Difference between revisions of "2007 AMC 8 Problems/Problem 22"
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− | 5 | + | Algebraic: The shortest segments would be perpendicular to the square. The lemming went <math>x</math> meters horizontally and <math>y</math> meters vertically. No matter how much it went, the lemming would have been <math>x</math> and <math>y</math> meters from the sides and <math>10-x</math> and <math>10-y</math> meters from the remaining two. To find the average, add the lengths of the four segments and divide by four: <math>\frac {x+10-x+y+10-y}{4} = 5</math> |
+ | ---- | ||
+ | Solution: 5 '''(C)''' |
Revision as of 19:55, 21 November 2011
Algebraic: The shortest segments would be perpendicular to the square. The lemming went meters horizontally and
meters vertically. No matter how much it went, the lemming would have been
and
meters from the sides and
and
meters from the remaining two. To find the average, add the lengths of the four segments and divide by four:
Solution: 5 (C)