When you say "primary school", do you mean Soviet primary school? Anyways, I'll need to remember myself a bit of geometry for this...

Thought I could solve this by taking the area of the triangled formed by the 5m stick and subtracting off the triangled formed by the 4m stick but then I realized that in terms of numerical values that'd be subtracting more from triangle 1 than just the intersection sigh....

>>929
I don't know which primary school textbook it originally came from. But the problem is easier for those primary school students because they know the solution method was covered in the chapter for which the problem is an exercise.

>>931
I'm sure it's something to do with the intersection of two segments or whatever. Is there any hint you could give for me to look into?

Wait... if it's primary school it's gotta be related to Pythagorean's theorem, right? That's the only geometry concept I remember learning that early, alongside area/perimeter.

>>932
Might be hard to give a hint without spoiling the aspect of the problem that isn't just plug and chug.

>>933
You need it for the way I did it but it's not the only concept you need. I can't rule out solutions I didn't think of, though. You might need some stuff you learned in early high school (or remember learning in early high school having forgotten it after the first time), since different curricula go at different rates. But you certainly don't need stuff you would typically first learn in trig or calculus, although I've seen some people get the solution with trig techniques; there's more than one way to do it.

h/4 + h/sqrt(7) = 1
this has to be the "primary school" method

>>935
Nice. This is a simpler solution than any of the others I've seen so far.

>>935
how do you even make this assumption

I don't have a pen and paper with me to do it but using the properties of 3:4:5 triangles you should be able to solve it.

Write the equations for the two lines:
y_1 = 5 - 5x/3
y_2 = 4x/3

Intersection when y_1 = y_2:
5 - 5x/3 = 4x/3
9x/3 = 5
9x = 15
x = 15/9 = 5/3

At x = 5/3:
y = 4/3 * 5/3 = 20/9 = h

This correct?

>>941
The equations are wrong, but the logic is right.

>>943
Oh right, I see my mistake. I used the length of the lines as their height, whereas you need to use Pythagoras to get those, so the equations should be y_1 = 4 - 4x/3 and y_2 = sqrt(7)/3. Fucking maths man.

>>937
Not that anon, but
△ABF is a right triangle, so (AB)² + 3² = 5² and AB = 4.
△EFB is a right triangle, so (EF)² + 3² = 4² and EF = √7.
△CDF is similar to △ABF, so DF/BF = CD/AB = h/4.
△CDB is similar to △EFB, so BD/BF = CD/EF = h/√7.
BD and DF together make BF, so DF/BF + BD/BF = h/4 + h/√7 = 1.
Thus h = 1/(1/4 + 1/√7) ≈ 1.5924467752259388 m.

An interesting application!
https://en.wikipedia.org/wiki/Crossed_ladders_problem#Application_to_paper_folding