Difference between revisions of "2008 AMC 10A Problems/Problem 10"
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<math>\mathrm{(A)}\ \frac{1}{2}\qquad\mathrm{(B)}\ 1\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ 4</math> | <math>\mathrm{(A)}\ \frac{1}{2}\qquad\mathrm{(B)}\ 1\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ 4</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | {{ | + | Since the area of the large square is <math>16</math>, the side equals <math>4</math> and if you bisect all of the sides, you get a square of side length <math>2\sqrt{2}</math> thus making the area <math>8</math>. If we repeat this process again, we notice that the area is just half that of the previous square, so the area of <math>S_{3} = 4 \longrightarrow \fbox{E}</math> |
| − | |||
==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 18:00, 15 May 2008
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 
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 | ||
