Difference between revisions of "1952 AHSME Problems/Problem 33"
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== Solution 1== | == Solution 1== | ||
| − | Assume that the side length of the square is 4 units. Then the perimeter for both shapes is is 4 * 4 = 16 units. Since the perimeter of a circle is 2πr, then 2πr =16 and r = 8/π, which is about 2.5. The area of a circle is πr^2 so (2.5^2)*π = 6.25π or about 19.6. This is more than the area of the square, which is 4^2 =16 | + | Assume that the side length of the square is 4 units. Then the perimeter for both shapes is is 4 * 4 = 16 units. Since the perimeter of a circle is 2πr, then 2πr =16 and r = 8/π, which is about 2.5. The area of a circle is πr^2 so (2.5^2)*π = 6.25π or about 19.6. This is more than the area of the square, which is 4^2 =16, so the answer is <math>\fbox{B}</math>. |
== See also == | == See also == | ||
Revision as of 21:33, 12 April 2020
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 * 4 = 16 units. Since the perimeter of a circle is 2πr, then 2πr =16 and r = 8/π, which is about 2.5. The area of a circle is πr^2 so (2.5^2)*π = 6.25π or about 19.6. This is more than the area of the square, which is 4^2 =16, so the answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
