Homework Problems, Errors, & Typos by Chapter

Introduction

The surfer/spotter problem of Geometry is answered on page 290 of the solutions manual.

Chapter 1

Chapter 1, Lesson 1, Number 19 -- the answer should include as an answer A-B-G instead of A-B-C.
Chapter 1, Lesson 1, Number 21 -- the answer should have AG instead of AC.

Chapter 1, Lesson 3, pg. 23, Set III, How to get an equilateral triangle by rearranging the polygons you cut out from the figure. (Number 5 in this set). Shown below on the left is the original shape as shown in the book, but the four polygons are labeled 1, 2, 3, and 4. Then on the right is shown the equilateral triangle you can make with those same polygons. Hopefully this shows you how to get an equliateral triangle out of the original shape in the book, but if you still have questions please Contact Us. We can help.

Original Shape from Book:                                     Equilateral Triangle you can form:

Chapter 2

Chapter 2, Lesson 6, Number 9 generates lots of questions as to why the demonstration does not prove the case. The issue is that in number 9 he has proved a SINGLE CASE, not all cases. If I tried to prove to you that the sun is "Always" to the west and walked outside in the afternoon - would you believe my evidence? You would tell me to try in the morning and you would show me I was wrong. Here Raoul demonstrated that the statement is true in a PARTICULAR case - but he did not PROVE it is so in ALL cases. It so happens it is true in all cases - but hopefully you can see from this that a simple example is NEVER proof.

Chapter 2, Lesson 6, Number 10 Similar to number 9 above, even 1000 examples showing me something does not PROVE anything. In fact all I need to DISPROVE 1000 examples is ONE example showing it does not work.

This is an important issue in science and in logic in general. Think about it. Even when you hear something on the news! How often do we find a particular example being used to argue truth? If someone steals from you have you proved ALL people are evil? No - you showed one example. Much of what people argue in society is from examples - which can be evidence of some larger truth - but not always. So like the word "TRUTH" the word "PROOF" is very powerful and has a very specific meaning. They do NOT mean sometimes or most of the time - they mean at ALL times. Now read John 17:17 and consider the word TRUTH and PROOF.

Chapter 3

Chapter 3, Lesson 2 Number 47

Chapter 3, Lesson 4, Number 36 has a typo in line 3 of the proof. It should say PC besects <APB.

Chapter 3, Lesson 4 Numbers 19-23. Download a pdf of this solution here. Or you can read it, right on the page:

Let’s start at the top.

19) This one is 45 degrees, because it is half of 90. You are right that half of 360 is 180, and you are also right that the figure is a square. You are further right that all the angles in a square equal 360. So congratulations! You have lots of things right. However, the question here is wanting to know about a specific angle. Let me see if I can draw you a picture.

Number 20

Here, they are continuing to bisect angles, only this time they are cutting in half angle ACD. Since we found in number 19 that angle ACD is 45 degrees, we know that angle FCD is 45/2, or 22.5 degrees.

Number 21) Here they are choosing a larger angle that while not a bisecting angle, IS an angle we can figure out based on what we have found in numbers 19 and 20 (Note that all of these problems are building on each other intentionally).

Notice that if angle FCD is half of angle ACD, then the other half: Angle ACF must be half of ACD also. This means that <FCD = <ACF, both are 22.5 degrees. Why is this important? Well look at this figure:

 Number 22) DCE is another addition angle. Meaning we have to add together smaller angles to figure it out (You could also subtract, I’ll show you that in a minute).

Option 1 :Let’s start by labeling these angles 1, 2, 3, and 4 for the sake of brevity.

DCE = <1 + <2 + <3 = 22.5 + 45 (based on what we’ve found in 19-21). That makes DCE = 67.5 degrees

Now I mentioned a second option, which is subtraction.

You know the “whole” angle (the original right angle) equals 90 degrees. We also know by process of deduction that <4 = 22.5 degrees. So we can subtract 90 – 22.5 = 67.5 degrees. Either way is correct.

Number 23) Last , but never least, is angle DFC. This one asks you to apply what you know about triangles.

Notice that Angle DFC is the top of a triangle. I will highlight the triangle here:

We know that the three inside angles of a triangle add together to equal 180 (as you so correctly pointed out originally). So we use that information here. We know that the corner angle of a square equals 90 degrees. Number 19 found for us that the small angle of this triangle, <FCD (aka <1) Is 22.5 degrees. We can now calculate the value of the remaining third angle, angle DFC.

180 – (22.5 + 90) = 180 – 112.5 = 67.5 degrees.

There are other ways to solve this problem, using many geometry theorems that will work. I have shown you the way that I would approach the problem. If you use another method, or have questions about alternate ways to work it, please let us know!

Part 1 Video Tutorial Geometry 3-6 Number 11

Part 2 Video Tutorial Geometry 3-6 Number 11

Midterm Review

Number 97

(Deals with concepts from Chapter 5) Learning about Pythagorean Theorem, and the extension to this theorem called Theorem 38 (Triangle Inequalities Theorem). These two concepts can seem to be at odds with each other, but in this explanation we show you how this question can be tricky, but you have the knowledge to out smart it!

Check out the explanation below:

What you are learning about here is called 'Theorem 38" or "The Triangle Inequalities Theorem" (pg. 201)

What it is telling you is that the sum of two sides of a triangle have to be greater than the third side.

In number 97 in the Midterm Review of Geometry, they are asking a trick question.

We know that the Pythagorean theorem says that


However, in this problem it specifically asks if we can have a triangle with side lengths a^2, b^2, and c^2, where the sum of a^2 and b^2 equals c^2. The answer is No. What?? Are we crazy?? Well, possibly, but that's not the answer. :) The distinction here is that the pythagorean theorem applies to side lengths, where "a" "b" and "c" represent the value, in length, of the side. If you take those lengths, and square them, then the relationship of
Makes sense.

What you have to be careful of, though, is that the SIDE LENGTHS for this hypothetical triangle in number 97 is not asking about what happens when you square the side lengths, it is asking what happens if the side length already ARE a squared value, then can the triangle exist? And the answer is no. Because if we add a^2 and b^2, we get c^2 (per the pythagorean theorem) so those side lengths are not possible in a triangle.

You can test this theory by trying to create a triangle with side lengths a^2, b^2 and c^2. (remember that the side length has to be squared values, not distances that you later square on paper).

Let's give it a try together. Here's an example that will help.

In this example problem they try out different side lengths and see if it is possible to form a triangle with them based on combining the simultaneous truths of the Pythagorean Theorem along side Theorem 38. I would encourage you to look specifically at the ones they say are "not possible" and try to draw them out on paper, using the measurements they indicate in the example. What happens? The ones that violate Theorem 38 are not possible to form.

Admittedly the question in the midterm review is tricky, but it was good hearted. They are trying to get you to think about what you're doing when you evaluate the side lengths of a triangle and realize that these relationships work together in complicated ways worthy of close attention :) I hope this helps you, but feel free to write again if we can help! (Oh, and there is a formulary in the back of the Geometry book that if you have not already found it might truly be helpful. They take all of the theorems and postulates and put them in one place. Starts on pg. 738. The glossary is cool too. )

Hope this helps! If you would like more clarification, or if you still need help with this problem. Contact Us. We can help.

Chapter 12

Chapter 12, Lesson 5, Number 43 has the wrong solution in older version of Teachers Guide Published by Freeman. The correct answer is "To other points on the major arc AB." Thanks to Charlie in Wisconsin for catching this error. Harold Jacobs indicates that this problem is corrected in some of the most recent printings of the Enhanced Teachers Guide. Not sure which guide you have? Ours is most current. Click Here to compare.

Chapter 12, Lesson 5, Number 54 has been corrected in the new Enhanced Teachers Guide to say "At other positions on the major arc." Again thanks to Harold Jacobs for pointing this out.

Chapter 14

Chapter 14, Lesson 2, Number 22, 23 The method implied in the teacher's guide solution is based on dividing the polygon into triangles by drawing diagonals from one vertex. This is illustrated by exercises #38-51 of Chapter 7, Lesson 1 (pages 262-263).

Chapter 14, Lesson 3, Number 34  Number 34

Tests

Test 1
Geometry Test 1 Problem 11

• If you missed problem 11 (under Review Section 1), you can use this template to print out that diagram, and fold the cube at home. Then you can see with your own eyes (and hands!) how that cube comes together. Print the Diagram Here. 

Geometry Test 1 Problem 17Geometry Test 1 Problem 18 (same as 17, see work for 17 above)

 


Resources
Trouble measuring angles? Use this online protractor to help!

• Note that the "teacher controls" on this website opens up more options so that you can create, and measure, all kinds of angles!