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?

$\textbf{(A)}\ 15 \qquad  \textbf{(B)}\ 20 \qquad  \textbf{(C)}\ 24 \qquad  \textbf{(D)}\ 30 \qquad  \textbf{(E)}\ 40$
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 $\frac{60}{2}=\boxed{\textbf{(D)}\ 30 }$.