The area of $\triangle \mathrm{ABC}$ with vertices $\mathrm{A}(3,0), \mathrm{B}(7,0)$ and $C(8,4)$ is (a) 14 sq units (b) 28 sq units (c) 8 sq units (d) 6 sq units
The area of $\triangle \mathrm{ABC}$ with vertices $\mathrm{A}(3,0), \mathrm{B}(7,0)$ and $C(8,4)$ is (a) 14 sq units (b) 28 sq units (c) 8 sq units (d) 6 sq units

The correct option is option (c) 8 sq units

The given points are $A(3,0), B(7,0)$ and $C(8,4)$.

$\left(x_{1}=3, y_{1}=0\right),\left(x_{2}=7, y_{2}=0\right)$ and $\left(x_{3}=8, y_{3}=4\right)$

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

$=\frac{1}{2}[3(0-4)+7(4-0)+8(0-0)]$

$=\frac{1}{2}[-12+28+0]$

$=\left(\frac{1}{2} \times 16\right)$

$=8$ sq. units