Precalculus (10th Edition)

$15,18,21$ units
By Heron's formula, the area $K$ can be computed $K=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{a+b+c}{2}.$ Hence here $s=\dfrac{5x+6x+7x}{2}=9x$ Thus, $K=\sqrt{9x(9x-5x)(9x-6x)(9x-7x)} \\K=\sqrt{9x(4x)(3x)(2x)} \\K=\sqrt{216x^4} \\K=6\sqrt6{\space x^2}$ Therefore $6\sqrt6\space x^2=54\sqrt6$, solve the equation above to obtain: \begin{align*} \dfrac{6\sqrt6\space x^2}{6\sqrt6}&=\dfrac{54\sqrt6}{6\sqrt6}\\ x^2&=9\\ x&=\pm3\end{align*} Since $x$ represents a length, it cannot be negative so $x=3$. Therefore, the length of the sides are: $5x=5(3)=15\\ 6x=6(3)=18\\ 7x=7(3)=21$