## 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.
 . . . . . FIGURE 1 0 000 000000 000000000 000000000000 0000000000000000 000000000000000000000 00000000000000000000000 222222222222222222222222II 222222222222222222222222III II I 222222222222222222222222II II II II I 222222222222222222222222III II II II II II 111111112222222222222222II II II II II II II III 111111112222222222222222III II II II II II II II II II 111111112222222222222222II II II II II II II II II II II II 111111112222222222222222III II II II II II II II II II II II II II 111111111111111111111111II II II II II II II II II II II II II II II II 111111111111111111111111III II II II II II II II II II II II II II II II III 111111111111111111111111II II II II II II II II II II II II II II II II II II III 111111111111111111111111III II II II II II II II II II II II II II II II II II II II I . . . . . . .figure 2 II III II I II II II II I III II II II II II II II II II II II II III III II II II II II II II II II II II II II II II II II II II II II III II II II II II II II II II II II II II II II II II II II II II II II II II II II II II III II II II II II II II II II II II II II II II III II II II II II II II II II II II II II II II II II II III III II II II II II II II II II II II II II II II II II II II I 11111111________2222222222232222222222220 11111111________222222222222222222222222000 11111111________222222222222222222222222000000 11111111________222222222222222222222222000000000 1111111111111111111111112222222222222222000000000000 11111111111111111111111122222222222222220000000000000000 1111111111111111111111112222222222222222000000000000000000000 111111111111111111111111222222222222222200000000000000000000000

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 Blog Entries: 9 Recognitions: Homework Help Science Advisor SLOPE THEY ARE NOT TRIANGLES Daniel.

 Quote by dextercioby SLOPE Daniel.
WRONG
That does not describe the source the extra area.

## Triangle with extra area???

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.

 here is a nice image:
 Recognitions: Gold Member Science Advisor Staff Emeritus Not to be a pompous @ or anything, but since gerben's linked to that picture... http://www.physicsforums.com/showpos...1&postcount=20
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Nice,Gokul,a page especially for u.Greg must have been in a very good mood that day... Daniel.
 Off topic, but dextercioby you can press that number that is to the right of "View dextercioby's Warnings" for example #7 and it will go to a single post on one screen. http://www.physicsforums.com/showpos...14&postcount=7

 Quote by Pseudopod just hard to do it in one word. . . . quadrilateral)
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.

 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.
 Recognitions: Gold Member Science Advisor Staff Emeritus 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).

 Quote by Gokul43201 SLOPES "HYPOTENUSE" QUADRILATERAL SIMILARITY
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

 "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."
 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

 Quote by Healey01 "Parallelogram" does not describe teh solution
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.

 Quote by RandallB 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.
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.

 Quote by Healey01 "There is no extra area."
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.