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Hard Math Middle School Hard Math for Middle School: Solutions for Workbook Glenn Ellison Chris Avery 1 Copyright © 2021 by Glenn Ellison and Chris Avery All right reserved ISBN-10: ISBN-13: 978- 2 Introduction Welcome to the solutions manual for the Hard Math Workbook! If you are reading this, we assume that you have just gotten a copy of the Hard Math for Middle School textbook and its accompanying workbook creatively titled Hard Math for Middle School: Workbook. If not, you should get them. This solution manual is super cheap, but it is of very little use without them. The Workbook supplements the text by presenting a page of problems on each of the 100+ sections of the textbook. Following the Hard Math philosophy, the problems in the workbook are intended to be hard because we think it is fun to work on things that are so hard that you can feel good if you get them right, not bad if you get them wrong. A difficulty with making the problems so hard, of course, is that almost all middle school math teachers also will not be able to do quite a number of them. It can be educational to work on a hard problem even if you don’t ever get it. But, working on a problem for a while and then having someone explain what you were missing is even better. We hope this solution manual will make the Workbook even more fun and useful. And we hope that it will make teachers and math team coaches more comfortable giving the Workbook to their students. It has taken multiple years to get this solution manual done. In part, this is because we have jobs. In part, it is because we slowed down after our daughters finished middle school. But, it is also because there are lots of problems in the Workbook and they really are hard. Writing out solutions took quite a while and made us realize that a number of answers in the first printing of the Workbook were wrong. We owe special thanks to all of the people who have helped us out by pointing out mistakes, especially Sam Sun. Enjoy! © 2021 Chris Avery and Glenn Ellison Meet #1 2.1 Basic Definitions 1. An obtuse angle has a measure is greater than 90°. The only such choice is 120°. 2. In the figure, the angle with the smallest degree measure is angle C and the angle with the largest degree measure is angle B. This implies that the measure of angle A is 40° and the measure of angle C is 20°. The positive difference is 20°. Here, one should be able to recognize which angles are larger by looking at how pointy they are. Another useful fact the largest angle in a triangle is the one opposite the longest side and the smallest angle is the one opposite the smallest side. Here the side opposite angle A (side BC) is clearly longer than the side opposite angle C (side AB), so angle A has a larger measure than angle C. 3. The measure of angle ABD is less than the measure of angle ABC. The largest multiple of 7 less than 40° is 35°. 4. The addition facts are that DAB + CDA = 200° and BCD + CDA = 210°. From the similarity between these expressions (or subtracting the first from the second), we find that BCD = DAB + 10°. Two of the four angles are obtuse and angle ABC is not obtuse. Hence, exactly two of DAB, BCD, and CDA are obtuse. Because BCD is larger than DAB we there are only two possibilities: (1) CDA and BCD are obtuse and DAB is acute; or (2) BCD and DAB are obtuse and CDA is acute. Consider each case: Case (1): One of BCD and CDA must have a measure of 100° so the other would have a measure of 110°. The measure of BCD cannot be 110°. If it were, then the measure of DAB would be 100° and there would be three obtuse angles. So in case (1) the only possibility is CDA = 110° and BCD = 100°. We can fill in the other angles and satisfy all conditions from here. DAB = BCD – 10° = 90° is not obtuse, and ABC = 360° – (CDA + BCD + DAB) = 360° – 300° = 60° is acute. The answer is 90°. If under time pressure you should stop after case (1), but to check your work it could be useful to do case (2) as well. Here, BCD cannot be the 100° angle because that would imply that DAB = 90° is not obtuse. So we would need to have DAB = 100°. However, the first addition fact, DAB + CDA = 200°, then tells us that CDA = 100°. This contradicts that BCD and DAB are the only two obtuse angles. Hence, case (2) does not lead to any valid solutions. A - 1 Meet 1 Solutions © 2021 Chris Avery and Glenn Ellison 2.2 Adding Up Rules 1. Angles are complements if their measures add up to 90°. 90° – 65° = 25°. 2. Angles BOD´ and D´OH are supplements so the D´OH = 180° – 155° = 25°. Similarly, BOG and GOH are supplements so GOH = 180° – 140° = 40°. The positive difference is 40° – 25° = 15°. The problem can be done more quickly by realizing that if one angle is 15° larger than another, then its supplement will be 15° smaller than the other’s supplement. 3. Let x be the measure of the angle in degrees. The text tells us in words that 180 – x = 2(90 – x) + 20 Expanding both sides we get 180 – x = 180 – 2x + 20. Cancelling the 180’s from the two sides and moving all the x’s to one side gives x = 20°. Students who are not comfortable with algebra could alternately make a table like the one shown below, trying out a couple nice round numbers to start, and then trying larger or smaller numbers to move toward the solution. Angle Supplement Complement 2×Complement+20 40 140 50 120 50 130 40 100 30 150 60 140 20 160 70 160 4. As in many problems like this, a good approach is to just keep looking for angles you can fill in with the facts you have (and have just discovered). First, because BMC and CMB are complements we fill in CMB = 90° – 53° = 37°. Next, we look to use this new fact. Since OMB and CMB are complements, we can fill in OMB = 90° – 37° = 53°. Finally, using that OMG and OMB are supplements we fill in OMG =180 – 53 = 127°. 5. Translating the words to algebra we have 180° – (90° – (180° – x)) = x – 7.2°. This simplifies to 270° – x = x – 7.2°, which gives 2x = 277.2° and x = 138.6°. To translate the words to algebra, some students will find it better to work back from the end step by step instead of trying to do it all at once: the supplement of the angle is 180 – x, the complement of the supplement is 90 – (180 – x) = x – 90, the supplement of this is 180 – (x – 90) = 270 + x. A - 2 Meet 1 Solutions © 2021 Chris Avery and Glenn Ellison 2.3 Equality Rules 1. DOG and PIG are corresponding angles on parallel lines. Corresponding angles on parallel lines have the same measure so PIG = DOG = 140°. 2. Angles DGF and COF are corresponding angles on parallel lines so COF = 48°. Angles COP and COA are complementary, so COA = 180° – 70° = 110°. Now by subtraction we have FOA = COA – COF = 110° – 48° = 62°. A F m C O n D G P 3. In triangle EMB we have EBM = ABM = 42° and BME = 180 – BMD = 90°. The angles in a triangle add up to 180°, so BEM = 180° – (42° + 90°) = 48°. Now, using that BEX and BEM are supplements we have BEX = 180° – 48° = 132°. Note that in this problem the lines OM and MC were just drawn in to make the diagram more complicated and make it less obvious which angles students should be trying to fill in. Often in angle chasing problems you do want to just blindly start filling in angles one by one, trusting that you’ll eventually get to the answer. But if you’re having a hard time getting started or the angle chasing is taking a while it is sometimes better to step back and think about where you’re trying to get. Here, I’m trying to find BEX. To find this I’ll probably want to first find AEX or BEM and then use supplements. I don’t see any way to get AEX so let me think about how I could get BEM. This will have to come from adding up in some triangle, … At which point, you hopefully come up with the first step of using adding up in triangle BEM. X O A B E M C G D A - 3 Meet 1 Solutions © 2021 Chris Avery and Glenn Ellison 2.4 Angles in Polygons 1. The angles in any quadrilateral (four sided polygon) add up to 360°. Here, the three angles that we are given add up to 130° + 70° + 100° = 300°. So, the fourth angle we are trying to find is 360° – 300° = 60°. 2. The exterior angles in an N-gon add up to 360 degrees. So exterior angle in a regular decagon is 360°/10=36°. The interior angle is the complement if this, 180° – 36° = 144°. 3. The exterior angle in a regular hexagon is 360°/6 = 60°. If the exterior angle in a regular N-gon is the complement of this, its measure is 90° – 60° = 30°.
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