The area of a triangle with vertices $ A(3,0), B(7,0) $ and $ C(8,4) $ is
(A) 14
(B) 28
(C) 8
(D) 6

Given: 

Vertices of a triangle $A (3,\ 0),\ B (7,\ 0)$ and $C (8,\ 4)$.

To do: 

We have to find the area of the triangle.

Solution:

The vertices of the given triangle $A (3,\ 0),\ B (7,\ 0)$ and $C (8,\ 4)$.

Here,

$x_1=3,\ y_1=0,\ x_2=7,\ y_2=0,\ x_3=8,\ y_3=4$

We know that,

Area of a triangle$=\frac{1}{2}[x_1( y_2-y_3)+x_2( y_3-y_1)+x_3( y_1-y_2)]$

Therefore,

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

$=\frac{1}{2}[-12+28-56]$

$=\frac{1}{2}[-40]$

$=-20$

$\because$ Area can't be negative.

Therefore, the area of the triangle is $20$ square units.