1. Know the postulates and theorems on pp. 770-2.
2. Go over the Chapter Reviews in the text.
3. Go through your quizzes.
4. There are extra practice problems starting on p.716. I would go over the following problems:
Chapter 1 (p.716): #18 - 37
#42 - 47
Chapter 2 : #1 - 28
Chapter 3: #1 - 13
#19 - 41
Chapter 4: #1 - 16
# 21 - 28
# 31-33
Chapter 5: #1 -9
# 13 - 20
# 33 - 42
Of course, you don't have to answer all of those but you can glance at each one and determine whether or not you know how to do it. If you're unsure, try the problem. Odd answers start on p.850.
I would say that most students had a little trouble with finding equations of lines. Make sure you know that material. Know how to find the slope of a line given two points, how to write the equation once you have a slope and how to tell whether lines are parallel or perpendicular.
If you have any questions, post them here or email me. I'll check in regularly.
Study!
Mrs. Dant
Saturday, December 18, 2010
Tuesday, December 7, 2010
Tips for New Proofs and Problems
You will have to use theorems from your notes for these problems. Problem #11 tells you to use Theorem 24, which is not in your book. Here it is:
Thm 24 - If two points are each equidistant from the endpoints of a segment, then the two points determine the perpendicular bisector of that segment.
Another useful theorem:
Thm 25 - If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of that segment.
TIPS:
4.) In order for PQ to be the perp. bis. of RS, P and Q must be equidistant from R and S. That means, PR = PS and QR = QS. You don't have to go from congruent segments to equal distances. If you have that PR congruent to PS, that means equidistant. Tip - draw radii. (4 step proof)
5.) You do not have to state that E is on the perpendicular bisector. Let's all agree that it is! Use Thm 25. (5 step proof)
6.) Use subtraction property. (3 step proof, with semicolons)
7.) Just answer it.
8.) Same.
9.) Again.
10.) a.) Ignore angle EBC. You can set up an equation with the two expressions. Make sure you use the right equation. Don't assume that segment AB is perpendicular to CD when you write the eq'n.
b.) Use the solution for x from part a to answer this.
11.) Given: PQ = PR (actually congruent but I don't have that key)
SQ bisects <PQR
______________ (you fill in this one)
Prove: PS perp QR
I used the Division property here, followed by another common theorem. You'll also use Thm. 24. (about 6 steps)
12.) There's a hint already there. To tell which segment is the perp. bisector, consider one point at a time. Once you find two points that are each equidistant to the endpoints of a segment, "connect them" to form the perp. bisector.
13.) Don't neglect to use the transitive property so that you can get all four segments congruent.
14.) Once you have the segments perpendicular, choose one angle to be right and then you can conclude the Prove.
15.) Number four acute angles. (Not the right angles.) You'll have to use Subtraction and 'If angles, then sides.' Theorem 24 is handy too.
16.) Make sure you DO NOT use congruent triangles. To prove something is an altitude, the step before that must include perpendicular segments. (4 step proof)
17.) Ok, on this one, you can prove one set of triangles to be congruent. Then use Thm 24.
18.) Again, what's the definition of altitude? Remember that the area of a triangle is 1/2 (base)(height). But, in a right triangle, the base and the height are the legs.
19.) a.) Figure out the coordinates of C and T.
b.) When you rotate point E 90 degrees clockwise, its coordinates are easier to figure out. (Only 1, 12, -1, or -12) are possible. Use that for a simpler sample and follow the pattern.
Thm 24 - If two points are each equidistant from the endpoints of a segment, then the two points determine the perpendicular bisector of that segment.
Another useful theorem:
Thm 25 - If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of that segment.
TIPS:
4.) In order for PQ to be the perp. bis. of RS, P and Q must be equidistant from R and S. That means, PR = PS and QR = QS. You don't have to go from congruent segments to equal distances. If you have that PR congruent to PS, that means equidistant. Tip - draw radii. (4 step proof)
5.) You do not have to state that E is on the perpendicular bisector. Let's all agree that it is! Use Thm 25. (5 step proof)
6.) Use subtraction property. (3 step proof, with semicolons)
7.) Just answer it.
8.) Same.
9.) Again.
10.) a.) Ignore angle EBC. You can set up an equation with the two expressions. Make sure you use the right equation. Don't assume that segment AB is perpendicular to CD when you write the eq'n.
b.) Use the solution for x from part a to answer this.
11.) Given: PQ = PR (actually congruent but I don't have that key)
SQ bisects <PQR
______________ (you fill in this one)
Prove: PS perp QR
I used the Division property here, followed by another common theorem. You'll also use Thm. 24. (about 6 steps)
12.) There's a hint already there. To tell which segment is the perp. bisector, consider one point at a time. Once you find two points that are each equidistant to the endpoints of a segment, "connect them" to form the perp. bisector.
13.) Don't neglect to use the transitive property so that you can get all four segments congruent.
14.) Once you have the segments perpendicular, choose one angle to be right and then you can conclude the Prove.
15.) Number four acute angles. (Not the right angles.) You'll have to use Subtraction and 'If angles, then sides.' Theorem 24 is handy too.
16.) Make sure you DO NOT use congruent triangles. To prove something is an altitude, the step before that must include perpendicular segments. (4 step proof)
17.) Ok, on this one, you can prove one set of triangles to be congruent. Then use Thm 24.
18.) Again, what's the definition of altitude? Remember that the area of a triangle is 1/2 (base)(height). But, in a right triangle, the base and the height are the legs.
19.) a.) Figure out the coordinates of C and T.
b.) When you rotate point E 90 degrees clockwise, its coordinates are easier to figure out. (Only 1, 12, -1, or -12) are possible. Use that for a simpler sample and follow the pattern.
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