The area of a triangle is $54\sqrt{6}$ square units. Find the lengths of the sides: $5x,6x,7x$

Question

I realize I can use Heron's formula for this question. I did $54\sqrt{6}=\sqrt{9x\cdot 3x\cdot 4x\cdot 2x}$ but from there I must have done something wrong. Thanks for the help.

Answer 1

Try using,

2*S= sum of lengths of all three sides (say, a , b, c)

Now use ,

$(Area of Triangle)^2$= $S(S-a)(S-b)(S-c)$

Answer 2

$$54\sqrt{6}=\sqrt{9x\cdot 3x\cdot 4x\cdot 2x}=\sqrt{216x^4}=6\sqrt{6x^4}=6x^2\sqrt{6}$$

$$\iff 54=6x^2\iff x^2=9\iff x=3$$

I used the fact that $\sqrt{x^4}=|x^2|=x^2$, since $x^2\ge 0, \forall x\in\mathbb R$
and $x^2=9\implies x=\pm 3\implies x=3$, since $x$ is a positive number.