Answer:
[tex]\frac{3\sqrt{13} }{2}[/tex]
Step-by-step explanation:
First we have to identify the parallel sides of the trapezium.
We know that the slopes are equal for parallel lines.
Slope of (x₁,y₁) and (x₂,y₂) is given by
[tex]m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
Slope of AB:
[tex]m_{AB} = \frac{3-5}{3-0}=-\frac{2}{3}[/tex]
Slope of BC:
[tex]m_{BC} = \frac{-2-3}{5-3}=-\frac{5}{2}[/tex]
Slope of CD:
[tex]m_{CD} = \frac{2+2}{-1-5}=-\frac{4}{6}=-\frac{2}{3}[/tex]
Slope of DA:
[tex]m_{DA} = \frac{2-5}{-1-0}=3[/tex]
We see that the slopes of AB and CD are equal, so, AB and CD are the parallel sides.
The length of the midsegment = (1/2)*(length of base1 + length of base2 )
Length of the bases can be calculated using distance formula,
[tex]d= \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}[/tex]
AB = [tex]\sqrt{(3-0)^{2}+(3-5)^{2}}= \sqrt{9+4} =\sqrt{13}[/tex]
CD = [tex]\sqrt{(-1-5)^{2}+(2+2)^{2}}= \sqrt{36+16} =\sqrt{52}=2 \sqrt{13}[/tex]
Length of the midsegment = (1/2) (√13 + 2√13) =3√13/2
