Question 1031166
One can start with coordinate geometry with the hypotenuse segment in quadrant 1, and find the triangle figure to be composed of a 2 by 2 square unit SQUARE figure and two right triangle.  These three parts form the main larger triangle.


The upper left triangle will have leg lengths 2 and y, and hypotenuse assigned h_sub_1;  the triangle on the right will have leg lengths x and 2, and hypotenuse assigned h_sub_2.


A list of equations can be found from the figure - you need to draw it or see it.
{{{system(h[1]+h[2]=sqrt(5),(2+y)^2+(2+x)^2=(sqrt(5))^2,y^2+2^2=h[1]*h[1],2^2+x^2=h[2]*h[2])}}}


Two formulas will come from that list, being radical forms for h_sub_1 and h_sub_2, leading to this simpler system of equations.


{{{system(sqrt(y^2+4)+sqrt(x^2+4)=sqrt(5),(y+2)^2+(x+2)^2=5)}}}--------This should be the system to work with to solve for x and y; but you must understand that these x and y are to be  values to ADD to the 2 and the other 2 for the drawing if imagined NOT on a cartesian grid system.


I have not solved further than this last listed system.  Yet, two equations in two unknowns.