Difference between revisions of "2000 AMC 10 Problems/Problem 5"
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(b) Obviously, the perimeter changes. | (b) Obviously, the perimeter changes. | ||
− | (c) The area clearly doesn't change, as the base and height remain the same. | + | (c) The area clearly doesn't change, as both the base <math>AB</math> and its corresponding height remain 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. | (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 <math>\boxed{\text{B}}</math>. | + | Only <math>1</math> quantity changes, so the correct answer is <math>\boxed{\text{B}}</math>. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2000|num-b=4|num-a=6}} | {{AMC10 box|year=2000|num-b=4|num-a=6}} |
Revision as of 10:03, 24 January 2009
Problem
Points and are the midpoints of sides and of . As moves along a line that is parallel to side , how many of the four quantities listed below change?
(a) the length of the segment
(b) the perimeter of
(c) the area of
(d) the area of trapezoid
Solution
(a) Clearly does not change, as . Since does not change, neither does .
(b) Obviously, the perimeter changes.
(c) The area clearly doesn't change, as both the base and its corresponding height remain the same.
(d) The bases and do not change, and neither does the height, so the trapezoid remains the same.
Only quantity changes, so the correct answer is .
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 |