$18 \cdot 65$ In a right triangle $ABC$, $BC=5, AC=12, AM=x, MN \perp AC$, $NP \perp BC$. If $y=MN+NP$ is the semi-perimeter of rectangle $MCPN$, then
(A) $y=\frac{1}{2}(5+12)$.
(B) $y=\frac{5 x}{12}+\frac{12}{5}$.
(C) $y=\frac{144-7 x}{12}$.
(D) $y=12$.
(E) $y=\frac{5 x}{12}+6$.
[Solution] By the property of similar triangles, we have $\frac{M N}{x}=\frac{5}{12}$, which means $M N=\frac{5}{12} x$.
Also, $\quad N P=12-x$,
then $y=N P+M N=\frac{144-7 x}{12}$.
Therefore, the answer is $(C)$.