25. A point is marked one quarter of the way along each side of a triangle, as shown.
What fraction of the area of the triangle is shaded?
A $\frac{7}{16}$
B $\frac{1}{2}$
C $\frac{9}{16}$
D $\frac{5}{8}$
E $\frac{11}{16}$
25. D Points $A, B, C, D, E, F$ on the perimeter of the triangle are as shown. Let $A D$ have length $x$ so that $D B$ has length $3 x$. Let the perpendicular from $C$ to $A B$ have length $4 h$. So, by similar triangles, the perpendicular from $E$ to $D B$ has length $h$. The area of triangle $A B C$ is $\frac{1}{2} \times 4 x \times 4 h=8 x h$. The area of triangle $D B E$ is $\frac{1}{2} \times 3 x \times h=\frac{3}{2} x h$. So the area of triangle $D B E$ is $\frac{3}{16}$ of the area of triangle $A B C$.

Similarly, by drawing perpendiculars to $C B$ from $A$ and from $F$, it may be shown that the area of triangle $F E C$ is $\frac{3}{16}$ of the area of triangle $A B C$.
So the fraction of the area of the triangle that is shaded is $1-\frac{3}{16}-\frac{3}{16}=\frac{5}{8}$.