Difference between revisions of "2008 AMC 10A Problems/Problem 10"
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==Solution 2== | ==Solution 2== | ||
Since the length ratio is <math>\frac{1}{\sqrt{2}}</math>, then the area ratio is <math>\frac{1}{2}</math>. This means that <math>S_2 = 8</math> and <math>S_3 = \boxed{\textbf{(E) }4}</math>. | Since the length ratio is <math>\frac{1}{\sqrt{2}}</math>, then the area ratio is <math>\frac{1}{2}</math>. This means that <math>S_2 = 8</math> and <math>S_3 = \boxed{\textbf{(E) }4}</math>. | ||
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==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=A|num-b=9|num-a=11}} | {{AMC10 box|year=2008|ab=A|num-b=9|num-a=11}} | ||
Revision as of 17:31, 19 November 2012
Contents
Problem
Each of the sides of a square 
with area 
is bisected, and a smaller square 
is constructed using the bisection points as vertices. The same process is carried out on to construct an even smaller square 
. What is the area of ?
Solution 1
Since the area of the large square is , the side equals 
and if you bisect all of the sides, you get a square of side length 
thus making the area 
. If we repeat this process again, we notice that the area is just half that of the previous square, so the area of 
Solution 2
Since the length ratio is 
, then the area ratio is 
. This means that 
and 
.
See also
| 2008 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 10 Problems and Solutions | ||
