Difference between revisions of "2014 AMC 8 Problems/Problem 6"
m (→Solution) |
(→Solution 2) |
||
| Line 10: | Line 10: | ||
==Solution 2== | ==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>. | 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>. | ||
| + | |||
| + | ==Solution 3== | ||
| + | The formula for a consecutive perfect squared sum is <math>S = \frac{n(n+1)(2n+1)}{6}</math>, where <math>n</math> is the number of terms from 1. | ||
| + | |||
| + | When multiplied by the constant length 2 this gives <math>S = \frac{n(n+1)(2n+1)}{3}</math>. | ||
| + | |||
| + | Plugging in <math>n = 6</math> gives <math>\boxed{\textbf{(D)}~182}</math>. | ||
==Video Solution (CREATIVE THINKING)== | ==Video Solution (CREATIVE THINKING)== | ||
Revision as of 00:38, 14 November 2024
Contents
Problem
Six rectangles each with a common base width of 
have lengths of 
, and 
. What is the sum of the areas of the six rectangles?
Solution
The sum of the areas is equal to 
. This is equal to 
, which is equal to 
. This is equal to our final answer of 
.
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 .
Solution 3
The formula for a consecutive perfect squared sum is 
, where 
is the number of terms from 1.
When multiplied by the constant length 2 this gives 
.
Plugging in 
gives .
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
https://youtu.be/SvjJETtxQnk ~savannahsolver
See Also
| 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. 
