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Triangle with extra area?

  1. Jan 26, 2005 #1
    Triangle with extra area???

    Four puzzle pieces; two right triangles and two six sided polygons with all right angles.
    The two six sided polygons can form a perfect 12 by 12 square as shown figure1.
    They are put together in two different ways to create triangular shapes of exactly the same height and width (fig 1 & 2).
    BUT, some extra uncovered area, a 4 by 4 square, shows up in fig 2!

    *Where does this extra space come from???
    *You can show you know the solution with one descriptive word.


    Without a drawing function it's hard to show in 'text drawing' but with the dimensions given above and figures below it should be easy to make four paper cutouts to help solve if needed.
     
    Last edited: Jan 26, 2005
  2. jcsd
  3. Jan 26, 2005 #2

    dextercioby

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    SLOPE

    THEY ARE NOT TRIANGLES

    Daniel.
     
  4. Jan 27, 2005 #3
    WRONG
    That does not describe the source the extra area.
     
  5. Jan 27, 2005 #4
    Sure it explains it, just hard to do it in one word. The hypotenuse of the red triangle must have a different slope than the hypotenuse of the black triangle. Specifically, the red triangle's hypotenuse must be a steeper slope than the black triangle's. This would make the first giant "triangle" (really a quadrilateral) have a concave hypotenuse and the lower "triangle" have a convex hypotenuse, accounting for the extra area.
     
  6. Jan 28, 2005 #5
    here is a nice image:
    HowCanThisBeTrue.jpg
     
  7. Jan 29, 2005 #6

    Gokul43201

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  8. Jan 29, 2005 #7

    dextercioby

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    Nice,Gokul,a page especially for u.Greg must have been in a very good mood that day... :tongue2:


    Daniel.
     
  9. Jan 29, 2005 #8
  10. Jan 29, 2005 #9
    Parallelogram

    Give credit to Pseudopod for coming close by mentioning the word quadrilateral.

    The two large Triangular Shapes, easily mistaken for triangles, are of course quadrilaterals. Overlaying the small quadrilateral on top of the large quadrilateral will expose another quadrilateral shape as a visible portion from the larger .

    Only this quadrilateral has two sides equal in length and both parallel to the hypotenuse of the small triangle and the other two are parallel with hypotenuse of the large triangle. Thus the best one word description of the source of the extra area:
    PARALLELOGRAM

    Which doesn’t directly give away the explanation,
    but at least a less advanced poster was able to give the detailed description.
     
  11. Jan 29, 2005 #10
    I disagree. Someone who did not understand the problem would not be helped by the word "parallelogram"--paralellogram where? A better answer is "hypotenuse." That at least identifies which region.
     
  12. Jan 29, 2005 #11

    Gokul43201

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    PARALLELOGRAM would be a poor choice of word. If you take any pair of congruent triangles (like what someone might interpret the two figue=res to be, if he hasn't cracked the problem), flip one triangle over, and stick it on the other, you get a parallelogram.

    I prefer SLOPES (the plural being important), "HYPOTENUSE" (within quotes to indicate it is not really), QUADRILATERAL (this is straightforward and requires no tricks), or even SIMILARITY (the fact that the three triangles - the two real ones, and the bogus one - are not similar is a sufficient observation).
     
  13. Jan 31, 2005 #12
    None of these directly describes “Where the extra space comes from” those are hints to or gives away the solution that a triangular shape is not necessarily a triangle. And all require more explanation, thus none was given as one word alone.

    However to just “show you know the solution” implies doing so without giving away the solution to others.
    Done to often by those that already know the answers and want to show off.

    Yet once they do solve, they couldn't doubt someone clearly knew that had given the one word.
    PARALLELOGRAM
    Yet it does the least to give away the subtle misdirection of the puzzle.

    PS: Gerben, thanks for the link
     
    Last edited: Jan 31, 2005
  14. Jan 31, 2005 #13
    "Parallelogram" does not describe teh solution or problem at all... You DO know what a parallelogram IS right?

    I think the best answer is "There is no extra area."
     
  15. Jan 31, 2005 #14
    Ratio....

    The hypotenuse length of the overall triangle (all shapes put together) is equal to the sum of the hypotenuse length on the II triangle and the base length on the 111 shape.

    Because the 00 triangle is 4 spaces shorter in hypotenuse length and height (as seen) than the II triangle, the 4x4 block must be inserted for the pieces to fit correctly, yielding the same measurements as the first whole triangle.

    ____________________________________________
    In seeking wisdom thou art wise; in imagining that thou hast attained it - thou art a fool.
    Lord Chesterfield
     
  16. Feb 1, 2005 #15
    Never said it described the solution! It describes how figure 2 is larger than figure 1.
    Figure 2 has to outline a larger area than figure 1 to allow for the 4x4 hole.
    Did you and Lord Chesterfield try cutting out the four shapes?
    Your both wrong, just trace them on paper if you have trouble seeing it.
     
  17. Feb 1, 2005 #16
    What do you mean Im wrong? I understand completely where the hole comes from. Neither of these two shapes are triangles, I don't see why people would have any trouble seeing this at all, unless the image that shows it is distorted or misleading.
     
  18. Feb 2, 2005 #17
    So your saying you finally did find the Parallelogram?
    You DO know what a parallelogram IS right?

    Follow the link for a better photo, but if you need to SEE the Parallelogram
    cut out the four shapes.
     
  19. Feb 2, 2005 #18

    Integral

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    A parallelogram, by definition has parallel sides that shape is NOT a parallelogram. Nor is it a Triangle. It is a Quadrilateral, It has four sides none of which are parallel to any other.
     
  20. Feb 2, 2005 #19

    Gokul43201

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    RandallB never claimed that the provided shape wa a parallelogram. I think what he said was that if the two shapes are superimposed, the bigger one will have an excess area that is the shape of a parallelogram. This excess parallelogram has the same area as the removed square.
     
  21. Feb 2, 2005 #20
    Ahh, I didn't follow what he was trying to say. I thought he meant somehow it all had to do with laws of parallelograms, not that it would generate that shape... I see now.
     
  22. Mar 25, 2006 #21
    If you want to physically confirm that some outside block has been pulled into the triangle by differences in slope, complete the rectangles using the empty space that corresponds to the two smaller triangles. Then complete the rectangle on the larger triangle and you will see that there are 16 blocks left outside the upper resulting shape and only 15 for the lower one (the 16th required block is inside the triangle).
     
    Last edited: Mar 25, 2006
  23. Mar 25, 2006 #22
    Wow Ubern0va you've dug an oldie. but one of my favorites:
    Lots of folks answered this one to quick.

    The point of the OP in this one is to show you understand the problem of how two “triangular shapes” of the same size (height & width) can have different areas.
    If they really are triangles then they should have the same area

    To show you understand you only need demonstrate that you know:
    “ *Where does this extra space come from???”
    And you can prove that know in one word because:
    “ *You can show you know the solution with one descriptive word.”

    So if your trying to give a solution, (I have no idea what your saying in those several lines) you only need give one and only one word to describe the shape of the source of that extra area.

    If you want to keep trying without looking at the other posts. I’ll let you know when you get it right.

    BUT here’s a tip, if you just ‘think’ you have the word; you likely do not.
    Because when you do get the word you will have no doubt.
    When you have no doubt, just browse though the rest of the thread.
     
  24. Mar 26, 2006 #23
    Actually I've already come to the realizations that are listed in this thread; I was just adding a simple way to confirm them, a way to "see" the extra area, for those who don't have the ability to approach this mathematically, or still aren't convinced for some reason.
     
    Last edited: Mar 26, 2006
  25. Mar 27, 2006 #24
    Then you know discribing something as a "block" in "(the 16th required block is inside the triangle)" dosn't go to the solution right.
     
  26. Apr 5, 2006 #25

    Mk

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    I always thought Gokul's post settled it, but hey I guess not.
     
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