Difference between revisions of "2010 AMC 12B Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | We find the area of the big rectangle to be <math>8 \times 5 = 40</math>, and the area of the smaller rectangle to be <math>(8 - 2) \times (5 - 2) = 18</math>. The answer | + | We find the area of the big rectangle to be <math>8 \times 5 = 40</math>, and the area of the smaller rectangle to be <math>(8 - 2) \times (5 - 2) = 18</math>. The answer is then <math>40 - 18 = 22</math> <math>(A)</math>. |
== See also == | == See also == | ||
| − | {{AMC12 box|year=2010|num-b= | + | {{AMC12 box|year=2010|num-b=1|num-a=3|ab=B}} |
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
| + | [[Category:Area Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 09:58, 4 July 2013
Problem 2
A big 
is formed as shown. What is its area?
![[asy] unitsize(4mm); defaultpen(linewidth(.8pt)); draw((0,0)--(5,0)--(5,2)--(2,2)--(2,8)--(0,8)--cycle); label("8",(0,4),W); label("5",(5/2,0),S); label("2",(5,1),E); label("2",(1,8),N); [/asy]](http://latex.artofproblemsolving.com/1/2/2/122a4290575eda1b63a0b302c47384ced6bb393e.png)
Solution
We find the area of the big rectangle to be 
, and the area of the smaller rectangle to be 
. The answer is then 
.
See also
| 2010 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 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 AMC 12 Problems and Solutions | |
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