find the length of midsegment YZ in trapezoid ABCD when A(-5,-6) B(-6,-2) C(-4,0) and D(3,2)

Answer:
Option C
Step-by-step explanation:
Since, Y and Z are the midpoints of sides AB and CD of the given trapezoid.
Segment YZ will the midsegment of trapezoid ABCD.
By the theorem of midsegment,
m(YZ) = [tex]\frac{1}{2}(AD+BC)[/tex]
By using expression for the length of a segment between two points,
Length of a segment = [tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Distance between two points A(-5, -6) and D(3, 2),
AD = [tex]\sqrt{(3+5)^2+(2+6)^2}[/tex]
AD = [tex]\sqrt{64+64}[/tex]
AD = [tex]\sqrt{128}[/tex]
AD = [tex]8\sqrt{2}[/tex]
Distance between B(-6, -2) and C(-4, 0)
BC = [tex]\sqrt{(-6+4)^2+(-2-0)^2}[/tex]
BC = [tex]\sqrt{8}[/tex]
BC = [tex]2\sqrt{2}[/tex]
Therefore, m(YZ) = [tex]\frac{1}{2}(8\sqrt{2}+2\sqrt{2})[/tex]
= [tex]\frac{1}{2} (10\sqrt{2})[/tex]
= [tex]5\sqrt{2}[/tex]
Option C will be the answer.
The length of the midsegment YZ is [tex]5\sqrt 2[/tex]
The coordinates are given as:
A(-5,-6) B(-6,-2) C(-4,0) and D(3,2)
Start by calculating distance AD and CB using:
[tex]d = \sqrt{(x_2 -x_1)^2 + (y_2 -y_1)^2[/tex]
So, we have:
[tex]AD = \sqrt{(3 +5)^2 + (2 +6)^2} = \sqrt{128}[/tex]
[tex]CB = \sqrt{(-6 +4)^2 + (-2 -0)^2} = \sqrt{8}[/tex]
The length of the midsegment is then calculated as:
[tex]YZ = \frac 12 * (AD + CB)[/tex]
This gives
[tex]YZ = \frac 12 * (\sqrt{128} + \sqrt 8)[/tex]
Simplify
[tex]YZ = \frac 12 * (8\sqrt{2} + 2\sqrt 2)[/tex]
[tex]YZ = \frac 12 * (10\sqrt 2)[/tex]
Simplify
[tex]YZ = 5\sqrt 2[/tex]
Hence, the length of the midsegment YZ is [tex]5\sqrt 2[/tex]
Read more about midsegments at:
https://brainly.com/question/7423948