Difference between revisions of "1970 AHSME Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | <math>\fbox{A}</math> | + | Let's say the circle has a circumference ( or perimeter of <math>4\pi</math>). Since the perimeter of the square is the same as the perimeter of the circle, the side length of the square is <math>\pi</math>. That means that the area of the square is <math>\pi^2</math>. The area of the circle is <math>4\pi</math>. So the area of the circle over the area of the square is <math>\frac{4\pi}{\pi^2} \Rightarrow \frac{4}{\pi} \Rightarrow</math> <math>\fbox{A}</math> |
== See also == | == See also == | ||
| − | {{AHSME box|year=1970|num-b=1|num-a=3}} | + | {{AHSME 35p box|year=1970|num-b=1|num-a=3}} |
[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 03:00, 12 March 2017
Problem
A square and a circle have equal perimeters. The ratio of the area of the circle to the are of the square is
Solution
Let's say the circle has a circumference ( or perimeter of 
). Since the perimeter of the square is the same as the perimeter of the circle, the side length of the square is 
. That means that the area of the square is 
. The area of the circle is . So the area of the circle over the area of the square is 
See also
| 1970 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
