Chapter 8.4, Problem 37WE

To determine

To find: the length of the median of trapezoid.

Answer to Problem 37WE

The length of medium of trapezoid is $\frac{j}{4}\left(3+\sqrt{3}\right)$ .

Explanation of Solution

Given:

The trapezoid with dimension is shown below:

  

Calculation:

Draw perpendiculars as shown below:

  

In triangle ABE,

  $\begin{array}{c}\sin \angle BAE=\frac{\text{perpendicular}}{\text{Hypotenuse}}\\ \sin 60°=\frac{BE}{AB}\\ \frac{\sqrt{3}}{2}=\frac{BE}{j}\\ BE\approx \frac{\sqrt{3}}{2}j\end{array}$

And,

  $\begin{array}{c}\cos \angle BAE=\frac{\text{Base}}{\text{Hypotenuse}}\\ \cos 60°=\frac{AE}{AB}\\ \frac{1}{2}=\frac{AE}{j}\\ AE\approx \frac{j}{2}\end{array}$

From above figure, BE is equal to FD.

Also, side FD and FC are equal as angles FDC and FCD are equal to 45 degree. Thus, Side FC is $\frac{\sqrt{3}}{2}j$ .

Side ED is calculated as follows:

  $\begin{array}{c}ED=AD-AE\\ =j-\frac{j}{2}\\ =\frac{j}{2}\end{array}$

From above figure, ED is equal to BF.

Thus,

  $BF=\frac{j}{2}$

Now by the midsegment theorem,

  $\begin{array}{c}PQ=\frac{1}{2}\left(AD+BC\right)\\ =\frac{1}{2}\left(j+BF+FC\right)\\ =\frac{1}{2}\left(j+\frac{j}{2}+\frac{\sqrt{3}}{2}j\right)\\ =\frac{j}{4}\left(3+\sqrt{3}\right)\end{array}$

Thus, the length of medium of trapezoid is $\frac{j}{4}\left(3+\sqrt{3}\right)$ .