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?

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, $x = \frac{24}{5}$.
