2014 AMC 8 Problems/Problem 6

Revision as of 10:10, 2 December 2015 by Ghghghghghghghgh (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 $\boxed{\textbf{(D)}~182}$.

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png