Difference between revisions of "1952 AHSME Problems/Problem 33"
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| − | Assume that the side length of the square is 4 units. Then the perimeter for both shapes is is 4 *3 = 12 units. Since the perimeter of a circle is 2 pi r, then 2 pi r =12 and r is 12/pi which is about 4. the area of a circle is pi r^2 so 4^2 pi =16 pi or about 48. This is more than the area of the square, which is 4^2 =16. So the answer is | + | Assume that the side length of the square is 4 units. Then the perimeter for both shapes is is 4 *3 = 12 units. Since the perimeter of a circle is 2 pi r, then 2 pi r =12 and r is 12/pi which is about 4. the area of a circle is pi r^2 so 4^2 pi =16 pi or about 48. This is more than the area of the square, which is 4^2 =16. So the answer is B. |
== See also == | == See also == | ||
Revision as of 12:55, 27 October 2019
Problem
A circle and a square have the same perimeter. Then:
Solution 1
Assume that the side length of the square is 4 units. Then the perimeter for both shapes is is 4 *3 = 12 units. Since the perimeter of a circle is 2 pi r, then 2 pi r =12 and r is 12/pi which is about 4. the area of a circle is pi r^2 so 4^2 pi =16 pi or about 48. This is more than the area of the square, which is 4^2 =16. So the answer is B.
See also
| 1952 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 32 |
Followed by Problem 34 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
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