2006 AMC 8 Problems/Problem 5

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]

$\textbf{(A)}\ 15 \qquad  \textbf{(B)}\ 20 \qquad  \textbf{(C)}\ 24 \qquad  \textbf{(D)}\ 30 \qquad  \textbf{(E)}\ 40$

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 $\frac{60}{2}=\boxed{\textbf{(D)}\ 30 }$.

See Also

2006 AMC 8 (ProblemsAnswer KeyResources)
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

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