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

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==Problem==
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==Solution==
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(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>.
 
(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>.
  
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Only <math>1</math> changes, so <math>\boxed{\text{B}}</math>.
 
Only <math>1</math> changes, so <math>\boxed{\text{B}}</math>.
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==See Also==
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{{AMC10 box|year=2000|num-b=4|num-a=6}}

Revision as of 19:33, 8 January 2009

Problem

Solution

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

See Also

2000 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions