Difference between revisions of "2014 AMC 8 Problems/Problem 6"

Line 5: Line 5:
  
 
==Solution==
 
==Solution==
 +
The sum of the areas is equal to <math>2*1+2*4+2*9+2*16+2*25+2*36</math>. This is simply equal to <math>2*(1+4+9+16+25+36)</math>, which is equal to <math>2*91</math>, which is equal to our final answer of \boxed{182}.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2014|num-b=5|num-a=7}}
 
{{AMC8 box|year=2014|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:30, 26 November 2014

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}.

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

Invalid username
Login to AoPS