Answer:
The answer is below
Step-by-step explanation:
What is the area of ABCD if the coordinates of the vertices are A (1 1/2, 1) = (3/2, 1), B(1 1/2, 6 1/3) = (3/2, 19/3), C(5 1/2, 6 1/3) = (11/2, 19/3) and D = (5 1/2, 1) = (11/2, 1). Hence A(3/2, 1), B(3/2, 19/3), C(11/2, 19/3) and D(11/2, 1)
Therefore from the vertices, we can see that ABCD is a rectangle.
Area of ABCD = AB * BC
The distance between two points [tex](x_1,y_1)\ a nd\ (x_2,y_2)[/tex] in the coordinate plane is given by:
[tex]Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Therefore:
[tex]AB=\sqrt{(\frac{3}{2}-\frac{3}{2} )^2+(\frac{19}{3}-1)^2} =\frac{16}{3} \ units\\\\BC=\sqrt{(\frac{11}{2}-\frac{3}{2} )^2+(\frac{19}{3}-\frac{19}{2} )^2} =4\ units[/tex]
Area of ABCD = AB * BC = 16/3 units * 4 units = 64/3 unit²
