$17 \cdot 198$ The three sides of a triangle are 11, 15, and $k\left(k \in Z^{+}\right)$, the number of $k$ values that make the triangle an obtuse triangle is
(A) 5.
(B) 7.
(C) 12.
(D) 13.
(E) 14.
(46th American High School Mathematics Examination, 1995)
[Solution] First, by the triangle inequality, we know that one side of a triangle is less than the sum of the other two sides and greater than their difference:
$$
4<k<26 \text{. }
$$

Since the triangle is an obtuse triangle, we have:
$11^{2}+15^{2}-k^{2}<0$ or $11^{2}+k^{2}-15^{2}<0$ or $15^{2}+k^{2}-11^{2}<0$.
Solving these inequalities, we get $k \leqslant 10$ or $k \geqslant 19$.
In summary, the values of $k$ that satisfy the problem's requirements are $5,6,7,8,9,10$ and $19,20,21,22,23,24,25$, a total of 13 values.
Therefore, the answer is $(D)$.