The lengths of the sides of a triangle are integers, and its area is also an integer. 
One side is $21$ and the perimeter is $48$. The shortest side is:
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 16$
Let $b$ and $c$ be the other two sides of the triangle.  The [perimeter](https://artofproblemsolving.com/wiki/index.php/Perimeter) of the triangle is $48$ units, so $c = 27-b$ and the semiperimeter equals $24$ units.
By [Heron's Formula](https://artofproblemsolving.com/wiki/index.php/Heron%27s_Formula), the area of the triangle is $\sqrt{24 \cdot 3(24-b)(b-3)}$.  Plug in the answer choices for $b$ and write the [prime factorization](https://artofproblemsolving.com/wiki/index.php/Prime_factorization) of the product to make sure it is a perfect square.
Testing $b = 8$ results in the area being $\sqrt{6 \cdot 4 \cdot 3 \cdot 16 \cdot 5} = \sqrt{2^7 \cdot 3^2 \cdot 5}$, so $8$ does not work.  However, testing $b = 10$ results in the area being $\sqrt{6 \cdot 4 \cdot 3 \cdot 14 \cdot 7} = \sqrt{2^4 \cdot 3^2 \cdot 7^2}$, so $10$ works.  The third side is $17$, and the sides satisfy the [Triangle Inequality](https://artofproblemsolving.com/wiki/index.php/Triangle_Inequality), so the answer is $\boxed{\textbf{(B)}}$.