Calculating the right angled triangle's cathetus

Question

We just started learning the Pythagorean theorem at school and we got a pretty difficult assignment.

5 meter tall bamboo broke and the top of it touched the floor 2 meters from the base of the bamboo. What height did it break at?

I've tried all sorts of different approaches, but I just can't get it right. I'm starting to think it's unsolvable.

Am I missing something?

Thanks!

Edit: I've tried using the Pythagorean theorem (blue is the bamboo after it broke, gray is where the bamboo was before it broke):

Answer 1

As you stated (I won't write the units of meters):

$$c+b=5, ~~~~~~~~~~~~~~~~~~~~~~ c^2=a^2+b^2.$$

You want to know either $c$ or $b$, so, you can write the first equation as $c=5-b$ or $b=5-c$. Now you can plug one of these into the second equation, and you would get an equation in terms of $a$ and $c$ or $b$, i.e. you CAN now solve for $b$ or $c$ and get the desired answer.

Answer 2

As you already observed, $$b+c=5, \tag{1}$$ $$c^2 = 4 + b^2.$$ From the second equation we can find that $$c^2 - b^2 = 4.$$ Factor the left-hand side: $$(c + b)(c - b) = 4.$$ This looks useful, since we already know that $c+b = b+c = 5.$ That is, we now know that $$5(c - b) = 4. \tag{2}$$

Now we have a system of two simultaneous linear equations in $b$ and $c$, namely the equations labeled $(1)$ and $(2)$. We can solve for $b$ and $c$ in whatever way we would usually solve such a system of equations; or in this case, we might notice that if we set $a = \frac{b+c}{2}$ and $d = \frac{c-b}{2}$, then $c = a + d$ and $b = a - d$. (This last bit is handy to have in your bag of tricks when you happen to know the sum and difference of two unknown values.)