If you are looking for details on for a transcript for Jacobs Geometry Course, this description should do the trick.
Geometry is a branch of mathematics that deals with 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 (aka precalculus) 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
Question from Kerry:
The answer key for the Geometry Midterm Review #74 says the answer is square root of 80, but shouldn’t it be the square root of 52? It says (4-0) squared + (8-2)squared = 16+64, but it should be = 16+36, shouldn’t it?
Answer from Dr. Callahan:
You are correct. The original Enhanced Teacher’s Guide for Geometry (ISBN: 978-0-7167-5607-1) had the correct answer. When it was republished as Answers to Exercises for Geometry (ISBN: 978-1-61999-116-3) they changed the answer.
The correct answer is SQRT(52).
I am doing Harold Jacobs Geometry, I am on chapter 11 lesson 3. I got to problems 39-41 and got stuck.
Dr Callahan Answered:
Let us start at #38.
We can see that <ABC = 120 so <ADB = 30. Therefore we know AB = BD based on Theorem 10 page 159.
AB = x because given
BD = x by substitution since AB = BD
<BDC = 30 (since a right triangle with one angle 60)
2BC = BD (Them 51 on page 442)
2BC = x by substitution
BC = x/2
same logic using thrm 51 again — of course starting from what we know from #39
CD = BC*SQRT(3) Thrm 51 on page 442
BC = x/2 from #39
CD = x *SQRT(3) / 2
We are to show how we get CD = 7/8 AB
AB = x
CD = x *SQRT(3) /2
We are to show CD/AB = 7/8 so we can divide what we have as AB and CD to get
CD/AB = (x *SQRT(3) /2 ) / x
CD/AB = SQRT(3) /2
Now you have to calculate SQRT(3) /2 to get about 0.866 and then do the math on 7/8 to get 0.875
They are not equal – but approximately equal.
Need some help with this problem – Geometry Ch. 5.1 #44:
The answer key for this question states that we can’t draw any definite conclusion about whether angle 3 and 4 are vertical angles.
We previously proved angles 3 and 4 to be equal; based on the drawing, how could they not be vertical, too?
Dr. Callahan Wrote
For #44 note they are asking if you can PROVE.
Look at the logic. If vertical angles, then <3 = <4 and <1 = <2. And since we know (given) that <2=<4 then we would have to conclude <1=<3 by substitution.
Note – that is NOT a given.
The books answer is that you are jumping to conclusions ASSUMING line AC is straight to the end after it intersects with BC. We do not know that. It is easy to be fooled by how a drawing looks. In proofs you have to stick with the logical conclusions you can make from the given facts and not what it looks like. So the book is correct – the answer is no.
Hope this helps.
Question from Kerry:
The answer key gives 5 statement answers, but the proof only has 4 statements. I don’t know which of the 5 in the answer key is extraneous. Assuming the last one should be dropped, the answer in the key for statement 2 is subtraction, but I think it is addition. There is no subtraction for statement 2. Could you clarify/verify the 4 correct reasons for the 4 steps?
Answer from Dr. Callahan:
This should be
1) Given2) Addition3) Given4) Substitution
The answer key you have is a reprint from the original (oddly the original has the correct answers) and is new. We will document any errors at askdrcallahan.com/support.
Question from Nathan:
In Lesson 6 Chapter 4 problem 3 of Harold Jacobs Geometry the answer was SSS. This is definitely accurate, however I answered with SAS and I believe that is also correct. Is it possible that both are correct? This happened again on problem 18 of the same lesson too.
Yes this is correct.
Often you can find multiple answers to be correct in these but be careful – you MUST be able to justify by a given or justification why each side or angle is equal. Sometimes it might appear you can use a given reason when you really have not been given enough detail.
Every now and then we have customers who have a bad DVD or one that is missing. How frustrating – right as you are ready to start and have energy, the key ingredient is missing. ARGH!!!
Here is what you need to do!
First, let us know. Send an email to us at firstname.lastname@example.org and tell us which disk is missing or defective (Ex. Geometry 3.1-5.4) and your name, address, and phone number. I know we probably have your address – but just in case. (You have moved. Or we may have lost all of our data due to an alien invasion – not that we have those often.) We will ship you a new disk.
Second, get started. I know your student might have been hoping for a moment of delay, but you can come to the rescue with the online videos!
Question: Can I use Geometry 2nd Edition by Harold Jacobs with the AskDrCallahan course?
Short Answer – No, not easily.
Longer Answer – Many have asked us if they could use the Second Edition of Geometry with our DVD set. The answer is no, at least not without a lot of grief to the teacher and student.
We initially were telling teachers and students they could use the 2nd edition. Several people told us they had been told the second edition was superior due to a better treatment of proofs. In support of those families, we attempted to develop the syllabus to develop a map between the two editions. We consulted with Harold Jacobs on the map. The resulting dialogue was interesting.
First, the two books cover basically the same material, but in different ways. But, the chapters and problem sets are different enough to cause a significant management issue for the teacher and student – even with a map.
Second, and more importantly, we came to believe that those who felt the 2nd edition was superior were wrong. This was further backed up by Jacobs when he wrote the following in an email back to us.
“There was more emphasis on two-column proofs in the second edition but, as is often the case with such proofs, the things to be proved were rarely surprising and almost always rather obvious. I endeavored in the third edition to build the exercises around more profound material and to give the student more freedom in developing proofs. For really challenging problems, this meant leading the student along in a Polya-like style. Forced into two-column format, some of the proofs in this edition would run 15-20 steps, much longer than the typical 4- or 5-step proofs in the earlier edition. In his foreword on page xii of the book, Andrew Gleason of Harvard University wrote that “This book restores the idea of proof to its rightful place in geometry. Whoever studies this text will know what a proof is, what has been proved, and what has not.” He, together with Dr. Peter Renz, who wrote the supplemental notes in the Teacher’s Guide, looked carefully at the content and structure of all of the exercises and gave his full approval. I feel very honored that he would take so much of his time to do this. Gleason, by the way, is a former president of the American Mathematical Society and one of our country’s most prestigious mathematicians. I hope this helps clarify what has changed between the second and third editions as well as why.”
We hope this clears up any misunderstanding. With the two books sitting side by side, we would clearly choose the 3rd edition for our children.