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 | + | 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== | ||
+ | https://youtu.be/SvjJETtxQnk ~savannahsolver | ||
==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}} |
Latest revision as of 09:14, 27 April 2022
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 .
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.