, 2016

Solution 1

Since square $PQRS$ has side length 2, then $PQ=QR=RS=SP = 2$.

Since $W$, $X$, $Y$, $Z$ are the midpoints of the sides of $PQRS$, then $PW = PZ = 1$.

Since $\angle ZPW = 90^\circ$, then $WZ = \sqrt{PW^2 + PZ^2} = \sqrt{1^2+1^2} = \sqrt{2}$.

Therefore, square $WXYZ$ has side length $\sqrt{2}$.

The area of square $WXYZ$ is $(\sqrt{2})^2 = 2$ and the area of square $PQRS$ is $2^2 = 4$.

The ratio of these areas is $2:4$ or $1:2$.

Solution 2

Join $W$ to $Y$ and $X$ to $Z$.

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Since $PQRS$ is a square and $W$, $X$, $Y$, and $Z$ are the midpoints of its sides, then $WY$ and $ZX$ divide the square into four identical squares.

Each of these four squares is divided into two triangles of equal area by its diagonal. (These diagonals are $WZ$, $WX$, $XY$, $YZ$.)

Square $WXYZ$ is made up of 4 of these triangles of equal area.

Square $PQRS$ is made up of 8 of these triangles of equal area.

Therefore, the ratio of these areas is $4:8$ or $1:2$.