91Ó°ÊÓ

The given diagram is:

The theorem 5-19 states that the median of a trapezoid is parallel to the bases and has a length equal to the average of the base lengths.

It is being given that $\stackrel{¯}{MN}$is the median of trapezoid $ZOID$.

From the given diagram it can be noticed the sides and ZO and DIare parallel sides, therefore the bases of the trapezoid ZOID are ZOand DI.

Therefore, by using the theorem 5-19 it can be noticed that the median $\stackrel{¯}{MN}$ is parallel to the bases ZO and DI and has a length equal to the average of the base lengths ZO and DI.

Therefore, $\stackrel{¯}{MN}âˆ\yen \stackrel{¯}{ZO}$.

Therefore, it can be said that $\stackrel{¯}{TN}âˆ\yen \stackrel{¯}{ZO}$.

As, $\stackrel{¯}{MN}$is the median of trapezoid ZOID, therefore it can be said that $\stackrel{¯}{TN}$is the median of the triangle $ZIO$.

The theorem 5-11 states that the segments that join the midpoints of two sides of a triangle are parallel to the third side and are half as long as the third side.

Therefore, as $\stackrel{¯}{TN}$is the median of the triangle $ZIO$, therefore by using the theorem 5-11 it can be said that $\stackrel{¯}{TN}âˆ\yen \stackrel{¯}{ZO}$and $\stackrel{¯}{TN}=\frac{1}{2}\left(\stackrel{¯}{ZO}\right)$.

Therefore, it can be obtained that:

$\begin{array}{c}\stackrel{¯}{TN}=\frac{1}{2}\left(\stackrel{¯}{ZO}\right)\\ =\frac{1}{2}\left(8\right)\\ =4\end{array}$

Therefore the length of $TN$is 4.

If $ZO=8$, then $TN=4$.