Answer:
The answer is below
Step-by-step explanation:
Distance between two points [tex]A(x_1,y_1)\ and\ B(x_2,y_2)\ on \ the\ coordinate\ plane[/tex]
[tex]|AB|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
The area of rectangle = length * breadth
The area of rectangle ABCD = |AB| * |BC|
[tex]|AB|=\sqrt{(6-0)^2+(4-4)^2}=6\\\\|BC|=\sqrt{(6-6)^2+(0-4)^2} =4[/tex]
The area of rectangle ABCD = |AB| * |BC| = 6 * 4 = 24
E(x, y) is midpoint of line DC. Its coordinate is:
x = (6 + 0) / 2 = 3; y = (0 + 0) / 2 =0
The coordinate of E = (3, 0)
F(a, b) is midpoint of line DA. Its coordinate is:
a = (0 + 0) / 2 = 0; b = (4 + 0) / 2 = 2
The coordinate of E = (0, 2)
The area of rectangle DEGF = |DE| * |DF|
[tex]|DE|=\sqrt{(3-0)^2+(0-0)^2}=3\\\\|DF|=\sqrt{(0-0)^2+(2-0)^2} =2[/tex]
The area of rectangle ABCD = |DE| * |DF| = 3 * 2 = 6
Therefore, area of rectangle DEGF is one fourth the area of rectangle ABCD
