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

Revision as of 19:58, 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 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 $\boxed{\text{B}}$ .

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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All AMC 10 Problems and Solutions

@@ Line 5: / Line 5: @@
 (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.
+(b) Obviously, the perimeter changes.
 (c) The area clearly doesn't change, as the base and height remain the same.