Difference between revisions of "1958 AHSME Problems/Problem 18"

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The area of a circle is doubled when its radius <math> r</math> is increased by <math> n</math>. Then <math> r</math> equals:
 
The area of a circle is doubled when its radius <math> r</math> is increased by <math> n</math>. Then <math> r</math> equals:
  
<math> \textbf{(A)}\ n(\sqrt{2} \plus{} 1)\qquad  
+
<math> \textbf{(A)}\ n(\sqrt{2} + 1)\qquad  
\textbf{(B)}\ n(\sqrt{2} \minus{} 1)\qquad  
+
\textbf{(B)}\ n(\sqrt{2} - 1)\qquad  
 
\textbf{(C)}\ n\qquad  
 
\textbf{(C)}\ n\qquad  
\textbf{(D)}\ n(2 \minus{} \sqrt{2})\qquad  
+
\textbf{(D)}\ n(2 - \sqrt{2})\qquad  
\textbf{(E)}\ \frac{n\pi}{\sqrt{2} \plus{} 1}</math>
+
\textbf{(E)}\ \frac{n\pi}{\sqrt{2} + 1}</math>
  
 
== Solution ==
 
== Solution ==

Revision as of 23:19, 13 March 2015

Problem

The area of a circle is doubled when its radius $r$ is increased by $n$. Then $r$ equals:

$\textbf{(A)}\ n(\sqrt{2} + 1)\qquad  \textbf{(B)}\ n(\sqrt{2} - 1)\qquad  \textbf{(C)}\ n\qquad  \textbf{(D)}\ n(2 - \sqrt{2})\qquad  \textbf{(E)}\ \frac{n\pi}{\sqrt{2} + 1}$

Solution

$\fbox{}$

See Also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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