2006 AMC 8 Problems/Problem 5: Difference between revisions
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Thus, the area of the smaller square in the picture is <math> \frac{60}{2}=\boxed{\textbf{(D)}\ 30 } </math>. | Thus, the area of the smaller square in the picture is <math> \frac{60}{2}=\boxed{\textbf{(D)}\ 30 } </math>. | ||
{{AMC8 box|year=2006|num-b=4|num-a=6}} | |||
Revision as of 14:59, 21 November 2011
Problem
Points 
and 
are midpoints of the sides of the larger square. If the larger square has area 60, what is the area of the smaller square?
![[asy]size(100); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle,linewidth(1)); draw((0,1)--(1,2)--(2,1)--(1,0)--cycle); label("$A$", (1,2), N); label("$B$", (2,1), E); label("$C$", (1,0), S); label("$D$", (0,1), W);[/asy]](http://latex.artofproblemsolving.com/e/7/8/e7899c538a174faf9963a60d09886b7fb12ae942.png)
Solution
If the side length of the larger square is 
, the side length of the smaller square is 
. Therefore the area of the smaller square is 
, half of the larger square's area, 
.
Thus, the area of the smaller square in the picture is 
.
| 2006 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
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| All AJHSME/AMC 8 Problems and Solutions | ||
