Find the perimeter of $\triangle CDE$ . Round your answer to the nearest hundredth. A trapezoid A B C D is plotted on a coordinate plane. A D represents the longer base and B C represents the shorter base. Vertex A lies at ordered pair negative 5 comma 4. Vertex B lies at ordered pair 0 comma 3. Vertex C lies at ordered pair 4 comma negative 1. Vertex D lies at ordered pair 4 comma negative 5. A line is drawn from vertex B and intersects line A D at point F plotted at ordered pair negative 2 comma 1. A line is drawn from vertex C and intersects the line A D at point E plotted at ordered pair 2 comma negative 3. The perimeter is about units.

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Answer:

The answer is below

Step-by-step explanation:

We are asked to find the perimeter of triangle CDE. The perimeter of a shape is simply the sum of all its sides, hence:

Perimeter of tiangle CDE = |CD| + |DE| + |CE|

Given that C(4, -1), D(4, -5), E(2, -3).

The distance between two points [tex]X(x_1,y_1)\ and\ Y(x_2,y_2)[/tex] is given as:

[tex]|XY|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

Therefore the lengths of the triangle are:

[tex]|CD|=\sqrt{(4-4)^2+(-5-(-1))^2} =4\ units\\\\|DE|=\sqrt{(2-4)^2+(-3-(-5))^2} =2.83\ units\\\\|CE|=\sqrt{(2-4)^2+(-3-(-1))^2} =2.83\ units[/tex]

Perimeter of CDE = 4 + 2.83 + 2.83 = 9.66 units

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