Question

What is the length of the midsegment of the trapezoid made by the vertices A(0, 5), B(3, 3), C(5, -2) and D(-1, 2). Show equations and all work that leads to your answer.

Answer 1

`(3√(13) )/(2)`

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

`m = (y_(2)-y_(1))/(x_(2)-x_(1))`

Slope of AB:

`m_(AB) = (3-5)/(3-0)=-(2)/(3)`

Slope of BC:

`m_(BC) = (-2-3)/(5-3)=-(5)/(2)`

Slope of CD:

`m_(CD) = (2+2)/(-1-5)=-(4)/(6)=-(2)/(3)`

Slope of DA:

`m_(DA) = (2-5)/(-1-0)=3`

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,

`d= \sqrt{(x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2)}`

AB =
`\sqrt{(3-0)^(2)+(3-5)^(2)}= √(9+4) =√(13)`

CD =
`\sqrt{(-1-5)^(2)+(2+2)^(2)}= √(36+16) =√(52)=2 √(13)`

Length of the midsegment = (1/2) (√13 + 2√13) =3√13/2