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

 
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Latest revision as of 11:03, 4 July 2013

Problem

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

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

Solution

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

See Also

2012 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 10 Problems and Solutions

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