Difference between revisions of "2008 AMC 10A Problems/Problem 10"
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==Solution 1== | ==Solution 1== | ||
| − | Since the area of the large square is <math>16</math>, the side length equals <math>4</math>. If all sides are bisected, | + | Since the area of the large square is <math>16</math>, the side length equals <math>4</math>. If all sides are bisected, the resulting square has 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> |
==Solution 2== | ==Solution 2== | ||
Revision as of 16:39, 4 June 2021
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 length equals 
. If all sides are bisected, the resulting square has 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 
(since the area ratio between two 2-dimensional objects is equal to the side ratio of those objects squared). 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
