Posted on Leave a comment

Algebra Ch5.6 #15

Ch5.6 #15

As before – all are similar so I will pick the harder one.



AB = 9(x-2)

CD = x + 3(x+5)


9(x-2) = x + 3(x+5)

Now multiply out and solve

9x – 18 = x + 3x + 15

9x – x – 3x  = 15 + 18

5x = 33

x = 6.6

Now go back and find the lengths

AB = 9(x-2) = 9(6.6 – 2) = 41.4

CD = AB = 41.4

CE = x = 6.6

ED = 3(x+5) = 3(6.6+5) = 34.8

Posted on Leave a comment

Algebra Ch5.6 #14

Now let’s look at Ch 5.6 #14.

Parts a-c are similar – so will do part c since the harder.

Area = x + 20 = (x-1)4    > I am just stating all things that are equal based on the given.

x+20 = (x-1)4

x+20 = 4x -4  > multiplied out

20 + 4 = 4x – x  > moved similar items to same side

24 =  3x

x = 8

Now we know x, we have to find the sides (ALWAYS GO BACK AND MAKE SURE YOU KNOW WHAT WAS ASKED)

x -1  = 8 – 1 = 7

4 = 4


Posted on Leave a comment

Algebra Ch5.6 #9

For problem 5.6, parts a-d are all pretty much the same. Here I will do part d)  since a hair more complex. The steps the same for the others.

Since AB = CD, we can substitute the following

AB is x + 6(x+2)

CD is 2(x+10)

To get

x + 6(x+2)  = 2(x+10)

So to find the length of AB and CD, I need to figure out x. So solve the equation.

x + 6(x+2)  = 2(x+10)

x + 6x + 12 = 2x + 20   – I just multiplied everything out

x + 6x – 2x = 20 – 12  > I moved all the x terms to the left and all the constants (numbers) to the right

5x = 8  > did the math

x = 8/5  = 1.6

Now I have x, but the problem asked for the lengths.

So now we can find the parts

AB = x + 6(x+2) = 1.6 + 6(1.6 + 2) = 1.6 + 6(3.6) = 23.2

AE = x = 1.6

EB = 6(x+2) = 21.6

CD = AB = 23.2



To do the others – follow the same process

Posted on Leave a comment

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

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



Posted on

Geometry Ch13.5 #15-18

For Chap 13, Lesson 6, #32, the basis for the answers is the figure for #6.  If we look carefully at the grid, applying the Pythagorean Theorem gives XY^2 = 2^2 + 2^2 = 8, so XY = sqrt8 = 2sqrt2.  YZ^2 = 1^2 + 1^2 = 2, so YZ =sqrt2.  Therefore XY = 2YZ.
Harold Jacobs



Posted on 2 Comments

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.



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 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.





Posted on

Geometry Ch8 Review #32

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.


Posted on

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.


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.


Posted on

Algebra 2 with Trig Course Description for Transcript

If you are looking for details on for a transcript for Algebra 2 with Trig,  this should do the trick.

Algebra 2 and Trigonometry covers build on basic algebra and geometry to prepare the student for calculus or other college-level mathematics courses.

Students who complete Algebra 2 with Trig should take Calculus next. Concepts from this course show up on the ACT and the SAT.

Topics Covered in this Course:

  • Algebraic Operations
  • Equations and Inequalities
  • Graphs and Functions
  • Polynomial and Rational Functions
  • Exponential and Logarithmic Functions
  • Trigonometric Functions
  • Trigonometric Identities and Conditional Equations
  • Additional Topics and Applications in Trigonometry
Posted on

Algebra Ch5.1 #10x

This is an error in the Algebra Solutions Manual

Mary asked,

In Algebra 1, chapter 5 lesson 1 problem 10x, the answer key says the answer is true for all positive integers.  But isn’t it true for all integers? 1 to the power of x =1.  If it is a negative power, it is still one, right?

Dr. Callahan Answer:
You are correct. Should be for all real numbers – not just integers. I tend to test these in a calculator to make sure though 😉
So that is incorrect in the Solutions manual.
Just for completeness sake – my goto math tool is Wolfram – so here is there answer.