Course Extras: Jacob’s Geometry

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).

Corrections on Schedule

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, 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 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.)

Picture of Problem

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.

Drawing of Problem

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?

geometry figure

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.

problem 46

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.

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 mm2. 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 mm2 x 1 m/1,000 mm x 1 m/1,000 mm = approximately 1 m2 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, Algebra Review, Problems 21-23

On problem 21, The left hand side of the Solutions Manual shows 1) multiplying both sides by Rr1r2 and distributing the multiplication. When you multiply the left side by that, you end up with r1r2; the right side Rr2 + Rr1. 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 Rr2 + Rr1 = r1r2. 3) Divide both sides by r1 + r2.

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 RR1r2 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 12, Lesson 4, Problem 189
This is an explanation of how this case could be proved in a similar way for those curious.

Geometry PictureWe’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 x1 and y1 (or as x2 and y2–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 x2 + y2 = 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 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.

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.)
    geometry figure