<s>[INST] An isosceles triangle with sides... An isosceles triangle with sides $|AB|=6$, $|BC|=|AC|=5$. Calculate the height of the triangle, against the side $BC$.
I drew the following:

By the figure I get that $|CD|=\sqrt{25-x^2}$ and $|BD|=5-\sqrt{25-x^2}.$ Using the pythagorean theorem on $\triangle ABD,$ I get $$36=x^2+\left(5-\sqrt{25-x^2}\right)^2 \Longleftrightarrow x^2 = 25 - \frac{49}{25}.$$
Here is where the trouble starts. I'm not allowed to have a calculator to solve this, but I have to simplify the RHS and check if it is a perfect square. What I did with the RHS was this:
$$25-\frac{49}{25}=\frac{25^2-7^2}{25}=\frac{(25-7)(25+7)}{25}=\frac{18\cdot 32}{25}=\frac{3\cdot3\cdot2\cdot2^5}{5^2}=\frac{3^2(2^3)^2}{5^2}=\left(\frac{24}{5}\right)^2.$$
Thus the only positive solution is $x=\frac{24}{5},$ which indeed is correct. However time is of the essence here, doing that last simplification on the RHS took a while, on probably on a test, I'd just give up thinking I'm wrong because it's not supposed to be that much of arithmetic on these questions.
Is there any simpler way to solve this problem?
 [/INST] An easy way to solve this problem is to compute the area of the triangle using Heron's formula. In particular, $s = (6+5+5)/2=8$, and $A = \sqrt{8(8-6)(8-5)^2}=12$. But $A=\frac{5x}{2}$. Therefore, $x = \frac{24}{5}$.
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