Chapter 11.3, Problem 20PSC

To determine

To prove: The median of a trapezoid is parallel to the bases and is equal to one-half of their sum.

Explanation of Solution

Given information : Trapezoid APTR with MN as median

Formula used : Slope of a line = $\frac{y_2-y_1}{x_2-x_1}$ and distance formula: $d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$ .

Where

“ d ” is the distance between the two points

Prove: Slope of line AP = $\frac{2c-2c}{2d-2b}=0$

Similarly, slope of line NM = $\frac{c-c}{d+a-b}=0$

Now, slope of line RT = $\frac{0-0}{2a-0}=0$

As, slope of AP, NM and RT is $0$ ,

  $\Rightarrow AP\parallel NM\parallel RT$

Now, according to distance formula,

AP = $\sqrt{{(2b-2d)}^2+{(2c-2c)}^2}$

  $\begin{array}{l}\Rightarrow \sqrt{{(2b-2d)}^2+{(2c-2c)}^2}\\ \Rightarrow \sqrt{{(2b-2d)}^2}\\ \Rightarrow 2b-2d\end{array}$

Similarly, NM = $\sqrt{{[b-(d+a)]}^2+{(c-c)}^2}$

  $\begin{array}{l}\Rightarrow \sqrt{{(b-d-a)}^2}\\ \Rightarrow b-d-a\end{array}$

And, RT = $\sqrt{{(0-2a)}^2+{(0-0)}^2}$

  $\begin{array}{l}\Rightarrow \sqrt{{(-2a)}^2}\\ \Rightarrow -2a\end{array}$

Now, we have to prove that, median is one-half of sum of bases of trapezoid.

  $\Rightarrow NM=\frac{1}{2}(AP+RT)$

Taking Right Hand Side,

  $AP+RT=2b-2d-2a$

Now, $\frac{1}{2}(AP+RT)=\frac{1}{2}(2b-2d-2a)=b-d-a$

Now, Left Hand Side = $NM=b-d-a$

  $\therefore$ Median of trapezoid is equal to one-half of sum of bases of trapezoid.