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 01:00, 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 | |
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All AHSME Problems and Solutions |
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