Let $ABC$ be a triangle such that $AB=4,BC=5$ and $CA=6$. Choose points $D,E,F$ on $AB,BC,CA$ respectively, such that $AD=2,BE=3,CF=4$. Then $\frac{\mathrm{area} \mathrm{\Delta }DEF}{\mathrm{area} \mathrm{\Delta }ABC}$ is

$\frac{area\left(\mathrm{\Delta }DEF\right)}{area\left(\mathrm{\Delta }ABC\right)}=$

Let area $\left(\mathrm{\Delta }ABC\right)=\mathrm{\Delta }$

$\frac{area\left(\mathrm{\Delta }BED\right)}{area\left(\mathrm{\Delta }ABC\right)}=\frac{\frac{1}{2}.6\sin B}{\frac{1}{2}.20\sin B}=\frac{3}{10}$

$\frac{area\left(\mathrm{\Delta }ADF\right)}{area\left(\mathrm{\Delta }ABC\right)}=\frac{\frac{1}{2}.4\sin A}{\frac{1}{2}.24\sin A}=\frac{1}{6}$

$\frac{area\left(\mathrm{\Delta }CEF\right)}{area\left(\mathrm{\Delta }ABC\right)}=\frac{\frac{1}{2}.8\sin C}{\frac{1}{2}.30\sin C}=\frac{4}{15}$

area $\left(\mathrm{\Delta }BED\right)=\frac{3}{10}\mathrm{\Delta }$

area $\left(\mathrm{\Delta }ADF\right)=\frac{\mathrm{\Delta }}{6}$

area $\left(\mathrm{\Delta }CEF\right)=\frac{4\mathrm{\Delta }}{15}$

area $\left(∆DEF\right)=\mathrm{\Delta }-\left(\frac{3}{10}+\frac{1}{6}+\frac{4}{15}\right)\mathrm{\Delta }$

$=\mathrm{\Delta }-\left(\frac{27+15+24}{90}\right)\mathrm{\Delta }$

$=\mathrm{\Delta }-\left(\frac{66}{90}\right)\mathrm{\Delta }=\frac{24\mathrm{\Delta }}{90}$

$\frac{area\left(\mathrm{\Delta }DEF\right)}{area\left(\mathrm{\Delta }ABC\right)}=\frac{24}{90}=\frac{4}{15}$