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Difference between revisions of "2008 AMC 10A Problems/Problem 10"

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==Solution 1==
 
==Solution 1==
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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>
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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

Problem

Each of the sides of a square $S_1$ with area $16$ is bisected, and a smaller square $S_2$ is constructed using the bisection points as vertices. The same process is carried out on $S_2$ to construct an even smaller square $S_3$. What is the area of $S_3$?

$\mathrm{(A)}\ \frac{1}{2}\qquad\mathrm{(B)}\ 1\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ 4$

Solution 1

Since the area of the large square is $16$, the side length equals $4$. If all sides are bisected, the resulting square has side length $2\sqrt{2}$, thus making the area $8$. If we repeat this process again, we notice that the area is just half that of the previous square, so the area of $S_{3} = 4 \longrightarrow \fbox{E}$

Solution 2

Since the length ratio is $\frac{1}{\sqrt{2}}$, then the area ratio is $\frac{1}{2}$ (since the area ratio between two 2-dimensional objects is equal to the side ratio of those objects squared). This means that $S_2 = 8$ and $S_3 = \boxed{\textbf{(E) }4}$.

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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