This problem appears in another thread:
The medians to the legs of a certain right triangle have lengths 13 and 19. What is the length of the hypotenuse of the triangle?
An analytical answer is given but lets see what we can do with geogebra.
1) Make a slider and range it from 0 to 25 with steps of .001
2) Create points A and B by entering into the input bar (0,0) and (a,0).
3) Draw line segment AB.
4) Use the circle with radius tool to create a circle with center at B and having a radius of 13.
5) Draw a line through A that is perpendicular to AB.
6) Get the intersection point with the circle and that perpendicular line using the intersect tool.
7) Hide the lower intersection point, the circle and the perpendicular line. You should now only have points A,B and D and the slider visible.
8) Draw line segment AD and get the midpoint of it E.
9) Draw another point by entering in the input bar (2a,0). Point F will be created.
10) Draw dotted line segments BD and FE. These are the medians.
11) Right click both these medians and in object properties check Show label - value. The lengths of the medians will appear. One will be 13 by construction and the other will be some arbitrary value.
12) Draw a big fat line from F to D, make it red and thick. This is the hypotenuse we seek.
Now we will try to get EF to be as close to 19 as is possible. The whole drawing is dependent on the slider which controls the horizontal motion of point B.
13) Start by moving the slider point marked a carefully with your mouse until you are close as you can get. I achieved,19.05.
14) That is good but not good enough. With the slider point highlighted use the shift right and shift left keys to get closer to 19. I got to g = 19.00010106815226 with an increment of .001.
Can we do better?
15) Go into the slider object properties and set the increment to 0.00001. I got to 19.000000987535433.
To do better we set the increment to .00000001 and try again. Now I get 18.999999999467065 that is the best we can do with the increment command.
Read off h, which is the length of FD, the hypotenuse. I am getting h = 20.591260281580603
Take that and go here:
enter into the simple lookup and you see a bunch of answers, the closest one is
sqrt(424), which is correct.
Not bad for no math at all!
In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.