Chapter 13.5, Problem 15WE

To determine

To find: the midpoints of the legs

Answer to Problem 15WE

The midpoints of the legs is $8$

Explanation of Solution

Given information: The length of the median of the trapezoid with vertices $C\left(-4,-3\right),D\left(-1,4\right),E\left(4,4\right)\&F\left(7,-3\right)$

Calculation:

  $M_1,M_2,M_3,M_4$ are midpoints of the legs

  $\begin{array}{l}{M}_1=\left(\frac{-4-1}{2},\frac{-3+4}{2}\right)\\ M_1=\left(\frac{-5}{2},\frac{1}{2}\right)\\ M_2=\left(\frac{4-1}{2},\frac{4+4}{2}\right)\\ M_2=\left(\frac{3}{2},\frac{8}{2}\right)=\left(\frac{3}{2},4\right)\\ M_3=\left(\frac{4+7}{2},\frac{-3+4}{2}\right)\\ M_3=\left(\frac{11}{2},\frac{1}{2}\right)\\ M_4=\left(\frac{7-4}{2},\frac{-3-3}{2}\right)\\ M_4=\left(\frac{3}{2},-3\right)\\ \text{median is }{M}_1{M}_3\\ M_1{M}_3=\sqrt{{\left(\frac{11}{2}-\left(-\frac{5}{2}\right)\right)}^2+{\left(\frac{1}{2}-\frac{1}{2}\right)}^2}\\ M_1{M}_3=\sqrt{{\left(8\right)}^2}=\sqrt{64}\\ M_1{M}_3=8\end{array}$