<s>[INST] Easier way to discover the area of a right triangle In the following right triangle: $y-x=5$ and the altitude to the hypotenuse is 12. Calculate its area.

I've managed to discover its area using the following method, but it ends up with a 4th degree equation to solve. Is there an easier way to solve the problem?
$ha=xy \Rightarrow 12 \sqrt{x^2+y^2} = xy$
Substitute $y=5+x$ and square both sides:
$144 (x^2 + (5+x)^2)=x^2 (5+x)^2 \Rightarrow x^4+10x^3-263x^2-1440x-3600=0$
Which only positive solution is $x=15$ and therefore $y=20$ and the area is $\frac{15 \cdot 20 }{2}=150$
Thanks in advance.
 [/INST] Square both sides of relationship $(x-y)=5$; then apply Pythagoras, giving
$\tag{1}x^2+y^2-2xy=25 \Leftrightarrow a^2-2xy=25$
Besides, the area $S$ of the triangle can be computed in two ways :
$\tag{2}S=\frac{xy}{2}=\frac{12a}{6}=6a$
Plugging the value of $a$ taken from (2) in (1), one gets a quadratic equation in variable $S$ which yields the looked for value for $S$. This equation is 
$$\left(\frac{S}{6}\right)^2-4S=25 \ \ \Leftrightarrow \ \ S^2-144S-900=0$$
whose roots are $S=150$ (the unique answer) and $S=-6$, this one having no geometrical meaning.
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