write a coordinate proof. Given: Rectangle ABCD has vertices A(0,4), B(6,4), C(6,0), D(0,0). E is the midpoint of line DC. F is the midpoint

Question

write a coordinate proof. Given: Rectangle ABCD has vertices A(0,4), B(6,4), C(6,0), D(0,0). E is the midpoint of line DC. F is the midpoint of line DA. Prove: The area of rectangle DEGF is one fourth the area of rectangle ABCD.

Please help I’ve been trying for a while ​

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Diễm Kiều 4 years 2021-08-14T19:05:47+00:00 1 Answers 81 views 0

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  1. minhkhue
    0
    2021-08-14T19:06:55+00:00
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    Answer:

    The answer is below

    Step-by-step explanation:

    Distance between two points A(x_1,y_1)\ and\ B(x_2,y_2)\ on \ the\ coordinate\ plane

    |AB|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

    The area of rectangle = length * breadth

    The area of rectangle ABCD = |AB| * |BC|

    |AB|=\sqrt{(6-0)^2+(4-4)^2}=6\\\\|BC|=\sqrt{(6-6)^2+(0-4)^2} =4

    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|

    |DE|=\sqrt{(3-0)^2+(0-0)^2}=3\\\\|DF|=\sqrt{(0-0)^2+(2-0)^2} =2

    The area of rectangle ABCD = |DE| * |DF| = 3 * 2 = 6

    Therefore, area of rectangle DEGF is one fourth the area of rectangle ABCD

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