Difference between revisions of "1999 AMC 8 Problems/Problem 14"
(Solution of AMC 8 Problem, 1999, #14) |
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− | + | There is a rectangle present, with both horizontal bases being <math>8</math> units in length. The excess units on the bottom base must then be 8. The fact that <math>AB</math> and <math>CD</math> are equal in length indicate, by the Pythagorean Theorem, that these excess lengths are equal. There are two with a total length of <math>8</math> units, so each is <math>4</math> units. The triangle has a hypotenuse of 5, because the triangles are <math>3-4-5</math> right triangles. So, the sides of the trapezoid are <math>8</math>, <math>5</math>, <math>16</math>, and <math>5</math>. Adding those up gives us the perimeter, <math>8 + 5 + 16 + 5 = 13 + 21 = 34</math> units. |
Revision as of 13:40, 12 November 2011
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.