Answers to Previously Asked Questions

No hw tonight.

This is very unlikely. I will not be picking them up until tomorrow. I will definitely have them back for you on Wednesday.

I am still working on it, but as of now, there are 3 proofs, one of which is a fill-in-the-missing-parts. There are 3 constructions, a few short answer, a few multiple choice, and a few true/false. It is a pretty long exam, but you will definitely have enough time to complete it.

#10: Use info from the last example I did in class on Thursday--the hexagon is very special as far as the angles formed by the radii (you get a 30-60-90 triangle). Also, you know that the apothem bisects a side of a regular polygon.

#11: This guy forms another special triangle (30-60-90) with the apothem, radius, and side. Use the properties of that triangle along with the fact that the apothem bisects the side.

See the 2nd to last post on this page.

See above.

We haven't discussed formulas for this (as far as I understand your question). This has to do with the perimeter/area of a cirlce, and we will discuss this on Tuesday.

Firgure this out by computing the areas of each triangle. The middle triangle is an equilateral triangle, and so you can compute the height by dropping an altitude and using the info about the 30-60-90 triangles. Additionally, the two outer triangles are 45-45-90 triangles, so you have information about them. You know the hypotenuse of each is 8*squareroot(2) (because this was the lenght of a side of the equilateral triangle). This will let you figure out the other legs of the triangles and then compute areas of each.

Compute the area of the rectangle and trapezoid separately (using the only formula we discussed for each). Also, you know XT = RS because RSTX is a parallelogram.

Chapter 6, and only the easier area formulas from section 7.1 and 7.2 (not 7.3).

#14: This is a parallelogram, so BC = AD. But 6 = CD = BC = AD. So now you have the base and height of the parallelogram. Just use the parallelogram formula now.

#17: Try letting x = length of second side. Then 76 = 2x + x + (x+12).

#21: For part (a), just plug this into the formula of theorem 7.2.6. For part (b), you need to first compute that ratio of s₁/s₂, by taking a side of the first and dividing it by the corresponding side of the second. THen apply theorem 7.2.6.

#9: Get a system of 2 equations in 2 unknowns. First equation: m(angle B) = 1/2 (m(arc ADC) - m(arc AC)). Second equation: m(arc ADC) + m(arc AC) = 360. After solving those, you should be able to get the rest.

#23: I'm not sure which parts you had trouble with, but you should use the formulas in the theorems of section 6.2. Remember: angle outside = 1/2 (difference of arcs); angle inside = 1/2 (sum of arcs); angle on circle = 1/2 (intercepted arc).

I'm not sure what you're talking about.

Use the formula: area = 1/2 perimeter * apothem. (You will need to calculate the perimeter for each question). Also, things may be simpler for you if you use a calculator for these questions.

From which section?

Use the fact that the diagonals of a parallelogram bisect each other, along with corollary 7.2.4.

Heron's thm will not be on Monday's test.

To find the measure of an angle inside the circle, use thm 6.2.2 (1/2 the sum of intercepted arc and arc intercepted by the vertical angle).

Use the formula given to us by theorem 6.3.6 (product of whole leg and outside bit = product of other leg and other outside bit).

#14: You need to figure out AD. But you know this is a parallelogram with BC = CD. Opposite sides of a parallelogram are congruent. So 6 = CD = BC (given) = AD (opposite side). Use formula for area of parallelogram with that info and the height (length of the altitude).

I'm not sure which other question you meant to ask about.

It will be one of: construct the circumcircle, construct the incircle, construct the orthocenter or construct the centroid.

The intersection of the medians is the centroid, so asking either is asking the same.

The centroid occurs 2/3 of the way down the triangle. Because this is an isosceles triangle, median from the vertex angle (at the top) to the base is the same line as the perpendicular bisector. This means that angle RZS is a right angle and that SZ is half of ST. Using information you have, you can use the Pythagorean Theoreom to calculate RZ. 2/3 of this number is RQ.

To compute SQ, you need to look at right triangle SQZ. You can calculate QZ because RZ = RQ + QZ. You've already calculated SZ. One more application of the Pythagorean theorem will give you SQ.

For 16, look at it as a rectangle and a triangle. You are given the dimensions for the rectangle, so it's just length times width. For the triangle, you are given the base (10), and you can calculate the height. The entired structure has a height of 13. The rectangle is only 9 ft tall... How tall (or what height) does that make the triangle?

For 17, you need to use the Pythagorean theorem to calculate that missing length... It's a base for both those triangles.

I understand your concern. Many have, in fact, come to me with this same request. Let me first address the issues behind removing the exam, and then I will let you know my current plans for Monday's test.

First of all, removing the exam entirely would leave 10% of your grade floating around. There would be a couple of options for this 10%. One is to absorb that 10% into the final. However, this would then make your final exam worth 30% of your grade, which, given the exam scores of most students,is too much risk. I could evenly distribute that 10% across the other portions of your grade; however, this eliminates using that 10% to really boost your grade (assuming you study). Both of these ideas are farfetched anyway because on the syllabus, I promised 10% for each of 5 exams.

Additionally, by studying for Monday's test, you are actually studying for the final as well. Not only will the material from Monday's test be on the final, but, in case you hadn't noticed, the new material in this course really builds on past material. Studying for this test will make you look back at previous ideas discussed.

Finally, you wonder why I don't make this a large quiz. To be honest, this goes back to the promised 5 exams each at 10%. I will not give a "large quiz" the same weight that I would the other closed note/book exams. This then means that people will be losing the opportunity to make up for a really rotten exam score that they may have "earned" along the way.

Now... What about Monday's exam... I am currently planning on making it easier than previous exams but still closed note/book. You still have to know the material, but you don't need to know it in as much depth as I've required in the past. Monday's exam will consist of problems relating to the following:

1. Basic circle vocabulary
2. Angles and circles (ie: formulas for angles formed inside, outside and on the circle)
3. Formulas for secant segments/chord segments (as in 6.3)
4. Inequalities in the circle
5. Construction/ideas of circumcircle/incircle/orthocenter/centroid
6. Basic area formulas from 7.1/7.2 (not 7.3)
7. No loci/No proofs

It is definitely still possible. From looking at people who have grades close to that, it looks like you would need to score high C's on the next exam and the final exam.

Yes... I talked about 31 on the slide "Inequalities in Circles" and 32 on the slide "Application of the Hinge Theorem to Circles".

While we were working on completing homework problems in class today, someone brought this request up. We had a discussion about it, explaining fully what each option means, and decided that it was best to leave the test on Monday.

I am still open to discussion on this issue, though, if you feel you have good reasons.

See above.

See above.

Hi!!! :)

6.3 #9: Try letting EC = x. Then how can you express ED in terms of x? Use these to set up the equations.

6.3 #13: Let x = AD. Set up the normal equation for the secant segments. Then use the information you found for AD along with the information given for AE in order to find DE.

This is a toughy... First, since m(angle S) = 15, we know that m(arc RV) = 2 * 15 = 30 since RV is the intercepted arc. But this means that the central angle associated with arc RV also measures 30. Ie: m(angle ROV) = 30.

But V was the midpoint of the arc. So m(arc SV) = m(arc RV) = 30, and so m(angle SOV) = 30 as well (you're going to need to drawn in line segment OS). This is now enough to get triangle ORT congruent to triangle OST by SAS (S = radii; A = we just got those congruent; S = common side OT). This means (by cpctc) that angle OTR and angle OTS are congruent; congruent adjacent angles are right angles, so we know m(angle OTR) = 90. Since m(angle ROT) = 30 and m(angle OTR) = 90, we know m(angle ORT) = 30. So you are dealing with a 30-60-90 triangle and you know one of the sides...

First, what equation do you get using the Intersecting Chords theorem we discussed in class today? Also, what does it mean to be a midpoint? Take the equation you get from the intersecting chords theorem (looking at chords AB and CE), and then substitute using the equation you get from the fact that D is a midpoint. Divide this new equation appropriately, and you should be left with the equation for the geometric mean.

I don't have a clue. My guess would be a D in college means a D in high school, but that's only a guess. Your high school counselor should know the answer as they are the ones who help transfer your grades over. (If you find out, let me know, and I'll post the answer up here.)

These questions get emailed to me, so I check them as often as I check my email. How often I check my email varies. I try to check several times a day.

To find MRP, you need to find MV. To get MV, use the fact that arc PMV = 180, and subtract from that arc MP. Then use the subtraction formula. To get angle 1, just apply the formula (you have enough info from the problem). To get angle 2, you need NP (look at MP - MN) and TV (look at PMV - (MP + MT)), then apply the formula.

No constructions.

Roughly same length as previous exams.

Suggested preparation for the exam along with the sections is listed on gradekeeper.

2 proofs. Test will be similar in style to all the other ones you've taken.

There will be 2 proofs from the problems I listed under suggested preparation for the exam.

As soon as I grade your final exams, those scores will be up on gradekeeper, so you will be able to know your grade immediately. As far as officially on your transcripts, I am not sure. They will show up on webadvisor, but I don't know if they mail them out to you. You can contact admissions and records to find out. Also, you can always obtain an official copy of your transcripts from them.

You are right. This class goes fast. We are condensing an entire year of high school algebra into only 23 class days. We need to get through all the material that would be taught in the normal class. Completing this class is definitely a challenge, and is not for everyone. It is my job to make sure we get through all the required content and to guide you through it. Ultimately, learning the material is up to you.

Success in this class is about 50% knowing the statements of the definitions/theorems cold and 50% being able to apply each of these. Memorizing is not something I can do for you. I do expect you to be able to recite to me definitions/theorems we've covered previously (which I test with the true/false questions). I can help you understand things, but if you are not understanding the material, you need to take initiative and come to office hours for extra help.

I am not sure how failing grades are recorded. My understanding is that a notation of "W" (for withdraw) on the transcripts is better than an F. Your best bet is to contact admissions and records and ask them.

We will be covering 1-7 and parts of 10.

For this one, try looking at the proportion AD/EH = AB/EF. You just need to calculate AD. A proportion similar to this one will help you get the other parts (FG and GH).

Basically, it's best to try to raise everything. If you're looking for one over the other to focus on, I would say the quizzes. The quizzes in total constitute the largest chunk on your grade. Plus, doing well on them will probably mean you are understanding more, and it will help you on the exams.

This doesn't mean stop studying for the exams. Just really make sure you understand the homework everyday. And you can always check your answers for the homework quizzes with me (either by email (you actually have to send an email because I don't post full answers on the website) or in person at my office hours).

Hopefully by tomorrow. I'm down to grading the last page.

Use Theorem 5.3.7 on page 224. Or did you mean that you don't understand how to interpret the results of that theorem?

I just finished the exam. It contains 14 True/false, a couple short answer, 1 construction, and 3 proofs.

See above.

See above.

See above.

See above. Or for content on the exam, look it up under http://thegradekeeper.com.

Constructions will be similar to ones you would find in section 3.4.

The main thing that factors into this is your hw quiz average. If you have a 90% going before said test, scoring better would only mean scoring within 10 percentage points of 90. That's all that could affect your grade in the positive. But if you have a 90%, you can score more than 10% less than a 90, which would lower your grade more than scoring 10% higher.

Nope!

Think about the definition of a square -- it has 2 consecutive sides congruent. What other object has that? What do you know about its diagonals?

For #2, see #29. For #6, see #28 or #30.

4.3 #28: Use the picture from the slide today entitled "Theorems About Rectangles" (of course your given/prove will be different). You need to show the parallelogram is a rectangle (ie: it has one right angle). First show the same triangles I talked about on that slide are congruent (this time by SSS). Then show that those two top angles are both congruent and supplementary (and therefore right angles).

4.3 #29: This time look at the slide called "Theorems About Rhombuses". You need to show the parallelogram is a rhombus (ie: has 2 consecutive sides congruent). Get the two triangles I talked about congruent (this time by SAS). The prove statement will follow after that.

4.3 #30: Put together the two theorems you just proved in #28 and #29

4.4 #23: See slide titled "Isosceles Trapezoid Theorems". Draw in two diagonals. Show the triangles formed are congruent by SAS; then use cpctc.

4.4 #25: Use theorem we had about 2 lines forming supplementary angles implies that the lines are parallel...

4.4. #26: This is very similar to the one I did for Thm 4.4.1 on the "Isosoceles Trapezoid Theorems" slide. Construct the same line. This time, though, instead of using the sides to get at the angles, you're going to use the angles to get to the sides... Looking at the triangle formed, you will be able to show the angles are congruent, and therefore, the sides opposite those angles are congruent.

These are just trying to get you to see geometry in the real world. For 4.2 #15, when the wheels drop, they are both must be the same distance from the ground (so they are congruent). Also they will be perpendicular.

4.2 #16: Since AM = MB and MC = MD, we must have that AB and CD bisect each other.

For 22, you can get the triangles congruent by SAS. Because MNPQ is a parallelogram we know angle Q = angle N, MQ = NP, and MN = QP. Use the fact that T and S are midpoints together with MN = QP to show that QS = TN (you also get MT = SP, which we need to show MSPT is a parallelogram). This will give you enough to show the triangles are congruent.

23 and 24 I went over plans for in class. You could ask me questions on what I did or you can come to my office hours (H-2 8:30), and I'll help you through them.

Use what we know about opposite angles in parallelograms, vertical angles, and the transitive property of congruence to put that one together.

This is a question to get you thinking about ideas to come. I think I'll tell you the answer and let you try to figure out why that might be true. DEC is a right triangle. Triangles ADE and BCE are both isosceles.

I practically did each of these in class today, so I can't do much more than that without actually doing it out for you (which I can do if you come to my office hours--H-2 at 8:30 am). Perhaps you could ask me questions on what I did, and maybe we can fill in the blanks that way.

That's a pretty low D.

Same place they always are... scroll down. Under Homework and Lectures...Under Week 3...Under Tuesday - 6/29/04: Quadrilaterals I.

I assure you that as soon as I make them, lecture notes will be online.

3.4 #18: Construct a 45 degree angle like you did for #15. Then construct a 60 degree angle like we did in class. Add those two angles together like you did for #11 or #12.

3.4 #21: Complement means construct the bit that will add to that angle to give you a right angle. Basically, construct a right angle and then subtract angle Q like you did for #10.

3.4 #31: To calculate the interior angle, refer to the formulas in section 2.5. You can either email me what you got, and I can help you from there, or you can try to put together 60 and 90 degree angles in a way that will give you what you want.

3.5 #18: You need to figure out if that 96 is a base angle or a vertex angle. Suppose it is a base angle. What does that mean at least 2 angles in the triangle measure? When you add those together, what do you get? Is that a problem?

3.5 #24: Write an equation for DC using the Segment Addition Postulate. Then use substitution to get the inequality you want.

3.5 #30: Change the statement to be of "If ... , then ..." form. Draw a picture, pick out your givens/proves. Let me know what you get for this, and I can help you with the rest.

3.5 #33: See #30.

3.4 #28: This one works out like the 3rd justification I did in class. First, concentrate on getting the big triangles (CAD and CBD) congruent. Then use the information you get from there to get the little triangles (CAM and CBM) congruent. Using cpctc, you will get your result.

That's a pretty involved problem. I'll assume you mean you want help from the beginning, so this is going to be a few step process. Let me break down the first part in steps: First you need to rephrase the statement into "if ..., then ..." form. After you've done that, make a picture to describe the hypothesis. Then pick out your given and prove statements using your picture. When you get all that, email me again, and I'll check them and help you with the proof.

It's quite likely for there to be at least 1...

Use the fact that VY is perpendicular to XZ to get an equation... What must angle 1 + angle 2 equal? Use that equation to figure out x, angles 1, 2, and 4. What relationship to angles 3 and 4 have (again... this is coming because VY is perpendicular to XZ)?

Use the parallel fact to get the rest. What are angles 1 and 5 called with respect to the parallel lines and the transversal? How do they relate? What about angles 6 and 7?

I'm working on them right now. They should be up very soon.

This one is completely dependent on the fact that the lines are parallel. Because AB || CD, we get that angles DCA and A have to be supplementary (interior angles on same side of transversal). We get a similar result from the other angles in question via the other parallel lines.

You can abbreviate; don't abbreviate any shorter than what I do in class.

It's about the same in length. There are about 7 true/false, 10 short answer, 1 construction, and 3 proofs.

Using exterior angles and sum of angles in a triangle, you should be able to get 2 equations. This will give you a system of 2 equations in two unknowns, which you can then turn to algebra to solve.

For #24, try to answer the following questions to get you started in the right direction: How much do angles A and C measure? How much do all the interior angles of the quadrilateral add up to? How much of that do you have already? And what's leftover?

It is possible that there will be constructions on tomorrow's test.

I agree; the homework is challenging. One of the main reasons for this is that this is an extremely condensed version of geometry, so you don't have the luxury of time to let the concepts sink in (the main drawback, I feel, to summer courses).

As far as empty seats, I happen to know there were at least 2 chairs not being used. Additionally, I can always ask for more desks.

It is hard because you have a lot to memorize. However, the definitions are as pared down as they can get. I suggest you go through and draw a picture to go with each; write a few words yourself--make it your own. As I suggested in the beginning of the class, making this list yourself will help you commit them to memory.

We can't get that angles 1 and 4 are congruent, but that's okay. We want to show the lines are parallel. The only way to do this (via info given) is to take a look at angles ABC and BCD. We need to show these are supplmentary, then you can use a theorem from this section to complete the proof.

How, though, can we show those angles are supplementary? That's where the givens come in. Using the fact that BE and CE are bisectors, you can get 2 equations. Also, by using the Angle Addition Postulate, you can get two more equations (from ABC and BCD). Try to put the equations you get together so that you can say that mABC + mBCD = 180.

For 21, if you can get angles 1 and 2 congruent, you can use one of the theorems from this section to finish the proof. Starting from the beginning, what does it mean when they say DE bisects CDA? What does that do to angles 2 and 3? How can you put that together with the other given?

For 29, there isn't really much I can say about that one other than to regurgitate what I said in class. Let me try to rephrase it, and if doesn't work for you, come to my office hours (H-2 at 8:30 am), and I'll take you through it step by step. Basically, your drawing will consist of 4 lines--3 lines that appear parallel and 1 transversal. Consider what you get from corresponding angles. Look at the drawing for exercise 18, considering only angles 1, 2, and 4. That is the setup you need.

For 21, if you can get angles 1 and 2 congruent, you can use one of the theorems from this section to finish the proof. Starting from the beginning, what does it mean when they say DE bisects CDA? What does that do to angles 2 and 3? How can you put that together with the other given?

For 29, there isn't really much I can say about that one other than to regurgitate what I said in class. Let me try to rephrase it, and if doesn't work for you, come to my office hours (H-2 at 8:30 am), and I'll take you through it step by step. Basically, your drawing will consist of 4 lines--3 lines that appear parallel and 1 transversal. Consider what you get from corresponding angles. Look at the drawing for exercise 18, considering only angles 1, 2, and 4. That is the setup you need.

Probably tomorrow morning before class.

Just 1.7 for Monday's quiz. Probably one proof with a couple other questions thrown in.

For 26, first of all, it is easiest to start by thinking about why you know AM = MB and why DN = NC. Once you've got that, the only missing part of the chain in the prove statement is to show MB = DN. Ask yourself how the measure of MB relates to the measure of AB, and likewise for how DN relates to DC (ie: What does it mean to be a midpoint?). Remember that DC = AB, and this should be enough to connect those chains in the prove statement via substitution.

For 28, working backwards, you are trying to show EBF is a right angle. What two angles is EBF made up of? What equation do you get from this fact? You are trying to show EBF is a right angle, so what sum are you trying to make equal to 90? Now we start working from the givens. Using each of the given statements, you can get at least one equation for each of them, just by applying definitions and postulates. Write the equations you get, and try to put them together by substitution.

For 29, the hint really gives you this problem; only minor adjustments are necessary. This means line 1 will change to supplementary, which will change lines 2 and 5 accordingly. Carry this change all the way through. The reasons will be the same.

We will cover Chapters 1-7 and 10.

I can definitely help. However, I need a little more specific information. What problems have you tried? What have you tried for those problem so far? Are you able to get the drawing and given/prove statements okay?

The assignments are consistent. Be sure to look at the correct dates. Homework I assigned on 6/15 would be on the 6/16 quiz, so it would show up under the assignment "6/16 Quiz - Sections 1.3, 1.4". thegradekeeper.com holds the assignments (the quizzes), not the homework.

For what I had under homework for today, look under Exam 1 on Gradekeeper.

I'm working on making them right now. They should be up in a few hours. After this week, the notes should be up at least a few days before the class.

No problem! :)

Thanks for the reminder; I totally forgot. At any rate, you need to have Adobe Acrobat Reader to view the notes. You can download this program for free at http://www.adobe.com/products/acrobat/readstep2.html. Once you've installed that program, you should be able to view the notes.

You need to just keep playing around with the compass until you find some way of holding it that works for you. The things you should keep in mind are: (1) keep the point in the same place and (2) keep the amount the compass is open constant.

As far as practice in class, there was a full hour from when lecture was over until I left (at 12:30). That would have been a very good time to sit with me and work on it. You can stop by my office hours in H-2 tomorrow at 8:30 am, and we can work on it together. I don't think I will dedicate more class time to this construction because there was plenty of time today to stick around and ask questions.