Difference between revisions of "2014 AMC 8 Problems/Problem 6"
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==Solution== | ==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>. | 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>. | ||
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| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/-oIKrq97ya0 | ||
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| + | ~Education, the Study of Everything | ||
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| + | |||
==Video Solution== | ==Video Solution== | ||
Revision as of 10:15, 2 July 2023
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 
.
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. 
