A [right triangle](https://artofproblemsolving.com/wiki/index.php/Right_triangle) $ABC$ with [hypotenuse](https://artofproblemsolving.com/wiki/index.php/Hypotenuse) $AB$ has side $AC=15$. [Altitude](https://artofproblemsolving.com/wiki/index.php/Altitude) $CH$ divides $AB$ into segments $AH$ and $HB$, with $HB=16$. The [area](https://artofproblemsolving.com/wiki/index.php/Area) of $\triangle ABC$ is:
$\mathrm{(A) \ }120 \qquad \mathrm{(B) \ }144 \qquad \mathrm{(C) \ } 150 \qquad \mathrm{(D) \ }216 \qquad \mathrm{(E) \ } 144\sqrt{5}$
$AHC\sim ACB$ by $AA$, so $\frac{AH}{AC}=\frac{AC}{AH+16}$. Since $AC=15$, we have $\frac{AH}{15}=\frac{15}{AH+16}$. Cross mutliplying, we have $AH(AH+16)=225$. Solving this [quadratic](https://artofproblemsolving.com/wiki/index.php/Quadratic) yields $AH=9$. Also, $AHC\sim CHB$, so $\frac{AH} {HC}=\frac{HC}{HB}$. Substituting in known values, we have $\frac{9}{HC}=\frac{HC}{16}$, so $HC^2=144$ and $HC=12$. 
The area of $\triangle ABC$ is $\frac{1}{2}(AB)(HC)=\frac{1}{2}(25)(12)=150, \boxed{\text{C}}$.