2000 AMC 10 Problems/Problem 5

Problem

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

(b) Obviously, the perimetar 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 $\boxed{\text{B}}$ .

2000 AMC 10 (Problems • Answer Key • Resources)
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