In trapezoid $ABCD$, the sides $AB$ and $CD$ are equal. The perimeter of $ABCD$ is

$\text{(A)}\ 27 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 34 \qquad \text{(E)}\ 48$
There is a rectangle present, with both horizontal bases being $8$ units in length. The excess units on the bottom base must then be $16-8=8$. The fact that $AB$ and $CD$ are equal in length indicate, by the Pythagorean Theorem, that these excess lengths are equal. There are two with a total length of $8$ units, so each is $4$ units. The triangle has a hypotenuse of $5$, because the triangles are $3-4-5$ right triangles. So, the sides of the trapezoid are $8$, $5$, $16$, and $5$. Adding those up gives us the perimeter, $8 + 5 + 16 + 5 = 13 + 21 = \boxed{\text{(D)}\ 34}$ units.