The area of a triangle with vertices A (3,0), | vedclass

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(B) The area of a triangle with vertices $A(x_1, y_1), B(x_2, y_2)$ and $C(x_3, y_3)$ is given by the formula:$	ext{Area} = \frac{1}{2} |x_1(y_2 - y_

Vedclass · Accepted Answer

$8$

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(B) The area of a triangle with vertices $A(x_1, y_1), B(x_2, y_2)$ and $C(x_3, y_3)$ is given by the formula:$	ext{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$Given vertices are $A(3, 0), B(7, 0)$ and $C(8, 4)$.Here,$x_1 = 3, y_1 = 0, x_2 = 7, y_2 = 0, x_3 = 8, y_3 = 4$.Substituting these values into the formula:$	ext{Area} = \frac{1}{2} |3(0 - 4) + 7(4 - 0) + 8(0 - 0)|$$	ext{Area} = \frac{1}{2} |3(-4) + 7(4) + 8(0)|$$	ext{Area} = \frac{1}{2} |-12 + 28 + 0|$$	ext{Area} = \frac{1}{2} |16|$$	ext{Area} = 8 	ext{ square units}$.Thus,the area of the triangle is $8$.

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

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