I am trying to draw a mathematical drawing in latex but I have difficulties with it.Could someone help me how to draw it as it is shown in picture below.

Best regards

Mathematical drawing with objects

Here is my code and how far I got. I would like to hear your suggestions and opinions if I am on the right way. Is there any easier solution or way to do this?

Since I am a beginner any help would be useful and welcome :)



    \draw[step=1cm,gray,very thin] (0,0) grid (11,6);

    \draw (1,1) circle (5pt); % lower
    \draw (10,5) circle (5pt); %upper
    \draw[line width=3pt] (1,1.17)--(2,5)--(9.83,5); %left,topside
    \draw[line width=3pt] (10,4.83)--(9,1)--(1.17,1); %right,bottomside

    \draw[line width=3pt] (1.16,1.07)--(6,5); %leftside t1
    \draw[line width=3pt] (6,5)--(9,1); %rightside t1

    \draw[line width=3pt] (2,5)--(5,1); %leftside t2
    \draw[line width=3pt] (5,1)--(9.91,4.85); %rightside t2



This is picture of mine how far I got.

My picture of parallelogram

  • 3
    Please show the code you have so far and ask a specific question about any place where you are stuck. Questions that just consist of an image an a request for the code are usually considered off topic. – David Carlisle Dec 14 '14 at 17:41
  • 1
    Welcome to TeX.SE. It would be helpful if you composed a fully compilable MWE including \documentclass and the appropriate packages that sets up the problem showing exactly which part you are having difficulty with. For instance, can you draw straight lines to draw the parallelogram. Then it is just two more straight lines for the two triangles. – Peter Grill Dec 14 '14 at 19:19
  • I have edited my post including my code.Could anyone of moderators just delete my answer and put it as an edit to my topic? – lesale Dec 14 '14 at 20:35

Here's a possibility using TikZ:

enter image description here

The code (with some explanatory comments):




    inner sep=1.5pt
% The paralellogram and the diagonals inside
% corners are labelled ``ll'' lower left, ``ur'' upper right,
% ``lr'' lower right and  ``ul'' upper left
% ``lm'' is the inner point in the lower base
% ``um'' is the inner point in the uper base
\draw[line width=1pt]
  (0,0) coordinate (ll) -- 
  ++(\Angle:\Side) coordinate (ul) -- 
  ++(0:\Base) coordinate (ur)-- 
  ++(180+\Angle:\Side) coordinate (lr) -- 
\coordinate (lm) at ([xshift=\Side]ll);
\coordinate (um) at ([xshift=-\Side]ur);
\path[draw,line width=1pt,name path=diag1]
  (ll) -- (um);  
\path[draw,line width=1pt,name path=diag2]
  (um) -- (lr);  
\path[draw,line width=1pt,name path=diag3]
  (ul) -- (lm);  
\path[draw,line width=1pt,name path=diag4]
  (lm) -- (ur);
% We find the intersection point between inner diagonals
\path[name intersections={of=diag1 and diag3, by={aux1}}];    
\path[name intersections={of=diag2 and diag4, by={aux2}}];
% We fill the inner cuadrilateral
\fill[pattern=north east lines,opacity=0.6]
  (lm) -- (aux1) -- (um) -- (aux2) -- cycle;    
% The arcs
  pic[draw,angle radius=\Side] {angle=lm--ll--ul};
  pic[draw,angle radius=\Side] {angle=um--ur--lr};
  pic[draw,angle radius=12pt,"\mybullet"] {angle=ul--aux1--ll};
  pic[draw,angle radius=12pt,"\mybullet"] {angle=lr--aux2--ur};
% The auxiliary lines with lengths
\begin{scope}[help lines,>=latex]
\draw (ll) -- ([yshift=15pt]ll|-ul);
\draw (ur) -- ++(0,15pt);
  ([yshift=7.5pt]ll|-ul) -- 
    node[fill=white] {\small$57$\,cm} 
  at ( $ (ll)!0.5!(lr) $ ) {};
% The dots at opposed corners
   at (ur) {};
   at (ll) {};
% The ``F'' label   
\node[fill=white,inner sep=0.5pt]
   at ( $ (ul)!0.5!(lr) $ ) {$F$};

| improve this answer | |
  • 1
    +1 for your broad definition of "starting point" :) – Pier Paolo Dec 14 '14 at 20:24
  • 2
    @PierPaolo Hehe. You are right ;) I've changed the wording a little. – Gonzalo Medina Dec 14 '14 at 20:31

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