Chapter 6.5, Problem 22PPS

To determine

Whether JKLM is a rhombus, a rectangle or a square.

Answer to Problem 22PPS

QRST is a rectangle, a rhombus and also a square.

Explanation of Solution

Given:

The vertices are:

  $J\left(-1,1\right),K\left(4,1\right),L\left(4,6\right),M\left(-1,6\right).$

Calculation:

Plot the vertices on the graph:

According to Rectangle Theorem, if the diagonals of a parallelogram are congruent, then it is a rectangle.

According to Rhombi Theorem, if the diagonals of a parallelogram are perpendicular, then it is a rhombus.

According to Square Theorem, if the quadrilateral is both, a rectangle and a rhombus, then it is a square.

Consider the diagonals of JKLM:

Finding the length of JL:

  $\begin{array}{l}JL=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ JL=\sqrt{{\left(4-\left(-1\right)\right)}^2+{\left(6-1\right)}^2}\\ JL=\sqrt{25+25}\\ JL=\sqrt{50}\end{array}$

Finding the length of KM:

  $\begin{array}{l}KM=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ KM=\sqrt{{\left(-1-4\right)}^2+{\left(6-1\right)}^2}\\ KM=\sqrt{25+25}\\ KM=\sqrt{50}\end{array}$

The diagonals are congruent, hence, QRST is a rectangle.

Finding the slope of JL:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{6-1}{4-\left(-1\right)}\\ m=\frac{5}{5}\\ m=1\end{array}$

Finding the slope of KM:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{6-1}{-1-4}\\ m=\frac{5}{-5}\\ m=-1\end{array}$

As, $1\left(-1\right)=-1$ , the diagonals are perpendicular. It is a rhombus.

Hence, QRST is a rectangle, a rhombus and also a square.