Let a,b,c,p,S be the length of 3 sides,the semiperimeter,the area of the triangle respectively,then :

\(\circledast a+b+c=160\) ; \(\circledast\dfrac{a}{5}=\dfrac{b}{13}=\dfrac{c}{12}\); \(\circledast p=\dfrac{160}{2}=80\)

Applying the property of sequence of equivalent ratios,we have :

\(\dfrac{a}{5}=\dfrac{b}{13}=\dfrac{c}{12}=\dfrac{a+b+c}{5+13+12}=\dfrac{160}{30}=\dfrac{16}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{16}{3}.5=\dfrac{80}{3}\\b=\dfrac{16}{3}.13=\dfrac{208}{3}\\c=\dfrac{16}{3}.12=64\end{matrix}\right.\)

Applying the Heron's formula,we have :

\(S=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)

\(S=\sqrt{80\left(80-\dfrac{80}{3}\right)\left(80-\dfrac{208}{3}\right)\left(80-64\right)}\)

\(=\sqrt{80.\dfrac{160}{3}.\dfrac{32}{9}.16}=\dfrac{2560}{3}=853\dfrac{1}{3}\)