Hi @The-Darkin-Blade!
We got the \( \sqrt{ 3}\) from finding the area of the 5-7-8 triangle using Heron's formula:
$$\text{semi-perimeter } = s = \frac{5+7+8}{2} = 10$$
By Heron's Formula, the area of a triangle with the sides \(a,\) \(b,\) \(c\) and semi-perimeter \(s\) is equal to:
$$ \text{Area } = \sqrt{s\cdot (s-a)\cdot (s-b)\cdot (s-c)}$$
So the area of the 5-7-8 triangle is equal to:
$$ \text{Area of 5-7-8} \triangle =\sqrt{10\cdot (10-5)\cdot (10-7)\cdot (10-8)}=\sqrt{10\cdot 5\cdot 3\cdot 2}=\sqrt{10\cdot 10\cdot 3}=\sqrt{10^2\cdot 3}=10\sqrt{3}$$
And, after that, using the area of this 5-7-8 triangle, we can then find its height:
$$10\sqrt{3}= \text{Area} =\frac{1}{2}\cdot 8\cdot \text{height}$$
