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2000 AMC 10 Problems/Problem 5

Revision as of 14:55, 7 January 2009 by BOGTRO (talk | contribs) (New page: (a) Clearly does not change, as <math>MN=\frac{1}{2}AB</math>. Since <math>AB</math> does not change, neither does <math>MN</math>. (b) Obviously, the perimetar changes. (c) The area cle...)
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(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 B.

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