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Is it possible to construct a non-rhombus but parallelogram and quadrilateral non-square, Non-rectangle by putting four 3-4-5 (Pythagorean triplet) triangles together and making the 90 degree angle not at the intersection of the two diagonals?

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  • $\begingroup$ There's some redundancy in your description: each of non-rhombus and non-rectangle already implies non-square, and parallelogram of course implies quadrilateral. $\endgroup$
    – Rand al'Thor
    Oct 29 at 13:02

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The answer is

yes

because

you can first put two triangles together to make a 3x4 rectangle (let's say the 3-sides are horizontal and the 4-sides are vertical) and then attach the other two at the sides, one each way up, so that you have two horizontal sides of length 6 and two neither-horizontal-nor-vertical sides of length 5.

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