Difference between revisions of "2012 AMC 10A Problems/Problem 2"

(Problem 2)
(Problem 2)
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<math> \textbf{(A)}\ \emph{2\ by\ 4}\qquad\textbf{(B)}\ \emph{\ 2\ by\ 6}\qquad\textbf{(C)}\ \emph{\ 2\ by\ 8}\qquad\textbf{(D)}\ \emph{4\ by\ 4}\qquad\textbf{(E)}\ \emph{4\ by\ 8} </math>
 
<math> \textbf{(A)}\ \emph{2\ by\ 4}\qquad\textbf{(B)}\ \emph{\ 2\ by\ 6}\qquad\textbf{(C)}\ \emph{\ 2\ by\ 8}\qquad\textbf{(D)}\ \emph{4\ by\ 4}\qquad\textbf{(E)}\ \emph{4\ by\ 8} </math>
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== Solution ==
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Cutting the square in half will bisect one pair of sides while the other side will remain unchanged. Thus, the new square is <math>\frac{8}{2}*8 = \qquad\textbf{(E)}\ 4*8</math>.

Revision as of 18:33, 8 February 2012

Problem 2

A square with side length 8 is cut in half, creating two congruent rectangles. What are the dimensions of one of these rectangles?

$\textbf{(A)}\ \emph{2\ by\ 4}\qquad\textbf{(B)}\ \emph{\ 2\ by\ 6}\qquad\textbf{(C)}\ \emph{\ 2\ by\ 8}\qquad\textbf{(D)}\ \emph{4\ by\ 4}\qquad\textbf{(E)}\ \emph{4\ by\ 8}$

Solution

Cutting the square in half will bisect one pair of sides while the other side will remain unchanged. Thus, the new square is $\frac{8}{2}*8 = \qquad\textbf{(E)}\ 4*8$.

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