Difference between revisions of "2007 AMC 8 Problems/Problem 16"
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The circumference of a circle is obtained by simply multiplying the radius by <math>2\pi</math>. So, the C-coordinate (in this case, it is the x-coordinate) will increase at a steady rate. The area, however, is obtained by squaring the radius and multiplying it by <math>\pi</math>. Since squares do not increase in an evenly spaced arithmetic sequence, the increase in the A-coordinates ( aka the y- coordinates) will be much more significant. The graph that satisfies these two conditions is '''graph A'''. | The circumference of a circle is obtained by simply multiplying the radius by <math>2\pi</math>. So, the C-coordinate (in this case, it is the x-coordinate) will increase at a steady rate. The area, however, is obtained by squaring the radius and multiplying it by <math>\pi</math>. Since squares do not increase in an evenly spaced arithmetic sequence, the increase in the A-coordinates ( aka the y- coordinates) will be much more significant. The graph that satisfies these two conditions is '''graph A'''. | ||
Revision as of 14:38, 31 January 2015
Problem
Amanda Reckonwith draws five circles with radii and . Then for each circle she plots the point , where is its circumference and is its area. Which of the following could be her graph?
Solution
The circumference of a circle is obtained by simply multiplying the radius by . So, the C-coordinate (in this case, it is the x-coordinate) will increase at a steady rate. The area, however, is obtained by squaring the radius and multiplying it by . Since squares do not increase in an evenly spaced arithmetic sequence, the increase in the A-coordinates ( aka the y- coordinates) will be much more significant. The graph that satisfies these two conditions is graph A.
See Also
2007 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.