The sides of a triangle are $30$, $70$, and $80$ units. If an altitude is dropped upon the side of length $80$, the larger segment cut off on this side is:
$\textbf{(A)}\ 62\qquad  \textbf{(B)}\ 63\qquad  \textbf{(C)}\ 64\qquad  \textbf{(D)}\ 65\qquad  \textbf{(E)}\ 66$
Let the shorter segment be $x$ and the altitude be $y$. The larger segment is then $80-x$.  By the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem)
, \[30^2-y^2=x^2 \qquad(1)\]
and \[(80-x)^2=70^2-y^2 \qquad(2)\]
Adding $(1)$ and $(2)$ and simplifying gives $x=15$. Therefore, the answer is $80-15=\boxed{\textbf{(D)}~65}$
~megaboy6679