Chapter 13.9, Problem 4WE

a.

To determine

To prove: median of a trapezoid is parallel to its bases.

Explanation of Solution

Given information:

  $MNOP$ is a trapezoid with vertices $M(d,c),N(b,c),O(0,0),P(a,0)$ .

  

Calculation:

Let us consider the mid points $M_1,M_2$ of the sides then,

  $\begin{array}{l}{M}_1=\left(\frac{b+0}{2},\frac{c+0}{2}\right)=\left(\frac{b}{2},\frac{c}{2}\right)\\ M_2=\left(\frac{a+d}{2},\frac{c+0}{2}\right)=\left(\frac{a+d}{2},\frac{c}{2}\right)\end{array}$

Now $MN\left|\right|OP$

  $\begin{array}{l}\text{Slope of }\overline{M_1{M}_2}=\frac{\frac{c}{2}-\frac{c}{2}}{\frac{a+d}{2}-\frac{b}{2}}=0\\ \text{Slope of }\overline{OP}=\frac{0-0}{a-0}=0\end{array}$

Which implies $\overline{M_1{M}_2}$ is parallel to $\overline{OP}$ . Hence median is parallel to its bases is proved.

b.

To determine

To prove: median of a trapezoid has a length equal to the average of the base lengths.

Explanation of Solution

Given information:

  $MNOP$ is a trapezoid with vertices $M(d,c),N(b,c),O(0,0),P(a,0)$ .

  

Calculation:

  $\begin{array}{l}\text{Length of base }\overline{OP}=a\\ \text{Length of base }\overline{MN}=\sqrt{{(d-b)}^2+{(c-c)}^2}=d-b\\ \text{Average of length of bases =(a-b+d)/2}\end{array}$

  $\text{Length of }\overline{M_1{M}_2}\text{ =}\sqrt{{\left(\frac{a+d}{2}-\frac{b}{2}\right)}^2+{\left(\frac{c}{2}-\frac{c}{2}\right)}^2}=\sqrt{{\left(\frac{a+d-b}{2}\right)}^2}=\frac{a+d-b}{2}=\text{ Average of length of bases}\text{.}$

Hence proved the result.