Difference between revisions of "1952 AHSME Problems/Problem 1"

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Let the radius of the circle be the common fraction <math>\frac{a}{b}.</math> Then the area of the circle is <math>\pi \cdot \frac{a^2}{b^2}.</math>
 
Let the radius of the circle be the common fraction <math>\frac{a}{b}.</math> Then the area of the circle is <math>\pi \cdot \frac{a^2}{b^2}.</math>
 
Because <math>\pi</math> is irrational and <math>\frac{a^2}{b^2}</math> is rational, their product must be irrational. The answer is <math>\boxed{B}.</math>
 
Because <math>\pi</math> is irrational and <math>\frac{a^2}{b^2}</math> is rational, their product must be irrational. The answer is <math>\boxed{B}.</math>
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==Solution 2==
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The phrasing of the problem makes it clear that the rule would apply to all rational radii. So let r be equal to 1. <math>\pi</math> times {1^2} is equal to <math>\pi</math> which is irrational. Therefore, the answer is <math>\boxed{b}.</math>
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~YJC64002776
  
 
==See also==
 
==See also==
 
{{AHSME 50p box|year=1952|before=First Question|num-a=2}}
 
{{AHSME 50p box|year=1952|before=First Question|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 10:32, 19 January 2021

Problem

If the radius of a circle is a rational number, its area is given by a number which is:

$\textbf{(A)\ } \text{rational}  \qquad \textbf{(B)\ } \text{irrational} \qquad \textbf{(C)\ } \text{integral} \qquad \textbf{(D)\ } \text{a perfect square }\qquad \textbf{(E)\ } \text{none of these}$

Solution

Let the radius of the circle be the common fraction $\frac{a}{b}.$ Then the area of the circle is $\pi \cdot \frac{a^2}{b^2}.$ Because $\pi$ is irrational and $\frac{a^2}{b^2}$ is rational, their product must be irrational. The answer is $\boxed{B}.$

Solution 2

The phrasing of the problem makes it clear that the rule would apply to all rational radii. So let r be equal to 1. $\pi$ times {1^2} is equal to $\pi$ which is irrational. Therefore, the answer is $\boxed{b}.$

~YJC64002776

See also

1952 AHSC (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
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All AHSME Problems and Solutions

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