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*1+2*4+2*9+2*16+2*25+2*36</math>. This is simply equal to <math>2*(1+4+9+16+25+36)</math>, which is equal to <math>2*91</math>, which is equal to our final answer of \boxed{182}. | + | The sum of the areas is equal to <math>2*1+2*4+2*9+2*16+2*25+2*36</math>. This is simply equal to <math>2*(1+4+9+16+25+36)</math>, which is equal to <math>2*91</math>, which is equal to our final answer of <math>\boxed{182}</math>. |
==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}} | ||
Revision as of 21:30, 26 November 2014
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 simply equal to 
, which is equal to 
, which is equal to our final answer of 
.
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. 
