# Course Description

The fundamental purpose of Mathematics II is to formalize and extend the mathematics that students learned in 9th grade.  Students will focus on quadratic expressions, equations, and functions but also be introduced to radical, piecewise, absolute value, inverse and exponential functions. They will extend the set of rational numbers to the set of complex numbers and link probability and data through conditional probability and counting methods. Finally, they will study similarity and right triangle trigonometry and circles with their quadratic algebraic representations. The course ties together the algebraic and geometric ideas studied. Students should experience mathematics as a coherent, useful, and logical subject that makes use of their ability to make sense of problem situations.

## Class Overview

This integrated Secondary Mathematics II course is based on the New Utah State Standards Initiative.

### Credit

This course is worth .25 credits, or nine weeks of Mathematics I. There are four Mathematics II quarter classes available. Taking all four will add up to one credit or one year of Mathematics I. In order to earn credit for each quarter, you must commit to following the EHS Honor Code: "As a student of the Electronic High School, I agree to turn in my assignments in a timely manner, do my own work, not share my work with others, and treat all students, teachers, and staff with respect." This course is for tenth grade students. After completing the work for the class, students must pass a proctored final exam to earn credit. There is not a paper-based textbook assigned for this course. If you find that having a textbook is useful, you can check out a textbook from most local libraries. You can also search for topics on the Internet to find many useful resources.

## Prerequisites

You should have successfully completed 9th grade math.

## Supplies needed

• Scientific or Graphing Calculator (You can download scientific and graphing calculator simulators or find online versions, but as you will need a graphing calculator for the rest of your high school career, you may consider buying one now.)
• Access to a printer to print the daily assignments is vital. Most assignments are NOT interactive and must be printed out to complete.
• Ability to scan or photograph a completed assignment to submit electronically.

## Organization of Secondary Math Level 2

Units: There are 11 units for the full credit of Secondary Math 2.

Quarter 1 has four units.
Quarter 2 has one unit.
Quarter 3 has three units.
​Quarter 4 has three units.

Schedule: When you enroll in a quarter class, you are given 10 weeks to finish all the requirements needed to earn the credit for a quarter. There are no "required" due dates for the assignments in this course. However, there is a pacing guide provided for you that will help you stay on track to being successful and finishing the course within the 10 week time frame. The pacing guide is located in the Syllabus in Module 1. Before you begin, go over the pacing guide to help you set up your own due dates for the assignments. Give your parents permission to nag you about it. You don't want to be one of those students who does a whole lot of work, but never finishes the course.

## This Quarter Class

The units in this class have lessons, assignments, quizzes and a unit test.

Lessons: Each lesson provides instruction on a given topic. Many include instructional videos (hosted on YouTube) and one or more assignments for independent practice.

## Proctored Final

Each quarter class has a proctored final exam and is worth 25% of the final grade.

### Information about the Final Exam

1. You must get approval from your teacher before you are allowed to take the final exam.
2. You must complete every assignment and have an overall grade of C in the course to be approved to take the final exam.
3. The final exam is a comprehensive exam that must be taken with an approved proctor.
4. You are allowed to have a page of notes. You also will need a calculator and scratch paper.
5. You must pass the final exam with a 60% in order to pass the class.
7. The exam is timed. You will have 2.0 hours to complete the exam. You must finish it in one attempt.

Assignments and quizzes are worth 75% of the final grade. The proctored final test is worth 25% of the final grade.

You earn a grade based on a modified total points percentage method. This means that the total number of points you earned is divided by the total number of points possible, times 100%. That will make up 75% of your final grade. The final exam is the remaining 25%. These scores are combined for a total percentage of the class. This percentage is translated into a grade based on this standard scale:

94-100% A
90-93% A-
87-89% B+
83-86% B
80-82% B-
77-79% C+
73-76% C
70-72% C-
67-69% D+
63-66% D
60-62% D-
0-59% no credit

### 00.00 *Student supplies for Secondary Math 2 (Math2)

 There is no textbook assigned for this course. If you find that having a textbook is useful, you will want to select a textbook that aligns with the Common Core State Standards Initiative. You can purchase textbooks online, at some bookstores (selection is limited), or you can check out a textbook from most local libraries. Supplies needed: Graph paper (this can also be downloaded) Tracing paper or transparency film (if you use transparencies you will need transparency markers). Straight edge (A ruler is an example of a straight edge. I have used a scrap of mat board. A piece of paper will not work! My favorite is a drafting ruler with the cork on the bottom; this gives a very clean line, and doesn't slide around.) Compass (There is a very wide range of compass prices. You don't need a drafting quality compass, but you will be happier if you don't buy the cheapest compass you can find. You want something that doesn't change size as you draw with it.) Graphing Calculator (You can download graphing calculator simulators, but as you will need a graphing calculator for the rest of your high school career, you may as well buy one now.) Geometric software.   Here is the College Board's 2014–15 List of Graphing Calculatorshttps://apstudent.collegeboard.org/apcourse/ap-calculus-ab/calculator-policy

### 00.01 Curriculum Standards (Math Level 2)

Overview information on the Utah Mathematics Level II Core is here.

### 00.01.01 Student Software Needs

 Students need access to a robust internet connection and a modern web browser. This class may also require the Apple QuickTime plug-in to view media. For students using a school-issued Chromebook, ask your technical support folks to download the QuickTime plug-in and enable the plug-in for your Chromebook. \$0.00 QuickTime for Chromehttps://support.google.com/chrome/answer/2445742?hl=enFirefox browserhttp://www.mozilla.org/en-US/firefox/new/ Recommended

### 00.02 About Me (Math Level 2)

 teacher-scored 10 points possible 10 minutes

About Me Assignment: This assignment gives me, as your teacher, a chance to get to know you better! To complete and submit this assignment copy the material between the asterisks into a blank word-processing document. Answer the questions using complete sentences, appropriate punctuation and sentence structure. Please write your answers in either BOLD or a $\dpi{100} \fn_phv {\color{Blue}DIFFERENT }$ $\dpi{100} \fn_phv {\color{Blue}COLOR }$. Save the document. Finally, select all, copy, then paste the entire document into the box that opens when you click to submit this assignment.

**********************************************************************************

1. What is your full name, what name do you prefer to go by, your parent's/guardian's names, and contact information for both you and your parents? (email addresses and phone numbers.)

2. What high school do you attend and what grade are you in? What is the name of the last math class you completed?

3. Why have you chosen to take this math class with EHS?

4. What is your counselor's full name and contact information?

5. Have you read the EHS Honor Code and do you commit to following it? EHS Honor Code "As a student of the Electronic High School, I agree to turn in my assignments in a timely manner, do my own work, not share my work with others, and treat all students, teachers, and staff with respect."

6. Are you committed to finishing the class within the 10 week time frame, completing your final exam in week 9?

7. Now tell me about you! What are your likes/dislikes etc. Please be sure to include anything you think I need to know as your teacher.

*************************************************************************

1. All requested information is included.

2. Complete sentences, correct punctuation and correct grammar are used.

Pacing: complete this by the end of Week 1 of your enrollment date for this class.

### 06.00 Geometry (Math Level 2)

 The goal of all math classes is for students to become mathematically proficient. In particular, students need to be able to: Make sense of problems and persevere in solving them. Reason abstractly and quantitatively. Construct viable arguments and critique the reasoning of others. Model with mathematics. Use appropriate tools strategically. Attend to precision. Look for and make use of structure. Look for and express regularity in repeated reasoning. As you work through this unit, keep these proficiencies in mind! This unit provides you a review of geometric terms and properities and the logic behind geometric proofs. You will also learn to construct formal proofs. At the conclusion of this unit you must be able to state without hesitation that: I understand the key geometric terms and properties. I understand the logic behind constructing geometric proofs. I can reason from a diagram to develop proof-like arguments. I can prove conjectures from a diagram and write formal proofs to prove the conjectures. I can prove statements about parallel lines. I understand the six parameters that uniquely determine a triangle. I can solve problems related to triangle congruency. I can use the Triangle Sum Theorem to solve problems. I can solve problems using knowledge of the perpendicular bisector of a line segment. I can solve problems using knowledge of the base angles of isosceles triangles. I can solve problems using knowledge of parallelograms’ basic properties. I can use my knowledge of triangle midsegments to solve problems. I can use my knowledge of triangle centroids and medians to solve problems.

### 06.01 Introduction to Proofs (Math Level 2)

 Understand the key geometric terms and properties and the logic behind constructing geometric proofs.

Geometry is actually one of the most ancient forms of math. After people discovered how to count, people developed geometry.

Geometry is basically the study of shapes. It comes from the Greek words for "earth" and "measure." Clearly, geometry was used to measure the Earth. The ancient Egyptians developed geometry to allocate and tax land. The ancient Babylonians solved quadratic equations in order to calculate land areas. Geometry was also used to measure the sky. Astronomy was a driving force behind the development of geometry in China and India.

Image by John Reid, http://commons.wikimedia.org/wiki/File:Similar-geometric-shapes.png

Geometry is usually associated with the ancient Greeks. This is partially because the Greek texts were available to the European mathematicians during the Renaissance. It is also because the Greeks elevated the study of geometry from calculating land areas to a true discipline, governed by rules of logic. Therefore, in this class we will also be learning about logic.

Let’s begin this lesson by watching a video:

### 06.01 Introduction to Proofs - External Video (Math Level 2)

 EngageNY: Lesson 9: Unknown Angle Proofs—Writing Proofshttp://youtu.be/o30UY_flFgM

Holmes examines a victim and makes deductions about the victim’s attacker.  He makes each deduction based on several pieces of evidence the victim provides.  The video clip sets the stage for the deductive reasoning you must use to write geometry proofs.  Each geometric conclusion must be backed up with a concrete reason, a fact that has already been established.

In this lesson we will discuss proofs. But before we do that, there are some important terms and properties you need to understand. Most should be very familiar.

### 06.01.01 - Introduction to Proofs - Definitions (Math Level 2)

The first three basic ideas are the point, the line and the plane.

A line is a one dimensional object. It continues infinitely in both directions. Lines in Euclidean geometry are straight. A line is represented by a double-ended arrow, as shown here. It is standard notation to label a line with a lowercase letter, or with 2 points on the line. We denote lines with a double-headed arrow over the points. The order of the points doesn't matter. This line shown is written as $\dpi{100} \fn_phv \overline{EF}$ or as $\dpi{100} \fn_phv \overline{FE}$. The second line is written as line h.

A plane is a two-dimensional object. In Euclidean space, a plane is a flat surface like a tabletop. We draw a plane as a parallelogram even though it extends infinitely in all directions. A plane is labeled with a capital letter and is determined by three non-collinear points. The plane below can either be called plane P or the plane defined by points L, M and N.

Distance is also an undefined term. Distance is how far apart two things are. The distance depends on the path we take. A line will give you the shortest distance between two points.

Geometry Vocabulary

Line segment: A line segment is a section of a line. The ends of the line section are points, so a line segment is the portion of a line that lies between two endpoints. We name line segments by their endpoints, with a bar on top, instead of a double arrow. Also, the order of the points doesn't matter, so the first line segment shown can be called either $\dpi{100} \overline{CB}$ or $\dpi{100} \fn_phv \overline{BC}$, and the second line segment shown can be called either $\dpi{100} \fn_phv \overline{PQ}$ or $\dpi{100} \fn_phv \overline{QP}$. The length of a line segment is the distance between the endpoints of the line segment. When describing lengths, you do not draw a line over the points. To compare the lengths of the line segments shown, we would say CB < PQ. If two line segments have the same length, we say that they are congruent.

Midpoint: The midpoint of a line segment is the point that divides the segment into two equal lengths. The point M is the midpoint of line segment $\dpi{100} \fn_phv \overline{CB}$, and the point N is the midpoint of line segment $\dpi{100} \fn_phv \overline{PQ}$. This means that length MB = CM and that PN = NQ. We can also say that $\dpi{100} \fn_phv \overline{MB}$ is congruent to $\dpi{100} \fn_phv \overline{CM}$. To denote congruence, we use the symbol $\dpi{100} \fn_phv \cong$. Therefore, we can write $\dpi{100} \fn_phv \overline{PN}\cong \overline{NQ}$.

Ray: A ray is a section of a line that has one endpoint and extends infinitely in the other direction. A ray is also denoted with two points, but the order does matter. The endpoint is listed first. The ray has a single arrow on top, so the rays shown here are denoted as $\dpi{100} \fn_phv \overrightarrow{JK}$ and $\dpi{100} \fn_phv \overrightarrow{TS}$.

Intersection: An intersection is when two objects cross each other. The image shows the intersection of the line $\dpi{100} \fn_phv \overline{EF}$ with ray $\dpi{100} \fn_phv \overrightarrow{JK}$. Planes and other objects also may intersect.

Parallel: If two lines lie in the same plane, and never intersect, these lines are parallel. If two planes never intersect, these planes are parallel. In the image, lines s and q are parallel; we write this 'line s || line q.'

Skew: Lines that do not intersect, which do not lie in the same plane, are called skew. In the image above, line m is skew to both line s and line q.

Circle: A circle is a figure defined by the set of points in a plane that are all the same distance from a set point called the center. Circles are denoted by the symbol $\bigodotC$$\dpi{100} \fn_phv \bigodot$ and with the midpoint. The circle shown is named $\dpi{100} \fn_phv \tiny \bigodot$$\dpi{100} \fn_phv C$ or just circle C.

Radius: The distance from the center to the circle is called the radius. In the circle above, $\dpi{100} \fn_phv \overline{JK}$ is a diameter.

Diameter: A line segment going through the center, and terminating at opposite edges of the circle is the diameter. The diameter bisects the circle or cuts it in half. In the circle above $\dpi{100} \fn_phv \overline{CK}$ is a radius.

Angle: An angle is formed by two rays with the same endpoint. The angle is the inclination between these two rays. The endpoint is called the vertex. Angles are denoted in several ways. Consider the angle shown below. This angle is made from the rays $\dpi{100} \fn_phv \overrightarrow{CB}$ and $\dpi{100} \fn_phv \overrightarrow{CD}$. An angle is denoted with the symbol $\dpi{100} \angle$. One way to name an angle is with the three points that define the rays. These are listed with the vertex in the center, so the first angle shown can either be written as $\angle$BCD or as $\angle$DCB, but the point C is always in the middle. If there is only one angle with that vertex, you can also denote the angle by just the vertex; so, in this case, we have $\angle$C and $\angle$L. Finally, you can name the angle with a label to identify that angle. Greek letters and numbers are both used to label angles. The angles shown can also be denoted as $\angle$θ and $\angle$1.

Vertex: The endpoint of an angle is called the vertex. Point L is the vertex of the angle above.

Measure of an Angle: The measure of the angle is the magnitude of the inclination. The measure is denoted as m$\angle$, so the measure of $\angle$BCD is denoted m$\angle$BCD, and the measure of $\angle$L is m$\angle$L.

Congruent Angles: If the measure of two angles is the same, those angles are congruent or $\cong$.

Right Angles: When the endpoint of a ray is on a line, two angles are formed. If the angles formed are congruent, then these angles are called right angles. You are probably more familiar with the idea that a right angle has a measure of $\dpi{100} \fn_phv 90^{\circ}$.

Perpendicular: When two lines intersect to form right angles, the lines are perpendicular. We use the symbol ⊥ to denote that 2 lines are perpendicular. The term orthogonal is also used to describe perpendicular lines. In the image above, angles ∠TQR and ∠SQT are both right angles. Therefore, $\dpi{100} \fn_phv \overline{SR}\perp \overline{QT}$.

Acute Angle: An acute angle is an angle whose measure is less than a right angle.

Obtuse Angle: An obtuse angle is an angle whose measure is greater than a right angle.

Complementary Angles: When the measures of two angles combine to form a right angle, the angles are complementary. This is easy to see if the angles are adjacent. $\angle$UTV is both adjacent and complementary to $\angle$VTW. $\angle$B and $\angle$A are not adjacent, but they are complementary.

Supplementary Angles: When the measures of two angles combine to form two right angles, the angles are supplementary. This is true in general, but it is particularly useful when the angles are adjacent. $\angle$G and $\angle$H are not adjacent, but they are supplementary. $\angle$CDF is both adjacent and supplementary to $\angle$FDE.

Linear Pairs: Linear pairs are the two supplementary adjacent angles formed by two intersecting lines. $\dpi{100} \fn_phv \angle CDF$ and $\dpi{100} \fn_phv \angle EDF$ above are examples of linear pairs. (wiktionary.org/wiki/linearpair).

Straight Angle: When the measure of one angle is equal to two right angles, this angle is called a straight angle. $\angle$$\dpi{100} \fn_phv CAB$ is a straight angle.

Vertical Angles: When two lines intersect as shown, the angles opposite one another are called vertical angles. In the image here, $\dpi{100} \fn_phv \angle LDH$ and $\dpi{100} \fn_phv \angle GDM$∠GDM are vertical angles. Also $\dpi{100} \fn_phv \angle GDL$ and $\dpi{100} \fn_phv \angle HDM$ are vertical angles.

Transversal: Remember that two lines are parallel if they lie in the same plane and never intersect. If a third line intersects both parallel lines, then eight angles are formed. The intersecting line is called a transverse line.  Line n is a transversal.

Interior Angles: Lines s and q are parallel. Line n transverses lines s and q. The eight angles formed are labeled 1-8 as shown. The four angles that lie between the two lines are called interior angles. Therefore, $\angle$3, $\angle$4, $\angle$5 and $\angle$6 are the interior angles in the figure above.

Exterior Angles: The four angles that lie outside of the two lines above are called exterior angles. Therefore, $\angle$1, $\angle$2, $\angle$7 and $\angle$8 are the exterior angles.

Alternate Interior Angles: Alternate interior angles are the interior angles that are opposite one another. So, $\angle$4 and $\angle$5 are alternate interior angles. Also $\angle$3 and $\angle$6 are alternate interior angles.

Alternate Exterior Angles: Alternate exterior angles are the exterior angles that are opposite one another. So, $\angle$1 and $\angle$8 are alternate exterior angles, as are $\angle$2 and $\angle$7.

Corresponding Angles: Corresponding angles are the angles that are located in the same position relative to the transverse line. Therefore, the pairs of corresponding angles are $\angle$1 and $\angle$5, $\angle$2 and $\angle$6, $\angle$3 and $\angle$7 and $\angle$4 and $\angle$8.

Figure: A figure is an enclosed shape that lies in a plane (see image below). It can be made of line segments, curves or combinations of these.

Sides: A figure is made of lines or curves that connect at points. The lines or curves are called the sides.

Points or Vertices: The points where the sides connect are called the vertices. If these are straight lines, they form an angle at the vertex.

Concave or Convex: Figures can be either concave or convex. A concave figure is one where part of the figure lies "inside" the figure. A convex figure is one where all the sides are "outside" of the figure. Figures B and D are concave. Figures A and C are convex.

Polygon: If a figure is made ONLY of straight-line segments, then Euclid called it rectilinear. We usually just call it a polygon. The figure A above is not rectilinear. Figures B, C and D are all rectilinear.

Regular Polygon: A polygon in which every side and every angle is equal is called a regular polygon. All of the polygons shown above are irregular polygons. Figures F-H below are regular polygons.

Triangle: A triangle is a polygon formed with three sides. Triangles are described by the relations between the sides and angles.

Equilateral Triangle: An equilateral triangle has three sides that are the same. An equilateral triangle is a regular polygon.

Isosceles Triangle: An isosceles triangle is one where two sides are the same so you may notice that an equilateral triangle is a special case of an isosceles triangle.

Scalene Triangle: A scalene triangle is one where none of the sides are the same.

Obtuse Triangle: If one of the interior angles is obtuse the triangle is an obtuse triangle. Obtuse triangles can be either scalene or isosceles.

Acute Triangle: If all of the interior angles of a triangle are acute it is an acute triangle. Acute triangles can be scalene, isosceles or even equilateral.

Square: A square is a quadrilateral with all four sides equal and all angles being right angles. This means that a square is a regular polygon.

Rectangle: A rectangle is a quadrilateral with all right angles so a square is a special case of a rectangle. Rectangles also have opposite sides that are equal and parallel.

Rhombus: A rhombus is a quadrilateral with all four sides equal but not necessarily the angles. Therefore, the square is also a special case of a rhombus. Rhombi also have opposite sides that are parallel.

Parallelogram: A parallelogram is a quadrilateral with opposite sides that are equal and parallel, but the angles do not need to be right. Therefore, squares, rectangles and rhombi are all special cases of parallelograms.

Trapezoid: A trapezoid is a quadrilateral with one pair of parallel sides.

Pentagon: A pentagon is a five-sided polygon.

Hexagon: A hexagon is a six-sided polygon.

Heptagon: A heptagon is a seven-sided polygon.

Octagon: An octagon is an eight-sided polygon.

n-gon: The remaining polygons are all named with the general form n-gon, where n is a prefix that indicates the number of sides. The prefixes are either Latin or Greek and the actual prefix is not always agreed upon.

Axiom: Axioms are the fundamental rules that the system uses to build everything else. An axiom is true, but it cannot be proven that it is true. Instead, it is assumed to be true.

Postulate: Postulates are a type of axiom.

Theorem: A theorem is a rule that can be proven to be true using axioms or postulates and definitions.

Here’s an example of a postulate that we will need in this lesson.

Linear Pair Postulate: The linear pair postulate states that if two angles form a linear pair, then they are supplementary. (stageometrych2.wikispaces.com)

We need to add one more list! You need to know and understand the following properties:

Basic Properties Reference Chart (Lesson 9 EngageNY)

 Property Meaning Geometry Examples Reflexive Property A quantity is equal to itself. $\dpi{100} \fn_phv A=A$ Transitive Property If two quantities are equal to the same quantity, then they are equal to each other. If $\dpi{100} \fn_phv AB = BC$ and $\dpi{100} \fn_phv BC = EF$, then $\dpi{100} \fn_phv AB = EF$. Symmetric Property If a quantity is equal to a second quantity, then the second quantity is equal to the first. If $\dpi{100} \fn_phv AB = BC$ then $\dpi{100} \fn_phv BC=AB$. Addition Property of Equality If equal quantities are added to equal quantities, then the sums are equal. If $\dpi{100} \fn_phv OA = AB$ then $\dpi{100} \fn_phv BC = CD$, then $\dpi{100} \fn_phv AB+BC=DF+CD.$ Subtraction Property of Equality If equal quantities are subtracted from equal quantities, the differences are equal. If $\dpi{100} \fn_phv AB+BC=DF+DE$ and $\dpi{100} \fn_phv BC=DE,$ then$\dpi{100} \fn_phv AB=CD.$ Multiplication Property of Equality If equal quantities are multiplied by equal quantities, then the products are equal. If $\dpi{100} \fn_phv DE+CD = CE$and $\dpi{100} \fn_phv m\angle ABC = m\angle XYZ$ then $\dpi{100} \fn_phv 2m\angle ABC = 2m\angle XYZ$ Division Property of Equality If equal quantities are divided by equal quantities, then the quotients are equal. If $\dpi{100} \fn_phv AB=XY$ then $\dpi{100} \fn_phv \frac{AB}{2} = \frac{XY}{2}$. Substitution Property of Equality A quantity may be substituted for its equal. If $\dpi{100} \fn_phv DE + CD = CE$ and $\dpi{100} \fn_phv CD = AB$, then $\dpi{100} \fn_phv DE + AB = CE$. Partition Property includes “Angle Addition Postulate,” “Segments add,” “Betweenness of Points,” etc. A whole is equal to the sum of its parts. If point $\dpi{100} \fn_phv C$ is on $\dpi{100} \fn_phv \overline{AB}$, then $\dpi{100} \fn_phv AC + CB + AB.$

Mathematical Logic Geometry

It is finally time to get back to Sherlock Holmes!

In order to prove things in Geometry we use logic as Holmes did. We also apply axioms and theorems we learn as we move through Geometry. Then we can construct formal mathematical proofs.

### 06.01.01 - Introduction to Proofs - Lesson 1 (Math Level 2)

 Oswego City School District Regents Exam Prep Center: Logichttp://www.regentsprep.org/regents/math/geometry/gp1/indexgp...MathisPower4u If-Then Statements and Converseshttps://www.youtube.com/watch?v=oEr27P1bX9o

Read all nine lessons and then do the practice problems.

### 06.01.01 - Introduction to Proofs - Lesson 2 (Math Level 2)

 Oswego City School District Regents Exam Prep Center: Related Conditionalshttp://www.regentsprep.org/regents/math/geometry/gp2/indexGP...MathisPower4u The Converse, Contrapositive, and Inverse of an If-Then Statementhttps://www.youtube.com/watch?v=IHd8jiUF3Lk

Read all the lessons Related Conditionals - Warm Up, Converse, Inverse, and Contrapositive and do the practice problems.

### 06.01.01 - Introduction to Proofs - Lesson 3 (Math Level 2)

 Oswego City School District Regents Exam Prep Center:Writing a Proof - Direct Euclidean Proofshttp://www.regentsprep.org/regents/math/geometry/gp3/indexGP...

Read the lessons Styles of Proof, and What is a Proof.  Do the Practice: Pre-Proofs Warm-Ups with Definitions.

### 06.01.01 - Introduction to Proofs - Lesson 4 (Math Level 2)

 Oswego City School District Regents Exam Prep Center: Indirect Proof - Proof by Contradictionhttp://www.regentsprep.org/regents/math/geometry/gp3b/indire...

If after completing this topic you can state without hesitation that...

• I understand the key geometric terms and properties.
• I understand the logic behind constructing geometric proofs.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

### 06.01.01 - Introduction to Proofs - Quiz (Math Level 2)

 computer-scored 22 points possible 35 minutes
Pacing: complete this by the end of Week 1 of your enrollment date for this class.

### 06.02 Develop Proof-like Arguments (Math Level 2)

 Reason from a diagram to develop proof-like arguments.

The material for this lesson is from the The Mathematics Vision Project under a Creative Commons Attribution Non Commercial Share Alike 3.0 Unported license.

You probably know that the sum of the interior angles of any triangle is 180°. If you didn’t know that, you do now.  But an important question to ask yourself is, “How do you know that?”

We know a lot of things because we accept it on authority—we believe what other people tell us; things such as the distance from the earth to the sun is 93,020,000 miles or that the population of the United States is growing about 1% each year. Other things are just defined to be so, such as the fact that there are 5,280 feet in a mile. Some things we accept as true based on experience or repeated experiments, such as the sun always rises in the east, or “I get grounded every time I stay out after midnight.” In mathematics we have more formal ways of deciding if something is true.

Experiment

Cut out a triangle and trace it onto a piece of paper. It will be helpful to color‐code each vertex angle of the original triangle with a different color. As new images of the triangle are produced during this experiment, color‐code the corresponding angles with the same colors. This is my example. Choose a different angle for your experiment.

$\angle$A is shaded green, $\angle$B is shaded red, and $\angle$C is shaded blue.

Locate the midpoints of each side of your cut out triangle by folding the vertices that form the endpoints of each side onto each other.

Rotate your triangle 180° about the midpoint of one of its sides. Trace the new triangle onto your paper and color‐code the angles of this image triangle so that corresponding image/pre‐image pairs of angles are the same color.

Now rotate the new “image” triangle 180° about the midpoint of one of the other two sides. Trace the new triangle onto your paper and color‐code the angles of this new image triangle so that corresponding image/pre‐image pairs of angles are the same color.

Look carefully at the three angles by the point C. One is blue, one is red, and one is green. $\angle$ACC’’ is a straight angle  = $\dpi{100} \fn_phv 180^{\circ}$. So, $\angle$A + $\angle$B + $\angle$C = $\dpi{100} \fn_phv 180^{\circ}$.

Take time to seriously ponder the answers to these questions!

1. What does this experiment reveal about the sum of the interior angles of the triangles you cut out, and how does it do so?
2. Do you think you can rotate all triangles in the same way about the midpoints of its sides, and get the same results?
3. Why or why not?

Make certain you memorize the following symbols before continuing!

 = Used when comparing numbers of equal value. $\dpi{100} \fn_phv m\angle C$ Indicates the measure of an angle.  It would be set equal to a number. $\dpi{100} \fn_phv GH$ The length of GH.  It would equal to a number. $\dpi{100} \fn_phv \triangle ABC$ Triangle ABC $\dpi{100} \fn_phv \perp$ Perpendicular – Lines, rays, segments, and planes can all be perpendicular $\dpi{100} \fn_phv \angle ABC$ Angle ABC – The middle letter is always the vertex of the angle. Refers to the infinite line GH.  Lines are not equal or congruent to other lines. $\dpi{100} \fn_phv \cong$ Congruent – Figures that are the same size and shape are said to be congruent. $\sim$ Similar – Figures that have been dilated are similar. $\dpi{100} \fn_phv \overline{GH}$ Line segment with endpoints G and H.  Line segments can be congruent to each other. You would not say they were equal. $\dpi{100} \fn_phv \overrightarrow{GH}$ Ray GH – The letter on the left indicates the endpoint of the ray. $\dpi{100} \fn_phv \left | \right |$ Parallel – used between segments, lines, rays, or planes. $\dpi{100} \fn_phv \pm$ Plus or minus – indicates two values, the positive value and the negative value. $\dpi{100} \fn_phv \left | x\right |$ Absolute value – it is always equal to the positive value of the number inside the lines.  It represents distance from zero.

There are also two other types of angles that we need to review, supplementary and complementary angles. This section of the lesson comes from NROC.

Supplementary and Complementary

Two angles whose measures add up to 180º are called supplementary angles. There’s also a term for two angles whose measurements add up to 90º, they are called complementary angles.

One way to remember the difference between the two terms is that “corner” and “complementary” each begin with c (a 90º angle looks like a corner), while straight and “supplementary” each begin with s (a straight angle measures 180º).

If you can identify supplementary or complementary angles within a problem, finding missing angle measurements is often simply a matter of adding or subtracting.

Example

Problem: Two angles are supplementary. If one of the angles measures 48º, what is the measurement of the other angle?

Two supplementary angles make up a straight angle, so the measurements of the two angles will be 180º.

You know the measurement of one angle. To find the measurement of the second angle, subtract 48º from 180º.

Answer: The measurement of the other angle is 132º.

Example

Problem: Find the measurement of $\fn_phv {\color{DarkGreen} \angle AXZ}$.

This image shows two intersecting lines, $\fn_phv \overline{AB}$ and $\fn_phv \overline{YZ}$. They intersect at point X, forming four angles.

Angles  and  are supplementary because together they make up the straight angle .

Use this information to find the measurement of $\fn_phv \angle AXZ$.

Answer: $\fn_phv {\color{DarkGreen} m\angle AXZ=150^{\circ}}$

Example

Problem: Find the measurement of $\fn_phv {\color{DarkGreen} \angle BAC}$.

This image shows the line $\fn_phv \overline{CF}$ and the rays $\fn_phv \overrightarrow{AB}$ and $\fn_phv \overrightarrow{AD}$ , all intersecting at point A. $\fn_phv \angle BAD$ is a right angle.

Angles  and  are complementary, because together they create $\fn_phv \angle CAD$.

Use this information to find the measurement of .

Answer: $\fn_phv {\color{DarkGreen} m\angle BAC=40^{\circ}}$

Example

Problem: Find the measurement of $\fn_phv {\color{DarkGreen} \angle CAD}$.

You know the measurements of two angles here: $\fn_phv \angle CAB$ and $\fn_phv \angle DAE$.You also know that $\fn_phv m\angle BAE=180^{\circ}$.

Use this information to find the measurement of $\fn_phv \angle CAD$.

Answer: $\fn_phv {\color{DarkGreen} m\angle CAD=80^{\circ}}$

Summary

Two angles whose measurements add up to 180º are said to be supplementary, and two angles whose measurements add up to 90º are said to be complementary. For most pairs of intersecting lines, all you need is the measurement of one angle to find the measurements of all other angles formed by the intersection.

If after completing this topic you can state without hesitation that...

• I can reason from a diagram to develop proof-like arguments.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

I highly recommend that you click on the links below and watch the videos before continuing.

### 06.02 Develop Proof-like Arguments - videos (Math Level 2)

 Properties of Angleshttp://www.montereyinstitute.org/courses/DevelopmentalMath/U...MathisPower4u Angle Basicshttps://www.youtube.com/watch?v=7iBc5bJdanIMathisPower4u Animation: Types of Angleshttps://www.youtube.com/watch?v=50eVno0s1DIMathisPower4u Complementary, Supplementary, and Vertical Angleshttps://www.youtube.com/watch?v=rjOjwcV79HMMathisPower4u Introduction to Proof Using Properties of Equalityhttps://www.youtube.com/watch?v=OO_nIBmicKMMathisPower4u Introduction to Proof Using Properties of Congruence https://www.youtube.com/watch?v=M6cbpQ_TUAQ

### 06.02 Develop Proof-like Arguments - Worksheet (Math Level 2)

 teacher-scored 35 points possible 35 minutes

Activity for this lesson

Complete the attached worksheet.

1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.

Pacing: complete this by the end of Week 1 of your enrollment date for this class.

### 06.03 Writing Formal Proofs (Math Level 2)

 Prove conjectures from a diagram and write formal proofs to prove the conjectures.

The material for this lesson is from the The Mathematics Vision Project under a Creative Commons Attribution Non Commercial Share Alike 3.0 Unported license.

Conjectures and Proofs

The diagram from the previous lesson has been extended by repeatedly rotating the image triangles around the midpoints of their sides to form a tessellation of the plane, as shown below.

Using this diagram, we will make some conjectures about lines, angles and triangles and then write proofs to convince ourselves that our conjectures are always true.

Vertical Angles

When two lines intersect, the opposite angles formed at the point of intersection are called vertical angles. In the diagram below, $\angle$1 and $\angle$3 form a pair of vertical angles, and $\angle$2 and $\angle$4 form another pair of vertical angles.

Examine the tessellation diagram above, looking for places where vertical angles occur. You may have to ignore some line segments and angles in order to focus on pairs of vertical angles. This is a skill we have to develop when trying to see specific images in geometric diagrams.

Based on several examples of vertical angles in the diagram, write a conjecture about vertical angles.

My conjecture:

The red angles are vertical angles. Because they are the same angle I hope you stated that vertical angles are congruent or something less mathematical than that. ☺

Exterior Angles of a Triangle

When a side of a triangle is extended, as in the diagram below, the angle formed on the exterior of the triangle is called an exterior angle. The two angles of the triangle that are not adjacent to the exterior angle are referred to as the remote interior angles. In the diagram, $\angle$4 is an exterior angle, and $\angle$1 and $\angle$2 are the two remote interior angles for this exterior angle.

Examine the tessellation diagram above, looking for places where exterior angles of a triangle occur. Again, you may have to ignore some line segments and angles in order to focus on triangles and their vertical angles.

Based on several examples of exterior angles of triangles in the diagram, write a conjecture about exterior angles.

My conjecture:

The exterior angle of this triangle is the sum of the red angle and the green angle.

Parallel Lines Cut By a Transversal

When a line intersects two or more other lines, the line is called a transversal line. When the other lines are parallel to each other, some special angle relationships are formed. To identify these relationships, we give names to particular pairs of angles formed when lines are crossed or cut by a transversal. In the diagram below, $\angle$1 and $\angle$5 are called corresponding angles, $\angle$3 and $\angle$6 are called alternate interior angles, and $\angle$4 and $\angle$6 are called same side interior angles.

Examine the tessellation diagram above, looking for places where parallel lines are crossed by a transversal line. Based on several examples of parallel lines and transversals in the diagram, write some conjectures about corresponding angles, alternate interior angles and same side interior angles.

My conjecture:

First, notice that $\dpi{100} \fn_phv \angle 1$= the sum of the red and blue angles. The same is true for $\dpi{100} \fn_phv \angle 8$. This means $\dpi{100} \fn_phv \angle 1\cong \angle 8$.

Second, notice that the same is true for $\dpi{100} \fn_phv \angle 4$ and $\dpi{100} \fn_phv \angle 8$. This means $\dpi{100} \fn_phv \angle 4\cong \angle 8$

Third, notice that angles 2, 3, 6, and 7 are all green angles. You should be able to make congruent statements about these angles as well.

Here are some of mine:

You can see that angle 1 and angle 5 are each the sum of red and blue and therefore $\dpi{100} \fn_phv \angle 1\cong \angle 5$.

You can see that angle 3 and angle 6 are both green and therefore $\dpi{100} \fn_phv \angle 3\cong \angle 6$.

You can see that angle 4 and angle 6 are equal to green + red + blue and are therefore equal to $\dpi{100} \fn_phv 180^{\circ}$ or supplementary.

Proving Our Conjectures

Finally, we get to act like Sherlock Holmes! In geometry, we follow a similar deductive thought process, much like Holmes’ used, to prove geometric claims.

For each of the conjectures you wrote above, write a proof that will convince you and others that the conjecture is always true. You can use ideas about transformations, linear pairs, congruent triangle criteria, etc. to support your arguments. A good way to start is to write down everything you know in a flow diagram, and then identify which statements you might use to make your case.

Vertical Angles Proof

You know that angles on a line sum to $180^{\circ}$.

Prove that vertical angles are equal in measure.

Make a plan:

What do you know about $\dpi{100} \fn_phv \angle w$ and x? $\angle$y and $\angle$x?

They are supplemental angles which means they sum to $\dpi{100} \fn_phv 180{^{\circ}}$.

What conclusion can you draw based on both pieces of knowledge?

m$\dpi{100} \fn_phv \angle w$  and m$\angle$y

Make certain you understand each step!

Exterior Angles of a Triangle

Given: $\dpi{100} \fn_phv \mathbf{m\angle 4=m\angle 1+m\angle 2}$

Parallel Lines Cut by a Transversal

If after completing this topic you can state without hesitation that...

• I can prove conjectures from a diagram and write formal proofs to prove the conjectures.
• I can solve problems involving vertical angles, supplementary angles and complementary angles.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

I highly recommend that you click on the links below and watch the videos before continuing.

### 06.03 Writing Formal Proofs - Worksheet (Math Level 2)

 teacher-scored 60 points possible 40 minutes

Activity for this lesson

Complete the attached worksheet.

1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.

Pacing: complete this by the end of Week 2 of your enrollment date for this class.

### 06.04 Parallel Lines Proofs (Math Level 2)

Before we begin this lesson, we need to review the angles formed when a transversal cuts through a pair of parallel lines.

Please view the following videos to review the angles formed when a transversal cuts through a pair of parallel lines and the angle relationships formed.

### 06.04 Parallel Lines Proofs-Review Lesson (Math Level 2)

It is time for another postulate.

The Corresponding Angles Congruence Postulate states that corresponding angles formed by a transversal are congruent. Corresponding angles are angles that are in the same relative positions in a transversal.

In figure 2, the corresponding pairs of angles are angles 1 and 5, 2 and 6, 3 and 7, 4 and 8. Since the Corresponding Angles Congruence Postulate is an axiom, it has no proof, but is taken to be true without proof.

Armed with this postulate we can prove that alternate exterior angles are congruent.

The Alternate Exterior Angles Theorem states that if two parallel lines are cut by a transversal, then each pair of alternate exterior angles is congruent.

Watch the following video before continuing:

### 06.04.01 Parallel Lines Proofs (Math Level 2)

 David McAdams: Alternate Exterior Angles are Congruenthttp://www.allmathwords.org/en/a/altexterioranglestheorem.ht...

Recap of the proof:

Given: line a || line b

Prove: $\angle$5 ≅ $\angle$3

Now, you should be take time to write the proof to show that ∠4 ≅ ∠6.

Next, the Alternate Interior Angles Theorem states that if two parallel lines are cut by a transversal, then each pair of alternate interior angles is congruent. See if you can identify these angles from the figure 2 above.

Watch the video below:

### 06.04.02 Parallel Lines Proofs (Math Level 2)

 David McAdams: Alternate Interior Angles are Congruenthttp://www.allmathwords.org/en/a/altinterioranglestheorem.ht...

Time for one more theorem: Consecutive Interior Angles are Supplementary

In figure 2 above, the following angles are consecutive interior angles: $\angle$1 and $\angle$8 as well as $\angle$2 and $\angle$7.

Given: line a $\left | \right |$ line b

Prove: $\angle$1 $\cong$ $\angle$8  and $\angle$2 $\cong$ $\angle$7

Try proving this yourself before watching the following video:

### 06.04.03 Parallel Lines Proofs - Video (Math Level 2)

 mathispower4u: Proof: Alternate Interior Angles Are Congruenthttp://www.youtube.com/embed/svprkO5bM88?wmode=transparent&r...mathispower4u: Example 1: Parallel Line Properties https://www.youtube.com/watch?v=rQcgfcqwv0Amathispower4u: Example 2: Parallel Line Properties https://www.youtube.com/watch?v=cyGOrEd2mRY

If after completing this topic you can state without hesitation that...

• I can prove statements about parallel lines.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

### 06.04.04 Parallel Lines Proofs - Worksheet (Math Level 2)

 teacher-scored 68 points possible 35 minutes

Activity for this lesson

Complete the attached worksheet.

1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.

Pacing: complete this by the end of Week 2 of your enrollment date for this class.

### 06.05 Review of Triangle Congruence (Math Level 2)

 Understand the six parameters that uniquely determine a triangle and solve problems related to triangle congruency.

This concept was developed in Secondary Math I. It is very important that you understand this material.

Here is a summary of the pertinent concepts:

• Each triangle is defined by six parameters: three sides and three angles.
• We need at least three parameters to uniquely define a triangle. Therefore, we need at least three parameters to determine if two triangles are congruent.
• Not every combination of three parameters will uniquely define a triangle. Therefore, not every combination of parameters will prove that two triangles are congruent.
• The sets of parameters that uniquely determine a triangle, and therefore prove congruence, are:

1. Side-Angle-Side, SAS

a. Proven by the axiom that if two things coincide, they are equal

2. Side-Side-Side, SSS

a. Proven by the axiom that if two things coincide, they are equal

Image from 3a-math-project.blogspot.com

3. Angle-Side-Angle, ASA

a. Equivalent to the SAA theorem
b. Proven by contradiction, and by SAS

4. Angle-Angle, AA

5. Side-Side-Angle, SSA

a. Also called the ambiguous case - with these parameters, the triangles might be congruent, or they might not; we cannot tell

6. There are logical limits on some of these combinations.

a. The sum of the angles cannot be greater than 180°.
b. The length of any side cannot be greater than the sum of the other two sides.
c. The SSA case has other limits that are less obvious.

Please take time to study the each of the following lessons!

 Determining Triangle Congruence, Part 1 Geometryhttps://share.ehs.uen.org/sites/default/files/Unit06Lesson3_...Determining Triangle Congruence, Part 2 Geometryhttps://share.ehs.uen.org/sites/default/files/Unit06Lesson4....Determining Triangle Congruence, Part 3 Geometryhttps://share.ehs.uen.org/sites/default/files/Unit06Lesson5_...mathispower4u: Introduction to Congruent Triangles https://www.youtube.com/watch?v=CA1TvVRApkQmathispowe4u: Example 1: Prove Two Triangles are Congruenthttps://www.youtube.com/watch?v=4giM5JT5QqYmathispower4u: Example 2: Prove Two Triangles are Congruent https://www.youtube.com/watch?v=12Ok_sKhQYY

If after completing this topic you can state without hesitation that...

• I understand the six parameters that uniquely determine a triangle.
• I can solve problems related to triangle congruency.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

### 06.05 Review of Triangle Congruence - Worksheet (Math Level 2)

 teacher-scored 88 points possible 40 minutes

Activity for this lesson

Complete the attached worksheet.

1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.

Pacing: complete this by the end of Week 2 of your enrollment date for this class.

### 06.06 Triangle Sum Theorem (Math Level 2)

 Use the Triangle Sum Theorem to solve problems.

The material for this lesson is from:

What if you wanted to classify the Bermuda Triangle by its sides and angles? You are probably familiar with the myth of this triangle; how several ships and planes passed through and mysteriously disappeared.

The measurements of the sides of the triangle from a map are in the image. What type of triangle is this? Using a protractor, find the measure of each angle in the Bermuda Triangle. What do they add up to? Do you think the three angles in this image are the same as the three angles in the actual Bermuda triangle? Why or why not? After completing this Concept, you'll be able to determine how the three angles in any triangle are related in order to help you answer these questions.

In polygons, interior angles are the angles inside of a closed figure with straight sides. The vertex is the point where the sides of a polygon meet.

Triangles have three interior angles, three vertices and three sides. A triangle is labeled by its vertices with a $\bigtriangleup$. This triangle can be labeled $\bigtriangleup$ABC; $\bigtriangleup$ACB; $\bigtriangleup$BCA; $\bigtriangleup$BAC; $\bigtriangleup$CBA or $\bigtriangleup$CAB. Order does not matter. The angles in any polygon are measured in degrees. Each polygon has a different sum of degrees, depending on the number of angles in the polygon. How many degrees are in a triangle?

Investigation: Triangle Tear-Up

Tools Needed: paper, ruler, pencil, colored pencils

1. Draw a triangle on a piece of paper. Try to make all three angles different sizes. Color the three interior angles three different colors and label each one, $\angle$1; $\angle$2; and $\angle$3.

2. Tear off the three colored angles, so you have three separate angles.

3. Attempt to line up the angles so their points all match up. What happens? What measure do the three angles add up to?

This investigation shows us that the sum of the angles in a triangle is $\dpi{100} \fn_phv 180^{\circ}$  because the three angles fit together to form a straight line. Recall that a line is also a straight angle and all straight angles are $\dpi{100} \fn_phv 180^{\circ}$.

The Triangle Sum Theorem states that the interior angles of a triangle add up to $\dpi{100} \fn_phv 180^{\circ}$. The above investigation is one way to show that the angles in a triangle add up to $\dpi{100} \fn_phv 180^{\circ}$. However, it is not a two-column proof. Here we will prove the Triangle Sum Theorem.

There are two theorems that we can prove as a result of the Triangle Sum Theorem and our knowledge of triangles.

Theorem #1: Each angle in an equiangular triangle measures $\dpi{100} \fn_phv 60^{\circ}$.

Theorem #2: The acute angles in a right triangle are always complementary.

Example A

What is the m$\angle$T?

From the Triangle Sum Theorem, we know that the three angles add up to $\dpi{100} \fn_phv 180^{\circ}$. Set up an equation to solve for T.

m$\angle$M+m$\angle$A+m$\angle$T = $\dpi{100} \fn_phv 180^{\circ}$

$\dpi{100} \fn_phv 82^{\circ}$ + $\dpi{100} \fn_phv 27^{\circ}$ + m$\angle$T= $\dpi{100} \fn_phv 180^{\circ}$

$\dpi{100} \fn_phv 109^{\circ}$ + m$\angle$T = $\dpi{100} \fn_phv 180^{\circ}$

m$\angle$T = $\dpi{100} \fn_phv 71^{\circ}$

Example B

Show why Theorem #1 is true.

$\bigtriangleup$ABC above is an example of an equiangular triangle, where all three angles are equal. Write an equation.

m$\angle$A + m$\angle$B + m$\angle$C = $\dpi{100} \fn_phv 180^{\circ}$

m$\angle$A + m$\angle$A+ m$\angle$A = $\dpi{100} \fn_phv 180^{\circ}$

$\dpi{100} \fn_phv 3\cdot$m$\angle$A = $\dpi{100} \fn_phv 180^{\circ}$

m$\angle$A = $\dpi{100} \fn_phv 60^{\circ}$

If m$\angle$A = $\dpi{100} \fn_phv 60^{\circ}$ , then m$\angle$B = $\dpi{100} \fn_phv 60^{\circ}$ and m$\angle$C = $\dpi{100} \fn_phv 60^{\circ}$.

Example C

Use the picture below to show why Theorem #2 is true.

m$\angle$O = $\dpi{100} \fn_phv 41^{\circ}$ and m$\angle$G = $\dpi{100} \fn_phv 90^{\circ}$ because it is a right angle.

m$\angle$D + m$\angle$O + m$\angle$G = $\dpi{100} \fn_phv 180^{\circ}$

m$\angle$D + $\dpi{100} \fn_phv 41^{\circ}$ + $\dpi{100} \fn_phv 90^{\circ}$ = $\dpi{100} \fn_phv 180^{\circ}$

m$\angle$D + $\dpi{100} \fn_phv 41^{\circ}$ = $\dpi{100} \fn_phv 180^{\circ}$

m$\angle$D = $\dpi{100} \fn_phv 49^{\circ}$

Notice that m$\angle$D+m$\angle$O = $\dpi{100} \fn_phv 90^{\circ}$ because G is a right angle.

If you need help understanding these three examples, click on the link below.

 cK12: Chapter 4 Triangle Sum Theorem Exampleshttp://youtu.be/Qf0gbch5Brk

Concept Problem Revisited

The Bermuda Triangle is an acute scalene triangle. The angle measures are in the picture to the right. Your measured angles should be within a degree or two of these measures. The angles should add up to 180. However, because your measures are estimates using a protractor, they might not exactly add up.

The angle measures in the picture are the actual measures, based off of the distances given, however, your measured angles might be off because the drawing is not to scale.

Guided Practice

1. Determine m 1 in this triangle:

2. Two interior angles of a triangle measure $\dpi{100} \fn_phv 50^{\circ}$ and $\dpi{100} \fn_phv 70^{\circ}$. What is the third interior angle of the triangle?

3. Find the value of x in the triangle below and the measure of each angle.

1. $\dpi{100} \fn_phv 72^{\circ}$ + $\dpi{100} \fn_phv 65^{\circ}$ + m$\angle$1 = $\dpi{100} \fn_phv 180^{\circ}$ .

Solve this equation and you find that m$\angle$1 = $\dpi{100} \fn_phv 43^{\circ}$.

2. $\dpi{100} \fn_phv 50^{\circ}$  + $\dpi{100} \fn_phv 70^{\circ}$ + x = $\dpi{100} \fn_phv 180^{\circ}$.

Solve this equation and you find that the third angle is $60^{\circ}$.

3. All the angles add up to $\fn_phv 180^{\circ}$.

$\dpi{100} \fn_phv (8x-1)^{\circ}+(3x+9)^{\circ}+(3x+4)^{\circ} = 180^{\circ}$

$\dpi{100} \fn_phv (14x+12)^{\circ} = 180^{\circ}$

$\dpi{100} \fn_phv (14x)^{\circ} = 168^{\circ}$

$\dpi{100} \fn_phv x= 12^{\circ}$

Substitute in 12 for x to find each angle.

$\dpi{100} \fn_phv (8\cdot 12) - 1 = 96-1 = 95^{\circ}$

$\dpi{100} \fn_phv (3\cdot 12) + 9 = 27+9 = 45^{\circ}$

$\dpi{100} \fn_phv (3\cdot 12) + 4 = 36+4 = 40^{\circ}$

Click on the links below for additional videos on the Triangle Sum Theorem.

### 06.06 Triangle Sum Theorem - Video Proofs (Math Level 2)

 mathispower4u:Animation: The Sum of the Interior Angles of a Trianglehttps://www.youtube.com/watch?v=Rt0TbTVbKhAmathispower4u:Proving the Triangle Sum Theorem https://www.youtube.com/watch?v=fBPTfmU6XaI

If after completing this topic you can state without hesitation that...

• I can use the Triangle Sum Theorem to solve problems.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

### 06.06 Triangle Sum Theorem - Worksheet (Math Level 2)

 teacher-scored 88 points possible 35 minutes

Activity for this lesson

Complete the attached worksheet.

1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.

Pacing: complete this by the end of Week 3 of your enrollment date for this class.

### 06.07 Perpendicular Bisectors and Isosceles Triangles Geometry (Math Level 2)

 Solve problems using knowledge of the perpendicular bisector of a line segment. Solve problems using knowledge of the base angles of isosceles triangles.

Perpendicular Bisectors and Isosceles Triangles

In previous classes, you may have constructed the perpendicular bisector for segments and for the sides of triangles.

A Perpendicular Bisector is a line that cuts through the exact middle, or the midpoint, of another line at a $90^{\circ}$ angle.

Theorem: The points on the perpendicular bisector of a line segment are equidistant from the endpoints.

You can view the proof of this theorem in the link "Proof: Perpendicular Bisector Theorem" below.

You will be able to determine if a point is on a perpendicular bisector to a line segment if it is the same distance from each endpoint of the line segment. You can do this by using the distance formula with the given points.

Distance Formula: $\fn_phv d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

EXAMPLE: In the image above, point A is located at (-3,8) and point B is located at (7,8). If point C is at (2,12), does point C line on the perpendicular bisector of AB?

To determine if point C is on the perpendicular bisector of AB, we will need to find out if the distance from point A to point C is the same as the distance from point B to point C. We can do that by using the distance formula to find the distance from A to C and from B to C.

Distance from point A to point C:

$\fn_phv d=\sqrt{(2-(-3))^{2}+(12-8)^{2}}$

$\fn_phv d=\sqrt{(2+3)^{2}+(12-8)^{2}}$

$\fn_phv d=\sqrt{(5)^{2}+(4)^{2}}$

$\fn_phv d=\sqrt{25+16}$

$\fn_phv d=\sqrt{41}$

Distance from point B to point C:

$\fn_phv d=\sqrt{(2-7)^{2}+(12-8)^{2}}$

$\fn_phv d=\sqrt{(-5)^{2}+(4)^{2}}$

$\fn_phv d=\sqrt{25+16}$

$\fn_phv d=\sqrt{41}$

The distances are equal and so yes, point C lines on the perpendicular bisector of AB.

The converse is: If a point is equidistant from the endpoints of a segment, this point lies on the perpendicular bisector of the segment.

Consider point P, which is equidistant from points, I and J. We can construct the midpoint of $\dpi{100} \fn_phv \overline{IJ}$ and construct the segment connecting this point and point P. This segment bisects $\dpi{100} \fn_phv \overline{IJ}$. All we need to do is prove that it is perpendicular to $\dpi{100} \fn_phv \overline{IJ}$.

Given point P, such that P is equidistant from the endpoints of $\dpi{100} \fn_phv \overline{IJ}$. Prove P lies on the perpendicular bisector of $\dpi{100} \fn_phv \overline{IJ}$.

Now let's move on to Isosceles Triangles.

The material for the rest of this lesson is from:

What if you were presented with an isosceles triangle and told that its base angles measure $\dpi{100} \fn_phv x^{\circ}$ and $\dpi{100} y^{\circ}$? What could you conclude about x and y? After completing this Concept, you'll be able to apply important properties about isosceles triangles to help you solve problems like this one.

Click on the links below and watch the two videos.

### 06.07 Perpendicular Bisectors and Isosceles Triangles Geometry - Videos (Math Level 2)

 cK12: Chapter 4 Isosceles Triangles A https://vimeo.com/45151570mathispower4u: How To Construct An Isosceles Trianglehttp://www.youtube.com/embed/HlFfbY821AE?wmode=transparent&r...mathispower4u: Proof: Perpendicular Bisector Theorem https://www.youtube.com/watch?v=ttZvFhEq6cg

Guidance

An isosceles triangle is a triangle that has at least two congruent sides. The congruent sides of the isosceles triangle are called the legs. The other side is called the base and the angles between the base and the congruent sides are called base angles. The angle made by the two legs of the isosceles triangle is called the vertex angle.

Investigation: Isosceles Triangle Construction

Tools Needed: pencil, paper, compass, ruler, protractor

Step 1: Using your compass and ruler, draw an isosceles triangle with sides of 3 in, 5 in and 5 in. Draw the 3 in side the  base horizontally 6 inches from the top of the page.

Step 2: Now that you have an isosceles triangle, use your protractor to measure the base angles and the vertex angle.

The base angles should each be $\dpi{100} \fn_phv 72.5^{\circ}$ and the vertex angle should be $\dpi{100} \fn_phv 35^{\circ}$.

We can generalize this investigation into the Base Angles Theorem.

Base Angles Theorem: The base angles of an isosceles triangle are congruent.

To prove the Base Angles Theorem, we will construct the angle bisector through the vertex angle of an isosceles triangle.

Given: Isosceles triangle $\dpi{100} \fn_phv \bigtriangleup DEF$ with $\dpi{100} \fn_phv \dpi{100} \fn_phv \overline{DE}\cong \overline{EF}$

Prove: $\dpi{100} \fn_phv \angle D \cong \angle F$

By constructing the angle bisector, $\dpi{100} \fn_phv \overline{EG}$, we designed two congruent triangles and then used CPCTC to show that the base angles are congruent. Now that we have proven the Base Angles Theorem, you do not have to construct the angle bisector every time. It can now be assumed that base angles of any isosceles triangle are always equal.

Let’s further analyze the picture from step 2 of our proof.

Because $\dpi{100} \fn_phv \bigtriangleup DEG\cong \bigtriangleup FEG$, we know that $\dpi{100} \fn_phv \angle EGD \cong \angle EGF$ by CPCTC. These two angles are also a linear pair, so they are congruent supplements, or 90° each. Therefore, $\dpi{100} \fn_phv \overline{EG} \perp \overline{DF}$. Additionally, $\dpi{100} \fn_phv \overline{DG} \cong \overline{GF}$ by CPCTC, so G is the midpoint of $\dpi{100} \fn_phv \overline{DF}$. This means that $\dpi{100} \fn_phv \overline{EG}$ is the perpendicular bisector of $\dpi{100} \fn_phv \overline{DF}$, in addition to being the angle bisector of $\dpi{100} \fn_phv \angle DEF$.

Isosceles Triangle Theorem: The angle bisector of the vertex angle in an isosceles triangle is also the perpendicular bisector to the base.

The converses of the Base Angles Theorem and the Isosceles Triangle Theorem are both true.

Base Angles Theorem Converse: If two angles in a triangle are congruent, then the opposite sides are also congruent.

So, for a triangle $\dpi{100} \fn_phv \bigtriangleup ABC$, if $\dpi{100} \fn_phv \angle A \cong \angle B$, then $\dpi{100} \fn_phv \overline{CB} \cong \overline{CA}$.  $\dpi{100} \fn_phv \angle C$ would be the vertex angle.

Isosceles Triangle Theorem Converse: The perpendicular bisector of the base of an isosceles triangle is also the angle bisector of the vertex angle.

In other words, if $\dpi{100} \fn_phv \bigtriangleup ABC$ is isosceles, $\dpi{100} \fn_phv \overline{AD} \cong \overline{CB}$ and $\dpi{100} \fn_phv \overline{CD} \cong \overline{DB}$, then $\dpi{100} \fn_phv \angle CAD\cong \angle BAD$.

Click on the link below and watch the video.

 mathispower4u: Using the Properties of Isosceles Triangles to Determine Values http://www.youtube.com/embed/judby1g8udY?wmode=transparent&r...

Example A

Which two angles are congruent?

This is an isosceles triangle. The congruent angles are opposite the congruent sides.

From the arrows we see that $\dpi{100} \fn_phv \angle S\cong \angle U$.

Example B

If an isosceles triangle has base angles with measures of $\dpi{100} \fn_phv 47^{\circ}$, what is the measure of the vertex angle?

Draw a picture and set up an equation to solve for the vertex angle, v.

If an isosceles triangle has a vertex angle with a measure of $\dpi{100} \fn_phv 116^{\circ}$, what is the measure of each base angle?

Draw a picture and set up and equation to solve for the base angles, b. Recall that the base angles are equal.

Click on the link below for help with the Examples above.

### 06.07 Perpendicular Bisectors and Isosceles Triangles Geometry - Videos 2 (Math Level 2)

 cK12: Help with Exampleshttp://vimeo.com/45151571

Guided Practice

1. Find the value of x and the measure of each angle.

2. Find the measure of x.

3. True or false: Base angles of an isosceles triangle can be right angles.

1. Set the angles equal to each other and solve for x:

$\dpi{100} \fn_phv 4x+12^{\circ} = 5x - 3^{\circ}$

$\dpi{100} \fn_phv 12^{\circ} = x - 3^{\circ}$

$\dpi{100} \fn_phv 15^{\circ} = x$

If $\dpi{100} \fn_phv x=15^{\circ}$, then the base angles are 4($\dpi{100} \fn_phv 15^{\circ}$) + $\dpi{100} \fn_phv 12^{\circ}$, or $\dpi{100} \fn_phv 72^{\circ}$. The vertex angle is $\dpi{100} \fn_phv 180^{\circ}-72^{\circ}-72^{\circ} = 36^{\circ}$.

2. The two sides are equal, so set them equal to each other and solve for x.

$\dpi{100} \fn_phv 2x-9 = x+5$

$\dpi{100} \fn_phv 2x = x+14$

$\dpi{100} \fn_phv x = 14$

3. This statement is false. Because the base angles of an isosceles triangle are congruent, if one base angle is a right angle then both base angles must be right angles. It is impossible to have a triangle with two right $\dpi{100} \fn_phv 90^{\circ}$ angles. The Triangle Sum Theorem states that the sum of the three angles in a triangle is $\dpi{100} \fn_phv 180^{\circ}$. If two of the angles in a triangle are right angles, then the third angle must be  and the shape is no longer a triangle.

The following videos can provide additional assistance with this concept as needed:

### 06.07 Perpendicular Bisectors and Isosceles Triangles Geometry - Videos 3 (Math Level 2)

If after completing this topic you can state without hesitation that...

• I can solve problems using knowledge of the base angles of isosceles triangles.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

### 06.07.01 Isosceles Triangles Geometry - Worksheet (Math Level 2)

 teacher-scored 82 points possible 35 minutes

Activity for this lesson

Complete the attached worksheet.

1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.

Pacing: complete this by the end of Week 3 of your enrollment date for this class.

### 06.08 Parallelogram Properties (Math Level 2)

 Solve problems using knowledge of parallelograms’ basic properties.

The material for this lesson is from:

• How can we use our knowledge of triangle congruence criteria to establish other geometry facts?
• For instance, what can we now prove about the properties of parallelograms?

To date, we have defined a parallelogram to be a quadrilateral in which both pairs of opposite sides are parallel. However, we have assumed other details about parallelograms to be true too. We assume that:

• Opposite sides are congruent.
• Opposite angles are congruent.
• Diagonals bisect each other.

Let us examine why each of these properties is true.

Example 1

If a quadrilateral is a parallelogram, then its opposite sides and angles are equal in measure. Complete the diagram and develop an appropriate Given and Prove for this case.

Given:_______________________________________

Prove:_______________________________________

Construction: Label the quadrilateral ABCD, and mark opposite sides as parallel. Draw diagonal $\dpi{100} \fn_phv \overline{BD}$.

Example 2

If a quadrilateral is a parallelogram, then the diagonals bisect each other. Complete the diagram and develop an appropriate Given and Prove for this case.

Given:_______________________________________

Prove:_______________________________________

Construction: Label the quadrilateral ABCD, and mark opposite sides as parallel. Draw diagonals AC and BD.

Now we have set the stage for proving why the properties of parallelograms are in fact true, by extension, these facts will hold for any type of parallelogram, including rectangles, squares, and rhombuses.

Let us look at one last fact concerning rectangles. We set the stage for establishing that the diagonals of general parallelograms bisect each other. Let us now demonstrate that a rectangle has congruent diagonals.

Example 3

If the parallelogram is a rectangle, then the diagonals are equal in length. Complete the diagram and develop an appropriate Given and Prove for this case.

Given:_______________________________________

Prove:_______________________________________

Construction: Label the rectangle GHIJ. Mark opposite sides as parallel, and add small squares at the vertices to indicate $90^{\circ}$ angles. Draw diagonal GI and HJ.

Converse Properties: Now we examine the converse of each of the properties we proved. Begin with the property and prove that the quadrilateral is in fact a parallelogram.

Example 4

If the opposite angles of a quadrilateral are equal, then the quadrilateral is a parallelogram. Draw an appropriate diagram, and provide the relevant Given and Prove for this case.

Given:_______________________________________

Prove:_______________________________________

Construction: Label the quadrilateral ABCD. Mark opposite angles as congruent. Draw diagonal BD. Label $\dpi{100} \fn_phv \angle A$ and $\dpi{100} \fn_phv \angle C$ as $\dpi{100} \fn_phv x^{\circ}$. Label the four angles created by BD as $\dpi{100} \fn_phv r^{\circ}$, $\dpi{100} \fn_phv s^{\circ}$, $\dpi{100} \fn_phv t^{\circ}$, and $\dpi{100} \fn_phv u^{\circ}$.

Example 5

If the opposite sides of a quadrilateral are equal, then the quadrilateral is a parallelogram. Draw an appropriate diagram, and provide the relevant Given and Prove for this case.

Given:_______________________________________

Prove:_______________________________________

Construction: Label the quadrilateral ABCD, and mark opposite sides as equal. Draw diagonal $\dpi{100} \fn_phv \overline{BD}$.

Example 6

If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. Draw an appropriate diagram, and provide the relevant Given and Prove for this case.

Given:_______________________________________

Prove:_______________________________________

Construction: Label the quadrilateral ABCD, and mark opposite sides as equal. Draw diagonals $\dpi{100} \fn_phv \overline{AC}$ and $\dpi{100} \fn_phv \overline{BD}$.

Example 7

If the diagonals of a parallelogram are equal in length, then the parallelogram is a rectangle. Complete the diagram, and develop an appropriate Given and Prove for this case.

Given:_______________________________________

Prove:_______________________________________

Construction: Label the quadrilateral GHIJ. Draw diagonals $\dpi{100} \fn_phv \overline{GI}$ and $\dpi{100} \fn_phv \overline{HJ}$.

Summary

A parallelogram has three basic properties.

1. Opposite sides are congruent
2. Opposite angles are congruent
3. Diagonals bisect each other

In addition, there is this special case:

4. Rectangles are parallelograms with congruent diagonals

Watch the following videos to see the proofs of the three properties. You will be asked to prove the special case in the assignment.

### 06.08 Parallelogram Properties - Videos (Math Level 2)

1.

Given: Parallelogram $\dpi{100} \fn_phv ABCD$, diagonal $\dpi{100} \fn_phv \overline{BD}$, with $\dpi{100} \fn_phv \overline{AB}$$\dpi{100} \fn_phv \left | \right |$$\dpi{100} \fn_phv \overline{CD}$ and $\dpi{100} \fn_phv \overline{BC}$$\dpi{100} \fn_phv \left | \right |$$\dpi{100} \fn_phv \overline{AD}$.

Prove$\dpi{100} \fn_phv \overline{AB}$ $\dpi{100} \fn_phv \cong$ $\dpi{100} \fn_phv \overline{CD}$$\dpi{100} \fn_phv \overline{BC}$ $\dpi{100} \fn_phv \cong$ $\dpi{100} \fn_phv \overline{AD}$$\dpi{100} \fn_phv \angle A$ $\dpi{100} \fn_phv \cong$ $\dpi{100} \fn_phv \angle C$ and $\dpi{100} \fn_phv \angle ABC$ $\dpi{100} \fn_phv \cong$ $\dpi{100} \fn_phv \angle CDA$

2.

Given: Parallelogram $\dpi{100} \fn_phv ABCD$, diagonals $\dpi{100} \fn_phv \overline{BD}$ and $\dpi{100} \fn_phv \overline{AC}$, with $\dpi{100} \fn_phv \overline{AB}$ $\dpi{100} \fn_phv \left | \right |$ $\dpi{100} \fn_phv \overline{CD}$ and $\dpi{100} \fn_phv \overline{BC}$ $\dpi{100} \fn_phv \left | \right |$ $\dpi{100} \fn_phv \overline{AD}$.

Prove: $\dpi{100} \fn_phv \overline{AE}$ $\dpi{100} \fn_phv \cong$ $\dpi{100} \fn_phv \overline{EC}$ and $\dpi{100} \fn_phv \overline{BE}$ $\dpi{100} \fn_phv \cong$ $\dpi{100} \fn_phv \overline{ED}$

3.

Given: Rectangle $\dpi{100} \fn_phv GHIJ$, diagonals $\dpi{100} \fn_phv \overline{GI}$ and $\dpi{100} \fn_phv \overline{HJ}$, with $\dpi{100} \fn_phv \overline{GH}$ $\dpi{100} \fn_phv \left | \right |$ $\dpi{100} \fn_phv \overline{IJ}$ and $\dpi{100} \fn_phv \overline{HI}$ $\dpi{100} \fn_phv \left | \right |$ $\dpi{100} \fn_phv \overline{GJ}$.

Prove: $\dpi{100} \fn_phv \overline{HK}$ $\dpi{100} \fn_phv \cong$ $\dpi{100} \fn_phv \overline{KJ}$ and $\dpi{100} \fn_phv \overline{GK}$ $\dpi{100} \fn_phv \cong$  $\dpi{100} \fn_phv \overline{KI}$

4.

Given: Quadrilateral $\dpi{100} \fn_phv ABCD$ and diagonal $\dpi{100} \fn_phv \overline{BD}$, with $\dpi{100} \fn_phv \angle A$ $\dpi{100} \fn_phv \cong$ $\dpi{100} \fn_phv \angle C$ $\dpi{100} \fn_phv \angle ABC$ $\cong$ $\dpi{100} \fn_phv \angle BCD$

Prove: Quadrilateral $\dpi{100} \fn_phv ABCD$ is a parallelogram ($\dpi{100} \fn_phv \overline{BC}$ $\dpi{100} \fn_phv \left | \right |$ $\dpi{100} \fn_phv \overline{AD}$ and $\dpi{100} \fn_phv \overline{AB}$ $\dpi{100} \fn_phv \left | \right |$ $\dpi{100} \fn_phv \overline{CD}$).

5.

Given: Quadrilateral $\dpi{100} \fn_phv ABCD$ and diagonal $\dpi{100} \fn_phv \overline{BD}$, with $\dpi{100} \fn_phv \overline{AB}$ $\cong$ $\dpi{100} \fn_phv \overline{BC}$ and $\dpi{100} \fn_phv \overline{CD}$ $\cong$ $\dpi{100} \fn_phv \overline{AD}$

Prove: Quadrilateral $\dpi{100} \fn_phv ABCD$ is a parallelogram ($\dpi{100} \fn_phv \overline{BC}$ $\dpi{100} \fn_phv \left | \right |$ $\dpi{100} \fn_phv \overline{AD}$ and $\dpi{100} \fn_phv \overline{AB}$ $\dpi{100} \fn_phv \left | \right |$ $\dpi{100} \fn_phv \overline{CD}$).

6.

Given: Quadrilateral $\dpi{100} \fn_phv ABCD$ and diagonals $\dpi{100} \fn_phv \overline{BD}$ and $\dpi{100} \fn_phv \overline{AC}$, with $\dpi{100} \fn_phv \overline{BC}$ $\cong$ $\dpi{100} \fn_phv \overline{AD}$ and $\dpi{100} \fn_phv \overline{AB}$ $\cong$ $\dpi{100} \fn_phv \overline{CD}$

Prove: Quadrilaterial $\dpi{100} \fn_phv ABCD$ is a parallelogram ($\dpi{100} \fn_phv \overline{BC}$ $\dpi{100} \fn_phv \left | \right |$ $\dpi{100} \fn_phv \overline{AD}$ and $\dpi{100} \fn_phv \overline{AB}$ $\dpi{100} \fn_phv \left | \right |$ $\dpi{100} \fn_phv \overline{CD}$).

7.

Given: Parallelogram $\dpi{100} \fn_phv GHIJ$ and diagonals $\dpi{100} \fn_phv \overline{GI}$ and $\dpi{100} \fn_phv \overline{HJ}$, with $\dpi{100} \fn_phv \overline{GI}$ $\cong$ $\dpi{100} \fn_phv \overline{HJ}$.

Prove: Parallelogram $\dpi{100} \fn_phv GHIJ$ is a rectangle.

If after completing this topic you can state without hesitation that...

• I can solve problems using knowledge of parallelograms’ basic properties.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

### 06.08 Parallelogram Properties - Worksheet (Math Level 2)

 teacher-scored 65 points possible 40 minutes

Activity for this lesson

Complete the attached worksheet.

1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.

Pacing: complete this by the end of Week 3 of your enrollment date for this class.

### 06.09 Midsegments (Math Level 2)

 Use knowledge of triangle midsegments to solve problems.

So far in this class we have only considered six 'parts' of a triangle: three vertices and three sides. However, there are other points and segments that are interesting. This lesson looks at one such segment: the midsegment. If we take any triangle such as $\dpi{100} \fn_phv \bigtriangleup KLM$, shown here, we can find the midpoint of each side (construct a perpendicular bisector). Label the intersections so that H is the midpoint of $\dpi{100} \fn_phv \overline{KL}$, I is the midpoint of  $\dpi{100} \fn_phv \overline{LM}$ and J is the midpoint $\dpi{100} \fn_phv \overline{KM}$.

Draw the segments connecting the midpoints. These are the midsegments of $\dpi{100} \fn_phv \bigtriangleup KLM$.

Notice how similar the small triangles look to the larger triangle. Notice how congruent the small triangles look. Hmm. I say that if $\dpi{100} \fn_phv \bigtriangleup HLI$ ∼ $\dpi{100} \fn_phv \bigtriangleup KLM$, then we should be able to prove this. Likewise, if $\dpi{100} \fn_phv \bigtriangleup HLI$ $\dpi{100} \fn_phv \cong$ $\dpi{100} \fn_phv \bigtriangleup JIM$, we should be able to prove this as well. Let's do these!

### 06.09 Midsegments - Proof (Math Level 2)

In a triangle the midsegment forms a smaller triangle that is similar to the larger triangle. We can prove this. Given ΔKLM, with mid-segment HI connecting sides KL and LM , as shown. Prove that ΔHLI ∼ ΔKLM.

 Statements Reasons 1. $\dpi{100} \fn_phv \overline{HI}$ is the midsegment connecting sides $\dpi{100} \fn_phv \overline{KL}$ and $\dpi{100} \fn_phv \overline{LM}$ in $\dpi{100} \fn_phv \bigtriangleup KLM$. 1. Given 2. H is the midpoint of $\dpi{100} \fn_phv \overline{KL}$, and I is the midpoint of $\dpi{100} \fn_phv \overline{LM}$. 2. Definition of a midsegment. 3. LH + HK = LK, LI + IM = LM, and LH = LK, LI = IM. 3. Definition of a midpoint. 4. LH + LH = LK, LI + LI = LM. 4. Substitution of equals. 5. 2 x LH = LK, 2 x LI = LM 5. Addition 6. $\dpi{100} \fn_phv \frac{LH}{LK}=\frac{1}{2}$, $\dpi{100} \fn_phv \frac{LI}{LM}=\frac{1}{2}$ 6. Division 7. $\dpi{100} \fn_phv \frac{LH}{LK}=\frac{LI}{LM}$ 7. If 2 things equal a third thing, they equal each other. 8. $\dpi{100} \fn_phv \overline{LH} : \overline{LK} = \overline{LI} : \overline{LM}$ 8. Equivalent statement to step 7. 9. $\dpi{100} \fn_phv \angle L \cong \angle L$ 9. An object is congruent to itself. 10. $\dpi{100} \fn_phv \bigtriangleup HLK \sim \bigtriangleup KLM$ 10. RAR

Also, all such smaller triangles are congruent. We can prove this as well. Given ΔKLM, with midsegment $\dpi{100} \fn_phv \overline{HI}$ connecting sides $\dpi{100} \fn_phv \overline{KL}$ and $\dpi{100} \fn_phv \overline{LM}$ and midsegment $\dpi{100} \fn_phv \overline{IJ}$ connecting sides $\dpi{100} \fn_phv \overline{KL}$ and $\dpi{100} \fn_phv \overline{LM}$, as shown. Prove ΔHLI ≅ ΔJIM.

 Statements Reasons 1. $\dpi{100} \fn_phv \bigtriangleup HLI$ and $\dpi{100} \fn_phv \bigtriangleup JIM$ are triangles formed by midsegments of $\dpi{100} \fn_phv \bigtriangleup KLM$. 1. Given 2. $\dpi{100} \fn_phv \bigtriangleup HLI \sim \bigtriangleup KLM$, $\dpi{100} \fn_phv \bigtriangleup JIM \sim \bigtriangleup KLM$. 2. In a triangle, a midsegment forms a triangle that is similar to the first triangle. 3. $\dpi{100} \fn_phv \angle L \cong \angle JIM, \angle LIH \simeq \angle M$. 3. For similar triangles, corresponding angles are congruent. 4. $\dpi{100} \fn_phv I$ is the midpoint of $\dpi{100} \fn_phv \overline{LM}$. 4. Definition of a midsegment. 5. $\dpi{100} \fn_phv \overline{IL} \cong \overline{MI}$ 5. Definition of a midpoint. 6. $\dpi{100} \fn_phv \bigtriangleup HLI \cong \bigtriangleup JIM$. 6. ASA

Cool. Can you prove that ΔHJI ≅ ΔHLI?

In summary,

1. Midsegment Theorem: The segment that joins the midpoints of a pair of sides of a triangle is:

a. parallel to the third side.
b. half as long as the third side.

2. In a triangle the midsegment forms a smaller triangle that is similar to the larger triangle.

3. All 4 smaller triangles created by the midsegments are congruent to each other.

Click on the links below and watch the video on midsegments.

### 06.09 Midsegments - Video (Math Level 2)

 mathispower4u: Introduction to the Midsegments of a Trianglehttps://www.youtube.com/watch?v=FmlkQQVOSs4&feature=youtu.bemathispower4u: Examples: Determining Unknown Values Using Properties of the Midsegments of a Triangle https://www.youtube.com/watch?v=P5SEb7xpTjE

If after completing this topic you can state without hesitation that...

• I can use my knowledge of triangle midsegments to solve problems.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

### 06.09 Midsegments - Worksheet (Math Level 2)

 teacher-scored 70 points possible 40 minutes

Activity for this lesson

Complete the attached worksheet.

1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.

Pacing: complete this by the end of Week 4 of your enrollment date for this class.

### 06.10 Centroid of a Triangle (Math Level 2)

 Use knowledge of triangle centroids and medians to solve problems.

The material for this lesson is taken from en.wikipedia.org/wiki/Median_%28geometry%29 under a Creative Commons Attribution-ShareAlike License.

In geometry, a median of a triangle is a line segment joining a vertex to the midpoint of the opposing side. Because the median goes to the midpoint of a side of triangle, it will divide that side in half evenly! Every triangle has exactly three medians: one running from each vertex to the opposite side. In the case of isosceles and equilateral triangles, a median bisects any angle at a vertex whose two adjacent sides are equal in length.

Each median of a triangle passes through the triangle's centroid, which is the center of mass of an object of uniform density in the shape of the triangle.1 Thus the object would balance on the intersection point of the medians. Each median divides the area of the triangle in half; hence the name.

The centroid divides each median into parts in the ratio 2:1, with the centroid being twice as close to the midpoint of a side as it is to the opposite vertex.

This is also know as the Concurrency of Medians Theorem: The medians of a triangle will intersect in a point that is two-thirds of the distance from the vertices to the midpoint of the opposite side.

Consider △ABC with midpoints of the sides located at F, D, and E and the point of concurrency of the medians at

the centroid,0.

The theorem states that CO = $\fn_phv \small \frac{2}{3}$CD, AO = $\fn_phv \small \frac{2}{3}$AE, and BO = $\fn_phv \small \frac{2}{3}$BF.

Example 1.

Use The Concurrency of Medians Theorem and the image above to find the lengths of the indicated segments in the following triangle that has medians AE,CD, and BF as indicated.

1. If AE = 12 , then AO = ____ and OE = ____.

AO = $\fn_phv \small \frac{2}{3}$ · 12 = 8

OE = $\fn_phv \small \frac{2}{3}$ · 12 = 4

2. If CO = 6 . then OD = ____ and CD = _____. We will start by finding CD.

CO = $\fn_phv \small \frac{2}{3}$CD

6 = $\fn_phv \small \frac{2}{3}$CD

9=CD

Now for OD,

OD = $\fn_phv \small \frac{1}{3}$CD

OD = $\fn_phv \small \frac{1}{3}$(9)

OD=3

Click on the link below to view the video.

### 06.10 Centroid of a Triangle - Video (Math Level 2)

 mathispower4u: The Medians of a Triangle https://www.youtube.com/watch?v=CUzC6rySeN0&feature=youtu.be

Concurrency of Medians Theorem: The medians intersect in a point that is two-thirds the distance from the vertices to the midpoint on the opposite sides.

Click on the link below to view the video.

### 06.10 Centroid of a Triangle - Video2 (Math Level 2)

 mathispower4u: Examples: Using the Properties of the Medians of a Triangle to Solve for Unknown Values https://www.youtube.com/watch?v=aaIX1rUdrgs&feature=youtu.be

If after completing this topic you can state without hesitation that...

• I can use my knowledge of triangle centroids and medians to solve problems.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

### 06.10 Centroid of a Triangle - Worksheet (Math Level 2)

 teacher-scored 58 points possible 35 minutes

Activity for this lesson

Complete the attached worksheet.

1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.

Pacing: complete this by the end of Week 4 of your enrollment date for this class.

### 06.10 Unit 6 review - Quiz (Math Level 2)

 computer-scored 85 points possible 40 minutes
Pacing: complete this by the end of Week 4 of your enrollment date for this class.

### 07.00 Similarity and Right Triangles (Math Level 2)

 The goal of all math classes is for students to become mathematically proficient. In particular, students need to be able to: Make sense of problems and persevere in solving them. Reason abstractly and quantitatively. Construct viable arguments and critique the reasoning of others. Model with mathematics. Use appropriate tools strategically. Attend to precision. Look for and make use of structure. Look for and express regularity in repeated reasoning. As you work through this unit, keep these proficiencies in mind! This unit focuses on the geometric concept of similarity and on the geometry of right triangles. At the conclusion of this unit you must be able to state without hesitation that: I can solve problems concerning dilations. I can answer questions about similar polygons. I can identify equilateral, isosceles, scalene, acute, right, and obtuse triangles. I can identify whether triangles are similar, congruent, or neither. I can identify corresponding sides of congruent and similar triangles. I can find the missing measurements in a pair of similar triangles. I can solve application problems involving similar triangles. I can use my knowledge of similar triangles to solve problems. I can solve problems using Angle-Angle similarity. I can prove the midsegment of a triangle is parallel to the third side. I can prove the medians of a triangle meet at a point. I can prove the Pythagorean Theorem.

### 07.01 Similarity and Dilations (Math Level 2)

The material for this lesson is taken from:

A transformation is an operation that moves, flips, or changes a figure to create a new figure. Transformations that preserve size are rigid and ones that do not are non-rigid. A dilation makes a figure larger or smaller, but has the same shape as the original. In other words, the dilation is similar to the original. All dilations have a center and a scale factor. The center is the point of reference for the dilation like the vanishing point in a perspective drawing and scale factor tells us how much the figure stretches or shrinks. A scale factor is typically labeled k and is always greater than zero. Also, if the original figure is labeled A, for example, the dilation would be A'. The ‘ indicates that it is a copy. This tic mark is said “prime,” so A’ is read “A prime.” A second dilation would be A’’, read “A double-prime.”

If the dilated image is smaller than the original, then the scale factor is 0 < k < 1.

If the dilated image is larger than the original, then the scale factor is k > 1.

Example A

The center of dilation is P and the scale factor is 3. Find Q’.

If the scale factor is 3 and Q is 6 units away from P, then Q’ is going to be 6 x 3 = 18 units away from P. Because we are only dilating a point, the dilation will be collinear with the original and center.

Example B

Using the picture above, change the scale factor to $\fn_phv \small \frac{1}{3}$. Find Q’’.

Now the scale factor is $\fn_phv \small \frac{1}{3}$, so Q’’ is going to be  the distance away from P as Q is. In other words, Q’’ is going to be $\fn_phv \small 6$ • $\fn_phv \small \frac{1}{3}$ = $\fn_phv \small 2$ units away from P. Q’’ will also be collinear with Q and center.

Example C

KLMN is a rectangle with length 12 and width 8. If the center of dilation is K with a scale factor of 2, draw K’L’M’N’.

End of lesson.

In summary, to dilate a figure, you multiply it's measurements by the scale factor. If you have the coordinates of the vertices, to find a dilation, you multiply both the x and y value of the coordinats by the scale factor.

SIMLIARITY

Just like congruence, similarity is a technical term. Two figures are similar if corresponding angles are congruent, and the lengths of corresponding sides have the same ratio. Does this sound like the result of a dilation? When we dilate a figure, the angles do not change, and all the sides are scaled by the same factor. So, yes, a dilation will produce a new figure that is, by definition, similar to the original figure.

Consider this example:

Figure ABCD $\sim$ A'B'C'D' The symbol $\sim$ states that figures are similar. This means the corresponding angles are congruent and the corresponding sides have the same ratio.

$\fn_phv \angle A = \angle A'$, $\fn_phv \angle B = \angle B'$, $\fn_phv \angle C = \angle C'$ and $\fn_phv \angle D = \angle D'$

$\fn_phv \frac{A'B'}{AB}=\frac{B'C'}{BC}=\frac{C'D'}{CD}=\frac{A'D'}{AD}$

Therefore, if we use rigid motions and a dilation to make two figures coincide, then these figures are similar. Again, let's look at a few examples.

The turquoise figure looks like it may be similar to the blue figure, so start by rotating the turquoise figure so that the long sides are parallel.

That looks very promising. Move the blue figure so that the bottom left vertices of each figure align. Now draw the lines connecting corresponding vertices.

The lines all go through the bottom left vertex. Good sign. The bases are collinear and so are the left sides. In addition, by construction, we can prove that the remaining corresponding line segments are parallel. These are the properties of a dilation, so without actually measuring the ratio of the segments, we can prove that the figures are similar. Yeah! Okay, I said "by construction." How do you prove lines are parallel by construction? Remember how to construct parallel lines? We are given lines a and b and I want to prove that line a is parallel to line b. Start by constructing line c parallel to a, through a point on line b. If line c is collinear to line b, then line b must be parallel to line a. I will use the lines through the vertices, since I already drew them.

All lines I constructed are parallel to the sides of the blue figure through points on the turquoise figure and are collinear with the turquoise figure; Therefore, these figures are similar. Disproving similarity is a bit easier. Consider the following.

Start by rotating the red pentagon so that the base of the red pentagon aligns with the base of the magenta pentagon.

So, none of these look right to me. Let's try reflecting each of these across the perpendicular bisector of the base segment.

Number 3 looks good. We got here by rotating the original figure by more than three fifths of a circle, then reflecting it across the perpendicular bisector of the new base. There is a single axis of reflection which will result in the same image, but we are not going to look for it.

Next, translate the red figure so that it is inside of the magenta figure and bottom left vertices align.

At this point, it should be obvious these figures are not similar. However, let's finish the "proof." Next draw the lines connecting the vertices. These should all go through the bottom left vertex.

These lines do not; Therefore, the red figure cannot be similar to the magenta figure. Are you ready to try this on your own?

Click on the link below and watch the video.

### 07.01.01 Similarity and Dilations (Math Level 2)

 cK12: Chapter 7 Dilationshttp://vimeo.com/45543577Using Similar Polygonshttps://vimeo.com/56959202

Guided Practice

1. Find the perimeters of KLMN and K’L’M’N’. Compare this ratio to the scale factor.

$\dpi{100} \fn_phv \bigtriangleup ABC$ is a dilation of $\dpi{100} \fn_phv \bigtriangleup DEF$. If P is the center of dilation, what is the scale factor?

3. Find the scale factor, given the corresponding sides. In the diagram, the black figure is the original and P is the center of dilation.

1. The perimeter of KLMN = 12 + 8 + 12 + 8 = 40. The perimeter of K’L’M’N’ = 24 + 16 + 24 + 16 = 80. The ratio is 80:40, which reduces to 2:1, which is the same as the scale factor.
2. Because $\bigtriangleup$ABC is a dilation of $\bigtriangleup$DEF, then $\bigtriangleup$ABC ~ $\bigtriangleup$DEF. The scale factor is the ratio of the sides. Since $\bigtriangleup$ABC is smaller than the original, $\bigtriangleup$DEF, the scale factor is going to be less than one, $\dpi{100} \fn_phv \frac{12}{20}=\frac{3}{5}$. If  $\bigtriangleup$DEF was the dilated image, the scale factor would have been $\dpi{100} \fn_phv \frac{5}{3}$.
3. Since the dilation is smaller than the original, the scale factor is going to be less than one $\dpi{100} \fn_phv \frac{8}{20}=\frac{2}{5}$.

Explore More

In the two questions below, you are told the scale factor. Determine the dimensions of the dilation. In each diagram, the black figure is the original and P is the center of dilation.

1. k = 4

2.     k = $\dpi{100} \fn_phv \frac{1}{3}$

In the question below, find the scale factor, given the corresponding sides. In the diagram, the black figure is the original and P is the center of dilation.

3.

4. Find the perimeter of both triangles in #1. What is the ratio of the perimeters?

5. Writing What happens if k = 1?

Construction We can use a compass and straight edge to construct a dilation as well. Copy the diagram below.

6. Set your compass to be CG and use this setting to mark off a point 3 times as far from C as G is. Label this point G’. Repeat this process for CO and CD to find O’ and D’.

7. Connect G’, O’ and D’ to make $\bigtriangleup$D'O'G'. Find the ratios, $\dpi{100} \fn_phv \frac{D'O'}{DO}$, $\dpi{100} \fn_phv \frac{O'G'}{OG}$, $\dpi{100} \fn_phv \frac{G'D'}{GD}$.

8. What is the scale factor of this dilation?

9. Describe how you would dilate the figure by a scale factor of 4.

10. Describe how you would dilate the figure by a scale factor of $\dpi{100} \fn_phv \frac{1}{2}$.

11. The scale factor between two shapes is 1.5. What is the ratio of their perimeters?

12. The scale factor between two shapes is 1.5. What is the ratio of their areas? Hint: Draw an example and calculate what happens.

13. Suppose you dilate a triangle with side lengths 3, 7, and 9 by a scale factor of 3. What are the side lengths of the image?

14. Suppose you dilate a rectangle with a width of 10 and a length of 12 by a scale factor of $\dpi{100} \fn_phv \frac{1}{2}$. What are the dimensions of the image?

15. Find the areas of the rectangles in #14. What is the ratio of their areas?

If after completing this topic you can state without hesitation that...

• I can solve problems concerning dilations.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

### 07.01.02 Similarity and Dilations - Worksheet (Math Level 2)

 teacher-scored 94 points possible 35 minutes

Activity for this lesson

Complete the attached worksheet.

1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.

Pacing: complete this by the end of Week 5 of your enrollment date for this class.

### 07.02 Similar Triangles (Math Level 2)

 Identify whether triangles are similar, congruent, or neither, corresponding sides of congruent and similar triangles, find the missing measurements in a pair of similar triangles, and solve application problems involving similar triangles. Use knowledge of similar triangles to solve problems.

The material for this lesson if from NROC Developmental Math Unit 7 Lesson 1 Topic 3 under a Creative Commons Attribution Non Commercial Share Alike 3.0 Unported license.

Identifying Congruent and Similar Triangles

Two triangles are congruent if they are exactly the same size and shape. In congruent triangles, the measures of corresponding angles and the lengths of corresponding sides are equal. Consider the two triangles shown below:

Since both $\angle$B and $\angle$E are right angles, these triangles are right triangles. Let’s call these two triangles $\bigtriangleup$ABC and $\bigtriangleup$DEF. These triangles are congruent if every pair of corresponding sides has equal lengths and every pair of corresponding angles has the same measure.

The corresponding sides are opposite the corresponding angles.

$\bigtriangleup$ABC and $\bigtriangleup$DEF are congruent triangles as the corresponding sides and corresponding angles are equal

Let’s take a look at another pair of triangles. Here are the triangles $\bigtriangleup$ABC and $\bigtriangleup$RST.

These two triangles are surely not congruent because $\bigtriangleup$RST is clearly smaller in size than $\bigtriangleup$ABC. But, even though they are not the same size, they do resemble one another. They are the same shape. The corresponding angles of these triangles look like they might have the same exact measurement, and if they did they would be congruent angles and we would call the triangles similar triangles.

Congruent angles are marked with hash marks, just as congruent sides are.

 Image showing angle measurements of both triangles. Image showing triangles ABC and RST using hash marks to show angle congruency.

We can also show congruent angles by using multiple bands within the angle, rather than multiple hash marks on one band.  Below is an image using multiple bands within the angle.

Image showing triangles ABC and RST using bands to show angle congruency.

Like we learned in lesson 07.01, if the corresponding angles of two triangles have the same measurements they are called similar triangles. This name makes sense because they have the same shape, but not necessarily the same size. When a pair of triangles is similar, the corresponding sides are proportional to one another. That means that there is a consistent scale factor that can be used to compare the corresponding sides. In the previous example, the side lengths of the larger triangle are all 1.4 times the length of the smaller. So, similar triangles are proportional to one another.

Just because two triangles look similar does not mean they are similar triangles in the mathematical sense of the word. Checking that the corresponding angles have equal measure is one way of being sure the triangles are similar.

Corresponding Sides of Similar Triangles

There is another method for determining similarity of triangles that involves comparing the ratios of the lengths of the corresponding sides.

If the ratios of the pairs of corresponding sides are equal, the triangles are similar.

Consider the two triangles below.

$\bigtriangleup$ABC is not congruent to $\bigtriangleup$DEF because the side lengths of $\bigtriangleup$DEF are longer than those of $\bigtriangleup$ABC. So, are these triangles similar? If they are, the corresponding sides should be proportional.

Since these triangles are oriented in the same way, you can pair the left, right, and bottom sides: $\dpi{100} \fn_phv \overline{AB}$ and $\dpi{100} \fn_phv \overline{DE}$, $\dpi{100} \fn_phv \overline{BC}$ and $\dpi{100} \fn_phv \overline{EF}$, $\dpi{100} \fn_phv \overline{AC}$ and $\dpi{100} \fn_phv \overline{DE}$. You might call these the two shortest sides, the two longest sides, and the two leftover sides and arrived at the same ratios. Now we will look at the ratios of their lengths.

$\dpi{100} \fn_phv \frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$

Substituting the side length values into the proportion, you see that it is true:

$\dpi{100} \fn_phv \frac{3}{9}=\frac{4}{12}=\frac{6}{18}$

If the corresponding sides are proportional, then the triangles are similar. Triangles ABC and DEF are similar, but not congruent.

Let’s use this idea of proportional corresponding sides to determine whether two more triangles are similar.

Problem: Determine if the triangles below are similar by seeing if their corresponding sides are proportional.

 First determine the corresponding sides, which are opposite corresponding angles. $\dpi{100} \fn_phv \overline{CA} \leftrightarrow \overline{FD}$, $\dpi{100} \fn_phv \overline{AB} \leftrightarrow \overline{DE}$, $\dpi{100} \fn_phv \overline{BC} \leftrightarrow \overline{EF}$ Write the corresponding side lengths as ratios. $\dpi{100} \fn_phv \frac{CA}{FD}=\frac{AB}{DE}=\frac{BC}{EF}$ Substitute the side lengths into the ratios, and determine if the ratios of the corresponding sides are equivalent. They are, so the triangles are similar. $\dpi{100} \fn_phv \frac{10}{5}=\frac{6}{3}=\frac{14}{7}$ $\dpi{100} \fn_phv {\color{Blue} \bigtriangleup }{\color{Blue} ABC}$ and $\dpi{100} \fn_phv {\color{Blue} \bigtriangleup }{\color{Blue} DEF}$ are similar. $\dpi{100} \fn_phv {\color{Blue} 2=2=2}$

The mathematical symbol ~ means “is similar to”. So, you can write $\dpi{100} \fn_phv \bigtriangleup ABC$ is similar to $\dpi{100} \fn_phv \bigtriangleup DEF$ as $\dpi{100} \fn_phv \bigtriangleup ABC$ ~ $\dpi{100} \fn_phv \bigtriangleup DEF$.

Determine whether the two triangles are similar, congruent, or neither.

A. $\dpi{100} \fn_phv \bigtriangleup ABC$ and $\dpi{100} \fn_phv \bigtriangleup DEF$ are congruent.

B. $\dpi{100} \fn_phv \bigtriangleup ABC$ and $\dpi{100} \fn_phv \bigtriangleup DEF$ are similar.

C. $\dpi{100} \fn_phv \bigtriangleup ABC$ and $\dpi{100} \fn_phv \bigtriangleup DEF$ are similar and congruent.

D. $\dpi{100} \fn_phv \bigtriangleup ABC$ and $\dpi{100} \fn_phv \bigtriangleup DEF$ are neither similar nor congruent.

A. $\dpi{100} \fn_phv \bigtriangleup ABC$ and $\dpi{100} \fn_phv \bigtriangleup DEF$ are congruent.

Incorrect. Congruent triangles have corresponding sides of equal length and corresponding angles of equal measure. They are the same exact size and shape.  is equilateral and  is isosceles, so they are not the same exact shape. The correct answer is $\dpi{100} \fn_phv \bigtriangleup ABC$ and $\dpi{100} \fn_phv \bigtriangleup DEF$ are neither similar nor congruent.

B. $\dpi{100} \fn_phv \bigtriangleup ABC$ and $\dpi{100} \fn_phv \bigtriangleup DEF$ are similar.

Incorrect. The ratios of the corresponding sides are not equal, so the triangles cannot be similar:

$\dpi{100} \fn_phv \frac{6.5}{5}=\frac{6.5}{3}\neq \frac{5}{5}.$

C. $\dpi{100} \fn_phv \bigtriangleup ABC$ and $\dpi{100} \fn_phv \bigtriangleup DEF$ are similar and congruent.

Incorrect. All congruent triangles are similar, but these triangles are not congruent. Congruent triangles have corresponding sides of equal length and corresponding angles of equal measure. $\dpi{100} \fn_phv \bigtriangleup ABC$ is equilateral and  $\dpi{100} \fn_phv \bigtriangleup DEF$ is isosceles, so they are not the same exact shape.

D. $\dpi{100} \fn_phv \bigtriangleup ABC$ and $\dpi{100} \fn_phv \bigtriangleup DEF$ are neither similar nor congruent.

Correct. The corresponding angle measures are not known to be equal as shown by the absence of congruence marks on the angles. Also, the ratios of the corresponding sides are not equal:$\dpi{100} \fn_phv \frac{6.5}{5}=\frac{6.5}{3}\neq \frac{5}{5}.$

Finding Missing Measurements in Similar Triangles

You can find the missing measurements in a triangle if you know some measurements of a similar triangle. Let’s look at an example.

Problem: $\dpi{100} \fn_phv \bigtriangleup ABC$ and $\dpi{100} \fn_phv \bigtriangleup DEF$ are similar triangles.

What is the length of side BC?

 In similar triangles, the ratios of corresponding sides are proportional. Set up a proportion of two ratios, one that includes the missing side. $\dpi{100} \fn_phv \frac{BC}{YZ}=\frac{AB}{XY}$ Substitute in the known side lengths for the side names in the ratio. Let the unknown side length be n. $\dpi{100} \fn_phv \frac{n}{2}=\frac{6}{1.5}$ Solve for n using cross multiplication. $\dpi{100} \fn_phv 2\cdot 6=1.5\cdot n$ $\dpi{100} \fn_phv 12=1.5\cdot n$ $\dpi{100} \fn_phv 8=n$ The missing length is 8 units.

This process is fairly straightforward—but be careful that your ratios represent corresponding sides, recalling that corresponding sides are opposite corresponding angles.

Solving Application Problems Involving Similar Triangles

Applying knowledge of triangles, similarity, and congruence can be very useful for solving problems in real life. Just as you can solve for missing lengths of a triangle drawn on a page, you can use triangles to find unknown distances between locations or objects.

Let’s consider the example of two trees and their shadows. Suppose the sun is shining down on two trees, one that is 6 feet tall and the other whose height is unknown. By measuring the length of each shadow on the ground, you can use triangle similarity to find the unknown height of the second tree.

First, let’s figure out where the triangles are in this situation! The trees themselves create one pair of corresponding sides. The shadows cast on the ground are another pair of corresponding sides. The third side of these imaginary similar triangles runs from the top of each tree to the tip of its shadow on the ground. This is the hypotenuse of the triangle.

If you know that the trees and their shadows form similar triangles, you can set up a proportion to find the height of the tree.

Problem: When the sun is at a certain angle in the sky, a 6-foot tree will cast a 4-foot shadow. How tall is a tree that casts an 8-foot shadow?

 The angle measurements are the same, so the triangles are similar triangles. Since they are similar triangles, you can use proportions to find the size of the missing side Set up a proportion comparing the heights of the trees and the lengths of their shadows. $\dpi{100} \fn_phv \frac{Tree 1}{Tree 2}=\frac{Shadow 1}{Shadow 2}$ Substitute in the known lengths. Call the missing tree height h. $\dpi{100} \fn_phv \frac{6}{4}=\frac{4}{8}$ Solve for h using cross-multiplication. $\dpi{100} \fn_phv 6\cdot 8 = 4h$ $\dpi{100} \fn_phv 48 = 4h$ $\dpi{100} \fn_phv 12 = h$ The tree is 12 feet tall.

The material for this portion of the lesson is from

Can you find any similar triangles in the picture below?

Click on the links below and watch the video:

Summary

Triangles are one of the basic shapes in the real world. Triangles can be classified by the characteristics of their angles and sides, and triangles can be compared based on these characteristics. The sum of the measures of the interior angles of any triangle is 180º. Congruent triangles are triangles of the same size and shape. They have corresponding sides of equal length and corresponding angles of the same measurement. Similar triangles have the same shape, but not necessarily the same size. The lengths of their sides are proportional. Knowledge of triangles can be a helpful in solving real-world problems.

If after completing this topic you can state without hesitation that...

• I can identify whether triangles are similar, congruent, or neither.
• I can identify corresponding sides of congruent and similar triangles.
• I can find the missing measurements in a pair of similar triangles.
• I can solve application problems involving similar triangles.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

### 07.02 Similar Triangles - Videos (Math Level 2)

 mathispower4u: Solving for Unknown Values Using the Properties of Similar Triangles https://www.youtube.com/watch?v=ueYQII8Lousmathispower4u: Indirect Measurement Using Similar Triangles https://www.youtube.com/watch?v=LhEe0kB4QIsmathispower4u: The Triangle Proportionality Theorem http://www.youtube.com/embed/YLtQmuTY5rI?wmode=transparent&r...mathispower4u: Using the Properties of the Triangle Proportionality Theorem to Solve for Unknown Values https://www.youtube.com/watch?v=5w59-7mJ07Ymathispower4u: The Triangle Angle Bisector Theoremhttps://www.youtube.com/watch?v=82qPgY5pYQgmathispower4u: Using the Properties of the Triangle Angle Bisector Theorem to Determine Unknown Values https://www.youtube.com/watch?v=kzjwpOy1j_8

Guidance

If two triangles are similar, then their corresponding angles are congruent and their corresponding sides are proportional. There are many theorems about triangles that you can prove using similar triangles.

1. Triangle Proportionality Theorem: A line parallel to one side of a triangle divides the other two sides of the triangle proportionally. This theorem and its converse will be explored and proved in Example A, Example B, and the practice exercises.
2. Triangle Angle Bisector Theorem: The angle bisector of one angle of a triangle divides the opposite side of the triangle into segments proportional to the lengths of the other two sides of the triangle. This theorem will be explored and proved in Example C.
3. Pythagorean Theorem: For a right triangle with legs a and b and hypotenuse c, a2 + b2 = c2. This theorem will be explored and proved in the Guided Practice problems.

Example A

Prove: $\dpi{100} \fn_phv \bigtriangleup ADE$ ~ $\dpi{100} \fn_phv \bigtriangleup ABC$

Solution: The two triangles share $\angle$A. Because $\dpi{100} \fn_phv \overline{DE}\left | \right |\overline{BC}$, corresponding angles are congruent. Therefore, . The two triangles have two pairs of congruent angles. Therefore, $\dpi{100} \fn_phv \bigtriangleup ADE$ ~ $\dpi{100} \fn_phv \bigtriangleup ABC$ by AA similarity.

Example B

Use your result from Example A to prove that $\dpi{100} \fn_phv \frac{AB}{AD} = \frac{AC}{AE}$. Then, use algebra to show that $\dpi{100} \fn_phv \frac{DB}{AD} = \frac{BC}{AE}$.

Solution: $\bigtriangleup ADE$$\bigtriangleup ABC$ which means that corresponding sides are proportional. Therefore, $\dpi{100} \fn_phv \frac{AB}{AD} = \frac{AC}{AE}$. Now, you can use algebra to show that the second proportion must be true. Remember that AB = AD + DB  and AC = AE + EC.

You have now proved the triangle proportionality theorem: a line parallel to one side of a triangle divides the other two sides of the triangle proportionally.

Example C

Consider $\dpi{100} \fn_phv \bigtriangleup ABC$ with $\dpi{100} \fn_phv \overline{AE}$ the angle bisector of $\dpi{100} \fn_phv \angle BAC$ and point $\dpi{100} \fn_phv D$ constructed so that $\dpi{100} \fn_phv \overline{DE}\left | \right |\overline{AE}$. Prove that $\dpi{100} \fn_phv \frac{EB}{BA}=\frac{BC}{CA}$.

Solution: By the triangle proportionality theorem, $\dpi{100} \fn_phv \frac{EB}{EC}=\frac{BA}{AD}$. Multiply both sides of this proportion by $\dpi{100} \fn_phv \frac{EC}{BA}$.

Now all you need to show is that AD = CA in order to prove the desired result.

• Because $\dpi{100} \fn_phv \overline{AE}$ is the angle bisector of $\dpi{100} \fn_phv \angle BAC$, $\dpi{100} \fn_phv \angle BAE\cong \angle EAC$.
• Because $\dpi{100} \fn_phv \overline{DC}\left | \right |\overline{AE}$, $\dpi{100} \fn_phv \angle BAE\cong \angle BDC$ (corresponding angles).
• Because $\dpi{100} \fn_phv \overline{DC}\left | \right |\overline{AE}$, $\dpi{100} \fn_phv \angle EAC\cong \angle DCA$ (alternate interior angles).
• Thus, $\dpi{100} \fn_phv \angle BDC\cong \angle DCA$ by the transitive property.

Therefore, $\bigtriangleup ADC$ is isosceles because its base angles are congruent and it must be true that $\dpi{100} \fn_phv \overline{AD}\cong \overline{CA}$. This means that AD = CA. Therefore:$\dpi{100} \fn_phv \frac{EB}{BA}\cong \frac{EC}{CA}$.

This proves the triangle angle bisector theorem: the angle bisector of one angle of a triangle divides the opposite side of the triangle into segments proportional to the lengths of the other two sides of the triangle.

Concept Problem Revisited

There are three triangles in this picture: $\dpi{100} \fn_phv \bigtriangleup BAC$, $\dpi{100} \fn_phv \bigtriangleup BCD$, $\dpi{100} \fn_phv \bigtriangleup CAD$. All three triangles are right triangles so they have one set of congruent angles (the right angle). $\dpi{100} \fn_phv \bigtriangleup BAC$ and $\dpi{100} \fn_phv \bigtriangleup BCD$ share $\dpi{100} \fn_phv \angle B$, so $\dpi{100} \fn_phv \bigtriangleup BAC$$\dpi{100} \fn_phv \bigtriangleup BCD$ by AA similarity. Similarly, $\dpi{100} \fn_phv \bigtriangleup BAC$ and $\dpi{100} \fn_phv \bigtriangleup CAD$ share $\dpi{100} \fn_phv \angle C$ so $\dpi{100} \fn_phv \bigtriangleup BAC$~$\dpi{100} \fn_phv \bigtriangleup CAD$ by AA similarity. By the transitive property, all three triangles must be similar to one another.

Vocabulary

A similarity transformation is one or more rigid transformations followed by a dilation.

Two figures are similar if a similarity transformation will carry one figure to the other. Similar figures will always have corresponding angles congruent and corresponding sides proportional.

The Pythagorean Theorem states that for a right triangle with legs a and b and hypotenuse c, a2 + b2 = c2.

An angle bisector divides an angle into two congruent angles.

An angle bisector divides an angle into two congruent angles.

Guided Practice

The large triangle above has sides a, b, and c. Side c has been divided into two parts: y and c – y. In the Concept Problem Revisited you showed that the three triangles in this picture are similar.

1. Explain why $\dpi{100} \fn_phv \frac{a}{c}=\frac{c-y}{a}$.
2. Explain why $\dpi{100} \fn_phv \frac{b}{c}=\frac{y}{b}$.
3. Use the results from #1 and #2 to show that $\dpi{100} \fn_phv a^{2}+b^{2}=c^{2}$.

When triangles are similar, corresponding sides are proportional. Carefully match corresponding sides and you see that $\dpi{100} \fn_phv \frac{a}{c}=\frac{c-y}{a}$.

When triangles are similar, corresponding sides are proportional. Carefully match corresponding sides and you see that $\dpi{100} \fn_phv \frac{b}{c}=\frac{y}{b}$.

Cross multiply to rewrite each equation. Then, add the two equations together.

You have just proved the Pythagorean Theorem using similar triangles!

Summary

Triangles are one of the basic shapes in the real world. Triangles can be classified by the characteristics of their angles and sides, and triangles can be compared based on these characteristics. The sum of the measures of the interior angles of any triangle is 180º. Congruent triangles are triangles of the same size and shape. They have corresponding sides of equal length and corresponding angles of the same measurement. Similar triangles have the same shape, but not necessarily the same size. The lengths of their sides are proportional. Knowledge of triangles can be a helpful in solving real-world problems.

If after completing this topic you can state without hesitation that...

• I can identify whether triangles are similar, congruent, or neither.
• I can identify corresponding sides of congruent and similar triangles.
• I can find the missing measurements in a pair of similar triangles.
• I can solve application problems involving similar triangles.
• I can use my knowledge of similar triangles to solve problems.

…you are ready for the assignment.  Otherwise, go back and review the material before moving on.

### 07.02 Similar Triangles - Worksheet (Math Level 2)

 teacher-scored 68 points possible 40 minutes

Activity for this lesson

Complete the attached worksheet.

1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.

Pacing: complete this by the end of Week 5 of your enrollment date for this class.

### 07.03 Angle-Angle Similarity (Math Level 2)

 Solve problems using Angle-Angle similarity.

We learned in Secondary Math 1 that two triangles are congruent if one of four conditions are met:

1. SSS – All three sides are congruent
2. SAS – Two side and their included angle are congruent.
3. ASA – Two angles and their included side are congruent.
4. AA – Two corresponding angles are congruent.

We are going to look at the fourth case as it relates to similarity.

http://en.wikibooks.org/wiki/Geometry/Similar_Triangles

Click on the link below for the rest of the lesson:

### 07.03 Angle-Angle Similarity - Extra Video (Math Level 2)

If after completing this topic you can state without hesitation that...

• I can solve problems using Angle-Angle similarity.

…you are ready for the assignment.  Otherwise, go back and review the material before moving on.

### 07.03 Angle-Angle Similarity - Quiz (Math Level 2)

 computer-scored 50 points possible 60 minutes
Pacing: complete this by the end of Week 6 of your enrollment date for this class.

### 07.04 Similarity Proofs (Math Level 2)

 Prove the midsegment of a triangle is parallel to the third side, the medians of a triangle meet at a point and the Pythagorean Theorem.

In this lesson we will finish our discussion of similarity by proving key theorems. You will be provided the needed information and your assignment will be to write formal proofs.

Proof 1: Midsegment of a Triangle is Parallel to the Third Side

jwilson.coe.uga.edu/emt725/midseg/midsegthm.html

phgeometryfinalexamreview.wikispaces.com/05+Midsegment+and+Midsegment+Theorem

The two images above should give you a starting place. The points D and E are midpoints of their respective segments, making $\dpi{100} \fn_phv \overline{DE}$ a midsegment. From the definition of midpoint, you can mark the congruent parts. From there, you should be able to determine that the $\dpi{100} \fn_phv \bigtriangleup ADE$~$\dpi{100} \fn_phv \bigtriangleup ABC$. So, $\dpi{100} \fn_phv \angle ADE$ and $\dpi{100} \fn_phv \angle ABC$ are corresponding parts of similar triangles. Therefore, $\dpi{100} \fn_phv \angle ADE\cong \angle ABC$. Use $\dpi{100} \fn_phv \overline{AC}$ as a transversal of $\dpi{100} \fn_phv \overline{DE}$ and $\dpi{100} \fn_phv \overline{BC}$ to prove that they are parallel.

Please view the following video showing more examples of proofs of Similar Triangle Theorems.

 David Yost: Similarity 6: Theorems Involving Similarityhttps://www.youtube.com/watch?v=dKRw2d-paXg

Proof 2: Prove Pythagorean Theorem by using similar triangles

This section of the lesson is from EngageNY under a under a Creative Commons Attribution Non Commercial Share Alike 3.0 Unported license.

You have used the Pythagorean Theorem in math classes before. Now, it is time to prove it using your knowledge of similar triangles.

Given, $\dpi{100} \fn_phv \bigtriangleup ABC$, prove that $\dpi{100} \fn_phv a^{2}+b^{2}=c^{2}$.

For the proof, we will draw a line from vertex C to a point D so that the line is perpendicular to side AB.

We draw this particular line, line CD, because it divides the original triangle into three similar triangles:

Let’s look at the triangles in a different orientation in order to see why they are similar. We can use our basic rigid motions to separate the three triangles. Doing so ensures that the lengths of segments and degrees of angles are preserved.

In order to have similar triangles, they must have two common angles, by the AA criterion. Which angles prove that $\dpi{100} \fn_phv \bigtriangleup ADC$ and $\dpi{100} \fn_phv \bigtriangleup ABC$ similar?

What must that mean about $\dpi{100} \fn_phv \angle C$ from $\dpi{100} \fn_phv \bigtriangleup ADC$ and $\dpi{100} \fn_phv \angle B$ from $\dpi{100} \fn_phv \bigtriangleup ACB$?

Which angles prove that $\dpi{100} \fn_phv \bigtriangleup ACB$ and $\dpi{100} \fn_phv \bigtriangleup CDB$ similar?

What must that mean about $\dpi{100} \fn_phv \angle A$ from $\dpi{100} \fn_phv \bigtriangleup ADC$ and $\dpi{100} \fn_phv \angle C$ from $\dpi{100} \fn_phv \bigtriangleup CDB$?

If $\dpi{100} \fn_phv \bigtriangleup ADC\sim \bigtriangleup ACB$ and $\dpi{100} \fn_phv \bigtriangleup ACB\sim \bigtriangleup CDB$, is it true that $\dpi{100} \fn_phv \bigtriangleup ADC\sim \bigtriangleup CDB$?   How do you know?

When we have similar triangles, we know that their side lengths are proportional. Therefore, if we consider $\dpi{100} \fn_phv \bigtriangleup ADC$ and $\dpi{100} \fn_phv \bigtriangleup ACB$, we can write:

$\dpi{100} \fn_phv \frac{\left | AC\right |}{\left | AB \right |}=\frac{\left | AD\right |}{\left | AC\right |}.$

By the cross-multiplication algorithm,

$\dpi{100} \fn_phv \left | BC \right |^{2}=\left | BA\right |\cdot \left | BD\right |.$

By considering $\dpi{100} \fn_phv \bigtriangleup ACB$ and $\dpi{100} \fn_phv \bigtriangleup CDB$ we can write:

$\dpi{100} \fn_phv \frac{\left | BA\right |}{\left | BC \right |}=\frac{\left | BC\right |}{\left | BD\right |}.$

Which again by the cross-multiplication algorithm,

$\dpi{100} \fn_phv \left | BC \right |^{2}=\left | BA\right |\cdot \left | BD\right |.$

By considering  and  we can write:

$\dpi{100} \fn_phv \left | AC \right |^{2}+\left | BC \right |^{2}=\left | AB\right |\cdot \left | AD\right |+\left | BA\right |\cdot \left | BD\right |.$

By using the distributive property, we can rewrite the right side of the equation because there is a common factor of |AB|. Now we have

$\dpi{100} \fn_phv \left | AC \right |^{2}+\left | BC \right |^{2}=\left | AB\right |(\left | AD\right |+\left | BD\right |)$

Keeping our goal in mind, we want to prove that a2 + b2 = c2; let’s see how close we are.

Using our diagram where three triangles are within one, (shown below), what side lengths are represented by $\dpi{100} \fn_phv \left | AC \right |^{2}+\left | BC \right |^{2}?$

Now, let’s examine the right side of our equation: $\dpi{100} \fn_phv \left | AB\right |(\left | AD\right |+\left | BD\right |)$. We want this to be equal to c2; does it?

If we add the lengths of AD and BD we get the entire length of AB; therefore, we have $\dpi{100} \fn_phv \left | AB\right |(\left | AD\right |+\left | BD\right |)=\left | AB\right |\cdot \left | AB\right |=\left | AB\right |^{2} = c^{2}$.

Good luck writing formal proofs for these!

If after completing this topic you can state without hesitation that...

• I can prove the midsegment of a triangle is parallel to the third side.
• I can prove the Theorems of Simliar Triangles.
• I can prove the Pythagorean Theorem.

…you are ready for the assignment.  Otherwise, go back and review the material before moving on.

### 07.04 Similarity Proofs - Worksheet (Math Level 2)

 teacher-scored 50 points possible 40 minutes

Activity for this lesson

Complete the attached worksheet.

1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.

Pacing: complete this by the end of Week 6 of your enrollment date for this class.

### 07.04 Unit 07 Review Quiz - (Math Level 2)

 computer-scored 43 points possible 30 minutes
Pacing: complete this by the end of Week 6 of your enrollment date for this class.

### 08.00 Trigonometry (Math Level 2)

 The goal of all math classes is for students to become mathematically proficient. In particular, students need to be able to: Make sense of problems and persevere in solving them. Reason abstractly and quantitatively. Construct viable arguments and critique the reasoning of others. Model with mathematics. Use appropriate tools strategically. Attend to precision. Look for and make use of structure. Look for and express regularity in repeated reasoning. As you work through this unit, keep these proficiencies in mind! This unit focuses on trigonometry. The six trigonometric functions are identified and used to solve problems. At the conclusion of this unit you must be able to state without hesitation that: I can identify the hypotenuse, adjacent side, and opposite side of an acute angle in a right triangle. I can determine the six trigonometric ratios for a given angle in a right triangle. I can recognize the reciprocal relationship between sine/cosecant, cosine/secant, and tangent/cotangent. I can use a calculator to find the value of the six trigonometric functions for any acute angle. I can use a calculator to find the measure of an angle given the value of a trigonometric function. I can explain and use the relationship between the sine and cosine of complementary angles. I can use the Pythagorean Identity to find values of the six trigonometric functions of an angle. I can use the Pythagorean Theorem to find the missing lengths of the sides of a right triangle. I can find the missing lengths and angles of a right triangle. I can find the exact trigonometric function values for angles that measure 30°, 45°, and 60°. I can solve applied problems using right triangle trigonometry.

### 08.01 Right Triangle Trigonometric Relationships Based on Similar Triangles (Math Level 2)

 Identify the hypotenuse, adjacent side, and opposite side of an acute angle in a right triangle, determine the six trigonometric ratios for a given angle in a right triangle, recognize the reciprocal relationship between sine/cosecant, cosine/secant, and tangent/cotangent, use a calculator to find the value of the six trigonometric functions for any acute angle, and use a calculator to find the measure of an angle given the value of a trigonometric function.

Identifying the Six Trigonometric Functions

Suppose you want to build a ramp for access to a loading dock that is 4 feet above ground level. You want to be able to easily push a cart on wheels up the ramp, and the angle of elevation should be no more than 20°. How long does the ramp have to be?

If after completing this topic you can state without hesitation that...

• I can identify the hypotenuse, adjacent side, and opposite side of an acute angle in a right triangle.
• I can determine the six trigonometric ratios for a given angle in a right triangle.
• I can recognize the reciprocal relationship between sine/cosecant, cosine/secant, and tangent/cotangent.
• I can use a calculator to find the value of the six trigonometric functions for any acute angle.
• I can use a calculator to find the measure of an angle given the value of a trigonometric function.

…you are ready for the assignment.  Otherwise, go back and review the material before moving on.

 NROC Algebra 1: Identifying the Six Trigonometric Functionshttp://www.montereyinstitute.org/courses/DevelopmentalMath/U...

Each topic is divided into sections that include the following:

• Warm Up - questions to answer to see if you are ready for the lesson.
• Presentation - high quality video with excellent illustrations that teaches the topic.
• Worked Examples - examples that are worked out step-by-step with narration.
• Practice - quiz problems on the topic covered.
• Review - practice test to check your knowledge before moving on.

You are not required to complete every section. However, REMEMBER the goal is to MASTER the material.

### 08.01 Right Triangle Trigonometric Relationships Based on Similar Triangles - Worksheet (Math Level 2)

 teacher-scored 52 points possible 40 minutes

Activity for this lesson

Complete the attached worksheet.

1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.

Pacing: complete this by the end of Week 7 of your enrollment date for this class.

### 08.02 Relationship between the Sine and Cosine of Complementary Angles (Math Level 2)

 Explain and use the relationship between the sine and cosine of complementary angles.

The material for this lesson is from:

Recall that the sine and cosine of angles are ratios of pairs of sides in right triangles.

• The sine of an angle in a right triangle is the ratio of the side opposite the angle to the hypotenuse.
• The cosine of an angle in a right triangle is the ratio of the side adjacent to the angle to the hypotenuse .

In the examples, you will explore how the sine and cosine of the angles in a right triangle are related.

Example A

Consider the right triangle below. Find the sine and cosine of angles A and B in terms of a, b, and c. What do you notice?

Solution: sin A = $\dpi{100} \fn_phv \frac{a}{c}$, sin B = $\dpi{100} \fn_phv \frac{b}{c}$, cos A = $\dpi{100} \fn_phv \frac{b}{c}$, cos B = $\dpi{100} \fn_phv \frac{a}{c}$.  Note that sin A = cos B and sin B = cos A.

Example B

Consider the triangle from Example A. How is $\angle$A related to $\angle$B?

Solution: The sum of the measures of the three angles in a triangle is $180^{\circ}$. This means that m$\angle$A + m$\angle$B + m$\angle$C = $\dpi{100} \fn_phv 180^{\circ}$. $\angle$C  is a right angle so m$\angle$C = $\dpi{100} \fn_phv 90^{\circ}$. Therefore, m$\angle$A + m$\angle$B = $\dpi{100} \fn_phv 90^{\circ}$. Angles A and B are complementary angles because their sum is $\dpi{100} \fn_phv 90^{\circ}$.

In Example A you saw that sin A = cos B and sin B = cos A. This means that the sine and cosine of complementary angles are equal.

Example C

Find sin $\dpi{100} \fn_phv 80^{\circ}$ and cos $\dpi{100} \fn_phv 10^{\circ}$. Explain the result.

Solution: sin $\dpi{100} \fn_phv 80^{\circ}\approx$ 0.985 and cos $\dpi{100} \fn_phv 10^{\circ}\approx$ 0.985. sin $\dpi{100} \fn_phv 80^{\circ}$ = cos $\dpi{100} \fn_phv 10^{\circ}$ because $\dpi{100} \fn_phv 80^{\circ}$ and $\dpi{100} \fn_phv 10^{\circ}$ are complementary angle measures. sin $\dpi{100} \fn_phv 80^{\circ}$ and cos $\dpi{100} \fn_phv 10^{\circ}$ are the ratios of the same sides of a right triangle, as shown below.

Concept Problem Revisited

$\bigtriangleup$ABC is a right triangle with  m$\angle$C = $\dpi{100} \fn_phv 90^{\circ}$ and sin A = k. What is cos B?

$\angle$A and $\angle$B are complementary because they are the two non-right angles of a right triangle. This means that sin A = cos B and sin B = cos A. If sin A = k, then cos B = k as well.

Click on the link below to watch the video.

 ONEEZE: Complementary Angle Theorem https://www.youtube.com/watch?v=gEPMmcDJSB8David Yost Trigonometry 3: Sine and Cosine of Complementary Angles https://www.youtube.com/watch?v=364W9bxBe84

If after completing this topic you can state without hesitation that...

• I can explain and use the relationship between the sine and cosine of complementary angles.

…you are ready for the assignment.  Otherwise, go back and review the material before moving on.

### 08.02 Relationship between the Sine and Cosine of Complementary Angles - Worksheet (Math Level 2)

 teacher-scored 54 points possible 30 minutes

Activity for this lesson

Complete the attached worksheet.

1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.

Pacing: complete this by the end of Week 7 of your enrollment date for this class.

### 08.03 Pythagorean Identity (Math Level 2)

 An identity is an equation that expresses a relationship that is always true, regardless of the value of the angle.  Before we move on, let's review a few of the basic trigonometric identities. The following are the Reciprocal Identities. Remember the reciprocal of a fraction  is the fraction [\tfrac{b}{a}] . That is, we find the reciprocal of a fraction by interchanging the numerator and the denominator, or flipping the fraction. These are the Quotient Identities.   The first and principle identity we will consider is the  Pythagorean identity which states that sin2 θ + cos2 θ =1. I should mention notation at this point. sin2 θ is the standard way to write sin θ2. It is written like this because it is cleaner to write sin2 θ than to write sin θ2. If we were to write sin θ2, this means that the term θ is squared, not the expression sin θ. Likewise, the expression cos θ is squared. If you see anything else, such as tan2 θ, this means that tan θ is squared. Okay, back to the Pythagorean identity. This is called the Pythagorean identity because it is a form of the Pythagorean theorem. We will use the Pythagorean theorem to prove it. cnx.org/resources/9cccf7fd65d661ce25e6bfefedb58e35/graphics1.jpg Now that we know the basic Pythagorean Identiy, we can use it, along with the other basic identities shown above to create several different versions of the Pythagorean Identity. sin2 θ + cos2 θ =1 If we divide everthing in this equation by cos2 θ, watch what happens: This portion of the lesson is from  High School Trigonometry/Relating Trigonometric Functions under the Creative Commons Attribution-ShareAlike License We can use this identity to find the value of the sine function, given the value of the cosine, and vice versa. We can also use it to find other identities. Example If cos θ = what is the value of sin θ? Assume that θ is an angle in the first quadrant. Solution: 1  1  1 Remember that it was given that θ is an angle in the first quadrant. Therefore, the sine value is positive, so I highly recommend that you click on the links below and watch the videos before continuing.

### 08.03 Pythagorean Identity - video (Math Level 2)

 Khan Academy: Lecture 13: Trigonometric Identitieshttp://freevideolectures.com/Course/2558/Trigonometry/13mathispower4u: Fundamental Trigonometric Identities: Reciprocal, Quotient, and Pythagorean Identities https://www.youtube.com/watch?v=OmJ5fxyXrfg

If after completing this topic you can state without hesitation that...

• I can use the Pythagorean Identity to calculate the value of sinθ, cosθ, tanθ, cscθ, secθ and cotθ.

…you are ready for the assignment.  Otherwise, go back and review the material before moving on.

### 08.03 Pythagorean Identity - Worksheet (Math Level 2)

 teacher-scored 60 points possible 40 minutes

Activity for this lesson

Complete the attached worksheet.

1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.

Pacing: complete this by the end of Week 7 of your enrollment date for this class.

### 08.04 Setting Up and Solving Right Triangles to Model Real World Contexts (Math Level 2)

 Use the Pythagorean Theorem to find the missing lengths of the sides of a right triangle, find the missing lengths and angles of a right triangle, find the exact trigonometric function values for angles that measure 30°, 45°, and 60°, and solve applied problems using right triangle trigonometry.

Suppose you have to build a ramp and don’t know how long it needs to be. You know certain angle measurements and side lengths, but you need to find the missing pieces of information.

There are six trigonometric functions, or ratios, that you can use to compute what you don’t know. You will now learn how to use these six functions to solve right triangle application problems.

There are many ways to find the missing side lengths or angle measures in a right triangle. Solving a right triangle can be accomplished by using the definitions of the trigonometric functions and the Pythagorean Theorem. This process is called solving a right triangle. Being able to solve a right triangle is useful in solving a variety of real-world problems such as the construction of a wheelchair ramp.

You can find the exact values of the trigonometric functions for angles that measure 30°, 45°, and 60°. You can find exact values for the sides in 30°, 45°, and 60° triangles if you remember that sin 45° = $\dpi{100} \fn_phv \frac{1}{\sqrt{2}}$ and sin 30°$\dpi{100} \fn_phv \frac{1}{2}$. For other angle measures, it is necessary to use a calculator to find approximate values of the trigonometric functions.

If sin 45° =  $\fn_phv \small \frac{1}{\sqrt{2}}$, that tells us that the "opposite side" across from the 45° angle is of length 1 and the hypotenuse is of length $\fn_phv \small \sqrt{2}$. We can find the length of the "adjacent side" using the Pythagorean Theorem: $\fn_phv a^{2}+b^{2}=c^{2}$.

$\fn_phv 1^{2}+b^{2}=\sqrt{2}^{2}$

$\fn_phv 1+b^{2}=2$ Now solve for b by subtracting 1 from both sides of the equation.

$\fn_phv b^{2}=1$  Now take the square root of both sides.

$\fn_phv \sqrt{b^{2}}=\sqrt{1}$

$\fn_phv b=1$  We now know that the adjacent side is also 1. This makes sense because the opposite and adjacent sides are both across from 45° angles and so they should be the same length. We can use these side lengths for any number.

In general:

Similarly, if sin 30° = $\fn_phv \small \frac{1}{2}$, that tells us that the "opposite side" across from the 30° angle is of length 1 and the hypotenuse is of length 2. Again we can use the Pythagorean Theorem to solve for the "adjacent side" of this right triangle.

$\fn_phv 1^{2} + b^{2}=2^{2}$

$\fn_phv 1 + b^{2}=4$

Solve for b by subtracting 1 from both sides of the equation.

$\fn_phv b^{2}=3$

Take the quare root of both sides.

$\fn_phv b=\sqrt{3}$

We now know that the adjacent side is $\fn_phv \sqrt{3}$.

In general:

If after completing this topic you can state without hesitation that...

• I can use the Pythagorean Theorem to find the missing lengths of the sides of a right triangle.
• I can find the missing lengths and angles of a right triangle.
• I can find the exact trigonometric function values for angles that measure 30°, 45°, and 60°.
• I can solve applied problems using right triangle trigonometry.

…you are ready for the assignment.  Otherwise, go back and review the material before moving on.

 NROC Algebra 1: Right Triangle Trigonometryhttp://www.montereyinstitute.org/courses/DevelopmentalMath/U...

Each topic is divided into sections that include the following:

• Warm Up - questions to answer to see if you are ready for the lesson.
• Presentation - high quality video with excellent illustrations that teaches the topic.
• Worked Examples - examples that are worked out step-by-step with narration.
• Practice - quiz problems on the topic covered.
• Review - practice test to check your knowledge before moving on.

You are not required to complete every section. However, REMEMBER the goal is to MASTER the material.

### 08.04 Setting Up and Solving Right Triangles to Model Real World Contexts - Worksheet (Math Level 2)

 teacher-scored 66 points possible 40 minutes

Activity for this lesson

Complete the attached worksheet.

1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.