Finding the area of equilateral $\triangle ABC$, given an interior point $I$ with $IA=8$, $IB=10$, $IC=6$

Question

Find the area of an equilateral $\triangle ABC$ if $I$ is inside point such that:

$$IA=8 \qquad IB=10 \qquad IC=6$$

My try:

First we note that: $10^{2}=8^{2}+6^{2}$

Take rotational $R\left(C,\frac{π}{3}\right)$

But I don't know what I get??

I think there other simple method without rotational because I don't understand it.

I have already to see your solution

Answer 1

Hint:

\begin{align} a&=|AB|=|BC|=|CA|=2\sqrt{25+12\sqrt3} ,\\ S_{ABC}&=\tfrac{\sqrt3}4\,a^2=36+25\sqrt3 . \end{align}

Answer 2

A pedestrian approach is to use the formula for the volume of a tetrahedron in terms of its side lengths. In this case, three sides have length $x$ and the other three are 6,8,10, so the volume is a constant times $$\left|\begin{matrix}0&1&1&1&1\\\1&0&36&64&100\\\1&36&0&x^2&x^2\\\1&64&x^2&0&x^2\\\1&100&x^2&x^2&0\end{matrix}\right|$$ This must be 0. So we get $x^4-200x^2+3088=0$. Solving $x^2=100\pm48\sqrt3$. $x$ is clearly greater than 10, so we want the positive sign: $$x^2=100+48\sqrt3$$ Now the area is $\frac{\sqrt3}{4}x^2=36+25\sqrt3$.