Difference between revisions of "2006 AMC 8 Problems/Problem 5"

Revision as of 23:41, 31 August 2018

Problem

Points $A, B, C$ and $D$ are midpoints of the sides of the larger square. If the larger square has area 60, what is the area of the smaller square?

$[asy]size(100); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle,linewidth(1)); draw((0,1)--(1,2)--(2,1)--(1,0)--cycle); label("$A$", (1,2), N); label("$B$", (2,1), E); label("$C$", (1,0), S); label("$D$", (0,1), W);[/asy]$

$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40 \qquad \textbf{(F)}\ suicide \qquad \textbf{(E)}\ it \qquad \textbf{(E)}\ would \qquad \textbf{(E)}\ be \qquad \textbf{(E)}\ awesome!$

Solution

Solution 1

Drawing segments $AC$ and $BD$ , the number of triangles outside square $ABCD$ is the same as the number of triangles inside the square. Thus areas must be equal so the area of $ABCD$ is half the area of the larger square which is $\frac{60}{2}=\boxed{\textbf{(D)}\ 30 }$ .

Solution 2

If the side length of the larger square is $x$ , the side length of the smaller square is $\frac{\sqrt{2} \cdot x}{2}$ . Therefore the area of the smaller square is $\frac{x^2}{2}$ , half of the larger square's area, $x^2$ .

Thus, the area of the smaller square in the picture is $\frac{60}{2}=\boxed{\textbf{(D)}\ 30 }$ .

2006 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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