2008 AMC 10A Problems/Problem 10

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 $S_3 = \boxed{\textbf{(E) }4}$ .

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.

Page

Toolbox

Search

2008 AMC 10A Problems/Problem 10

Contents

Problem

Solution 1

Solution 2

See also