# 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 12:40, 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.