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.
 A: 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.
A: 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 $A$:
$$
\left(\frac A6\right)^2-4A-25=0
$$
A: you have to ways to arrive at the area.
$A = \frac 12 xy = 6\sqrt{(x^2 + y^2)}$
and you know: $(x-y) = 5$
$(x-y)^2 = 25\\
x^2 + y^2 - 2xy = 25\\
x^2 + y^2 = 25 + 4A$
Plug this into the equation above for areas.
$A = 6\sqrt{25 + 4A}$
square both sides and solve the quadratic
$A^2 = 36(25 + 4A)\\
A^2 - 144A - 900 = 0\\
(A - 150)(x+6) = 0\\
A = 150$
A: Due to similarity of two parts of the right triangle 
$$ (h/x)^2 + (h/y)^2 = 1 ; \, ( 1/x^2 + 1/y^2)  = 1/144 $$
It is given $ y = x + 5 $
Solve for $x,y$ choosing the  $\pm $ sign properly and then find $xy/2.$
