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

(Solution 2)
(Solution 2)
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==Solution 2==
 
==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. <math>\pi</math> times {1^2} is equal to <math>\pi</math> which is irrational. Therefore, the answer is <math>\boxed{B}.</math>
+
The phrasing of the problem makes it clear that the rule would apply to all rational radii. So let r be equal to one. <math>\pi</math> times {1^2} is equal to <math>\pi</math> which is irrational. Therefore, the answer is <math>\boxed{B}.</math>
  
 
~YJC64002776
 
~YJC64002776

Revision as of 11:33, 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 one. $\pi$ 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
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 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

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