**This page is designed to help provide support for Kate’s Jacob’s Geometry eCourse,** a video supplement that walk students through *Jacob’s Geometry* in an engaging, visual/auditory way! View samples on MasterBooksAcademy.com.

### Helpful Resources

**Free Online Graphing Calculators**– These calculators can be used instead of a graphing calculator for this course. (It is recommended that college-bound students get a graphing calculator, however, so they are already familiar with using one.)- Desmos (suggested, due to ease of identifying values and zooming in and out) – Note that you have to press the keyboard symbol in the bottom left to get a keyboard. Press y = and the values y equals, using x to stand for the independent variable. Use ^ to show exponents (use your right arrow to get back from an exponent to a regular value). Click on the graph of the curve itself to see the value at any particular point. Zoom in or out using the plus or minus signs. Add a second equation by clicking underneath where the first is typed and typing another one.
- Math Is Fun – Use ^ to show exponents; use the first line for inputting the first equation, and the second for the second.

## Corrections/Clarifications

As I get questions about problems or discover errors, I will post clarifications or corrections here. If your question isn’t here and you’re taking the eCourse, please email me so I can get back with you on it (and add it here for others to learn from too).

**Chapter 1, Summary and Review, Problem 2**

Students may think the answer is planes, and all the polygons are indeed in different planes. However, in geometry, it’s really important to read very carefully and give precise answers, as we then build on those answers. The question was what kind of geometric *figure*. A plane is not a figure–if you look at page 9, you’ll see it make a point about how while it’s pictured as having edges, it really has no boundaries. So while the sides are *in* different planes, the *figures* they are are polygons. Make sure the understands that a plane doesn’t have boundaries.

**Chapter 1, Summary and Review, Problem 9 **

Students may wonder why RO is being counted as an edge; the key here is in reading very carefully. The problem explains that the pattern can be used to form a cube. So in answering the question, it’s looking it at not as a flat figure (in which case RO wouldn’t be an edge) but as a cube (which you’d get if you folded it up).

**Chapter 2, Lesson 4, General Note **Solving problems using the thinking of indirect proofs can be tricky at first. Stress understanding the logic of it, not writing it out as a formal proof.

Notice in the example on page 56 that the assumption made is the opposite of a possible conclusion (think *answer*)–not something that immediately contradicts information given. Think of the conclusions as the possible answers; you need to assume the opposite of what you think the answer is going to be. The whole idea in an indirect proof is to assume a possible answer and see if it is right by seeing if you get a contradiction; the possible answer without a contradiction is the right answer.

**Chapter 2, Algebra Review, Problem 26**

Students who start by dividing both sides by x may think the problem is unsolvable, getting *x* + 7 = *x*. Dividing both sides by *x* works fine provided the unknown does not equal 0. Since we can’t divide by 0, it wouldn’t work to divide both sides by *x* if *x* = 0. Typically math books operate with the assumption that the unknown is not 0 and allow for dividing by an unknown. Here, though, was one of those rare exceptions where they made *x* equal to 0. If you ever come across an unsolvable result or a different answer than the *Solutions Manual* after dividing by an unknown, try solving another way (or plugging in 0 for the unknown) to see if perhaps the unknown is 0.

**Chapter 3, Lesson 2, Problems 38-40**

You may be tempted to write AC = BD as the answer to 38. Although that is a true statement, it’s not true *because *A-B-C. When the book is asking because A-B-C, it’s asking you to think through what you can conclude *based on that information alone. *A-B-C is a way of writing that B is between A and C. On page 86, we learned If A-B-C, then AB + BC = AC. So that’s why we can conclude that AC = AB + BC *because *A-B-C. The same thing for problem 39–we need to write down what we can conclude *because *B-C-D. Make sure you’re taking notes on what you learn so you can see what you can conclude *because *of something else. Geometry isn’t just about listing a true statement as is the case in other math courses–it’s about showing logically how that statement was arrived at, thus teaching critical thinking and deductive reasoning.

On #40, the question is why can we say that AB + BC equals BC + CD. Look closely at what happened and the properties on page 79. In the addition property, we’re concluding that if a = b, then a+ c = b + c. In other words, we have 2 equal quantities, and adding an equal quantity to both sides will not change that equality. It’s just a fancy way of stating what you’ve been doing in algebra already. But in the case of AB + BC = BC + CD, we’re starting with the answers to 38 and 39: AC = AB + BC and BD = BC + CD. We were told at the beginning that AC = BD. So we’d use substitution to substitute AB + BC for AC and BC + CD for BD.

Note that throughout this course, there are a lot of questions that build on the previous one. The curriculum is trying to break down thinking through logic step by step. But if you struggle on one part, you may end up getting the others wrong too. That’s why I don’t have the exercises count in grading–it’s assumed students will get a lot of the exercises wrong as they get a handle on the material. The goal is to learn so by the tests you’ve got it.

**Chapter 3, Lesson 7, Problem 12**

Don’t look at the lines that are next to each other, as they are not perpendicular. The picture here shows the 5 sets of perpendicular lines. (The green , pink ones, red ones, penciled ones, and blue ones are all perpendicular.)

**Chapter 3, Summary and Review, Problem 19 **Notice in the figure where I’ve marked all the information given (which is the first step to solving the problem). OB’s coordinate would be found by taking AOD minus BOD, not minus AOC. That gives 110-80, which equals 30.

**Chapter 4, Lesson 6, Problem 35**

Note that this is a tough problem, so I’m giving a hint here.

This problem is based on several that come before it. In order to reach these conclusions, as you go through the preceding problems, mark the figure with whatever information you learn at each step. So when you learn certain triangles are congruent, mark their corresponding angles and sides as congruent. Using different colors will help in long problems like this where the tick marks could get to be way too many otherwise. (Rather than using 2 tick marks for the second pair of angles you mark, just use a different color pencil.)

Now look at your notes (which are super important to be taking!) and go through everything you know about triangles, seeing what you can tell about these triangles. You can see in the markings that there are two angles in DEB that are green (so the same) and two in AEC that are red (and thus the same). Looking at your notes, what can you then conclude?

**Chapter 4, Lesson 6, Problem 46**

The image shows how AB was found using the Pythagorean theorem. The other measurements were obtained the same way.

**Chapter 5, Lesson 2, Problem 47**

If you are wondering why angle APC is greater than angle AXC, it’s because in problem 46 he extended line AP, essentially drawing AX and forming 3 triangles. Angle APC is an exterior angle to triangle PXC, which means it’s greater than both the interior angles, one of which is angle PXC. Angle PXC is an exterior angle to triangle AXB, which means that it is greater than both the interior angles in that triangle, one of which is angle B. By the transitive property, then, Angle APC is greater than angle B.

**Chapter 5, Lesson 4, Problem 32**

This is an alternate valid proof from a student (way to go, Isaac!): After the first three steps, you can say that AB + BC + CD + AC is greater than AC + AD using the Addition Theorem of Inequality. Then use subtraction (of AC) to prove that AB + BC + CD is greater than AD.

**Chapter 5, Summary and Review, Problem 50**

Here is an alternate valid proof based on one from a student (good job, Isaac):

1) AX + XB is greater than AB – Triangle Inequality Theorem

2) XB = XC – Given

3) AX + XC is greater than AB – Substitution

4) AX + XC = AC – Betweenness of Points Theorem

5) AC is greater than AB – Substitution

**Chapter 6, Lesson 2, Problem 11**

Students who list linear pair as the answer are correct; however, that answer doesn’t help us prove what we’re trying to prove in the next few problems. The questions are building to try to a conclusion in problem 12 that can only be established if we know that they are supplementary–that they add to 180 degrees–because one way we know that lines are parallel is that supplementary interior angles on the same side of a transversal (problem 13).

The *Solutions Manual* is using whichever definition is used in the next point it’s making–supplementary is used as part of the way to know lines are parallel in this case, so it’s being used.

In a proof, writing a linear pair here would be incorrect, as just knowing it’s a linear pair does not automatically lead to concluding the lines are parallel–you’d have to first say that a linear pair forms supplementary angles. The whole idea in geometry is to build step by step. While all linear pairs have supplementary angles, the definition of linear pair emphasizes the idea of a common side and opposite rays–that definition would be used in proofs where that’s what we need to show or where we plan on using another theorem or definition that uses linear pair. We’d use supplementary angle when trying to describe the angle sizes or where we plan on using another theorem or definition that uses supplementary angles, as was the case in this problem. While a linear pair answer can be counted as correct here, if you saw it on a test, in the future, when you get to a follow on question (like question 13) and see what the book was trying to prove, go back to your answer and adjust to supplementary.

**Test 6, Problem 7c, Clarification **

*that uses the information given in the preceding steps to prove the problem*–it’s as if you’re writing out the steps of a proof, so your answer should be a

*logical conclusion of the previous steps alone*. What was already shown in steps a and b would lead to proving that the triangles congruent based on ASA; now, if we added other steps, we could have used a different way to prove it, but the idea is to view the question as showing the next step in a proof (so it needs to build on the previous statements). Note that you may find it helpful on problems like this to redraw the two triangles seperately, marking them with the information given in the problem.

**Chapter 7, Lesson 5, Problems 46-51 **

To those who answered “ABCD is a trapezoid” for number 50, here’s a clarification. In an indirect proof, you assume the opposite of what you’re trying to prove (that’s your hypothesis) and look for a contradiction that shows that can’t be true. Here, in problem 46, we’re supposing that AC and DB bisect each other, which is the opposite of what we’re trying to prove. That’s our hypothesis. So while it is true that the answer to 49 contradicts what we were given that ABCD is a trapezoid, it also means that the hypothesis we assumed in 46 is incorrect, which is what we care about in this case. In an indirect proof, we’re assuming the opposite of what we’re trying to prove and then looking to prove that can’t be the case. We’ve shown that ABCD is not a trapezoid in problem 49; now, we’re stating that means that the hypothesis we had is false…which means that the opposite of it must be true.

**Midterm Review, Problem 97**

97 should read as follows: “Could *this *triangle have sides of lengths a^2, b^2, and c^2? Explain.”

Since we know that a^2 + b^2 = c^2 for this triangle, and we know the sum of any 2 sides has to be greater than the 3rd, we can’t have sides that are a^2, b^2 and c^2, as two sides would then equal the third. In other words, we can’t make a = a^2, b = b^2, and c = c^2.

**Chapter 9, Lesson 2, Problem 6**The answer is really square units, not just units.

**Chapter 9, Lesson 4, Schedule**The schedule should say problems 1-53 on pages 360-364.

**Chapter 9, Summary & Review, Problem 45**

The problem should really include that the altitudes for triangles BDE and CFD as well and say that they are equal to BH and EG. If we knew that, then the *Solution Manual’s* steps would explain the answer.

**Chapter 10, Lesson 1, Problem 47**

Seeing the geometric mean is tricky when you’re dealing with fractions. However, since we know when simplified that F/G = 9/8 and G/A = 9/8, we know that F/G = G/A…which would mean that G is the geometric mean. We’re NOT saying that 9 or 8 is the geometric mean–the geometric mean is G, which is 2/3. We were able to figure out that it’s the geometric mean by seeing that the ratios formed a proportion, and since G was in the denominator of one and numerator of the other of two equal ratios, it’s the geometric mean. Note that if we were to insert the original values for F/G = G/A, we’d have 3/4/2/3 = 2/3/16/27. You can see then that 2/3 is the geometric mean. But we had to simplify the fractions down to 9/8 to see that they were indeed equal fractions.

**Chapter 10, Lesson 2, Number 44**

In problem 41, we set up a ratio between the sides of AO and the sides of A1, getting 1188/*x* = *x*/594 and solving the proportion in 42 to see that *x* was 840 mm. We could set up that proportion because we told that the sizes were similar.

We find the area of a rectangle by multiplying length times width, so we multiplied 1188 by 840, the value we found for *x*. We get 1188 x 840 = 997,920 mm^{2}. That’s the answer in square milimeters, but the question asked for it in square meters. 1 meter = 1,000 milimeters, so we have to convert. 997,920 mm^{2} x 1 m/1,000 mm x 1 m/1,000 mm = approximately 1 m^{2} when you round.

**Chapter 10, Lesson 4, Number 50 **

*Alternate Proof for Students Who Already Know About SSS Theorem from Other Material (Thank you, Isaac!)*

MO = 1/2 AC; MN = 1/2 BA; ON = 1/2 BC Reason: The midsegment of a triangle is 1/2 as long as the 3rd side

MO/AC = 1/2; MN/BA = 1/2; ON/BC = 1/2 Reason: Division

MO/AC = MN/BA = ON/BC Reason: Substitution

triangle MNO ~ triangle ABC Reason: SSS Theorem

**Chapter 10, Lesson 6, Problems 1-9**

These particular questions require thinking about a problem a little differently. You don’t know actual dimensions of the triangles, but you know they’re all congruent. So you that triangle 1 is congruent to triangle 2 and 4. You know that corresponding parts of congruent triangles are congruent, so BC equals the corresponding line segment in triangle 2 and 4…and those segments added together equal DE. So that means the ratio between BC and DE, or BD/DE is going to be 1/2–1 segment divided by 2 of those same segments.

Again, you don’t know the actual dimensions. But whatever they are, you know they’d simplify down to 1/2 because you know from congruent triangles that BC is half of DE.

Problem 2 uses your answer to problem 1 and Theorem 48. That theorem says the ratio of the areas of two similar polygons (and we know these triangles are similar polygons since all their angles are the same) is equal to the square of the ratio of the corresponding sides (which is what we found in problem 1). So we square the answer to problem 1.

The other problems follow similar logic to those first two.

**Chapter 10, Algebra Review, Problems 21-23**

On problem 21, The left hand side of the Solutions Manual shows 1) multiplying both sides by Rr_{1}r_{2} and distributing the multiplication. When you multiply the left side by that, you end up with r_{1}r_{2}; the right side Rr_{2} + Rr_{1}. 2) Here they’ve swapped sides of the equation to put R on the left AND factored out an R at the same time. They skipped showing the step of writing Rr_{2} + Rr_{1} = r_{1}r_{2}. 3) Divide both sides by r_{1} + r_{2}.

Problem 22 follows similar steps to those above.

Problem 23, they multiplied both sides by vur and distributed the multiplication, continuing from there.

The general idea is that we multiply both sides by a common denominator, which we find by multiplying each of the denominators together (so RR_{1}r_{2} in problems 21-22). Then we distributed the multiplication and simplify.

This skill was covered on pages 614-617 in *Elementary Algebra*. Do not stress over this if you used a different course that did not cover this. It is covered in Algebra 2 in more depth. Feel free to skip these problems if you don’t have *Elementary Algebra* to look back at and the above explanation is not sufficient.

**Chapter 11, Lesson 6, Problem 63**

The answer can be found algebraically. You can take the reciprocal of both sides of an equation without changing the value, as shown in pic 1 below.

If you’re wondering why you can take the reciprocal of both sides, it’s really the same as multiplying both sides by the denominators and then dividing them both by the numerators, as shown in pic 2 below.

**Chapter 12, Lesson 4, Problem 18**9

*This is an explanation of how this case could be proved in a similar way for those curious.*

We’d draw PO (2 points determine a line) and continue it through point B.

< APB = <1 + <2 (Betweenness of rays theorem)

<APB = 1/2 mAC + 1/2mCB = 1/2 (mAC + 1/2mCB) (substitution based on what was proved in Case 1)

mAC + mCB = mAB (arc addition postulate)

<APB = 1/2mAB (substitution)

**Chapter 12, Lesson 6, Problem 29 (Number 39 Is Similar) **

*This is an explanation of how the answer is obtained. Note that problem 39 follows similar reasoning.*

Look at the graph drawn in the Solution manual for problem 28. The distance formula is to find the distance between two points–if you look it up in the Index, you’ll find it is explained on page 134. Here, we know that the distance between the center point (which is 0,0) and

*any*point on the circle is 5, as we’re told the radius is 5. Thus,

*d*in the formula is 5, and we can plug in 0,0 as

*x*

_{1}and

*y*

_{1}(or as x

_{2}and y

_{2}–it really wouldn’t matter) and

*x*and

*y*as the others since those are the symbols we’re using in this case. That leaves us with √[(

*x*– 0)

^{2}+ (

*y*– 0)

^{2}] = 5 OR √[(0 –

*x*)

^{2}+ (0 –

*y*)

^{2}] = 5. In either case, if we square both sides, we get that

*x*

^{2}+

*y*

^{2}= 25.

**Chapter 13, Lesson 3, Number 14**

You could also use SAS to prove the triangles congruent. Note on page 244 that SAS is part of the proof of HL.

**Chapter 13, Lesson 3, Number 19 **The answer is based on the answer to problem 16, where we saw that the sums of the lengths of the quadrilateral’s opposite sides must be equal. The bases are two sides of a trapezoid and the legs the other two sides. We’re basically just restating problem 16’s conclusion for a trapezoid.

**Chapter 13, Lesson 3, Numbers 34-36**The answer to 35 can also be found using the Reflexive Property with PC and PB and still applying AAS. The answer to 36 is based on the answer to 35. If you look at the triangles shown congruent in 35, PF and PG are corresponding parts of one of the pair of triangles, and PG and PH are corresponding parts of the other pair. Corresponding parts of congruent triangles are congruent. And since both PF and PH equal PG, they must equal each other as well.

**Chapter 13, Lesson 4, Number 51**

The answer mentions that the altitude is the third altitude. The idea of it being the third altitude is not important to have in the answer. The other two altitudes were already identified in 49, and each triangle has 3 altitudes, so that’s why it said the third. It would have been the first if it was the first one drawn. The important part of the answer is that AF is the altitude and an understanding as to how we know that.

**Chapter 13, Lesson 5, Set IV (Optional Exercise; Explanation) **For this problem, we are told the area of the frame equals the area of the picture, which is 4^2, or 16. We can think of the frame as 4 rectangles with width x and the same length as the length of the picture, which we know is 4 cm, plus the 4 corners, which are going to be squares with sides x. That gives us this:

4(4x + x^2) = 16 We have 4 cm times x, so 4x plus the corner area of x^2, and there are four sides and four corners, so we’re multiplying all that by 4.

We can then distribute the multiplication to get this: 16x + 4x^2 = 16

Rearrange to be in quadratic form:

4x^2 + 16x – 16 = 0

Factor out a 4:

4(x^2 + 4x – 4) = 0

Solve the quadratic using the quadratic formula to get -2 plus or minus 2 times the square root of 2. The solution left off the minus sign as with geometry, we’re looking for the positive answer (we’re not going to have a negative length).

**Chapter 13, Lesson 6, Problems 1-3**

A key to understanding these problems is to remember that on a coordinate graph, unlike in regular geometry, we take into account direction and thus have positive and negative. We consider down and left as negative, and right and up as positive.

The slope of DA is -3/2 because if you start at D and look at the rise/run, you’re going down 3 (rise) and to the right 2 (run), getting -3/2 (the 3 is negative since it is in the downward direction). For CB, if you start at C, you’re going down 3 and to the right 2, so -3/2 as well.

The product of the slopes of two consecutive sides would be 2/3 x -3/2, which simplifies to -1. We saw in problem 1 the two sides that have a slope of 2/3; the other two (that are consecutive to those) have -3/2, as seen in problem 2.

Problem 3 is asking why it equals -1. It’s referring back to a theorem covered on page 463.

**Chapter 14, Lesson 3, Problems 34 and 35**

Explanation of Answer: We know from 31 that we’re dealing with a 30-60 right triangle. We know based on Theorem 51 that PB is r/2 and *a* is r/2(SQRT3). (Answers to 32 + 33.) 35 asks for the area of the larger triangle (AOB) in terms of *r*. We find the area by taking 1/2 times the base times the height, as A = 1/2(bh). The height is *a*, which we saw in 33 that we can write as r/2(SQRT3). The base is 2 times PB, which we can saw in 32 that we can write PB as r/2. Multiplying that by 2 would just yield r, as r/2(2) = r. So plugging in r for the base and r/2(SQRT3) for the height in the area equation gives us A = 1/2(SRT3))r, which equals r^2/4(\sqrt3), which could also be written the way it is in the Solutions Manual as \sqrt3/4(r^2). (Algebraically, they mean the same thing.) 35 is asking for the area of the hexagon, which we find by multiplying the area of triangle ABO by 6. So we’d multiply \qsrt3/4(r^2) times 6 to get the answer given in 35.

**Chapter 15, Lesson 7, Number 45**

In the Solution Manual, there should be an A instead of a V = and the answer should be mi^2 not mi^3.

### Schedule

**Day 124 should be**

Watch Algebra Review: Graphing Linear Equations.

Complete exercises 1-20 on page 482.

Continue reviewing for test.

**Day 125 should be**

Take Test 11 on pages 81-84 (TG).

Review test results.

Help! I’m Struggling.

If you’re struggling through the first several chapters, take heart! It should get easier. Geometry requires training the brain in a lot of logic. Keep in mind that the problems are designed to be exercises and to stretch you! It is okay if you get some wrong–use the tests as your assessment. Here are few things to make sure you’re doing:

**Make sure you’re taking notes**in a way that you can easily find everything you know about a specific topic! More info on this is in the eCourse Downloads under Topical Notes.**Make flashcards.**(Whether on Index cards or using a site like Quizlet.com.) Becoming familiar with the definitions/postulates/theorems as you go is really important so you don’t get overwhelmed.**Use the answers given in the back of the textbook**to help you. Check those problems as you go and use them to help you on the other problems.- Parents, you can
**give hints**on some problems if needed. Avoid having the student look at the*Solution Manual*when trying to solve unless absolutely necessary. Hints would be more helpful in building their critical thinking skills. **Use colored pencils**to mark every piece of information you learn as you learn it. That way, when you’re asked to draw a conclusion, you can see visually all you know already. An example is given below from Chapter 4, Lesson 6, problem 35. (Rather than using 2 tick marks for the second pair of angles you mark, just use a different color pencil.)

- On
**true and false questions**, remember you’re trying to decide if the statement is*always*true. If it’s not always true, then it’s a false statement, as you can’t make that conclusion all of the time. I know these questions are tough. Don’t give up and try to remember you’re trying to really think through the definitions and the theorems/postulates related to see if you can make the statements definitively or not. Often we’re using geometry to measure things (like objects in outer space) we can’t physically measure, so we need to know if we can safely make conclusions or not.