The correct option is option (c) 6

Let $A\left(x_{1}=5, y_{1}=0\right), B\left(x_{2}=8, y_{2}=0\right)$ and $C\left(x_{3}=0, y_{3}=4\right)$ be the vertices of the triangle. Then,

$\text { Area }(\Delta \mathrm{ABC})=\frac{1}{2}\left[\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right]$

$=\frac{1}{2}[5(0-4)+8(4-0)+8(0-0)]$

$=\frac{1}{2}[-20+32+0]$

$=6$ sq. units