# 2006 AMC 8 Problems/Problem 5

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

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