# 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

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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.

Updated on 10-Oct-2022 13:28:26