Pythagorea Level 22.17 Answer Solution

Pythagorea Rhombuses Level 22.17 Solution/Answer

Pythagorea Rhombuses Level 22.17 New Version Game Answers,  detailed solutions, Tips, and Walkthrough. Scroll below to find answer to this level.

Pythagorea is android/iOS app developed by Horis International Limited. Solutions hints and answers to pythagorea are available in this post scroll down to find solutions to all the levels.

This game is mostly focused on geometric puzzles and construction. The work space is divided into grids to draw lines. You should know all the basic Math operations. All lines and shapes are drawn on a grid whose cells are squares. Most of the game levels can be answered using natural intuition and by some basic laws of geometry.

Click Here for All other Pythagorea Levels: http://puzzlegamemaster.com/pythagorea/

  • Pythagorea Level 22.17: Inscribe a rhombus in the triangle so that they share the common angle A.

If you have any doubt regarding Pythagorea answers given here you can watch video below or you can comment on this post-

 

 

 

8 thoughts on “Pythagorea Level 22.17 Answer Solution

  • February 24, 2023 at 04:30
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    This is a mess and a half

    Reply
  • January 10, 2020 at 21:44
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    Has anyone got this to work and/or got a simpler method?

    Reply
    • January 19, 2021 at 10:18
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      I’ve developed a nice solution in 7 lines.

      Reply
    • March 26, 2022 at 00:34
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      I cannot get this solution to work either.

      Reply
      • July 13, 2024 at 23:16
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        Can be completed in nine moves. Don’t need to do the segment on the bottom right, and the next two that connect to it up top as well as the segment that connects them.

        The two lines that are parallel are needed. they form an intersection inside the acute point. draw a line from the original angles point to the new angles point to bisect the right-top face. Draw the line from this new point in the middle of the short right-top face to the second node on the top of the grid, counting left from the right corner. Draw the long line on the left side of the grid L1R-L6L, and the short line in L2-L2. This creates the intersection necessary for the horizontal lines. Use the existing lines, and the few that have been added to make the horizontal line that completes the necessary nodes for the rhombus. Draw the remaining lines to close the gap and it should be satisfied.

        Reply
        • July 13, 2024 at 23:18
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          In that, I counted from the right, and then from the left on top, but failed to mention the shift in direction in the original post. the L1 part is counting left to right, the prior is right to left.

          Reply
  • November 6, 2019 at 02:49
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    Helllo, can anyone please explain 22.17 solution?

    Reply
    • February 9, 2020 at 08:09
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      seach for the video on Youtube, there’s detailed explanation. It’s quite long.

      Reply

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