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Difference between revisions of "2021 GMC 10B Problems/Problem 2"

(Created page with "==Problem== The radius of a circle that has an area of <math>\frac{\pi}{\sqrt{2}}</math> is <math>r</math>. Find <math>r^{2}</math> <math>\textbf{(A)} ~\frac{1}{2} \qquad\tex...")
 
m (→‎Solution)
 
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\boxed{\textbf{(B)}~\frac{\sqrt2}{2}} &= r^2.
 
\boxed{\textbf{(B)}~\frac{\sqrt2}{2}} &= r^2.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 +
~pineconee

Latest revision as of 14:17, 6 March 2022

Problem

The radius of a circle that has an area of $\frac{\pi}{\sqrt{2}}$ is $r$. Find $r^{2}$

$\textbf{(A)} ~\frac{1}{2} \qquad\textbf{(B)} ~\frac{\sqrt{2}}{2} \qquad\textbf{(C)} ~\sqrt{2} \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~4$

Solution

By the formula for the area of a circle, \[\frac{\pi}{\sqrt{2}} = \pi r^2.\]

Solving this equation, we get \begin{align*} \frac{\pi}{\sqrt{2}} &= \pi r^2 \\ \frac{1}{\sqrt{2}} &= r^2 \\ \boxed{\textbf{(B)}~\frac{\sqrt2}{2}} &= r^2. \end{align*} ~pineconee

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