25. The diagram shows a parallelogram $W X Y Z$ with area $S$. The diagonals of the parallelogram meet at the point $O$. The point $M$ is on the edge $Z Y$. The lines $W M$ and $Z X$ meet at $N$. The lines $M X$ and $W Y$ meet at $P$. The sum of the areas of triangles $W N Z$ and $X Y P$ is $\frac{1}{3} S$. What is the area of quadrilateral $M N O P$ ?

A $\frac{1}{6} S$
B $\frac{1}{8} S$
C $\frac{1}{10} S$
D $\frac{1}{12} S$
E $\frac{1}{14} S$
25. D The area of parallelogram $W X Y Z$ is $S$. Therefore the area of triangle $W X M$, which has the same base and height, is $\frac{1}{2} S$. Hence the sum of the areas of triangle $W M Z$ and triangle $X Y M$ is also $\frac{1}{2} S$. The sum of the areas of triangle $W N Z$ and triangle $X Y P$ is given as $\frac{1}{3} S$ and therefore the sum of the areas of triangle $Z N M$ and triangle $M P Y$ is $\frac{1}{2} S-\frac{1}{3} S=\frac{1}{6} S$. The area of triangle $Z O Y$, which has the same base as the parallelogram but only half the height is $\frac{1}{2} \times \frac{1}{2} S=\frac{1}{4} S$. Therefore the area of quadrilateral MNOP is $\frac{1}{4} S-\frac{1}{6} S=\frac{1}{12} S$.