question

I was soooo close to getting it I messed up on the square roots! Also I kind of forgot where did the sqrt 3 come from in 5 times sqrt 3/ 2 or is there a question here already that I can use to see why?

Hi @TheDarkinBlade!
We got the \( \sqrt{ 3}\) from finding the area of the 578 triangle using Heron's formula:
$$\text{semiperimeter } = s = \frac{5+7+8}{2} = 10$$
By Heron's Formula, the area of a triangle with the sides \(a,\) \(b,\) \(c\) and semiperimeter \(s\) is equal to:
$$ \text{Area } = \sqrt{s\cdot (sa)\cdot (sb)\cdot (sc)}$$
So the area of the 578 triangle is equal to:
$$ \text{Area of 578} \triangle =\sqrt{10\cdot (105)\cdot (107)\cdot (108)}=\sqrt{10\cdot 5\cdot 3\cdot 2}=\sqrt{10\cdot 10\cdot 3}=\sqrt{10^2\cdot 3}=10\sqrt{3}$$
And, after that, using the area of this 578 triangle, we can then find its height:
$$10\sqrt{3}= \text{Area} =\frac{1}{2}\cdot 8\cdot \text{height}$$