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Difference between revisions of "1978 AHSME Problems/Problem 2"

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==Solution 1==
 
==Solution 1==
 
Creating equations, we get <math>4\cdot\frac{1}{2\pi r} = 2r</math>. Simplifying, we get <math>\frac{1}{\pi r} = r</math>. Multiplying each side by <math>r</math>, we get <math>\frac{1}{\pi} = r^2</math>. Because the formula of the area of a circle is <math>\pi r^2</math>, we multiply each side by <math>\pi</math> to get <math>1 = \pi r^2</math>.
 
Creating equations, we get <math>4\cdot\frac{1}{2\pi r} = 2r</math>. Simplifying, we get <math>\frac{1}{\pi r} = r</math>. Multiplying each side by <math>r</math>, we get <math>\frac{1}{\pi} = r^2</math>. Because the formula of the area of a circle is <math>\pi r^2</math>, we multiply each side by <math>\pi</math> to get <math>1 = \pi r^2</math>.
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Therefore, our answer is <math>\boxed{\textbf{(C)  }1}</math>
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Therefore, our answer is <math>\boxed{\textbf{(C)  }1}</math>  
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~awin
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==See Also==
 +
{{AHSME box|year=1978|num-b=1|num-a=3}}
 +
{{MAA Notice}}

Latest revision as of 10:59, 13 February 2021

Problem 2

If four times the reciprocal of the circumference of a circle equals the diameter of the circle, then the area of the circle is

$\textbf{(A) }\frac{1}{\pi^2}\qquad \textbf{(B) }\frac{1}{\pi}\qquad \textbf{(C) }1\qquad \textbf{(D) }\pi\qquad \textbf{(E) }\pi^2$

Solution 1

Creating equations, we get $4\cdot\frac{1}{2\pi r} = 2r$. Simplifying, we get $\frac{1}{\pi r} = r$. Multiplying each side by $r$, we get $\frac{1}{\pi} = r^2$. Because the formula of the area of a circle is $\pi r^2$, we multiply each side by $\pi$ to get $1 = \pi r^2$. Therefore, our answer is $\boxed{\textbf{(C) }1}$

~awin


See Also

1978 AHSME (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
All AHSME Problems and Solutions

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