Difference between revisions of "2008 AMC 10A Problems/Problem 10"
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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> | 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> | ||
==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> (since the area ratio between two 2-dimensional objects is equal to the side ratio of those objects squared). This means that <math>S_2 = 8</math> and <math>S_3 = \boxed{\textbf{(E) }4}</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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:45, 9 June 2017
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
(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 |
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