1952 AHSME Problems/Problem 33: Difference between revisions
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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. So the answer is <math>\fbox{B}</math>. | |||
Assume that the side length of the square is 4 units. Then the perimeter for both shapes is is 4 * | |||
== See also == | == See also == | ||
Revision as of 21:30, 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 | |
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