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.

Slightly easier: Write the unknown area as $A:=xy/2$. From $12 \sqrt{x^2+y^2} = xy$ deduce $$x^2+y^2=(A/6)^2.$$ Substitute this into $x^2-2xy+y^2=25$. This gets you a quadratic for $A$:
$$
\left(\frac A6\right)^2-4A-25=0
$$
