Question

Trapezoid ABCD is graphed in a coordinate plane.

Answer 1

Answer:

The area of ABCD is 12 square units. Option b is correct.

Step-by-step explanation:

From the given figure it is clear that the vertices of triangle are A(-5,-2), B(-1,2), C(0,-1), D(-2,-3).

The area of ABCD is the sum of area of triangle ABC and ACD.

Area of a triangle is

$A=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

Area of triangle ABC is

$A=\frac{1}{2}|-5(2+1)-1(-1+2)+0(-2-2)|$

$A=\frac{1}{2}|-16|$

$A=\frac{16}{2}$

$A=8$

The area of triangle ABC is 8 square units.

Area of triangle ACD is

$A=\frac{1}{2}|-5(-1+3)+0(-3+2)-2(-2+1)|$

$A=\frac{1}{2}|-8|$

$A=\frac{8}{2}$

$A=4$

The area of triangle ABC is 4 square units.

The are of ABCD is

$Area(ABCD)=Area(ABC)+Area(ACD)$

$Area(ABCD)=8+4=12$

Therefore the area of ABCD is 12 square units. Option b is correct.