Difference between revisions of "1999 AMC 8 Problems/Problem 14"
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draw((0,0)--(4,3)--(12,3)--(16,0)--cycle); | draw((0,0)--(4,3)--(12,3)--(16,0)--cycle); |
Latest revision as of 20:16, 24 March 2022
Contents
Problem
In trapezoid , the sides and are equal. The perimeter of is
Solution
There is a rectangle present, with both horizontal bases being units in length. The excess units on the bottom base must then be . 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 , because the triangles are right triangles. So, the sides of the trapezoid are , , , and . Adding those up gives us the perimeter, units.
Video Solution
https://youtu.be/DKoTd9S4y-Q Soo, DRMS, NM
See Also
1999 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
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