<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] 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
$A^2 = 36(25 + 4A)\\
A^2 - 144A - 900 = 0\\
(A - 150)(x+6) = 0\\
A = 150$
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