2014 AMC 8 Problems/Problem 6

Revision as of 16:11, 28 November 2014 by Chinmaiss (talk | contribs) (Solution)

Problem

Six rectangles each with a common base width of $2$ have lengths of $1, 4, 9, 16, 25$, and $36$. What is the sum of the areas of the six rectangles?

$\textbf{(A) }91\qquad\textbf{(B) }93\qquad\textbf{(C) }162\qquad\textbf{(D) }182\qquad \textbf{(E) }202$

Solution

The sum of the areas is equal to $2*1+2*4+2*9+2*16+2*25+2*36$. This is simply equal to $2*(1+4+9+16+25+36)$, which is equal to $2*91$, which is equal to our final answer of $\boxed{182 : (D)}$.

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AJHSME/AMC 8 Problems and Solutions

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