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

Revision as of 01:04, 15 September 2024

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 $\boxed{\textbf{(D)}~182}$ .

Video Solution (CREATIVE THINKING)

https://youtu.be/-oIKrq97ya0

~Education, the Study of Everything

Video Solution

https://youtu.be/SvjJETtxQnk ~savannahsolver

2014 AMC 8 (Problems • Answer Key • Resources)
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.

@@ Line 6: / Line 6: @@
 ==Solution==
 The sum of the areas is equal to <math>2\cdot1+2\cdot4+2\cdot9+2\cdot16+2\cdot25+2\cdot36</math>. This is equal to <math>2(1+4+9+16+25+36)</math>, which is equal to <math>2\cdot91</math>. This is equal to our final answer of <math>\boxed{\textbf{(D)}~182}</math>.
+==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 <math>\boxed{\textbf{(D)}~182}</math>.
 ==Video Solution (CREATIVE THINKING)==

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