$17 \cdot 151$ In the right triangle $ABC$ with hypotenuse $AB$, the side $AC=15$. The altitude from $C$ to the hypotenuse $AB$ divides $AB$ into segments $AH$ and $HB$, where $HB=16$. The area of triangle $ABC$ is
(A) 120.
(B) 144.
(C) 150.
(D) 216.
(E) $144 \sqrt{5}$.
[Sol] Let $A H=x$, then by the given condition,

i.e. $\square$
$$
225=x(x+16) \text {, }
$$

or
$$
\begin{array}{l}
x^{2}+16 x-225=0, \\
(x+25)(x-9)=0 .
\end{array}
$$

Thus, $x=9$. (The negative value has been discarded)
Therefore, $A B=25, \quad C H=\sqrt{15^{2}-9^{2}}=12$.
So the area of $\triangle A B C$ is $\quad \frac{1}{2} \cdot 25 \cdot 12=150$. Hence, the answer is $(C)$.