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Geometry 6th Ed. (since July 2015)
Corrections and Clarifications - read carefully.
IMPORTANT! Make liberal use of the Geometry Student Solutions Guide, not so much to just see the answer but to study and understand the method(s) to reach a solution.

PROOFS are special kinds of problems and require special attention. The expected outcome in completing proofs is an understanding and appreciation of logical thought (and not to memorize a bunch of postulates and theorems). It is important to realize that there are many right ways to structure a proof. The answer in the solutions guide is just one correct method. In general, proofs do not appear on standardized tests. While it is important to attempt proofs, it is more important to understand the sequential thinking that goes into a solution. Please utilize technical support as needed.

In grading a student's homework look for a reasonable and logical flow of proof. While working with proofs, encourage the student to have handy a list of definitions, postulates, and theorems for reference and verify that all of the items in the REASONS column are definitions, postulates, theorems , or a close variation of one.

Don't hesitate to consult the Student Solutions Guide and compare the method in the solutions guide with the method used by the student. If a student continually gets stuck with proofs, stop for a day or so and try this: Go back one or two sections and consider 4-5 proofs that were attempted before. Read over each proof problem and look carefully at the flow of the proof in the solutions guide. The idea is to understand the logic flow, not to memorize the proof. Then close the solutions guide and try the problems from scratch.

One strategy that may help you in determining the steps involved is to think backwards through the problem. That is, think about what information you need in order to conclude the last step. Then think about the step that needs to precede that, and so on.

If the formal statements and reasons format is intimidating, have the student write the "logic path" for the proof in a kind of narrative form before using the two-column format. Be flexible in grading. Allow for creativity. Learning proofs is a process. Valuable assets in the learning process include patience, think backwards, realize there are several correct ways to complete a proof, give yourself some latitude with informality in the REASONS column, make liberal use of the solutions guide when you get stuck, positive attitude, and keep your sense of humor.

It is neither advised nor necessary for a student to memorize theorems, postulates and definitions. The student should make a list of each and add to the lists as the course progresses and refer to the lists frequently when completing a proof. Complete lists can be found at the end of the textbook. Objectives in learning proofs include:
1. Understanding and improving the skill of logical thinking.
2. Developing appreciation and skill in forming a well conceived "argument".
3. Improving problem solving skills, both personal and mathematical.

Allowing lists to be used advances the additional benefit of teaching the student the value and use of reference tools in study and problem solving.


Corrections and Clarifications

Geomtery 6th edition

Section 1.4, problem 4b. In the solutions guide, the answer should be "adjacent."

Chapter 1 Test, problem 19. In the solutions guide, the construction is not correct. A correct illustration has points at A and B, the endpoints of the given segment, and the crossing arcs above and below the given segment with a line drawn that contains the points of intersection of the crossing arcs. The line is perpendicular to the segment.

Section 1.6, problem 23. The proof in the solutions guide does not correspond with the problem. The correct proof is below.

Section 1.6, problem 25. The proof in the solutions guide does not correspond with the problem. The correct proof is below.

Chapter 3 Review, #3. The proof in the solutions guide does not correspond with the problem. The correct proof is below.


Geomtery 5th edition

Section 1.7, problem 22. In the solutions guide, the measure of angle 2 should be 70 degrees rather than 110 degrees.

Chapter 2 Review, problems 4 and 5. The solutions require that segments AB and CD are parallel and that information is not given. Please omit the problem.

Chapter 2 Test, problem 6. The construction in the solutions guide does not correspond with the problem in the textbook. The correct construction has the following steps.

1. With the point of the compass at point A and using a fixed opening greater than the distance to the line, draw an arc that intersects the line at two points. Label the points B and C. [Only the intersection points need to be shown.]
2. With the point of the compass at B and using an opening greater than half the distance between B and C, draw an arc below the line and approximately below point A. 3. Using the same opening, place the point of the compass C and cross the previous arc. Label the crossing point D.
4. Draw line AD. [Point may be labeled differently or not at all. They are used for clarity in this description.]

Section 3.1, problem 34. In the solutions guide, the answer should be yes. The proof is:

Section 3.2, problem 4. In the solutions guide, in Statement 2 the first part should say angle M is congruent to angle R (rather than angle Q).

Section 3.2, problem 12. In the solutions guide, Reason #2, the second "if" should be replaces with "it".

Section 3.3, problem 10. In the solutions guide, the answer should be, the two sets are equivalent.

Chapter 3 Review, #3. The proof in the solutions guide does not correspond with the problem. The correct proof is below.

Section 5.1, problem 2(c). In the solutions guide, the answer should be 5/8 rather than 8/5.

Section 5.5, problem 2. In the solutions guide, both answers should be "a" rather than 1.

Chapter 6, Mid-Chapter Review, problem 23(f). There is a misprint in the solutions guide. In the first step of 3 steps, the sign between the arc measures should be plus rather than minus.

Section 6.3, problem 9. In the solutions guide, when all the terms of the quadratic equation are on one side of the equal sign, the constant should be +48 rather than -48.

Section 6.4, problem 23. In the solutions guide, the answers in both parts are incorrect. By the way, those answers are correct in the back of the textbook.

Section 7.2, problem 4. In the solutions guide, the correct answer is orthocenter.

Section 9.1, problem 6. In the solutions guide, the unit used to measure its lateral area is square inches and the unit used to measure its volume is cubic inches.

Section 9.3, problem 6. In the solutions guide, the replacement for the radius in the formula should be 1.5 rather than 1.05. The correct value is used in the calculation so the answer is correct.

Chapter 9 Review, problem 20. In the solutions guide, the answer should be 1/9 rather than 1/2.

Chapter 10 Test, problem 2. In the solutions guide, point D is plotted incorrectly. The correct location for point D is 9 units above the origin (on the y-axis).


Chalk Dust Special 2nd Edition
Geomtery 4th edition - Since October 2007

Section 1.3, problem 17. In the solutions guide, the first step should read 2x + 1 + 3x + 2 = 6x - 4.

Section 1.4, problem 4b. In the solutions guide, the answer should be "adjacent."

Section 1.4, problem 26. In the solutions guide the expression for Part C is actually the simplification of the expression in Part B. Part C should be
90 - (2x + 5y)
90 - 2x - 5y

Section 1.5, problem 29. In the solutions guide, step 5 should be "Addition" and step 6 can be "Division Property."

Section 1.6, problem 12.

Section 1.6, problem 26.

Clarification of statements in the proof

The steps starting with statement 4...

4. 0 < x < 90

5. x + m angle 2 = 90

6. m angle 2 = 90 - x

7. -x < 0 < 90 - x

This is from statement 4,
0 < x < 90
Subtract x in all three parts of the compound inequality to get statement 7.

Following statement 4, there are two things that must be addressed. One is the manipulation in steps 5 and 6 that allows for a substitution later in the proof and the other is a continuation with the compound inequality. It looks like a couple of skipped steps, but if steps 5 and 6 are not in place there, then they will have to be added later in the proof and, again, it would look like a skip in the flow.

8. 90 - x < 90 < 180 - x

This is the result of adding 90 in all parts of statement 7.

9. 0 < 90 - x < 90

Here is the real dilemma. It looks like statement 9 must come from statement 8 and if you assume that then statement 9 does not make sense because to clear -x in the outside parts of the inequality in statement 8, x has to be added to all parts and that would make the middle part 90 + x rather than 90 - x.

However, the author of the solutions guide did not come up with statement 9 by manipulating statement 8. Rather, two previous steps were put together (transitive property) to make statement 9. Specifically, part of statement 7 is 0 < 90 - x and part of statement 8 is 90 - x < 90. Put them together to get statement 9.

Further, the reason given in the solutions guide is "Transitive Property of Inequality" but statement 9 does not represent the normal usage of the transitive property because normally the left side is made less than the right side and the middle part is left out. However, the validity of the statement should be clear in spite of the unusual use of the transitive property.

I hope this helps clear things up.

Section 1.7, problem 20. In the solutions guide, the measure of angle 2 should be 70 degrees rather than 110 degrees.

Section 2.1, problem 26. In the solutions guide, omit the "||" symbol in the third reason.

Section 2.1, problem 27. The labeling of the illustration is incorrect in the solutions guide. Please swap the numbers 1 and 2.

Section 2.2, problem 13. In the solutions guide, the notation used in the figure is not consistent with the notation used in the textbook. Please omit the problem.

Section 2.2, problem 14. The solution in the solutions guide does not match the problem in the textbook. Please omit this problem.

Section 2.3, problem 18. In the solutions guide, the last statement and reason should be labeled "6" rather than "3".

Section 2.3, problem 21. The proof in the solutions guide does not correspond with the problem. The correct proof follows:

Statements Reasons
1. Ray DE bisects angle CDA 1. Given
2. Angle 2 is congruent to Angle 3 2. Definition of angle bisector
3. Angle 3 is congruent to Angle 1 3. Given
4. Angle 2 is congruent to Angle 1 4. Transitive Property (no substitution)
5. Segment ED parallel to Segment AB 5. Theorem 2.3.2

Section 2.4, problem 41.

Section 2.4, problem 44.

Section 2.5, problem 27.

Section 2.5, problem 40.
In the solutions guide, the segment to be drawn should be BD rather than BC.

Section 2.6, problem 4.
The letter "I" also has point symmetry.

Chapter 2 Review, problem 14. In the solutions guide, some of the steps in the solution are incorrect. Here is one way to work the problem.

From the given information, two equations emerge.

Since a || b,
Equation 1: 2x - y = 3x + 2y

In the left-side triangle, the sum of the measures of the angles is 180. So,
Equation 2: (2x - y) + 100 + x = 180

There are many ways to solve the system of equations. Begin by simplifying both equations.

(1)
2x - y = 3x + 2y
2x - 3x = 2y + y
-x = 3y
x = -3y

(2)
2x - y + 100 + x = 180
3x - y = 80

Use substitution and replace x with -3y in the second equation.

(2)
3(-3y) - y = 80
-9y - y = 80
-10y = 80
y = -8

Use either equation to find x.

(1)
x = -3y
x = -3(-8)
x = 24

Chapter 2 Review, problem 17.
In the solutions guide, step 2, the right side of the equation should be 111 rather than 222.

Chapter 2 Review, problem 18.
In the solutions guide, the third equation is supposed to be the simplification of the first so the first term should be 8x rather than 3x.

Chapter 2 Test, problem 6. The construction in the solutions guide does not correspond with the problem in the textbook. The correct construction has the following steps:

1. With the point of the compass at point A and using a fixed opening greater than the distance to the line, draw an arc that intersects the line at two points. Label the points B and C. [Only the intersection points need to be shown.]
2. With the point of the compass at B and using an opening greater than half the distance between B and C, draw an arc below the line and approximately below point A.
3. Using the same opening, place the point of the compass at C and cross the previous arc. Label the crossing point D.
4. Draw line AD.

[Point may be labeled differently or not at all. They are used for clarity in this description.]

Section 3.1, problem 39. In the solutions guide, the angles listed in step 7 are not correct. The angles should be angle ABC and angle DEC.


Section 3.2, problem 29. In the solutions guide, the angle in the first statement should be SRU rather than RSU.

Section 3.3, problem 10. In the solutions guide, the solution is incorrect. The sets are equivalent.

Section 3.3, problem 11. In the solutions guide, replace "S" with "O".

Section 3.4, problem 30.

Chapter 3 Review, problem 3.

Section 4.1, problem 25.

Section 4.1, problem 36. In the solutions guide the solution is not consistent with the problem in the textbook. The answer is, "The bisectors should bisect each other."

Section 4.2, problem 27. In the solutions guide, reason #11, the symbol for angles should be replaced with the word "segments."

Section 4.2, problem 35.
In the solutions guide, statement 13, segment ND should be segment BC.

Section 4.3, problem 22.

Section 4.3, problem 26. In the solutions guide, the answer should be (b).

Section 4.3, problem 35. In the solutions guide, statement 4 should be "segment AE is congruent to segment CE."

Section 4.3, problem 36. In the solutions guide, there are two mistakes in the proof. Statement 7 should be, segment BE is congruent to segment BE. Reason 8 should be SAS.

Section 4.4, problem 30. In the solutions guide, in statements 5, 7, and 8, the parallel symbol should be replaced with the congruence symbol. Also, the reason for statement 9 is incorrect. It should be, "If the legs of a trapezoid are congruent, the trapezoid is an isosceles trapezoid."

Section 5.1, problem 2. In the solutions guide, the answers for parts (c) and (d) should be swapped.

Section 5.2, problem 24. In the solutions guide, the first numerator is the length of side AD so the numerator should be 4 rather than x. The mistake is corrected in the next step.

Section 5.3, problem 22. In the solutions guide, the last item is incorrect. It should read R6. CSSTP

Section 5.4, problem 43. The proof in the solutions guide contains a typographical error. Rather than "angle CDB is congruent to angle ACD", it should be "angle CDB is congurent to angle ADC."

Section 5.5, problem 2. In the solutions guide, the answers should be:

a. AC = a    b. BC = a

Section 5.5, problem 36b. In the solutions guide, the length of CF should be 2x rather than 12.

Section 5.6, problem 30. In the solutions guide, statement 2, the denominator on the right side of the equation should be SY.

Section 6.1, problem 19. In the solutions guide, there are several errors. Please omit this problem.

Section 6.3, problem 9. In the solutions guide, the constant term in the quadratic equation should be +48 rather than -48.

Section 6.3, problem 30. In the solutions guide, the ratio of RT / RS is calculated incorrectly.
If TO = x, then RO = x and RS = 2x. Since RT = x sqrt 2,
RT / RS = x sqrt 2 / 2x,
RT / RS = sqrt 2 / 2

Section 6.4, problem 20.The problem is not valid. It appears the author used inscribed angles when central angles were inrended. Please omit the problem.

Section 6.4, problem 23.In the solutions guide the inequality in part "a" should be "greater than".

Section 6.4, problem 33. In the solutions guide, the tangent requested should be drawn.

Section 6.6, problem 4. In the solutions guide, the answer should be Orthocenter rather than Circumcenter.

Chapter 6 Review, problem 10. In the solutions guide, the measure of arc AD should be 180 degrees rather than 80 degrees.

Chapter 6 Review, problem 33.
The solutions guide does not provide enough information for one to know where the listed equation comes from and it is not intuitively obvious.

Make a sketch of the problem — a right triangle with an inscribed circle. In order for my description to match your diagram, place the right angle of the triangle at the lower left with acute angles above and to the right of the right angle. The hypotenuse should be slanting from upper left to lower right.

From the center of the circle, draw segments to the three sides of the triangle at the point of tangency. Notice a right angle is formed by the segments drawn to the horizontal and vertical legs of the triangle and a square is formed with one vertex at the right angle of the right triangle. Label 6 on all sides of the square.

Draw segments from the center of the circle to the vertices of the acute angles of the triangle. Notice a little pair of congruent right triangles formed at each acute angle.

Move around the outer edge of the original right triangle and label every little segment. Starting at the right angle and moving up and then clockwise, the first segment is the 6 labeled earlier.

Label x the segment from the point of tangency on the vertical side to the upper left vertex.

Label x the segment along the hypotenuse from the upper left vertex to the point of tangency on the hypotenuse.

Since the hypotenuse is 29 cm, the segment from the point of tangency on the hypotenuse to the vertex at the lower right is (29 - x).

Also label the segment from the vertex at the lower right of the figure to the point of tangency on the horizontal leg (29 - x).

The segment from the point of tangency on the hoizontal leg of the triangle to the vertex of the right angle is already labeled 6.

Now take a look at the original big right triangle and consider the lengths of the sides.

The vertical left side is (x + 6).

The hypotenuse is 29.

The horizontal bottom side is 6 + 29 - x or (35 - x).

Using those side lengths, apply the Pythagorean Theorem and you have the equation in the solutions guide.

Section 7.1, problem 34. In the solutions guide, the solution for this problem is missing; they duplicated the solution for problem 38 instead.

There are 3 similar right triangles in the figure. Consider the small one (with hypotenuse "a") and the large one (with hypotenuse "c"). Similar trianlges have the characteristic that corresponding sides are proportional. That is, ratios of corresponding sides can be set equal to one another.

One such equation can be formed using the pattern,

(long leg / hypotenuse) = (long leg / hypotenuse)
     [small triangle]        [large triangle]

                 h / a  =  b / c  Multiply both sides by "a" to get...
                     h  =  ab / c
		         


Section 7.1, problem 49.

Section 7.3, problem 12. In the solutions guide, since the area and perimeter are equal, they cancel in the last step given in the solutions guide. Therefore, a = 2.

Section 7.5, problem 18. In the solutions guide, while calculating the area of the sector, the radius is incorrect. The radius should be 10 / sqrt 3 rather than just 10. The change results in an answer of (100pi - 75sqrt3)/9.

Section 7.5, problem 23. In the solutions guide, the diagram does not correspond with the problem in the textbook. However, the calculation of the area of the circle is correct.

To find the radius of the circle, draw a vertical segment from the top vertices down to the bottom side. Since the trapezoid is isosceles, the base of the right triangles formed (on both sides) is 5. Calculate the length of the vertical leg using the Pythagorean Theorem.

Section 8.1, problem 32. In the solutions guide, the factor 2/3 should be 1/3 (because 4 inches in 1/3 foot). The correct answer is 13 1/3 cubic yards.

Section 8.2, problem 35.
In the solutions guide, the answers given are correct but part (c) is not included, the method of solution for parts (a) and (b) are not clear and the figure does not seem to match the situation.

Since the regular tetrahedron in the text is composed of 4 congruent equilateral triangles with sides of length e, begin with the sketch of an equilateral triangle. Draw the triangle with a horizontal base at the bottom of the figure. Draw a segment from the top vertex vertically to the base.

The segment drawn is a height of the equilateral triangle and the point of intersection with the bottom side bisects the bottom side. Therefore, the length of each half is e/2.

Two right triangles were formed by the construction of the height of the equilateral triangle. Each right triangle has a leg of length e/2 and hypotenuse e. Find the length of the vertical leg using the Pythagorean Theorem.

a^2 + b^2 = c^2
(e/2)^2 + b^2 = e^2
e^2 / 4 + b^2 = e^2
b^2 = e^2 - e^2 / 4

Create a common denominator on the right by multiplying e^2 by 4/4...

b^2 = 4e^2 / 4 - e^2 / 4
b^2 = 3e^2 / 4
b = sqrt (3e^2 / 4)
b = e sqrt 3 / 2

(a) Since b is the height of the little right triangles, the area of each of them is...

A = (1/2)bh
A = (1/2)(e/2)(e sqrt 3 / 2)
A = (e^2 sqrt 3) / 8

Since there are two such right triangles forming the equilateral triangle, the area of the equilateral triangle is

2 (e^2 sqrt 3) / 8
(e^2 sqrt 3) / 4

(b) Since there are 4 congruent triangles forming the overall figure, the total area is 4 times the area of one triangle...

Total area = 4 (e^2 sqrt 3) / 4
Total area = e^2 sqrt 3

(c) When e = 4
Total Area = 4^2 sqrt 3
Total Area = 16 sqrt 3

Section 8.2, problem 36. In the solutions guide there is a typo in the third step. The fraction 1/2 should be 1/12.

Chapter 8 Review, problem 20. In the solutions guide, the answer to the first part should be 1/9 rather than 1/2.

Section 9.3, problem 24. In the solutions guide, the equation in the third to last step is 2st = -3st. It should be 2st = -2st.

Chapter 9 Test, problem 2. In the solutions guide, point D should be located 9 units above the origin on the y-axis.

Section 10.4, problem 39. The solution in the solutions guide does not match the problem. Draw the segment Q, perpendicular to side MN, intersecting MN at point R. In the right triangle MQR,
sin y = QR/a So
QR = a sin y

The area of the parallelogram is found by multiplying the base times the height.
Area = base x height

Since the base of the paeallelogram is b and the height is QR,
A = b(QR)

Replace QR with a sin y to get
A = b(a sin y) or
A = ab sin y

Chapter 10 Test, problem 3(b). In the solutions guide, replace "tan" with "sin".

Geometry - Traditional Program
Comments, Corrections and Clarifications
Caution!! The first chapter, Volume 1, is a brief review of many key topics discussed in more detail in later chapters. Do not let your child become confused with the pace of the Volume 1 review.

At the end of each section, all of the problems under Guided Practice and Independent Practice should be completed. Problems under Integrated Review are optional and may be omitted when they involve skills which must be remembered from Algebra. Omit all problems under Exploration and Extension, Chapter Exploration, Using a Computer Drawing Program, and Communications about Geometry. The last section of each section, entitled Extended Application, may be considered optional.

There should be considerable flexibility in choosing assignment problems in Geometry. Since the number of problems following each section is relatively limited (often less than 50) compared to the number in other texts (often over 100), we recommend that the students work both even and odd-numbered problems up to a point. That point changes from section-to-section but it mostly involves the beginning of the "real-life" problems. Have the student attempt only one or two of those real-life problems per section but do expect to refer often to the solutions manual for guidance. The "real life" problems are quite challenging and only the most gifted student will accomplish them without difficulty. Also, problems suggesting the use of a computer drawing program are optional and may be omitted.

IMPORTANT! Make liberal use of the Geometry Solutions Guide, not so much to just see the answer but to study and understand the method(s) to reach a solution.

PROOFS are special kinds of problems and require special attention. The expected outcome in completing proofs is an understanding and appreciation of logical thought (and not to memorize a bunch of postulates and theorems). It is important to realize that there are many right ways to structure a proof. An answer in the solutions guide is just one correct method.

In grading a student's homework look for a reasonable and logical flow of proof. While working with proofs, encourage the student to have handy a list of definitions, postulates, and theorems for reference and verify that all of the items in the REASONS column are definitions, postulates, theorems , or a close variation of one.

Don't hesitate to consult the Complete Solutions Guide and compare the method in the solutions guide with the method used by the student. If a student continually gets stuck with proofs, stop for a day or so and try this: Go back one or two sections and consider 4-5 proofs that were attempted before. Read over each proof problem and look carefully at the flow of the proof in the solutions guide. The idea is to understand the logic flow, not to memorize the proof. Then close the solutions guide and try the problems from scratch.

One strategy that may help you in determining the steps involved is to think backwards through the problem. That is, think about what information you need in order to conclude the last step. Then think about the step that needs to precede that, and so on.

If the formal statements and reasons format is intimidating, have the student write the "logic path" for the proof in a kind of narrative form before using the two-column format. Be flexible in grading. Allow for creativity. Learning proofs is a process. Valuable assets in the learning process include patience, think backwards, realize there are several correct ways to complete a proof, give yourself some latitude with informality in the REASONS column, make liberal use of the solutions guide when you get stuck, positive attitude, and keep your sense of humor.

It is neither advised nor necessary for a student to memorize theorems, postulates and definitions. The student should make a list of each and add to the lists as the course progresses and refer to the lists frequently when completing a proof. Complete lists can be found at the end of the textbook. Objectives in learning proofs include:
1. Understanding and improving the skill of logical thinking.
2. Developing appreciation and skill in forming a well conceived "argument".
3. Improving problem solving skills, both personal and mathematical.

Allowing lists to be used advances the additional benefit of teaching the student the value and use of reference tools in study and problem solving.

Corrections and Clarifications

Problems of (1) Exploration and Extentision, (2) Mixed Review, and (3) Integrated Review may be considered optional or omitted. Exploration and Extension problems are designed to explore concepts and are often a long reach for most students(and not necessary for what we want to accomplish in this course). The time spent with those problems is not worth the marginal gains in knowledge. The other topics are algebra based and often require the student to dig through Algebra 1 books for the answers. They offer an association to algebra but do not broaden the students understanding of Geometry.

Section 1.1, problem 11. Answers in the solutions guide are incorrect because segments k and n are exactly the same length and segments l and m are also the same length.

Section 1.2, page 12, problem 2. "L" is not a quadrilateral because the sides must be straight, as is true with squares, triangles, parallelograms, etc.

Section 1.3 Guided Practice, page 19-21, problems 9-12. It sounds like the student expects the video program to show an example of every kind of problem the student will face in the homework. That is not the case. Generally, the video program will talk about ideas and sometimes strategy for solving problems, using for illustration a few specific examples. Those ideas and strategies are to be used on other problems, some very similar, some not, in the problem set.
Problems 9-12 are designed to give students practice in recognizing patterns. All the student had to do is (1) trace the figures, then (2) cut them into congruent parts. The solutions manual shows where the cuts should be.

Section 1.3, "computer drawing program". The problems in Geometry involving drawing programs are considered optional because they fall into the "enrichment" category. Therefore, you may elect to omit those problems. If you are interested in purchasing a drawing program anyway, a few possibilities are listed below.

     Software                                   Company
Geometer's Sketchpad              	Key Curriculum Press
Geometric SuperSupposer          	Sunburst
Geometry Inventor                       LOGAL Software      
General interest and purchase information can be found on each company's website (use a search engine to find addresses) along with system requirements. Both Geometric SuperSupposer and Geometry Inventor can be found on the website for Sunburst which is a catalog company selling technology products and software.

Section 1.6, page 39, problems 4D and 5.
#4 D. The area of a triangle is found using the formula A = bh where b is a base and h is the related height. When the triangle is a right triangle, the base and height are the two sides of the triangle that form the right angle, so... A = (4)(4) and A = 8 square units.
#5. To find the area of the entire figure you add the areas of the individual figures found in #4. See Solutions Guide. For the perimeter, remember that perimeter is the distance around the outer edge of a figure so you would not just add all the perimeters together because they would contain the lengths of some sides within the overall figure. If I start with the top side of figure A and proceed clockwise I would add the following sides to find the perimeter:
P = 8 + 2 + 2 + 5 + 4rt2 + 3pi + 5 + 2 + 4 + 2 + 2
P = 32 + 4rt2 + 3pi

Section 1.7, page 47, problem 4. Notice that it says to draw an angle of ABOUT 60 degrees. It does not have to be exact. She can use the reference of a 90 degree angle and figure its 2/3 of that right angle. If you are having trouble now you can expect to experience considerable frustration with proofs. I suggest that you review your "READ ME FIRST" letter shipped with the program (and on our website at www.chalkdust.com) for information about proofs and for reference to many previously answered problems in geometry.

Section 2.1, problem 10. This problem is quite a stretch for any student to work correctly so I recommend that it be skipped. The correct answer is given in the solutions guide so just use that expression to help with #11.

Section 2.1, page 60, problem 12. You do not need to figure out a pettern for this one. All you have to do is add a few items from the table in number 9. Notice the directions say to "show the sum of n triangular numbers," and triangular numbers are found in number 9. For example, to find the sum of the first 4 triangular numbers, just add the 1, 3, 6, and 10 (from the table in number 9) to get 20.
A nice way to set up the problem is to add a third line to the chart on #9. Label that line "Sum of n Triangular Numbers." Then you just add numbers from the previous line up to the position in the third line. Here is how it goes:
For n=1, "add" one number (1) from the table in number 9. Enter 1.
For n=2, add two numbers (1 + 3) from the table in number 9. Enter 4
For n=3, add three numbers (1 + 3 + 6) from the table in number 9. Enter 10 .
For n=4, add four numbers (1 + 3 + 6 + 10) from the table in number 9. Enter 20.
For n=5, add five numbers (1 + 3 + 6 + 10 + 15) from the table in number 9. Enter 35.
For n=6, add six numbers (1 + 3 + 6 + 10 + 15 + 21) from the table in number 9. Enter 56.

Section 2.1, page 61, problems 25 and 26. The reference to a segment is used just to be able to describe the notions of "same"side and "different" sides of line AB. The important thing is to understand the given definitions and how they use the segment idea to clarify "same" or "different" side. Make sure you understand the difference between line CD and segment CD. If points C and D are on the same side of AB, then the line containing C and D (line CD) may very well intersect line AB, but segment CD cannot. On the other hand, if C and D are on opposite sides of line AB, then segment CD must intersect line AB.

Section 2.4, page 83, Mixed Review Problem 10, Solutions Guide. The misprint in the solutions guide is that DC should be DQ. The calculation for DQ = 2 can be accomplished several ways but the SM explanation is sketchy at best. The easiest way to see it is from the diagram. Notice that since AP = 2, the other "halves" of the side of the square are the same length so QC = 2. Since triangle DQC is isosceles (because it is a right triangle with one acute angle of 45 degrees so the other acute angle must be 45 degrees), the sides opposite the equal angles are equal so DQ = QC = 2.

Chapter 3 Test. You may use the books outline of postulates and theorems and definitions on your test.

Chapter 3 Test, problem 12. In the solutions guide the correct answer is, "You can conclude that we will celebrate."

Chapter 3 Review, #39. Omit this problem. Part of the given information in the solutions guide is not given directly in the text and that essential information cannot be derived because isosceles triangles have not been studied yet.

Section 3.2, problem 44. In the Solutions Guide (page 75) the coordinates of the point where 12 intersects the mirror are reversed. The coordinates should be (1/16, -1/2). The problem is worked correctly, however, because the y-coordinate of -1/2 is used in the solution.

Section 3.3, page 125, problems 26-29. The problems should be omitted because the figure is mis-labeled.

Section 3.7, problem 32. In the solutions manual, the vector should be AD and not AC.

Chapter 4. What is an included angle/side? A formal definition would be more complicated than an informal explanation. Imagine (or draw) a triangle with vertices labeled A, B, and C. The triangle contains three sides and three angles. For any two sides, like sides AB and BC, there is an angle "between" the sides called the included angle which is the angle formed by those two sides, in this case it is angle B (which is the same as saying angle ABC). And for any two angles, like angle A and angle B, there is a side in between the angles (called the included side), it is a side shared by the two angles; in this example the included side is AB.

Section 4.5, problem 17. The triangles can be proven congruent by a, b, or c.

Section 4.6, problems 20 and 21. In the solutions guide, the solutions for these two problems are reversed.

Section 4.7, problems 5 and 33. The problems are vague and should be omitted. For example, Section 4.7, #33. The intention is for the student to notice that the ratio of segment AM to segment AB is golden (from problem 32), triangle ABM is isosceles with base angles of 72 degrees (problem 30), the diagonals and sides of pentagon ALBCD also form isosceles triangles with base angles of 72 degrees, and therefore the ratio of the diagonals to the sides of the pentagon is likely golden as well.

Section 4.7, problem 27. Assume that O is the center of the circle.

Chapter 4 Test, problem 15, page 213. In the solutions guide, statement 5 should read "Triangle ADB is congruent to triangle ACE".

Section 5.5, page 228, problem 23. Your question was can a median be an angle bisector and not a perpendicular bisector? It's not possible for a median to be an angle bisector and not a perpendicular bisector. In the drawing, the angle is ALMOST bisected because the angle at the midpoint is ALMOST a right angle. One way to see it is to draw an angle bisector then take something like a toothpick or popsicle stick, mark a midpoint, and try to put the midpoint on the angle bisector so that the ends of the stick just touch the rays of the angle. You'll find that the only way to do it is for the angle formed at the midpoint of the stick is a right angle.

Section 5.5, Mixed Review, page 251, problem 6. The answer is incorrect in the solutions guide. The correct answer is x^4/y^2, and NOT x^3/y^2.

Section 5.6, Independent Practice, problem 9. Since S is on both circles, the only way the two circles do not intersect in two points is if A, B, and S are collinear in some fashion [S between A and Bor A and B on the same side of S, but in either case "in line" with one another]. So it is remotely possible that the two circles intersect at only one point and the position of the ship could, indeed, be determined. However, that possibility is extraordinarily remote in the context of the problem. Notice the figure in the text to the right of #12. I think that is a clue as to a more likely configuration of the points A, B, and S. NOTE : Problems 9-14may be skipped altogether. These problems are designed to let the student know how the geometry content MAY be used in real life. But they can be presented in a way which is too complicated to be beneficial. These are "enrichment" category and are non-essential.

Chapter 5 test, page 263, problems 12, 15 and 17.
Problem 12. The request is for a single segment rather than the sum of segments.
Problem 13. There is not enough information given in the figure to determine the relative lengths of the sides of the triangle ABC. If you consider the measure of the angle A to be fixed at 61 degrees and the lengths of AB and AC to be variable, then point A can be moved around so that the relative lengths of AB and AC change.
Problem 15. Omit this problem.
Problem #17. The answer should be the Converse of the Hinge Theorem.

In the problem, the legs of a tripod are being spread apart. [We can assume that the lengths of the tripod legs are not changing in that process.] The problem is asking what theorem guarantees that the angle between the legs is getting larger as the legs of the tripod are spread apart.

Intuition tells us that when the legs of a tripod are spread apart two things are happening at the same time. One, the angle between the legs is growing larger and, two, the distance between the ends of the legs is getting longer, and each can be thought of as a consequence of the other. That is, the legs get farther apart because the angle between the legs gets bigger or the angle between them gets bigger because the legs get farther apart.

It is somewhat basic and may be obvious to you, but it's important to realize that in the context of the theorems we are considering only two of the legs of the tripod at a time and the "triangles" formed are composed of two tripod legs and the segment connecting the ends of those legs. Also, the two triangles are: the one formed as described above BEFORE spreading the legs of the tripod and the other is the triangle formed as described above AFTER spreading the legs of the tripod. Further, either of those triangles can be regarded as the "first" triangle, and the other one would be regarded as the "second."

Because we want to conclude that the angles are getting larger, the premise implied by the spreading of the tripod legs is that the distance between the legs are getting longer. That is, in the context of Hinge Theorem and converse, the length of the third side of a triangle is getting longer.

Now consider the Hinge Theorem and its converse.

HINGE THEOREM
Premises:
1. Two sides of one triangle are congruent to two sides of another triangle.
2. The included angle of the first is larger than the included angle of the second.
Conclusion:
The third side of the first triangle is longer than the third side of the second triangle.

CONVERSE OF THE HINGE THEOREM
Premises:
1. Two sides of one triangle are congruent to two sides of another triangle.
2. The third side of the first is longer than the third side of the second.
Conclusion:
The included angle in the first triangle is larger than the included angle in the second.

Since the Converse of the Hinge Theorem is concluding a change in angle when given a change in length of the third side, that is the one most applicable to #17.

Chapters 1-6 Cumulative Review, problem 39. Please omit this problem because it does not provide enough additional understanding to justify the amount of time necessary to work with it.

Chapter 7 Mid-Chapter Self Test, problem 11. Rotational symmetry: 90 degrees and 180 degrees. Line symmetry: 4 lines; vertical, horizontal and two diagonals.

Chapter 7 Review, page 367, problem 32. Omit because the proof in the solutions manual uses a theorem which is not introduced until section 8.6.

Section 8.5, Golden Rectangle. The Golden Rectangle is just a rectangle with rather special proportions. That is, any rectangle having a particular ratio of length to width (or width to length) is a Golden Rectangle. That particular ratio is given on page 405 and it seems a bit complicated. I think the actual ratio is of less importance than the fact that a Golden Rectangle can be generated within itself!!
Look at the figure on page 405. I'm not certain that the figure is drawn exactly to scale, but there is the general idea from a geometric standpoint. Rectangle ABCD is a Golden Rectangle. Now take side AD and imagine "attaching" it at point A but allow the segment to swing down so that point D meets side AB. You can actually do this with a compass. D will intersect AB at point E. Now imagine segment AE, and swing it so that point E is fixed and point A moves to intersect CD. The point of intersection in the figure is F. It turns out that BCFE is a Golden Rectangle and it was generated geometrically from the original. I think that's more important than the algebraic calculation of the actual lengths of sides but the author is asking you to complete a table in order to see the same relationship just described.
Anyway, the ancient Greeks thought the golden Rectangle to be the most pleasing to the eye and much old Greek architecture reflects that feeling. Indeed, the overall dimensions of the Parthenon (top to bottom, left to right) is a Golden Rectangle; the overall column structure is a golden Rectangle; the rectangles formed column to column are Golden Rectangles; ... see what I mean.
Also, painters would begin with a canvas in the dimensions of a Golden Rectangle, then generate rectangles within itself again and again, often working toward the middle of the original canvas (like you see in the figure on page 405 but continued a few more steps) and then begin the painting by putting an important feature of the art, like the eye of a cat, inside the innermost golden Rectangle.

Chapter 8 Test, problem 11, clarification.
In the #8 triangle, the sum of the squares of the legs is equal to the square of the hypotenuse, AND it is therefore true that the triangle is indeed a right triangle, but it turns out that the numbers 2,3, and square root 13 are not a Pythagorean triple because by definition the numbers are supposed to be integers. That said, though, since the problem is so tricky go ahead and mark it correct if you chose #8 to be a Pythagorean triple.

Chapter 8 Test, page 423, problem 14.
Notice in the two trapezoids that AB corresponds with BC and the ratio is 2.8/1.4 or 2 to 1. That implies that if the two trapezoids are similar, all corresponding sides would have the same ratio. Which means that all the sides of the large trapezoid are double the corresponding sides of the small one or, put another way, all the sides of the small trapezoid are half the length of the corresponding sides of the large one. Now notice that AG corresponds with BF; so if the figures are similar, then BF must be half the length of AG. Since AG is 5, BF would have to be 2.5.
Now consider the overall triangle AGD. BF is parallel to AG and if BF is really of length 2.5 then BF is a midsegment of triangle AGD (because it is parallel to and half the length of AG). However, if it is a midsegment, then it means that B is the midpoint of AD which means that AB would have to be equal to BD; but that is not the case because of the lengths given.
To put it another way, since B is not the midpoint of AD, BF cannot be the midsegment of AGD. Therefore BF is not 2.5. But if the two trapezoids containing AF and BF are similar, then BF would have to be 2.5. That is a contradiction so the two trapezoids are not similar.

Section 9.2, page 440, problems 2 and 16.
Problem #2. Rationalizing means removing the radical from the denominator. To do that you multiply the fraction by a number that will cause you to have the square root of a square in the denominator, but yet you don't want to change the value of the overall fraction, so you need to multiply by a form of one. In problem 2, you should multiply by sqrt7/sqrt7 , a form of one because it is a number over itself. In the denominator you get the square root of 49 which becomes 7; in the numberator you get 14 (sqrt7) so the fraction looks like 14 (sqrt7) / 7. Then the 14 and 7 cancel leaving 2 (sqrt7) for the answer.
Problem #16. You might want to get out your Algebra 1 book and review factoring. The idea is to rewrite the expression as the product of two binomials. The factored form is shown in the solutions manual but from your question about 1 - or a + I can tell that you really don't remember the concept. The idea of factoring comes from the idea of multiplying polynomials and the topics are to comprehensive for me to try to explain via email. There are several sections on multiplying polynomials and several sections on factoring that are needed to understand the whole idea. I am sure you had this before and a couple of examples in a book would probably cause the process to come rushing back in your memory.

Cumulative Review for Chapters 7-9, text page 474, solutions guide page 319, problem 7. The angle measure is incorrect. It should read: Rotational symmetry (72 degrees, 144 degrees, 216 degrees, and 288 degrees), 5 lines of symmetry.

Chapter 10 test, problem 12. Add point N to the list.

Section 11.4, problem 29. In the solutions guide, the fifth step from the bottom includes a factor 2.5 which should be 25.

Chapter 11 test, problem 8, solutions guide. In the solutions guide, the calculation of perimeter contains a fraction with a numerator of 2.7. That numerator should be 14.

Section 13.2, #20. In the solutions guide, the answer for the 3-dimensional part should be the same as the answer for the first part since the points must be 2 inches from the parallel lines.

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