Chapter 13.1, Problem 28WE

To determine

To show that the quadrilateral diagonals are congruent

Explanation of Solution

Given information:

The quadrilateral has vertices

  $T\left(4,6\right),A\left(6,-4\right),U\left(-4,-2\right)\text{ and }L\left(-2,4\right)$

Proof:

The quadrilateral diagonal are congruent can be show as

Let, a rectangle is a quadrilateral that must have congruent diagonals.

So, use distance formula to show that $TU=AL$

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Let,

  $\begin{array}{l}T\left(x_1,y_1\right)\text{ be }T\left(4,6\right),A\left(x_2,y_2\right)\text{ be }A\left(6,-4\right)\\ U\left(x_3,y_3\right)\text{ be }U\left(-4,-2\right)\text{ and }L\left(x_4,y_4\right)\text{ be }L\left(-2,4\right)\end{array}$

Therefore,

  $TU=\sqrt{{\left(x_3-x_1\right)}^2+{\left(y_3-y_1\right)}^2}$

Put the values of $x_1=4,y_1=6,x_3=-4\text{ and }{y}_3=-2$ on the above equation

  $\begin{array}{l}TU=\sqrt{{\left(-4-4\right)}^2+{\left(-2-6\right)}^2}\\ TU=\sqrt{{\left(-8\right)}^2+{\left(-8\right)}^2}\\ TU=\sqrt{64+64}\\ TU=\sqrt{128}\\ TU=8\sqrt{2}\end{array}$

And,

  $AL=\sqrt{{\left(x_4-x_2\right)}^2+{\left(y_4-y_2\right)}^2}$

Put the values of $x_2=6,y_2=-4,x_4=-2\text{ and }{y}_4=4$ on the above equation

  $\begin{array}{l}AL=\sqrt{{\left(-2-6\right)}^2+{\left(4-\left(-4\right)\right)}^2}\\ AL=\sqrt{{\left(-2-6\right)}^2+{\left(4+4\right)}^2}\\ AL=\sqrt{{\left(-8\right)}^2+{\left(8\right)}^2}\\ AL=\sqrt{64+64}\\ AL=\sqrt{128}\\ AL=8\sqrt{2}\end{array}$

Hence,

  $TU=AL$ so, quadrilateral diagonals are congruent.