Given:
In trapezoid [tex]ABCD, AD\parallel BC, MN[/tex] is the mid-segment of [tex]ABCD, AD=30x-10,MN=31x+1,BC=30x+28[/tex].
To find:
The length of [tex]AD[/tex].
Solution:
We know that the length of the mid-segment of a trapezoids is half of the sum of lengths of two parallel sides of the trapezoid.
In trapezoid [tex]ABCD[/tex],
[tex]MN=\dfrac{AD+BC}{2}[/tex]
[tex]31x+1=\dfrac{30x-10+30x+28}{2}[/tex]
[tex]31x+1=\dfrac{60x+18}{2}[/tex]
[tex]31x+1=30x+9[/tex]
Isolate variable terms.
[tex]31x-30x=9-1[/tex]
[tex]x=8[/tex]
Now,
[tex]AD=30x-10[/tex]
[tex]AD=30(8)-10[/tex]
[tex]AD=240-10[/tex]
[tex]AD=230[/tex]
Therefore, the length of [tex]AD[/tex] is 230.
