smc

Use vectors. Place an origin at $A$, with $B=p,D=q,C=p+q$. We know that $\Vert p\times q\Vert =10$, and also $E=\frac{2}{3}p,F=p+\frac{2}{5}q,G=\frac{2}{5}q$, and now we can find the area of $EFHG$ by dividing it into two triangles and using cross-products (the expressions simplify using the fact that the cross-product distributes over addition, it is anticommutative, and a vector crossed with itself gives zero). $\boxed{C}$ # Solution 2 We note that $ABFG$ is a parallelogram because $AG=BF=2$ and $AG\parallel BF$. Using the same reasoning, $GFCD$ is also a parallelogram. Assume that the height of parallelogram $ABFG$ with respect to base $AB$ is $x$. Then, the area of parallelogram $ABFG$ is $AB\ast x$. The area of triangle $EFG$ is $\frac{AB\ast x}{2}$, which is half of the area of parallelogram $ABFG$. Likewise, the area of triangle $FGH$ is half the area of parallelogram $GFCD$. Thus, $[EFHG]=[EFG]+[FGH]=1/2[ABFG]+1/2[GFCD]=1/2[ABCD]=1/2(10)=\boxed{5}$