Difference between revisions of "1970 AHSME Problems/Problem 2"

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== Solution ==
 
== Solution ==
<math>\fbox{A}</math>
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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}}   
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{{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

$\text{(A) } \frac{4}{\pi}\quad \text{(B) } \frac{\pi}{\sqrt{2}}\quad \text{(C) } \frac{4}{1}\quad \text{(D) } \frac{\sqrt{2}}{\pi}\quad \text{(E) } \frac{\pi}{4}$

Solution

Let's say the circle has a circumference ( or perimeter of $4\pi$). Since the perimeter of the square is the same as the perimeter of the circle, the side length of the square is $\pi$. That means that the area of the square is $\pi^2$. The area of the circle is $4\pi$. So the area of the circle over the area of the square is $\frac{4\pi}{\pi^2} \Rightarrow \frac{4}{\pi} \Rightarrow$ $\fbox{A}$

See also

1970 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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