Difference between revisions of "2000 AMC 10 Problems/Problem 5"

Revision as of 18:03, 8 January 2009

(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 $\boxed{\text{B}}$ .

Revision as of 13:55, 7 January 2009 (view source) 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...)		Revision as of 18:03, 8 January 2009 (view source) 5849206328x (talk \| contribs) m Newer edit →
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	(d) The bases <math>AB</math> and <math>MN</math> do not change, and neither does the height, so the trapezoid remains the same.		(d) The bases <math>AB</math> and <math>MN</math> do not change, and neither does the height, so the trapezoid remains the same.

−	Only <math>1</math> changes, so B.	+	Only <math>1</math> changes, so <math>\boxed{\text{B}}</math>.