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How to align text above triangle figure



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Rotate a node but not its content: the case of the ellipse decorationHow to define the default vertical distance between nodes?TikZ scaling graphic and adjust node position and keep font sizealign text in tikz figureNumerical conditional within tikz keys?Relating tree nodes in forest to content in a tableTikZ: Node position in draw environmentLatex and Game Theory: Combining an Extensive and Normal Form for a Three Players GameDrawing graph with Tikz: Link it with main text without overlapping with textTikZ: define arrow starting position based on style and format node label










3















I managed to align my hypotenuse text with the hypotenuse side of my triangle, but I feel like it was done inefficiently using a lot of ~~~~~~ in this line node[above] $sqrt1+x^2$~~~~~~~ (B) --.



Is there a better way to get the same alignment that I have now without the excessive use of ~?



enter image description here



documentclass[hidelinks,14pt, letterpaper]extarticle
usepackageamsmath, amssymb, tikz

newcommandpythagwidth3cm
newcommandpythagheight2cm

begindocument
beginfigure[h]
centering

begintikzpicture[scale=1.25]
coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
draw
(A) --
node[above] $sqrt1+x^2$~~~~~~~ (B) --
node[right] ? (C) --
node[below] ?
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);

endtikzpicture
captionCaption
labelfig:my_label
endfigure
enddocument









share|improve this question






















  • I realise you already have an answer for this, but I thought I would mention that you can also use xshift for this.

    – bradrn
    4 hours ago















3















I managed to align my hypotenuse text with the hypotenuse side of my triangle, but I feel like it was done inefficiently using a lot of ~~~~~~ in this line node[above] $sqrt1+x^2$~~~~~~~ (B) --.



Is there a better way to get the same alignment that I have now without the excessive use of ~?



enter image description here



documentclass[hidelinks,14pt, letterpaper]extarticle
usepackageamsmath, amssymb, tikz

newcommandpythagwidth3cm
newcommandpythagheight2cm

begindocument
beginfigure[h]
centering

begintikzpicture[scale=1.25]
coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
draw
(A) --
node[above] $sqrt1+x^2$~~~~~~~ (B) --
node[right] ? (C) --
node[below] ?
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);

endtikzpicture
captionCaption
labelfig:my_label
endfigure
enddocument









share|improve this question






















  • I realise you already have an answer for this, but I thought I would mention that you can also use xshift for this.

    – bradrn
    4 hours ago













3












3








3


1






I managed to align my hypotenuse text with the hypotenuse side of my triangle, but I feel like it was done inefficiently using a lot of ~~~~~~ in this line node[above] $sqrt1+x^2$~~~~~~~ (B) --.



Is there a better way to get the same alignment that I have now without the excessive use of ~?



enter image description here



documentclass[hidelinks,14pt, letterpaper]extarticle
usepackageamsmath, amssymb, tikz

newcommandpythagwidth3cm
newcommandpythagheight2cm

begindocument
beginfigure[h]
centering

begintikzpicture[scale=1.25]
coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
draw
(A) --
node[above] $sqrt1+x^2$~~~~~~~ (B) --
node[right] ? (C) --
node[below] ?
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);

endtikzpicture
captionCaption
labelfig:my_label
endfigure
enddocument









share|improve this question














I managed to align my hypotenuse text with the hypotenuse side of my triangle, but I feel like it was done inefficiently using a lot of ~~~~~~ in this line node[above] $sqrt1+x^2$~~~~~~~ (B) --.



Is there a better way to get the same alignment that I have now without the excessive use of ~?



enter image description here



documentclass[hidelinks,14pt, letterpaper]extarticle
usepackageamsmath, amssymb, tikz

newcommandpythagwidth3cm
newcommandpythagheight2cm

begindocument
beginfigure[h]
centering

begintikzpicture[scale=1.25]
coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
draw
(A) --
node[above] $sqrt1+x^2$~~~~~~~ (B) --
node[right] ? (C) --
node[below] ?
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);

endtikzpicture
captionCaption
labelfig:my_label
endfigure
enddocument






tikz-pgf






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 8 hours ago









Evan KimEvan Kim

1553




1553












  • I realise you already have an answer for this, but I thought I would mention that you can also use xshift for this.

    – bradrn
    4 hours ago

















  • I realise you already have an answer for this, but I thought I would mention that you can also use xshift for this.

    – bradrn
    4 hours ago
















I realise you already have an answer for this, but I thought I would mention that you can also use xshift for this.

– bradrn
4 hours ago





I realise you already have an answer for this, but I thought I would mention that you can also use xshift for this.

– bradrn
4 hours ago










2 Answers
2






active

oldest

votes


















7














Something like this? I use node[midway,above left=0pt,inner sep=0.5pt] $sqrt1+x^2$, where inner sep=0.5pt controls the distance.



documentclass[hidelinks,14pt, letterpaper]extarticle
usepackageamsmath, amssymb, tikz

newcommandpythagwidth3cm
newcommandpythagheight2cm

begindocument
beginfigure[h]
centering

begintikzpicture[scale=1.25]
coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
draw
(A) --
node[midway,above left=0pt,inner sep=0.5pt] $sqrt1+x^2$ (B) --
node[right] ? (C) --
node[below] ?
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);

endtikzpicture
captionCaption
labelfig:my_label
endfigure
enddocument


enter image description here



ADDENDUM: Just for fun: an even simpler and shorter code with TikZ...



documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture[scale=1.25]
draw (-1.5,-1) coordinate [label=left:$A$] (A) --
node[midway,above,sloped] $sqrt1+x^2$
(1.5,1) coordinate [label=above:$B$] (B) --
node[right] ?
(1.5,-1)coordinate [label=below right:$C$] (C) --
node[below] ? cycle;
draw ([xshift=-0.25cm]C) |- ([yshift=0.25cm]C);
endtikzpicture
enddocument


enter image description here






share|improve this answer

























  • yes that is it thanks! It seems like simply having node [midway,above left=0pt] $sqrt1+x^2$ (B) -- does the trick too without using inner_sept

    – Evan Kim
    8 hours ago



















2














Just for fun: with pstricks, a very short code to have this figure:



 documentclassarticle
usepackagepst-eucl%,
usepackageauto-pst-pdf

begindocument

beginpostscript
pssetunit=2, linejoin=1, PointSymbol=none,
pstTriangle(-1.5,-1)A(1.5,1)B(1.5,-1)C
ncline[linestyle=none]ABnaput*[nrot=:U]$ sqrt1 + x^2$
pssetPointName=none
pstMiddleABACIuput[d](I)?
pstMiddleABBCJuput[r](J)?
endpostscript

enddocument


enter image description here






share|improve this answer























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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7














    Something like this? I use node[midway,above left=0pt,inner sep=0.5pt] $sqrt1+x^2$, where inner sep=0.5pt controls the distance.



    documentclass[hidelinks,14pt, letterpaper]extarticle
    usepackageamsmath, amssymb, tikz

    newcommandpythagwidth3cm
    newcommandpythagheight2cm

    begindocument
    beginfigure[h]
    centering

    begintikzpicture[scale=1.25]
    coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
    coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
    coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
    draw
    (A) --
    node[midway,above left=0pt,inner sep=0.5pt] $sqrt1+x^2$ (B) --
    node[right] ? (C) --
    node[below] ?
    (A);
    draw
    (1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);

    endtikzpicture
    captionCaption
    labelfig:my_label
    endfigure
    enddocument


    enter image description here



    ADDENDUM: Just for fun: an even simpler and shorter code with TikZ...



    documentclass[tikz,border=3.14mm]standalone
    begindocument
    begintikzpicture[scale=1.25]
    draw (-1.5,-1) coordinate [label=left:$A$] (A) --
    node[midway,above,sloped] $sqrt1+x^2$
    (1.5,1) coordinate [label=above:$B$] (B) --
    node[right] ?
    (1.5,-1)coordinate [label=below right:$C$] (C) --
    node[below] ? cycle;
    draw ([xshift=-0.25cm]C) |- ([yshift=0.25cm]C);
    endtikzpicture
    enddocument


    enter image description here






    share|improve this answer

























    • yes that is it thanks! It seems like simply having node [midway,above left=0pt] $sqrt1+x^2$ (B) -- does the trick too without using inner_sept

      – Evan Kim
      8 hours ago
















    7














    Something like this? I use node[midway,above left=0pt,inner sep=0.5pt] $sqrt1+x^2$, where inner sep=0.5pt controls the distance.



    documentclass[hidelinks,14pt, letterpaper]extarticle
    usepackageamsmath, amssymb, tikz

    newcommandpythagwidth3cm
    newcommandpythagheight2cm

    begindocument
    beginfigure[h]
    centering

    begintikzpicture[scale=1.25]
    coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
    coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
    coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
    draw
    (A) --
    node[midway,above left=0pt,inner sep=0.5pt] $sqrt1+x^2$ (B) --
    node[right] ? (C) --
    node[below] ?
    (A);
    draw
    (1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);

    endtikzpicture
    captionCaption
    labelfig:my_label
    endfigure
    enddocument


    enter image description here



    ADDENDUM: Just for fun: an even simpler and shorter code with TikZ...



    documentclass[tikz,border=3.14mm]standalone
    begindocument
    begintikzpicture[scale=1.25]
    draw (-1.5,-1) coordinate [label=left:$A$] (A) --
    node[midway,above,sloped] $sqrt1+x^2$
    (1.5,1) coordinate [label=above:$B$] (B) --
    node[right] ?
    (1.5,-1)coordinate [label=below right:$C$] (C) --
    node[below] ? cycle;
    draw ([xshift=-0.25cm]C) |- ([yshift=0.25cm]C);
    endtikzpicture
    enddocument


    enter image description here






    share|improve this answer

























    • yes that is it thanks! It seems like simply having node [midway,above left=0pt] $sqrt1+x^2$ (B) -- does the trick too without using inner_sept

      – Evan Kim
      8 hours ago














    7












    7








    7







    Something like this? I use node[midway,above left=0pt,inner sep=0.5pt] $sqrt1+x^2$, where inner sep=0.5pt controls the distance.



    documentclass[hidelinks,14pt, letterpaper]extarticle
    usepackageamsmath, amssymb, tikz

    newcommandpythagwidth3cm
    newcommandpythagheight2cm

    begindocument
    beginfigure[h]
    centering

    begintikzpicture[scale=1.25]
    coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
    coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
    coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
    draw
    (A) --
    node[midway,above left=0pt,inner sep=0.5pt] $sqrt1+x^2$ (B) --
    node[right] ? (C) --
    node[below] ?
    (A);
    draw
    (1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);

    endtikzpicture
    captionCaption
    labelfig:my_label
    endfigure
    enddocument


    enter image description here



    ADDENDUM: Just for fun: an even simpler and shorter code with TikZ...



    documentclass[tikz,border=3.14mm]standalone
    begindocument
    begintikzpicture[scale=1.25]
    draw (-1.5,-1) coordinate [label=left:$A$] (A) --
    node[midway,above,sloped] $sqrt1+x^2$
    (1.5,1) coordinate [label=above:$B$] (B) --
    node[right] ?
    (1.5,-1)coordinate [label=below right:$C$] (C) --
    node[below] ? cycle;
    draw ([xshift=-0.25cm]C) |- ([yshift=0.25cm]C);
    endtikzpicture
    enddocument


    enter image description here






    share|improve this answer















    Something like this? I use node[midway,above left=0pt,inner sep=0.5pt] $sqrt1+x^2$, where inner sep=0.5pt controls the distance.



    documentclass[hidelinks,14pt, letterpaper]extarticle
    usepackageamsmath, amssymb, tikz

    newcommandpythagwidth3cm
    newcommandpythagheight2cm

    begindocument
    beginfigure[h]
    centering

    begintikzpicture[scale=1.25]
    coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
    coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
    coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
    draw
    (A) --
    node[midway,above left=0pt,inner sep=0.5pt] $sqrt1+x^2$ (B) --
    node[right] ? (C) --
    node[below] ?
    (A);
    draw
    (1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);

    endtikzpicture
    captionCaption
    labelfig:my_label
    endfigure
    enddocument


    enter image description here



    ADDENDUM: Just for fun: an even simpler and shorter code with TikZ...



    documentclass[tikz,border=3.14mm]standalone
    begindocument
    begintikzpicture[scale=1.25]
    draw (-1.5,-1) coordinate [label=left:$A$] (A) --
    node[midway,above,sloped] $sqrt1+x^2$
    (1.5,1) coordinate [label=above:$B$] (B) --
    node[right] ?
    (1.5,-1)coordinate [label=below right:$C$] (C) --
    node[below] ? cycle;
    draw ([xshift=-0.25cm]C) |- ([yshift=0.25cm]C);
    endtikzpicture
    enddocument


    enter image description here







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 7 hours ago

























    answered 8 hours ago









    marmotmarmot

    118k6153288




    118k6153288












    • yes that is it thanks! It seems like simply having node [midway,above left=0pt] $sqrt1+x^2$ (B) -- does the trick too without using inner_sept

      – Evan Kim
      8 hours ago


















    • yes that is it thanks! It seems like simply having node [midway,above left=0pt] $sqrt1+x^2$ (B) -- does the trick too without using inner_sept

      – Evan Kim
      8 hours ago

















    yes that is it thanks! It seems like simply having node [midway,above left=0pt] $sqrt1+x^2$ (B) -- does the trick too without using inner_sept

    – Evan Kim
    8 hours ago






    yes that is it thanks! It seems like simply having node [midway,above left=0pt] $sqrt1+x^2$ (B) -- does the trick too without using inner_sept

    – Evan Kim
    8 hours ago












    2














    Just for fun: with pstricks, a very short code to have this figure:



     documentclassarticle
    usepackagepst-eucl%,
    usepackageauto-pst-pdf

    begindocument

    beginpostscript
    pssetunit=2, linejoin=1, PointSymbol=none,
    pstTriangle(-1.5,-1)A(1.5,1)B(1.5,-1)C
    ncline[linestyle=none]ABnaput*[nrot=:U]$ sqrt1 + x^2$
    pssetPointName=none
    pstMiddleABACIuput[d](I)?
    pstMiddleABBCJuput[r](J)?
    endpostscript

    enddocument


    enter image description here






    share|improve this answer



























      2














      Just for fun: with pstricks, a very short code to have this figure:



       documentclassarticle
      usepackagepst-eucl%,
      usepackageauto-pst-pdf

      begindocument

      beginpostscript
      pssetunit=2, linejoin=1, PointSymbol=none,
      pstTriangle(-1.5,-1)A(1.5,1)B(1.5,-1)C
      ncline[linestyle=none]ABnaput*[nrot=:U]$ sqrt1 + x^2$
      pssetPointName=none
      pstMiddleABACIuput[d](I)?
      pstMiddleABBCJuput[r](J)?
      endpostscript

      enddocument


      enter image description here






      share|improve this answer

























        2












        2








        2







        Just for fun: with pstricks, a very short code to have this figure:



         documentclassarticle
        usepackagepst-eucl%,
        usepackageauto-pst-pdf

        begindocument

        beginpostscript
        pssetunit=2, linejoin=1, PointSymbol=none,
        pstTriangle(-1.5,-1)A(1.5,1)B(1.5,-1)C
        ncline[linestyle=none]ABnaput*[nrot=:U]$ sqrt1 + x^2$
        pssetPointName=none
        pstMiddleABACIuput[d](I)?
        pstMiddleABBCJuput[r](J)?
        endpostscript

        enddocument


        enter image description here






        share|improve this answer













        Just for fun: with pstricks, a very short code to have this figure:



         documentclassarticle
        usepackagepst-eucl%,
        usepackageauto-pst-pdf

        begindocument

        beginpostscript
        pssetunit=2, linejoin=1, PointSymbol=none,
        pstTriangle(-1.5,-1)A(1.5,1)B(1.5,-1)C
        ncline[linestyle=none]ABnaput*[nrot=:U]$ sqrt1 + x^2$
        pssetPointName=none
        pstMiddleABACIuput[d](I)?
        pstMiddleABBCJuput[r](J)?
        endpostscript

        enddocument


        enter image description here







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 7 hours ago









        BernardBernard

        176k778210




        176k778210



























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