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> (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>.
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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 similar 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}}

Latest revision as of 15:15, 1 July 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 similar 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 (ProblemsAnswer KeyResources)
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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