1999 AMC 8 Problems/Problem 14
Revision as of 20:34, 16 June 2011 by Kingofmath101 (talk | contribs) (Solution of AMC 8 Problem, 1999, #14)
Okay. There is a rectangle present, with both horizontal bases being 
units in length. The excess units on the bottom base must then be 8. The fact that 
and 
are equal in length indicate, by the Pythagorean Theorem, that these excess lengths are equal. There are two with a total length of units, so each is 
units. The triangle has a hypotenuse of 5, because the triangles are 
right triangles. So, the sides of the trapezoid are , 
, 
, and . Adding those up gives us the perimeter, 
units.
