An isosceles right triangle with legs of length $8$ is partitioned into $16$ congruent triangles as shown.  The shaded area is

$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 64$
Solution 1
Because the smaller triangles are congruent, the shaded area take $\frac{10}{16}$ of the largest triangles area, which is $\frac{8 \times 8}{2}=32$, so the shaded area is  $\frac{10}{16} \times 32= \boxed{\text{(B)}\ 20}$.

Solution 2
Each of the triangle has side length of $\frac{1}{4} \times 8=2$, so the area is $\frac{1}{2} \times 2 \times 2=2$. Because there are $10$ triangles is the shaded area, its area is $2 \times 10 =\boxed{\text{(B)}\ 20}$.