$17 \cdot 67$ The three sides of a certain triangle are 30, 70, and 80. A perpendicular is drawn from the vertex opposite the side of length 80, to this side. The longer segment of the side of length 80 that is cut by the perpendicular is
(A) 62.
(B) 63.
(C) 64.
(D) 65.
(E) 66.
(9th American High School Mathematics Examination, 1958)
[Solution] Let the length of the longer segment be $d$. By the Pythagorean theorem, we have
$$
\begin{array}{l}
\quad 70^{2}-d^{2}=h^{2}=30^{2}-(80-d)^{2} \text {, } \\
\text { Solving for } \quad d=65, \\
\text { Therefore, the answer is }(D) \text {. }
\end{array}
$$