Geometry (A): READ: Classifying Quadrilaterals Reading

Classifying Quadrilaterals

Learning Objectives

Identify and classify a parallelogram.
Identify and classify a rhombus.
Identify and classify a rectangle.
Identify and classify a square.
Identify and classify a kite.
Identify and classify a trapezoid.
Identify and classify an isosceles trapezoid.
Collect the classifications in a Venn diagram.
Identify how to classify shapes on a coordinate grid.

Introduction

There are many different classifications of quadrilaterals. In this lesson, you will explore what defines each type of quadrilateral and also what properties each type of quadrilateral has. You have probably heard of many of these shapes before, but here we will focus on things we’ve learned about other polygons—the relationships among interior angles, and the relationships among the sides and diagonals. These issues will be explored in later lessons to further your understanding.

Parallelograms

A parallelogram is a quadrilateral with two pairs of parallel sides. Each of the shapes shown below is a parallelogram.

As you can see, parallelograms come in a variety of shapes. The only defining feature is that opposite sides are parallel. But, once we know that a figure is a parallelogram, we have two very useful theorems we can use to solve problems involving parallelograms: the Opposite Sides Theorem and the Opposite Angles Theorem.

Opposite Angles in Parallelogram Theorem: The opposite angles of a parallelogram are congruent.

Opposite Sides of Parallelogram Theorem: The opposite sides of a parallelogram are congruent.

Rhombi

A rhombus (plural is rhombi or rhombuses) is a quadrilateral that has four congruent sides. While it is possible for a rhombus to have four congruent angles, it’s only one example. Many rhombi do NOT have four congruent angles.

Theorem: A rhombus is a parallelogram

By this theorem we know that all the properties of a parallelogram also apply to a rhombus.

Rectangle

A rectangle is a quadrilateral with four congruent angles. Since you know that any quadrilateral will have interior angles that sum to $360^\circ$ (using the expression $180(n-2)$ ), you can find the measure of each interior angle.

$360 \div 4 = 90$

Rectangles will have four right angles, or four angles that are each equal to $90^\circ$ .

Square

A square is both a rhombus and a rectangle. A square has four congruent sides as well as four congruent angles. Each of the shapes shown below is a square.

Kite

A kite is a different type of quadrilateral. It does not have parallel sides or right angles. Instead, a kite is defined as a quadrilateral that has two distinct pairs of adjacent congruent sides. Unlike parallelograms or other quadrilaterals, the congruent sides are adjacent (next to each other), not opposite.

Trapezoid

A trapezoid is a quadrilateral that has exactly one pair of parallel sides. Unlike the parallelogram that has two pairs, the trapezoid only has one. It may or may not contain right angles, so the angles are not a distinguishing characteristic. Remember that parallelograms cannot be classified as trapezoids. A trapezoid is classified as having exactly one pair of parallel sides.

Isosceles Trapezoid

An isosceles trapezoid is a special type of trapezoid. Like an isosceles triangle, it has two sides that are congruent. As a trapezoid can only have one pair of parallel sides, the parallel sides cannot be congruent (because this would create two sets of parallel sides). Instead, the non-parallel sides of a trapezoid must be congruent.

Example 1

Which is the most specific classification for the figure shown below?

A. parallelogram

B. rhombus

C. rectangle

D. square

The shape above has two sets of parallel sides, so it is a parallelogram. It also has four congruent sides, making it a rhombus. The angles are not right angles (and we can’t assume we know the angle measures since they are unmarked), so it cannot be a rectangle or a square. While the shape is a parallelogram, the most specific classification is rhombus. The answer is choice B.

Example 2

Which is the most specific classification for the figure shown below? You may assume the diagram is drawn to scale.

A. parallelogram

B. kite

C. trapezoid

D. isosceles trapezoid

The shape above has exactly one pair of parallel sides, so you can rule out parallelogram and kite as possible classifications. The shape is definitely a trapezoid because of the one pair of parallel sides. For a shape to be an isosceles trapezoid, the other sides must be congruent. That is not the case in this diagram, so the most specific classification is trapezoid. The answer is choice C.

Using a Venn Diagram for Classification

You have just explored many different rules and classifications for quadrilaterals. There are different ways to collect and understand this information, but one of the best methods is to use a Venn Diagram. Venn Diagrams are a way to classify objects according to their properties. Think of a rectangle. A rectangle is a type of parallelogram (you can prove this using the Converse of the Interior Angles on the Same Side of the Transversal Theorem), but not all parallelograms are rectangles. Here’s a simple Venn Diagram of that relationship:

Notice that all rectangles are parallelograms, but not all parallelograms are rectangles. If an item falls into more than one category, it is placed in the overlapping section between the appropriate classifications. If it does not meet any criteria for the category, it is placed outside of the circles.

To begin a Venn Diagram, you must first draw a large ellipse representing the biggest category. In this case, that will be quadrilaterals.

Now, one class of quadrilaterals are parallelograms—all quadrilaterals with opposite sides parallel. But, not all quadrilaterals are parallelograms: kites have no pairs of parallel sides, and trapezoids only have one pair of parallel sides. In the diagram we can show this as follows:

Okay, we are almost there, but there are several types of parallelograms. Squares, rectangles, and rhombi are all types of parallelograms. Also, under the category of trapezoids we need to add isosceles trapezoids. The completed Venn diagram is like this:

You can use this Venn Diagram to quickly answer questions. For instance, is every square a rectangle? (Yes.) Is every rhombus a square? (No, but some are.)

Strategies for Shapes on a Coordinate Grid

You have already practiced some of the tricks for analyzing shapes on a coordinate grid. You actually have all of the tools you need to classify any quadrilateral placed on a grid. To find out whether sides are congruent, you can use the distance formula.

Distance Formula: Distance between points $(x_1, y_1)$ and $(x_2, y_2) = \sqrt{(x_2-x_1)^2 + (y_2 - y_1)^2}$

To find out whether lines are parallel, you can find the slope by computing $\mathrm{slope} = \frac{\mathrm{rise}} {\mathrm{run}} = \frac{y_{2} - y_{1}} {x_{2} -x_{1}}$ . If the slopes are the same, the lines are parallel. Similarly, if you want to find out if angles are right angles, you can test the slopes of their lines. Perpendicular lines will have slopes that are opposite reciprocals of each other.

Example 4

Classify the shape on the coordinate grid below.

First identify whether the sides are congruent. You can use the distance formula four times to find the distance between the vertices.

For segment $\overline{AB}$ , find the distance between $(-1,3)$ and $(1,9)$ .

$AB &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ &=\sqrt{(1 - (-1))^2 + (9-3)^2} \\ &=\sqrt{(1 + 1)^2 + (9 - 3)^2} \\ &=\sqrt{(2)^2 + (6)^2} \\ &=\sqrt{4 + 36} \\ &=\sqrt{40}$

For segment $\overline{BC}$ , find the distance between $(1, 9)$ and $(3, 3)$ .

$BC &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ &=\sqrt{(3 - 1)^2 + (3-9)^2} \\ &=\sqrt{(2)^2 + (-6)^2} \\ &=\sqrt{4 + 36} \\ &=\sqrt{40}$

For segment $\overline{CD}$ , find the distance between $(3, 3$ ) and $(1, -3)$ .

$CD &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ &=\sqrt{(1 - 3)^2 + ((-3)-3)^2} \\ &=\sqrt{(-2)^2 + (-6)^2} \\ &=\sqrt{4 + 36} \\ &=\sqrt{40}$

For segment $\overline{AD}$ , find the distance between $(-1, 3)$ and $(1, -3)$ .

$AD &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ &=\sqrt{(1 - (-1))^2 + ((-3)-3)^2} \\ &=\sqrt{(2)^2 + (-6)^2} \\ &=\sqrt{4 + 36} \\ &=\sqrt{40}$

So, the length of each segment is equal to $\sqrt{40}$ , and the sides are all equal. At this point, you know that the figure is either a rhombus or a square. To distinguish, you’ll have to identify whether the angles are right angles. If one of the angles is a right angle, they all must be, so the shape will be a square. If it isn’t a right angle, then none of them are, and it is a rhombus.

You can check whether two segments form a right angle by finding the slopes of two intersecting segments. If the slopes are opposite reciprocals, then the lines are perpendicular and form right angles.

The slope of segment $\overline{AB}$ can be calculated by finding the “rise over the run”.

$\text{slope } \overline{AB} &= \frac{y_2 - y_1}{x_2 - x_1}\\ &= \frac{9-3} {1-(-1)} \\ &= \frac{6} {2} \\ &= 3$

Now find the slope of an adjoining segment, like $\overline{BC}.$

$\text{slope } \overline{BC} &= \frac{y_2 - y_1}{x_2 - x_1}\\ &= \frac{3-9} {3-1} \\ &= \frac{-6} {2} \\ &= -3$

The two slopes are $-3$ and $3$ . These are opposite numbers, but they are not reciprocals. Remember that the opposite reciprocal of $-3$ would be $\frac{1} {3}$ , so segments $\overline{AB}$ and $\overline{BC}$ are not perpendicular. Since the sides of $ABCD$ do not intersect a right angle, you can rule out square. Therefore $ABCD$ is a rhombus.

Lesson Summary

In this lesson, we explored quadrilateral classifications. Specifically, we have learned:

How to identify and classify a parallelogram.
How to identify and classify a rhombus.
How to identify and classify a rectangle.
How to identify and classify a square.
How to identify and classify a kite.
How to identify and classify a trapezoid.
How to identify and classify an isosceles trapezoid.
How to collect the classifications in a Venn diagram.
How to identify and classify shapes using a coordinate grid.

It is important to be able to classify different types of quadrilaterals in many different situations. The more you understand the differences and similarities between the shapes, the more success you’ll have applying them to more complicated problems.

The following questions are for your own review. The answers are listed below to help you check your work and understanding.

Review Questions

$x =\underline{\;\;\;\;\;\;\;}, y =\underline{\;\;\;\;\;\;\;}$
$w =\underline{\;\;\;\;\;\;\;}, z =\underline{\;\;\;\;\;\;\;}$
$a =\underline{\;\;\;\;\;\;\;}, b =\underline{\;\;\;\;\;\;\;}$

Use the diagram below for exercises 4-7:

Find the slope of $\overline{QU}$ and $\overline{DA}$ , and find the slope of $\overline{QD}$ and $\overline{UA}$ .
Based on $4$ , what can you conclude now about quadrilateral $QUAD$ ?
Find $QD$ using the distance formula. What can you conclude about $UA$ ?
If $m\angle{Q} = 53^\circ$ , find $m \angle{U}$ and $m\angle{A}$ .
Prove the Opposite Angles Theorem: The opposite angles of a parallelogram are congruent.
Draw a Venn diagram representing the relationship between Widgets, Wookies, and Wooblies (these are made-up terms) based on the following four statements:
1. All Wookies are Wooblies
2. All Widgets are Wooblies
3. All Wookies are Widgets
4. Some Widgets are not Wookies
Sketch a kite. Describe the symmetry of the kite and write a sentence about what you know based on the symmetry of a kite.

Review Answers

$x = 57^\circ, y = 123^\circ$
$w = 90^\circ, z = 9 \;\mathrm{m}$
$a = 97^\circ, b = 89^\circ$
The slope of $\overline{QU}$ and the slope of $\overline{DA}$ both $= 0$ since the lines are horizontal. For $\overline{QD}$ , $\;\mathrm{slope} \ \overline{QD} = \frac{3 - (-1)} {-2 - (-5)} = \frac{3 + 1} {-2 + 5} = \frac{4} {3}$ . Finally for $\overline{UA}$ , $\;\mathrm{slope}\ \overline{UA} = \frac{3 - (-1)} {6 - 3} = \frac{3 + 1} {3} = \frac{4} {3}$
Since the slopes of the opposite sides are equal, the opposite sides are parallel. Therefore, $QUAD$ is a parallelogram
Using the distance formula,
$QD&=\sqrt{((-2)-(-5))^2 + (3 - (-1))^2} \\ &=\sqrt{(-2 + 5)^2 + (3 + 1)^2} \\ &=\sqrt{(3)^2 + (4)^2} \\ &=\sqrt{9 + 16} \\ &=\sqrt{25} \\ &=5$

Since $QUAD$ is a parallelogram, we know that $UA = QD = 5$
$m \angle{U} = 127^\circ$ and $m \angle{A} = 53^\circ$

First, we convert the theorem into “given” information and what we need to prove: Given: Parallelogram $ABCD$ .

Prove: $\angle{A} \cong \angle{C}$ and $\angle{D} \cong \angle{B}$

Statement	Reason
1. $ABCD$ is a parallelogram	1. Given
2. Draw auxiliary segment $\overline{AC}$ and label the angles as follows	2. Line Postulate

3. $\overline{AB} \\| \overline{DC}$	3. Definition of parallelogram
4. $\angle{1} \cong \angle{3}$	4. Alternate Interior Angles Theorem
5. $\overline{AD} \\| \overline{BC}$	5. Definition of parallelogram
6. $\angle{2} \cong \angle{4}$	6. Alternate Interior Angles Theorem
7. $\overline{AC} \cong \overline{AC}$	7. Reflexive property
8. $\triangle ADC \cong \triangle CBA$	8. ASA Triangle Congruence Postulate
9. $\angle{D} \cong \angle{B}$	9. Definition of congruent triangles (all corresponding sides and angles of congruent triangles are congruent)
10. $m \angle{DAB} = m \angle{1} + m \angle{4}$	10. Angle addition postulate
11. $m \angle{DCB} = m \angle{2} + m \angle{3}$	11. Angle addition postulate
12. $\angle{DAB} \cong \angle{DCB}$	12. Substitution

Now we have shown that opposite angles of a parallelogram are congruent

See below. The red line is a line of reflection. Given this symmetry, we can conclude that $\angle{I} \cong \angle{E}$ .

Last modified: Monday, June 28, 2010, 4:20 PM