Question from Joy:
We are in Geometry 3 rd edition. Chapter 14 lesson 2 number 22-23. Page 582.
How do you get 108?
The answer in key says 3 x 180/5? Why? I think I missed something important yet simple. I was thinking divide 360/5. What an I doing wrong? Thanks!
Answer from Dr. Callahan:
I have to admit – I could not find this in the book. I think it SHOULD be there – but could not find it.
The answer is the a triagle has interior angles that add up to 180 (we know that – right)
So each time we add a side (triangle to square) we add 180.
See here https://www.mathsisfun.com/geometry/interior-angles-polygons.html
So in #22, he is saying we have 180 3 times (triangle plus 2 sides adding 2 more 180s) as the total angles and then dived by the number of sides to get 108.
Again – I do not see this in the book.
Hope this helps
The 3rd edition of the Harold Jacobs Geometry textbook has gone through three different printings with 3 different covers and ISBNs. If you are picking a text and looking at used options, it can be confusing. So let me try to help.
First, all 3 versions are basically the same – as long as they say 3rd edition. They look like the photos below.
The Three Geometry Textbook Versions that Still Work
The original published by Freeman – ISBN: 978-0-7167-4361-3 and the corresponding Teachers Guide which contains the solutions to the problems. This book had a separate Test Bank for tests.
The second printing published by My Father’s World – ISBN: 978-1-61999-109-5 and the corresponding Teachers Guide which contained the tests and the Answer Key.
The third printing published by Master Books – ISBN: 9787-1-68344-020-8. This is the latest and if you are buying new, this is what you are getting. The tests are in the Teacher’s Guide and the answers are in the Solutions Manual. You will need all 3 books for the course.
The Geometry Videos that Support the above Books
All 3 of the printings will work with the AskDrCallahan videos. The various printings have minor changes in page numbers, drawings, and some problems have changed – but the basic content remains the same. But, just to be more confusing, the videos come in 3 different packages. All are the same content but they might look different. All require the student to have one of the 3 textbooks, the tests, and a set of solutions.
1 – The original DVDs published by us (AskDrCallahan) -ISBN:978-0-615-27810-0
2 – The latest version of the DVDs was published by Master Books – UPC: 713438-110236-8
3 – The online version, offered by us only, have the same content and bypass the need for DVDs and a DVD player. They are the same content as on the DVD. The online videos come in two options.
- Monthly option – Pay for the course monthly and cancel anytime by logging into your AskDrCallahan account (or email us to cancel). This option is perfect if you think you will need just a few months.
- Lifetime access – If you need a full year, or have siblings who might use the course later, then this is your option. One price one time.
Question from Jennifer:
Chap 8 review
We aren’t sure how the answer would be 24mm.
This is not explained well in the textbook.
First – the translation is the movement of the bat from position B to position C (see chap 8, Lesson 2) A translation is two rotations around a parallel line – or in other words, we moved it over from one place to another.
So he is asking, how far did it move. Or, what is the length of the black line moving from B to C.
Since we can measure the bat to be about 42 mm (with a ruler), the length of the black line is about 24 mm using the same ruler.
Question from Courtney
I’m having trouble with Chapter 5, Lesson 1, Problem 29.
I have worked the problem, I know what the answer should be, but given the figure, I don’t understand how that could be false.
Without the figure, I understand it.
You can never assume anything. You only know what you are given in the definition. Don’t trust the figure. If it doesn’t tell you, you don’t know it. So for problem # 29 we know the following:
It’s a line
AB is less than BC
BC is less than CD
CD is less than DE
While the lengths in the figure look similar (or even equal), we don’t know that they really are similar or equal. Don’t trust anything but the defined statements in the problem and marked items in a figure.
Here is the question to think about. Given the definitions AB < BC < …. can you draw the figure such that the last DE is much larger.
Put some numbers in it and think inches.
AB = 1
BC = 2
CD = 3
DE = 500
Does that fit the definition of the problem? (Not the figure – the definition?)
Again – figures are often only ONE example that fits the problem – but they do not show every example.
Look for the following material or heading in the contents.
- Reasoning and Proof
· Deductive reasoning
· Direct and indirect proofs
· Parallel and Perpendicular Lines
· ASA and SAS
· Ratio and proportion
· Similar figures
- Right Triangle Trigonometry
· Pythagorean Theorem
· Tangent, Sine, and Cosine
- Surface Area and Volume
· Geometric solids
· Rectangular solids
Notes for use: One of the keys of geometry is learning deductive reasoning or how to do proofs. This can be a challenge to teach, so getting a teachers manual will really help here. The second main idea of geometry is getting used to thinking in space with shapes and the relationships between them. Lots of figures should be drawn.
The book has a problem
(x-3)(x^2 – 7x – 2)
which results in an answer of
x^3 – 10x^2 – 2x +19x + 6
The solution is incorrect in the new solutions manual ( the one with a bee on the cover)
However, it is correct in the old solutions manual -the one with a chess set on the cover.
Question: how do we work the problem on Chapter 14 test #4.
Answer from Harold Jacobs:
Since all 16 white regions are identical, we can find the area of one of them and multiply it by 16.
This figure represents one of the 16 small squares in the figure and one of the white regions.
Since the diameter of the large circle in the figure shown on the test page is 4 units, the side of the square of one of the small squares is 1 unit.
We can find the area of region B by finding the area of regions B & C combined (a quarter of a circle) and subtracting the area of region C (a triangle):
> The area of B & C combined is1/4 (pi r squared) or 1/4 (pi 1 squared) = 1/4 pi
> The area of C is1/2 r squared or 1/2 1 squared = 1/2So B = (B + C) – C = 1/4 pi – 1/2
and the total area of the 16 white regions is16(1/4 pi – 1/2) = 4 pi – 8.
In Chp. 10, Lesson 4, Set III Question 4 the textbook asks “Why is QD/AC = DB/CD?”
Shouldn’t it be DB/CB instead of DB/CD?
Dr. Callahan Answered:
Again you are correct. This is an error in the new printing of the 3rd edition by My Father’s World.
If you are looking for details on for a transcript for Geometry Seeing, Doing, Understanding by Harold Jacobs this should do the trick.
Geometry is a branch of mathematics that deals the understanding of the world around us by using measurement, properties, and relationships of points, lines, angles, surfaces, and solids. This course covers Euclidean geometry – which is a mathematical system where we assume a small set of truths (axioms) and from them we deduce through logic many other propositions (theorems). This course emphasizes the concept of proofs, but in a reachable way that helps the student learn both the logic and the geometrical mathematics. Students who complete Geometry should take Algebra II with Trig next. Concepts from this course show up on the ACT and the SAT.
Topics Covered in this Course:
- Deductive Reasoning
- Lines and Angles
- Parallel Lines
- The Right Triangle
- Regular Polygons and the Circle
- Geometric Solids