Let a rational right triangle have legs $A,B$, hypotenuse $C$, and area $AB/2=D$. After interchanging the legs if necessary, set
$$
x=\frac{A+C}{B},
\qquad
y=\frac{2x}{B}.
$$
Using $C^2=A^2+B^2$ and $AB=2D$ gives
$$
x^2-1=\frac{2A(A+C)}{B^2}=\frac{4Dx}{B^2}=\frac{Dy^2}{x},
$$
so $P=(x,y)$ lies on the <congruent number elliptic curve> $E_D:Dy^2=x^3-x$. The triangle is nondegenerate, so $y\ne0$ and $2P\ne O$.

Conversely the three displayed lengths in the question satisfy
$$
\left(\frac{x^2-1}{y}\right)^2+\left(\frac{2x}{y}\right)^2
=\left(\frac{x^2+1}{y}\right)^2,
$$
and their area is
$$
\left|\frac{x(x^2-1)}{y^2}\right|=D.
$$
Thus every such triangle is $\Delta_P$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 125 / 5 / a / Solution

{title2=$E_D:Dy^2=x^3-x$}
{wiki=Congruent_number#Connection_with_elliptic_curves}

For a positive square-free integer $D$, rational points on $E_D:Dy^2=x^3-x$ with $2P\ne O$ correspond to rational right triangles of area $D$. The point $(x,y)$ gives side lengths
$$
\left|\frac{x^2-1}{y}\right|,
\qquad
\left|\frac{2x}{y}\right|,
\qquad
\left|\frac{x^2+1}{y}\right|.
$$


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 Incoming links: Congruent number elliptic curve