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 18:33, 8 January 2009
Problem
Solution
(a) Clearly does not change, as 
. Since 
does not change, neither does 
.
(b) Obviously, the perimetar changes.
(c) The area clearly doesn't change, as the base and height remain the same.
(d) The bases and do not change, and neither does the height, so the trapezoid remains the same.
Only 
changes, so 
.
See Also
| 2000 AMC 10 (Problems • Answer Key • Resources) | ||
| 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 | ||
