The area of a triangle with vertices A(5, 0), B(8, 0) and C(8, 4) in square units is

Let A(x1 = 5, y1 = 0), B(x2 = 8, y2 = 0) and C(x3 = 8, y3 = 4) be the vertices of the triangle. Then

Area $(\Delta A B C)=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

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

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

$=6$ sq. units

Hence, the correct answer is option (c).