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Geometry: Note on Missing Chapters

AskDrCallahan Jacobs Geometry

Geometry Course

AskDrCallahan Jacobs GeometryChapters 13 and 16 were originally omitted from our Geometry course for time constraints. The concepts in those chapters are advanced, though not difficult. Students will not likely see them on the ACT/SAT or even in college, except for specialized fields. When needed in a college track, they will be covered anew. They are often skipped at the high school level.

Geometry Chapter 13 and Chapter 163rd Edition Jacobs Geometry

However, we now recommend Chapter 13 be covered. Chapter 14 briefly refers to concepts introduced there and some students need the information. Others will be okay just referring back. Most of our students still skip chapter 13. We still consider Chapter 16 optional for everyone.
We have lectures for both chapters now on our website for FREE.
 Remember that our video instruction, online or DVD, can be used with any printing of the 3rd Edition of Jacobs Geometry.
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Homework Help Complex Problems – Moving Deeper into Concepts

As you move deeper into each textbook (Algebra, Geometry, Calculus, etc), the problems will contain several concepts you’ve learned earlier – all in one problem.  This complexity can make it difficult to find the “how-to” or examples in the book to help us understand how to work it.

Recently, we received this question from our student support page: There doesn’t seem to be any examples or teaching that I can find that helps me solve this problem.

This is a common question – not this particular problem, but in general, as the problems develop into including several simple concepts stacked into compound calculations. However, the explanations are there, they just may be back a few, or several, chapters.

For example, Jacobs Algebra Chapter 12 Summary and Review Problem 14h.

Concepts include but are not limited to:

  • Chapter 12: Square Roots. Simplify radical as much as possible. Example of this step page 480-481
  • Chapter 5: Equations in One Variable. Specifically for this review problem, Equivalent Equations (Lesson 3) page 162-163.
  • Chapter 12: Square Roots: Radical Equations. Page 505 has examples of squaring both sides to eliminate the radical.

Math builds on itself. As you learn more and more concepts, the problems reach back and build on calculating and analyzing skills learned earlier in the book or even in an earlier course. These complex problems can be hard to find “how-to” when we just can’t see it! The solutions manual is a good resource to help with steps, but sometimes even with those steps, we need to see where it was explained or taught.

We are here to help! Send us your homework questions and let us help. Filling out this form makes it easy to be sure you’ve told us what we need to know to help you, but you can also send an email to support@askdrcallahan.com.

Be sure to tell us:

  1. the course,
  2. the chapter,
  3. the lesson,
  4. the problem,
  5. YOUR issue as best as you can explain it.

We love to help.

 

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Geometry Ch11 test #7b

Question from Anya 
Course: Geometry
Chapter/Lesson/Problem11 test7b
Issue: I’m supposed to find Angle APM. To find the angle, I used tan APM=20/70, which gives 15.9 degrees. The solutions manual used sin APM=20/70, which gives 16.6 degrees. Sine is opposite/hypotenuse. However, it seems to me that the given measurements of the problem are opposite side (20) to the adjacent side (70). Which is correct? Thank you!
Answer by Dr.Callahan
Yes, you are correct. That should be tangent as you stated.
tan(APM) = 20/70
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Geometry 14.2 #22-23

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

dwc

 

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The Different Geometry Textbooks and Videos

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.

3rd Edition Jacobs Geometry

 

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.

Jacobs Geometry Curriculum

 

 

 

 

 

 

 

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 online videos, offered by AskDrCallahan, contain the same instruction, 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.

AskDrCallahan Jacobs Geometry

 

 

 

 

 

 

2 – The latest version of the DVDs is published by Master Books – UPC: 713438-110236-8

 

 

 

 

 

 

 

 

 

3 – The original DVDs published by us (AskDrCallahan) and no longer available new.  -ISBN:978-0-615-27810-0

 

 

 

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Geometry Ch8 Review #32

Question from Jennifer:
Chap 8 review
#32
We aren’t sure how the answer would be 24mm.
Answer:
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.

 

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Geometry Ch5.1 #29

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.

Answer:

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.

 

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Geometry Textbook Evaluations

Look for the following material or heading in the contents.

  • Reasoning and Proof
    · Proofs
    · Deductive reasoning
    · Direct and indirect proofs
  • Lines
    · Parallel and Perpendicular Lines
    · Angles
  • Triangles
    · Congruent
    · Isosceles
    · Equilateral
    · ASA and SAS
  • Quadrilaterals
    · Parallelograms
    · Rectangles
    · Squares
    · Trapezoids
  • Area
    · Squares
    · Rectangles
    · Triangles
  • Similarity
    · Ratio and proportion
    · Similar figures
  • Right Triangle Trigonometry
    · Pythagorean Theorem
    · Proportions
    · Tangent, Sine, and Cosine
  • Surface Area and Volume
    · Geometric solids
    · Rectangular solids
    · Spheres
  • Circles
    · Radius
    · Chords
    · Tangents
  • Transformations
    · Reflections

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.

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Geometry Ch14 Test #4

 

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.

 

2016 04 04 12 59 30

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.