Is there a trapezoid with side-lengths $3$, $5$, $7$, $11$? 
Is it possible for a trapezoid to have the side lengths of 3, 5, 7, and 11?
  Explain completely.

I tried using multiple shapes to construct this trapezoid, but it didn't really work. I think it might be a right trapezoid. Does anyone have any idea how to prove this trapezoid's existence?
 A: I just want to expand on the answer given by @KBDave. Let's say the trapezoid has side lengths $a$, $b$, $c$, $d$, such that sides $a$ and $c$ are parallel. From the corners of the shorter side, say $a$, draw perpendiculars onto $c$. You have a rectangle, and two triangles. If you slide the two triangles together, you get a new triangle, with sides $d$, $b$, and $c-a$.
Now going back to a problem like this, if you can find the above numbers that obey triangle inequalities, you can construct the trapeze. In your case you can choose $a=7$ and $c=11$. Then $3$, $5$, and $4=11-7$ form the sides of a right angle triangle. 
However, it's not the only solution. You can also choose $a=5$, $c=11$. Then $3$, $6=11-5$, and $7$ also can form a triangle, and therefore you can get a trapeze. Some combinations will not work. $a=3$, $c=5$ means that $c-a=2$. But $2+7<11$, so it does not obey triangle inequality. Therefore you cannot construct such a trapeze.
A: There's one that dissects into a $3$-by-$7$ rectangle and a $3$-$4$-$5$ right triangle.
A: Yes. ${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$

