Monday, 20 February 2017

Pythagorea Parallels Complete Level (2.1-2.19) Answers

Pythagorea Parallels Complete Level 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 1.14 2.15 2.16 2.17 2.18 2.19 solution walkthrough 


game details: Study geometry while playing on squared paper.

> 330+ tasks: from very simple to really geometric puzzles
> 25 subjects to explore
> 76 geometric terms in a glossary
> Easy to use
> Friendly interface
> Train your mind and imagination

*** About ***
Pythagorea is a collection of geometric puzzles of different kind that can be solved without complex constructions or calculations. All objects are drawn on a grid whose cells are squares. A lot of levels can be solved using just your geometric intuition or by finding natural laws, regularity, and symmetry.

Pythagorea Level 2.1 Answer :

Pythagorea Level 2.2 Answer :


Pythagorea Level 2.3 Answer :


Pythagorea Level 2.4 Answer :


Pythagorea Level 2.5 Answer :


Pythagorea Level 2.6 Answer :


 Pythagorea Level 2.7 Answer :


Pythagorea Level 2.8 Answer :


Pythagorea Level 2.9 Answer :



Pythagorea Level 2.10 Answer :



Pythagorea Level 2.11 Answer :


Pythagorea Level 2.12 Answer :


Pythagorea Level 2.13 Answer :



Pythagorea Level 2.14 Answer :



Pythagorea Level 2.15 Answer :


Pythagorea Level 2.16 Answer :



Pythagorea Level 2.17 Answer :



Pythagorea Level 2.18 Answer :



Pythagorea Level 2.19 Answer :


Feel free to comment below if you have any doubts regarding solution. I will try to help you guys. all other levels are posted on this blog please visit them too.

 Video link to these solutions in case you want to know the detailed solutions: 

Posted by at
Labels: , ,

6 comments:

  1. Unknown26 July 2017 at 02:09

    Re 2.16. Sorry, I can't follow your construction! I used simple ratios x=5 & y=3 therefore x=2.5 & y=1.5 therefore it is easy to construct diagonals in square Row 1, Column 5 to give the square's centre point matching 2.5 & 1.5. Can you explain your construction pleae?

    ReplyDelete
    Replies
    1. Unknown28 July 2017 at 08:39

      This comment has been removed by the author.

      Delete
    2. Unknown28 July 2017 at 08:42

      Hello
      Also had problems with the suggested answer.
      But here is a possible solution:
      As Tim stated the slope y/x of the given line is 3/5.
      Normally: If you want to make a parallel through a given point A for a given line with given slope you only have to step x to the right and y upwards from A. Then draw the line through A and the constructed point.
      That's not possible here as the grid is too small. You can go only 4 max to the right and 2 up.
      So let's choose 3 to the right. But how far up?
      Here comes the auxiliary line. We draw it with slope 1/5 (5 right, 1 up). Then look at the intersection between the line and the second vertical grid line from the right. It also represents slope 1/5, but with 1/5 from the uppermost grid line.
      So from Point A this intersection point is 3 right and 9/5 up. And so the slope (9/5)/3 makes 3/5 as wished. So draw the line between the point A and the intersection point. QED

      Delete
  2. Unknown26 July 2017 at 02:11

    RE 2.19, again, sorry, i can't follow your construction, please can you clarify? Thanks

    ReplyDelete
  3. Unknown28 July 2017 at 11:54

    The problem for me was how to construct a gradient of x=5, y=6 whilst staying on the grid. That was until I realised that a diagonal drawn in 6 linear squares will give x=1,y=1-1/6 then x=2,y=2/6 (1/3) then x=3,y=3/6 (1/2)and so on. This gives us the ability to construct the required point as shown in the left hand column in the diag above. We're looking for a point 1/3 of a square in from the y axis on the 32nd line up from the x axis. Silly me!

    ReplyDelete
  4. Unknown28 July 2017 at 11:55

    Sorry, for x=1, y=1-1/6 please read y=1/6!

    ReplyDelete

Subscribe to: Post Comments (Atom)