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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.
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