It is elegant and advanced for beginners. Good exercise to study the code.
                – Sigur
                15 hours ago










              • @Sigur Conceptually it is the same as your nice post. I slightly modified the parametrization to make my answer, which came later, not completely pointless.
                – marmot
                15 hours ago










              • You can define the polar command using as two arguments the length of semi-diagonals, as we did before. The computation is simpler and no tan and sqrt needed. Anyway, nice constructions.
                – Sigur
                14 hours ago
















              Very interesting.
              – Sigur
              16 hours ago




              Very interesting.
              – Sigur
              16 hours ago












              @Sigur I'll be happy to remove my answer.
              – marmot
              16 hours ago




              @Sigur I'll be happy to remove my answer.
              – marmot
              16 hours ago












              Why? It is elegant and advanced for beginners. Good exercise to study the code.
              – Sigur
              15 hours ago




              Why? It is elegant and advanced for beginners. Good exercise to study the code.
              – Sigur
              15 hours ago












              @Sigur Conceptually it is the same as your nice post. I slightly modified the parametrization to make my answer, which came later, not completely pointless.
              – marmot
              15 hours ago




              @Sigur Conceptually it is the same as your nice post. I slightly modified the parametrization to make my answer, which came later, not completely pointless.
              – marmot
              15 hours ago












              You can define the polar command using as two arguments the length of semi-diagonals, as we did before. The computation is simpler and no tan and sqrt needed. Anyway, nice constructions.
              – Sigur
              14 hours ago




              You can define the polar command using as two arguments the length of semi-diagonals, as we did before. The computation is simpler and no tan and sqrt needed. Anyway, nice constructions.
              – Sigur
              14 hours ago











              4














              A PSTricks solution just for fun purposes.



              documentclass[pstricks,12pt]{standalone}
              usepackage{pst-eucl}
              begin{document}
              foreach a in {0,10,...,350}{%
              pspicture(-7,-7)(7,7)
              pstGeonode(0,0){O}(2;a){A}([offset=6]{A}O){B}
              pstRotation[RotAngle=-90]{O}{A}[D]
              pstRotation[RotAngle=90]{O}{B}[C]
              psline(A)(B)(C)(D)(A)(C)(D)(B)
              endpspicture}
              end{document}


              enter image description here



              Algorithm




              • Define two points O and A.

              • Define point B such that OA is perpendicular to OB.

              • Define C as the image of rotating point B about O counter-clockwise.

              • Define D as the image of rotating point A about O clockwise.

              • Draw the lines.






              share|improve this answer




























                4














                A PSTricks solution just for fun purposes.



                documentclass[pstricks,12pt]{standalone}
                usepackage{pst-eucl}
                begin{document}
                foreach a in {0,10,...,350}{%
                pspicture(-7,-7)(7,7)
                pstGeonode(0,0){O}(2;a){A}([offset=6]{A}O){B}
                pstRotation[RotAngle=-90]{O}{A}[D]
                pstRotation[RotAngle=90]{O}{B}[C]
                psline(A)(B)(C)(D)(A)(C)(D)(B)
                endpspicture}
                end{document}


                enter image description here



                Algorithm




                • Define two points O and A.

                • Define point B such that OA is perpendicular to OB.

                • Define C as the image of rotating point B about O counter-clockwise.

                • Define D as the image of rotating point A about O clockwise.

                • Draw the lines.






                share|improve this answer


























                  4












                  4








                  4






                  A PSTricks solution just for fun purposes.



                  documentclass[pstricks,12pt]{standalone}
                  usepackage{pst-eucl}
                  begin{document}
                  foreach a in {0,10,...,350}{%
                  pspicture(-7,-7)(7,7)
                  pstGeonode(0,0){O}(2;a){A}([offset=6]{A}O){B}
                  pstRotation[RotAngle=-90]{O}{A}[D]
                  pstRotation[RotAngle=90]{O}{B}[C]
                  psline(A)(B)(C)(D)(A)(C)(D)(B)
                  endpspicture}
                  end{document}


                  enter image description here



                  Algorithm




                  • Define two points O and A.

                  • Define point B such that OA is perpendicular to OB.

                  • Define C as the image of rotating point B about O counter-clockwise.

                  • Define D as the image of rotating point A about O clockwise.

                  • Draw the lines.






                  share|improve this answer














                  A PSTricks solution just for fun purposes.



                  documentclass[pstricks,12pt]{standalone}
                  usepackage{pst-eucl}
                  begin{document}
                  foreach a in {0,10,...,350}{%
                  pspicture(-7,-7)(7,7)
                  pstGeonode(0,0){O}(2;a){A}([offset=6]{A}O){B}
                  pstRotation[RotAngle=-90]{O}{A}[D]
                  pstRotation[RotAngle=90]{O}{B}[C]
                  psline(A)(B)(C)(D)(A)(C)(D)(B)
                  endpspicture}
                  end{document}


                  enter image description here



                  Algorithm




                  • Define two points O and A.

                  • Define point B such that OA is perpendicular to OB.

                  • Define C as the image of rotating point B about O counter-clockwise.

                  • Define D as the image of rotating point A about O clockwise.

                  • Draw the lines.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 13 hours ago

























                  answered 13 hours ago









                  God Must Be CrazyGod Must Be Crazy

                  6,05511039




                  6,05511039






















                      M. C. is a new contributor. Be nice, and check out our Code of Conduct.










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                      M. C. is a new contributor. Be nice, and check out our Code of Conduct.
















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