2002 AMC 12A Problems/Problem 7
- The following problem is from both the 2002 AMC 12A #7 and 2002 AMC 10A #7, so both problems redirect to this page.
Contents
Problem
A arc of circle A is equal in length to a arc of circle B. What is the ratio of circle A's area and circle B's area?
Solutions
Solution 1
Let and be the radii of circles and, respectively.
It is well known that in a circle with radius , a subtended arc opposite an angle of degrees has length .
Using that here, the arc of circle A has length . The arc of circle B has length . We know that they are equal, so , so we multiply through and simplify to get . As all circles are similar to one another, the ratio of the areas is just the square of the ratios of the radii, so our answer is .
Solution 2
The arc of circle is that of circle .
The circumference of circle is that of circle ( is the larger circle).
The radius of circle is that of circle .
The area of circle is that of circle .
Solution 3
We can use a very cheap way to find the answer based on the choices. Firstly, if a higher angle measure equals a lower angle measure, that means that the ratio of areas must be lower, cutting D and E. Then we realize that the ratio has to be with square numbers, as there are no square roots. Therefore, it is
-dragoon
Solution 4
- Written by a 6th grader, please don't mind*
Set Circle A's arc to 45 degrees. In a full circle, there is 360 degrees, so 360/45 = 8. If we set the arc to be pi degrees, then in total there will be 8 pi in the circumference of this circle, yielding 4 as the radius and 16pi as the area.
Since the arc of 30 degrees in Circle B is also pi, and there are 12 30 degree arcs in a circle, there will be 12pi in the circumference of Circle B, yielding 6 as the radius and 36pi as the area.
Now, we put 16pi/36pi as our ratio, yielding 16/36=4/9. Therefore, it is 4/9.
~MathKatana
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
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 12 Problems and Solutions |
2002 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.