Chapter 9.4, Problem 31PSC

a.

To determine

To find: The type of the quadrilateral QUAD .

Answer to Problem 31PSC

Quadrilateral QUAD is a parallelogram.

Explanation of Solution

Given information:

Quadrilateral QUAD has vertices at $Q=\left(-7,1\right),U=\left(1,16\right),A=\left(9,10\right)$ and $D=\left(1,-5\right)$ .

Formula used:

Slope of a line:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=slope\end{array}$

  $(x_1,y_1)=$ coordinates of first point in the line.

  $(x_2,y_2)=$ coordinates of second point in the line.

The distance between two points by using the distance formula, which is an application of the Pythagoreantheorem.

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

Calculation:

First we draw these four points.

  

We now check the slopes of the sides of the quadrilateral.

Slope of $\overline{QU}=\frac{16-1}{1-(-7)}=\frac{15}{8}$

Slope of $\overline{DA}=\frac{10-(-5)}{9-1}=\frac{15}{8}$

Slope of $\overline{QD}=\frac{-5-1}{1-(-7)}=\frac{-6}{8}=\frac{-3}{4}$

Slope of $\overline{AU}=\frac{10-16}{9-1}=\frac{-6}{8}=\frac{-3}{4}$

Observe that the adjacent sides QD and AD are not perpendicular, so it is not a rectangle or square.

Also, we have two pairs of parallel sides. QD is parallel to UA , UQ is parallel to AD .

So, it is a parallelogram or a rhombus.

To distinguish between them,we need to check the length of the sides.

Distance $\left(Q,D\right)$

  $\begin{array}{l}=\sqrt{{\left(-5-1\right)}^2+{\left(1-\left(-7\right)\right)}^2}\\ =\sqrt{36+64}\\ =\sqrt{100}\\ =10\end{array}$

Distance $\left(A,D\right)$

  $\begin{array}{l}=\sqrt{{\left(-5-10\right)}^2+{\left(1-9\right)}^2}\\ =\sqrt{225+64}\\ =\sqrt{289}\\ =17\end{array}$

Since the length of the sides are not equal, the quadrilateral is a parallelogram.

b.

To determine

Tocalculate:The perimeter of the quadrilateral QUAD .

Answer to Problem 31PSC

The perimeter of the quadrilateral QUAD is 54.

Explanation of Solution

Given information:

Quadrilateral QUAD has vertices at $Q=\left(-7,1\right),U=\left(1,16\right),A=\left(9,10\right)$ and $D=\left(1,-5\right)$ .

Formula used:

The distance between two points by using the distance formula, which is an application of the Pythagoreantheorem.

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

Calculation:

  

Distance $\left(Q,D\right)$

  $\begin{array}{l}=\sqrt{{\left(-5-1\right)}^2+{\left(1-\left(-7\right)\right)}^2}\\ =\sqrt{36+64}\\ =\sqrt{100}\\ =10\end{array}$

Distance $\left(A,D\right)$

  $\begin{array}{l}=\sqrt{{\left(-5-10\right)}^2+{\left(1-9\right)}^2}\\ =\sqrt{225+64}\\ =\sqrt{289}\\ =17\end{array}$

Perimeter is sum of all sides.

Perimeter of QUAD :

  $\begin{array}{l}P=AD+UQ+AU+DQ\\ P=17+17+10+10\\ P=54\end{array}$