2000 AMC 10 Problems/Problem 5

Revision as of 19:58, 8 January 2009 by Waffle (talk | contribs) (Solution)

Problem

Solution

(a) Clearly does not change, as $MN=\frac{1}{2}AB$. Since $AB$ does not change, neither does $MN$.

(b) Obviously, the perimeter changes.

(c) The area clearly doesn't change, as the base and height remain the same.

(d) The bases $AB$ and $MN$ do not change, and neither does the height, so the trapezoid remains the same.

Only $1$ changes, so $\boxed{\text{B}}$.

See Also

2000 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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All AMC 10 Problems and Solutions