Chapter 6, Problem 39SGR

To determine

To calculate:

Given $▭\mathrm{Q}\mathrm{R}\mathrm{S}\mathrm{T}$ is rhombus, rectangle or square.

Answer to Problem 39SGR

Since the shape is rhombus and a rectangle, it has four right angles and four congruent sides, so by definition it is square.

Explanation of Solution

Given information:

  $\mathrm{Q}\left(12,0\right),\mathrm{R}\left(6,-6\right),\mathrm{S}\left(0,0\right),\mathrm{T}\left(6,6\right)$

Using distance formula,

  $\mathrm{d}=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}$

Length of each segment: $\mathrm{Q}\mathrm{T},\mathrm{T}\mathrm{S},\mathrm{S}\mathrm{R},\mathrm{R}\mathrm{Q}$ by using the point in distance formula.

  $\begin{array}{l}\mathrm{Q}\mathrm{T}=\sqrt{{\left(6-12\right)}^2+{\left(6-0\right)}^2}=\sqrt{6^2+{\left(6\right)}^2}=\sqrt{36+36}=\sqrt{72}\\ \mathrm{T}\mathrm{S}=\sqrt{{\left(6-0\right)}^2+{\left(6-0\right)}^2}=\sqrt{6^2+{\left(6\right)}^2}=\sqrt{36+36}=\sqrt{72}\\ \mathrm{S}\mathrm{R}=\sqrt{{\left(6-0\right)}^2+{\left(-6-0\right)}^2}=\sqrt{6^2+{\left(-6\right)}^2}=\sqrt{36+36}=\sqrt{72}\\ \mathrm{R}\mathrm{Q}=\sqrt{{\left(12-6\right)}^2+{\left(0-\left(-6\right)\right)}^2}=\sqrt{6^2+{\left(-6\right)}^2}=\sqrt{36+36}=\sqrt{72}\end{array}$

It is observed that that this appears to be a square, which is also a rhombus (four sides the same length) and a rectangle (four right angles).

To prove that four sides are congruent (rhombus), use the distance formula on each side. $\mathrm{m}=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}$

it is a rectangle by showing that all angles are right angles. Need to find slopes of all of the lines and show that they are perpendicular (negative reciprocal slopes).

By using formula for slope to find the slopes of each line.

Slope of

  $\begin{array}{l}\mathrm{Q}\mathrm{T}=\frac{6-0}{6-12}=\frac{6}{-6}=-1\\ \mathrm{T}\mathrm{S}=\frac{0-6}{0-6}=\frac{-6}{6}=1\\ \mathrm{S}\mathrm{R}=\frac{-6-0}{6-0}=\frac{-6}{6}=-1\\ \mathrm{R}\mathrm{Q}=\frac{6-12}{-6-0}=\frac{-6}{-6}=1\end{array}$

  $\mathrm{Q}\mathrm{T}$ and $\mathrm{T}\mathrm{S}$ have negative reciprocal slopes, they are perpendicular.

  $\mathrm{T}\mathrm{S}$ and $\mathrm{S}\mathrm{R}$ have negative slopes, so they are perpendicular.

  $\mathrm{S}\mathrm{R}$ and $\mathrm{R}\mathrm{Q}$ have negative slopes, so they are perpendicular.

  $\mathrm{R}\mathrm{Q}$ and $\mathrm{Q}\mathrm{T}$ have negative slopes, so they are perpendicular.

This the angles formed are all right angles so the shape in a rectangle.

Since the shape is rhombus and a rectangle, it has four right angles and four congruent sides, so by definition it is square.