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
[tex]A=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|[/tex]
Area of triangle ABC is
[tex]A=\frac{1}{2}|-5(2+1)-1(-1+2)+0(-2-2)|[/tex]
[tex]A=\frac{1}{2}|-16|[/tex]
[tex]A=\frac{16}{2}[/tex]
[tex]A=8[/tex]
The area of triangle ABC is 8 square units.
Area of triangle ACD is
[tex]A=\frac{1}{2}|-5(-1+3)+0(-3+2)-2(-2+1)|[/tex]
[tex]A=\frac{1}{2}|-8|[/tex]
[tex]A=\frac{8}{2}[/tex]
[tex]A=4[/tex]
The area of triangle ABC is 4 square units.
The are of ABCD is
[tex]Area(ABCD)=Area(ABC)+Area(ACD)[/tex]
[tex]Area(ABCD)=8+4=12[/tex]
Therefore the area of ABCD is 12 square units. Option b is correct.
