2005 AMC 8 Problem 19
Attempt Problem 19 of the 2005 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
19.
What is the perimeter of trapezoid
Answer: A
Small hint:
Drop perpendiculars to split the trapezoid into right triangles and a rectangle.
Big hint:
Use the and right triangles.
Solution:
Drop the altitude from to meet at . The existing altitude from meets at .
By the Pythagorean theorem, and . Also .
Thus , and the trapezoid perimeter is .
Thus, A is the correct answer.
Problem 19 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
- 1988 AMC 8
- 1989 AMC 8
- 1990 AMC 8
- 1991 AMC 8
- 1992 AMC 8
- 1993 AMC 8
- 1994 AMC 8
- 1995 AMC 8
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- 1997 AMC 8
- 1998 AMC 8
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