Today, we started the class by correcting the homework of pages 366 and 367 - just for a change. Many exercises brought questions; Ms. Saatkamp's adivices for each + its solution are:

13 to 16 - we are supposed to find the parallel lines and their transversal and start analysing the diagram by looking at them. (13- DF / 14-AC / 15-GB / 16-IE).

.

19 -The exercise says AE=3/5x, which means

, if we consider that AB=x

Now, the exercise asks if its possible for

, however

BE:EA is not 3:2, but 2:3. So, it is false

20 - *The yellow part is a parallelogram. y is the opposite side of the side that measures 6. That means y=6 as well.

*To find x you can compare the two smaller triangles ( per example, side that measures x with the one that measures 6 and the one that measures 25 with the one that measures 10) or the bigger triangle with the medium one (x+6 with x and 35 with 25).

21 - *To find y, just compare the two segments that have it (3y-9 = 2y+6). In the end you should find y=15.

*To find x, base wourself on Theorem 7-6. In the end you should find x=21

27 - The key for the solution of this problem is given on it: 2DE=BC. That means D is midpoint for AB and E is midpoint for AC. All you have to do is use the midpoint formula. You should get to B (5,-2) and C (11,2).

29 - Just as in problem 27, the key point in this problem is the fact A, B and C are midpoints. That brings us to the conclusion that DE, for example, has its total number (18) divided equally by DB and BE (2x9). That plus Theorem 7-6, which states AC is parallel to DE, brings us to the conclusion that AC is also 9. Following the same line of thought with the other sides, you should get: AB=7, BC=18 and AC=9.

######################

After that, we had a little revision on the subject of chapter 5-1, which is Special Segments in Triangles (Yey!). So, we reviewed what each of these is:

Median: Segment from a vertex to the midpoint of the opposite side.


Altitude: Segment from a vertex to the oposite side of the triangle forming 90º.

Angle Bisector: Segment that bisects an angle and ends at the opposite side of the triangle.

Perpendicular Bisector: Line that bisects one side of a triangle and is perpendicular to it.

As soon as we had finished reviewing, Ms. Saatkamp led us through a "logical reasoning":

Let's say triangles ABC and DEF are similar. That means that all angles are congruent AND that

.

As we already learned, the perimeters of the triangles are also equal to that. So let's say AB:DE=1:3. That means (perimeter of ABC):(perimeter of EDF) = 1:3 as well. Because the altitudes, angle bisectors and medians are speciel SEGMENTS in proportional triangles, they have the same ratio from one triangle to the other. That means (altitude of ABC):(altitude of DEF) is also = 1:3.

Because perpendicular bisectors are NOT segments, but lines, they don't fit into these cathegories, and have a different kind of Theorem (which are about the review).

######################

And now... THEOREMS! \o/

[It's very important to repeat what Ms. Saatkamp said: we are having these theorems, not for proving (because we are not doing provings this semester) but for understandment of the subject]

>Today we had from Theorem 7-7 until Theorem 7-11.

>>I'll use triangles ABC, which is proportional to DEF from 7-7 to 7-10:

*Theorem 7-7: Proportional Perimeters

if: ABC~DEF

then: perimeters have the same scale factor (ratio) as corresponding sides.

*Theorem 7-8:

if: ABC~DEF
then: altitudes have the same scale factor (ratio) as corresponding sides.

*Theorem 7-9:
if: ABC~DEF

then: angle bisectors have the same scale factor (ratio) as corresponding sides.

*Theorem 7-10:
if: ABC~DEF

then: medians have the same scale factor (ratio) as corresponding sides.

**Theorem 7-11: Angle Bisector Theorem

if: AD is an angle bisector

then: BD:DC = AB:AC

#######

HW of today: Page 373 #s 12-39 ALL

Impotant Dates:

• Monday we have a quiz


• Tuesday we have a review


• Wednesday we have a... TEST \o


• And on Thursday be sure to brng your scientific calculators because we are starting a new chapter

Ms, sorry for taking so long to post, but I had some issues with my computer...