2014 AMC 8 Problems/Problem 6

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 $2\cdot1+2\cdot4+2\cdot9+2\cdot16+2\cdot25+2\cdot36$. This is equal to $2(1+4+9+16+25+36)$, which is equal to $2\cdot91$. This is equal to our final answer of $\boxed{\textbf{(D)}~182}$.


Solution 2

we can just multiply the common width 2 by each of the lengths 1 by 1, the sum would be 182. This is slow and grouping the lengths is easier to. The answer is still $\boxed{\textbf{(D)}~182}$.

Solution 3

The formula for a consecutive perfect squared sum is $S = \frac{n(n+1)(2n+1)}{6}$, where $n$ is the number of terms from 1.

Multiplying by the constant length 2 for area gives $S = \frac{n(n+1)(2n+1)}{3}$.

Plugging in $n = 6$ gives $\boxed{\textbf{(D)}~182}$.

~PeterDoesPhysics

Video Solution (CREATIVE THINKING)

https://youtu.be/-oIKrq97ya0

~Education, the Study of Everything


Video Solution

https://youtu.be/SvjJETtxQnk ~savannahsolver

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