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F
2/1/08
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HW due: Read §9.7; write §9.7 #1-5 all (work optional), 8-9
all (work required, as usual), choice of 20 or 21.
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M
2/4/08
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HW due: Read §9.8 and learn the terms face, edge, diagonal (of a rectangular solid), base, vertex, altitude, slant height, and
lateral edge (of a regular square
pyramid). Segment PK in the regular square pyramid on p. 413 or p. 414 is
called a lateral edge. Then write
#2, 3, 5, 8, and perform the supplementary activities indicated below as you
do them.
#2: Be sure to include a sketch.
#3(a): Compute altitude as requested.
#3(b): Compute slant height as requested.
Add #3(c): Compute length AB.
Add #3(d): What is AB called?
#5: Copy the diagram and answer parts (a) through (e).
Add #5(f): Compute RD.
Add #5(g): Compute PD.
Add #5(h): What is PD called?
#8: In addition to the measure of arc AC as requested, compute  .
In class: Review and additional problems.
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T
2/5/08
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Test (100
pts.) on §§9.1–9.8. Once again, there
will be a 4-point “immunity challenge” available for students who do a good
job on their review problems. (In other words, you are immunized against 4
points of loss on the test if your review problems are well done.) The
following problems on pp. 429-are recommended: #1-11 odd, 12-19 all, 21, 24,
27-31 all. That is a total of 21 questions. If time is short, you may skip a
few, but make sure that you understand all of the concepts addressed by the questions.
Here are some hints (please peek only if you need to):
1.(a) Use the first altitude-on-hypotenuse theorem:  . Plug in HG and EG; solve for GF.
3. You need to know four families of triples by heart (3-4-5, 5-12-13,
8-15-17, and 7-24-25) and the two families of “special degrees” (45-45-90 and
30-60-90).
5. The altitude to the base of any isosceles triangle bisects the base.
7. Make a sketch and look for a family solution. (No real work is needed.)
9. This is like problem #5, but in reverse. Remember that equilateral
triangles are the parents of twin 30-60-90 triangles.
11. Drop an auxiliary perpendicular from R down to the base.
12. As in #11, drop a perpendicular from X to a point (call it F) that is
collinear with V, Z, and W. Since TXFZ is a rectangle, ZF = 8, which makes VZ
= FW = 2. Now use 30-60-90 facts on  .
13. Use the 3-dimensional distance formula we developed in class.
14. Use the 3-4-5 family to get RS = 15, from which MK = 30. The answer
should now be clear. The problem would have been better if it had asked for
PS (slant height) and PK (lateral edge length) as well, since you are
responsible for those, too.
15. Similar to #13. You will need to use a calculator for this one unless you
are good at estimating.
16. Use the 8-15-17 family to get BC = 8. Then, since DE = 4, you should use
the Midline Theorem (p. 296) or similar triangles. (You are still responsible
for these.)
17. Similar to #13 and #15, except easier since this is only 2-dimensional.
18. Half SAD (half the difference of arcs since angle vertex is outside
circle).
19(b) Half SAD again.
(c) You can use Half SAD on the 110° arc BD to solve
this in one step. However, there is an alternate method you should try as a
cross-check. Let X denote the intersection just to the right of O. Then since
arc ADE is 100°,  = 80° by Half SAD
(half the sum of arcs since is inside circle).
That is fortunate, since 80° is the proper value to make  have the proper number
of degrees.
21.(a, b) Half SAD again (inscribed angle is half of the 180° arc DEF).
(c) From previous classes, you know that the circle
has circumference  . Make the appropriate adjustment since arc DEF is a
semicircle.
24.(a) By altitude-on-hypotenuse “Near Chunk Whole Chunk” theorems, 102
= 8(8 + x), which you should be
able to solve for x. Half credit on
the test if you provide a proper equation.
(b) Similarly, 62 = y(y + 9). Then, solve
for y by using your knowledge of quadratic equations from Algebra I.
(If that knowledge has left you, then leave the equation as it is for half
credit.)
27. The rightmost triangle is 5-12-13 by inspection, making the leftmost
triangle’s sides 4, x, and 8.5. You
should recognize that as a member of the 8-15-17 family (all sides have been
divided by 2).
28. Net change is 24 paces east, 45 paces north. This is now very much like
#27, except that the 8-15-17 family member has been multiplied through by 3.
29.(a) Since the diagonals of a kite are perpendicular (by PBT), and since
triangles KIT and KET are right, you can apply altitude-on-hypotenuse
theorems. (PE)2 = KP · TP = 36, which you can solve for PE. Then
use property of kites (or PBT again) to figure out what IE must be.
(b) Regardless of whether you solved (a) or not, let a denote the length of PE. Then you
can say 
by using the “conceptual approach” I have been stressing this week. Final
answer: 2KE + 2TE, which I would accept for full credit unless simplification
is explicitly required.
30.(a) E = (9, –3) by inspection
(b) Arectangle
= height · width = 5 · 14 = 70 square units [note: work is required for full credit]
(c) I would not ask for the nearest tenth. You should
be able to get  as the answer.
31. You can use the distance formula to show that both pairs of opposite
sides are congruent (QU = AD =  , and UA = DQ =  ), which proves that QUAD is a parallelogram. Then, show by
finding opposite reciprocal slopes that at least one vertex has a right
angle.
A faster method for #31 is probably to compute all the slopes and establish
that you have 4 right angles. That saves having to use the distance formula,
which is dangerous for many students. (Ask yourself if you can successfully
compute QU, AD, UA, and DQ above with the distance formula.)
Slope of 
Similarly, the slope of  is –3, the slope of  is 1/3, and the slope
of  is –3. Since the
slopes of consecutive sides are opposite reciprocals, QUAD contains 4 right
angles and must be a rectangle. (Q.E.D.)
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W
2/6/08
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No additional HW due today. Bring your binder, however,
in case some older problems are checked a second time.
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Th
2/7/08
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HW due: Write §10.1 #2, 7, 9, 11, 12. Reading notes
(normally required every day) are optional this time.
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F
2/8/08
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HW due: Read §10.3; write §10.2 #4, 5, 8, 11, 12; §10.3
#1-4 all. Reading notes (normally required every day) are optional this time.
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M
2/11/08
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HW due: Write a §10.3 summary for your notes (see
suggestion below) plus §10.3 #5-12 all, 15, 16. For #5-7, a short paragraph
proof is acceptable. For example, here is how I would do #6:
6. [diagram]
Given: Circle D with  .
Prove: 
By ITT, we know  . Since chords of a circle are  iff their arcs are , as claimed. (Q.E.D.)
Note: A two-column proof is
expected for #8 and #16. Also, you need to know that there are addition and
subtraction properties that work for arcs, just as there are corresponding
properties for segments that we learned last fall. For example, in the
diagram for #8, you can show that arcs AB and CD are congruent. After doing
that, you can say that since arc BC is congruent to itself (reflexive), the
addition property for arcs gives arc AC congruent to arc BD. (That is not the
conclusion of the proof, but it is certainly helpful!)
Suggestion: Since no reading notes
were required for the assignment due Friday 2/8, it is important that you
summarize the key theorems in this section in some suitable way. You should
use your own words if you can, or if you get stuck you may copy my words.
Here is what I would say:
“In a circle, or in congruent circles, central angles are iff their arcs are iff chords are . In other words, all three situations are equivalent:
congruent central angles, congruent arcs, congruent chords.”
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T
2/12/08
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HW due: Read §10.4, especially the common-tangent procedure
on p. 462; write #9, 10, 14.
If you cannot do #14, then copy (yes, copy) the entire Problem 3 on p. 462:
statement of the problem, initial diagram, marked-up diagram, work, and
answer. You may skip the copying of Problem 3 on p. 462 if you do #14
correctly.
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W
2/13/08
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HW due: Read §10.5; write §10.4 #5, 6, 11, 13, 14 (re-do),
22, 27. When you re-do #14, carefully follow the instructions on p. 462 and
the example below.
Example similar to #14: The
centers of two circles with radii 9 and 11 are 32 units apart. Find the
length of a common internal tangent.
Solution: First, make a diagram
showing the common internal tangent.
Second, add the line joining the centers, Q and R. Also add two perpendicular
radii, and label all known lengths.
Finally, extend  to join a line
through Q that is parallel to  . Do you see that this creates rectangle WTVQ?
Note: As we discussed in class, a right triangle has now
been created with its hypotenuse along
the line that joins the centers. This triangle,  , overlaps the rectangle. This is different from what we
saw in the external tangent procedure, in which the triangle and rectangle
had a side in common but did not overlap.
The goal of the problem is to find WT, the length of the common internal
tangent.
By properties of rectangles, TV = WQ = 9. Also note that if we could find QV,
we would have the answer to the problem, since WT = QV. Apply the Pythagorean
Theorem shortcut taught in class to get  . Since WT = QV by properties of rectangles, final answer =
 .
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Th
2/14/08
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HW due: Read §10.6; write §10.4 #22, §10.5 #5de, 6abcd, 10,
15, 18, and any one puzzle from the Angle-arc
puzzles or More angle-arc
puzzles collection. (Copy diagram and givens onto your regular homework
paper; do not simply mark up the printed copy.)
Regarding #22, we learned the “easy” way of doing the problem by means of the
walk-around technique. However, I would like you to re-do it the “hard” way,
by using the Pythagorean Theorem. In class, after we let x = either external tangent length from Y, we found that WX = 24
– x, and XY = 4 + x. (If you missed class, you may take
those as givens.) The hypotenuse, WY, is given to be 20. Therefore, write
(WX)2 + (XY)2 = (WY)2, substitute, and solve
for x.
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F
2/15/08
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No school (faculty professional meetings).
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M 2/18/08
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No school (holiday).
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T
2/19/08
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HW due: Reread the internal tangent procedure (see 2/13
calendar entry), and read the material below. No additional written work is
due. Enjoy your long weekend and get plenty of sleep!
Quiz (10 pts.) is likely on the
internal tangent procedure (as demonstrated in 2/13 calendar entry) or the
external tangent procedure (as demonstrated in the book’s example and in the
space below). You should work through all of these examples, as well as
similar examples of your own invention, in order to gain proficiency and
confidence.
Example external tangent problem: Find the length of belt needed to
connect two pulleys whose diameters are 10 and 2 inches if their centers are
8 inches apart.
Solution: First, make a sketch. If the diameters are 10 and 2, the
radii are 5 and 1 as shown. By symmetry, trapezoid ACQP is congruent to
trapezoid TUQP, a fact that will be useful later.
The second step is to add rectangle ACQB in such a way that the segment
joining the centers P and Q is kept as a hypotenuse.
Students often make a mistake on this step, attempting (hopelessly) to make
BCQP be a rectangle. (That is impossible, since then  ACB would be a right triangle with 2 right angles!) All you
have to do to avoid this mistake is to keep the hypotenuse on the line that
joins the centers of the circles. Here is the correct way to add rectangle
ACQB:
Now, look at right triangle BQP, which has a leg of 4 and a hypotenuse of 8. By
the Pythagorean Theorem shortcut,  . However, if  has sides of 4,  , and 8, then must be a 30°-60°-90° triangle, since only a 30°-60°-90°
triangle can have sides in the pattern of x,
 , and 2x.
Therefore,  = 60°.
By the congruence of trapezoids ACQP and TUQP,  = 60° also. Since  = 90° + 30° = 120°,
we know  = 120° also.
Therefore, arc CMU is 240°, forcing minor arc CU to be 120°, or 1/3 of a
circle. (This is important, since the belt connecting the pulleys must cover
minor arc CU.)
Notice that in circle P, minor arc ALT is 120°, forcing major arc TZA to be
240°, or 2/3 of a circle.
By a property of rectangles, AC = BQ = , and by the congruence of trapezoids ACQP and TUQP, TU = also. We now have
all the ingredients for the belt that connects the pulleys:
- AC =
- minor arc CU = 1/3 of the circumference of
circle Q =
- UT =
- major arc TZA = 2/3 of the circumference of
circle P =
Add them all up:  36.895 inches.
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W
2/20/08
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HW due: Get plenty of sleep, and be on your toes for the
quiz originally planned for yesterday. (See yesterday’s calendar entry for
details.)
If you bomb the quiz, you can earn partial credit by showing evidence that
you did some practice problems correctly (including neatly drawn diagrams).
If you cannot dream up any additional practice problems on your own, please
try #14 on p. 507 and #29 on p. 601.
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Th
2/21/08
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HW due: Read middle of p. 487 (Theorem 93 and its proof)
and §10.8; write §10.7 #9, 10, 11, 19, §10.8 #1, 6.
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F
2/22/08
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HW due: Read §10.9; write §10.8 #8-11 all, 13, 15, 16, 17.
Because of the snow day today, you will have to be your own teacher.
Carefully read through the lesson plan,
and work through the problems assigned for Monday. On Monday we will review
for Tuesday’s test. If you get stuck or need help, be sure to call me at 703-599-6624.
Leave a message with a phone number and a time to call you back, and I will
get in touch with you.
However, please don’t call to complain about having work to do on a snow day.
Your parents (and I) are investing a lot of effort in your education, and it
is reasonable for you to set aside an extra hour or so on your day off to
work through the detailed lesson plan provided for you. It is best if you
stay home and stay off the roads.
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M
2/25/08
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HW due: §10.9 #3, 4, 5, 8, 10, 17. For #5, give answer two
ways: (1) as requested by the textbook instructions and (2) rounded to the
nearest hundredth. Use a calculator for assistance.
Note: In addition to checking this
HW assignment, I will also be evaluating how well you followed Friday’s lesson plan. Remember, since you have to
serve as your own teacher for Friday’s class, you need to read carefully and
learn from each mistake. Mark your changes, preferably in ink or in a different
color of pencil. There is extra credit available if you prove the lemma
needed in #17 on p. 497.
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T
2/26/08
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Test on
Chapter 10. You should definitely
do more to prepare for the test than the several problems listed below, but
if you do these, you can earn an Immunity Challenge bonus of up to 4 points.
Bring your problems (your own work—no copying allowed) to class and turn them
in before the test begins.
p. 505 #2 extended (find all angles and arcs, not only the two angles
requested)
p. 506 #7
p. 506 #10
p. 506 #12 (simplify without using calculator)
p. 507 #14
p. 507 #16
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W
2/27/08
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HW due: Read §11.1. Also, if you have not already done so,
read through Friday’s lesson plan from beginning
to end and correct your homework completely.
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Th
2/28/08
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HW due: Read §§11.2, 11.3; write §11.1 #8, 9, §11.2 #1-9
odd, 10, 14.
Note: Students in period A may omit
the written work in §11.1. Thank you, period A students, for your superior attention
and focus in class on Wednesday.
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F
2/29/08
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HW due: Read §11.4; write §11.3 #1-6 all, 9, 13.
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