1999 AMC 8 Problems/Problem 14

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Okay. There is a rectangle present, with both horizontal bases being $8$ units in length. The excess units on the bottom base must then be 8. The fact that $AB$ and $CD$ are equal in length indicate, by the Pythagorean Theorem, that these excess lengths are equal. There are two with a total length of $8$ units, so each is $4$ units. The triangle has a hypotenuse of 5, because the triangles are $3-4-5$ right triangles. So, the sides of the trapezoid are $8$, $5$, $16$, and $5$. Adding those up gives us the perimeter, $8 + 5 + 16 + 5 = 13 + 21 = 34$ units.

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