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

14

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28

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8

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6

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Solution

The correct option is B $8$
The area of the $Δ A B C$ with vertices
$A (x_{1}, y_{1}) = (3, 0), B (x_{2}, y_{2}) = (7, 0) & C (x_{3}, y_{3}) = (8, 4)$ is
$A r . Δ A B C = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$
$= \frac{1}{2} {3 (0 - 4) + 7 (4 - 0) + 8 (0 - 0)}$ sq.units $= 8$ sq. units.
Hence, option C.

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