, 2019

Suppose that $SU=UW=WR=b$ and $PS=h$. Since the width of rectangle $PQRS$ is $3b$ and its height is $h$, then its area is $3bh$. Since $SU = b$ and the distance between parallel lines $PQ$ and $SR$ is $h$, then the area of $\triangle STU$ is $\frac{1}{2}bh$. Similarly, the area of each of $\triangle UVW$ and $\triangle WXR$ is $\frac{1}{2}bh$. Therefore, the fraction of the rectangle that is shaded is $\dfrac{3 \times \frac{1}{2}bh}{3bh}$ which equals $\dfrac{1}{2}$.