- Euler segment conjecture
- Euler line construction again if you didn't get it from this weekend
#4.1
- (1-18) all
don't forget to post your answer to conjecture 14 on here!
Chen Ying will be offering free tutoring sessions in the library tomorrow morning at 7:05. You will have to locate him. If you don't know what he looks like, find Dana or maybe Stefan. Dana met Chen today and knows what he looks like. Stefan plays tennis with him. Hint-hint: he's Chinese.
posted by msichan @ 11:41 AM
27 Comments:
Can someone help me with pretty much all of 3.6?! I seriously need help ASAP! WITH ALOT OF DETAILS!
Hola! Can someone tell me what happened in class to day?
-JJ
we went over the Eulers line and we did 3.7 and 3.8
02/12/07--
*falls off chair!*
Chen?! XD Mister Heather "Bawt-tun" the Chinese Wonder?! Bwaahaaa! Yaaay! I can finally get some help with those inscribed circles inside of obtuse triangles! *collapses on knees and rejoices!*
Ack, well. I've got to fold the laundry right about now. But, before I go, here's what I had gotten for Conjecture Fourteen:
Conjecture Page 162--
The centroid divides the Euler segment into two parts so that the smaller part is half as long as the larger part.
I'm sorry if I made a mistake. Any confirmation or opinions would be greatly appreciated. XD Thank you in advance.
British, what problems do you need help with in section 3.6? And Jessica, we basically went over our weekend homework, our work for 3.7 and 3.8 [including investigations], and we covered investigations 4.1 and 4.2 [or was it 4.1.2? XD]
Sincerely,
--Mary.
Hey if ne one wants to im me my im is presdnt1@aol.com. Just gve your name first do I will no who you are.
Keenan
Everything on 3.6
Conjecture 14:
The centroid divides the Euler segment into two parts so that the smallest part is half the larger part.
can someone tell me how they did their circumcenter for the Euler Line thing....!!
thats the only thing i didnt do..=/
thanx...
-juan moreno
The centroid divides the Euler segment into two parts so that the smallest part is half the larger part.
well i hope this is right-
the centroid divides the euler segment into two parts so that the smallest part is half the larger part.
-
dana
ok...ummm...i got...
the centroid divides the Euler segment into two parts so that the smallest part is half the larger part.
British, I honestly don't know what to tell you since pretty much all of the questions in section 3.6 involve constructing. Wait. Do you need help constructing the problems or with something else...? Either way, I'll try to explain each one as best as I can--if I can get my scanner to work, I can e-mail you some of my constructions for reference.
Section 3.6 #1-14, #16
Number One:
Given lengths--
MA .____________________.
AS ._________________________.
MS .____________________________.
Construct--Triangle MAS.
--Duplicate MS
--Radius setting: SA; arc from point S to construct side SA.
--Radius setting: MA; arc from point M
--Point A is where the two arcs intersect; draw lines.
Number Two:
Given--
Angle O.
DO .___________________.
OT ._____________________.
Construct--Triangle DOT.
--Duplicate OT.
--Duplicate < O at O; duplicate DO from < O.
--Draw DT.
Number Three:
Given--
Angle Y
Angle I
IY .______________.
Construct--Triangle IGY
--Duplicate IY
--Duplicate < Y and Angle I [I had to write that out since the comment box thought I typed in a broken tag.] from respective points.
--Draw G
Number Four:
I think this problem is rather self-explanatory. ^-^ Just make sure your angles are congruent to the given triangle, and you'll be fine, British.
Number Five:
Self-explanatory as well.
Number Six:
Given--
X ._______.
Y .___________________.
Construct-- Isosceles Triangle CAT with Y the perimeter and X the length of the base.
--Subtract X from Y.
--After subtracting X to find the new length of Y, bisect Y. [Divide it into two equal parts; end point to midpoint a congruent side.]
--Duplicate X as base.
--Radius Setting: bisected Y; arc from endpoints of base.
--Draw lines; label Triangle CAT.
Number Seven:
Given--
Z .__________.
Construct--An isosceles right triangle with z the length of each of the two congruent sides. [I'm labeling my triangle ABC.]
--Duplicate Z as base. [End points A and B.]
--Construct perpendicular segment from point A.
--Radius Setting: Z [or AB]; construct point C from point B.
Number Eight:
This should be simple enough to do. After you construct your triangle, just find the median and angle bisector they're asking for.
Number Nine:
This one is okie dokies, as well. XD I'm too lazy to type out an explanation. If you still need help with this one, call me after my nap. Ahaha!
Numbers Ten-Fourteen:
Gyaaah. XD The answers will probably vary too much, but just remember to mark off those double-hash marks to represent congruency.
Number Sixteen:
If you look back to number eight in section 1.4 on January 15, 2007, you'll see the term and value chart for "diagonals from one vertex". Just use the same nth term [n-3] to get your answer. ^-^
Wowwies. XD That was a doozy to type up. *headscratch* I hope I was able to help you. Ooh, and Juan? What do you want to know about the circumcenter? You mean how to find the circumcenter? And yaaay! XD A consistency of conjecture fourteen. Phew. *wipes sweat off forehead!* I had thought I was missing something since there were three points to choose from and so many things you could compare. XD
Sincerely,
--Mary.
ok so for investigation 3.8.3 do we like draw the four points of concurreny for c-13? for c-14?
or do we just complete the conjecture?
i just want to make sure.
-
Dana
Thanx Mary.
Hmm. Well Dana, I suppose technically our weekend homework was to complete and construct for conjecture thirteen, seeing as we're doing the same thing. So, if you construct the Euler line with the instructions Ms. Chan posted on Friday, then you would have already located the four points of concurrency.
And by doing that, you can just use the exact same construction for conjecture fourteen since the three points on the Euler line create a Euler segment. So, yeah. XD You'll have to complete the construction [I think only once! Just as long as you do Friday's assignment], as well as the conjectures.
Sincerely,
--Mary.
P.S. Ooh! You're welcome, British. ^-^
thank you so much mary =]
This si what i got:The centroid divides the Euler segment into two parts so that the smaller part of the segment is half as long as the larger part. i think...
For conjecture 14, I basically have what everyone else has. The centroid divides the Euler segment into two parts so that the smaller part is half as long as the larger part.
This is what I got for C-14: The centroid divides the Euler segment into two parts so that the smaller part is half the larger part.
I need help with #4,5,9,11, and 15 on tonite's homework.
I'm only on number fourteen myself; I need a bit of advice for number eleven as well, so if anyone can lend a helping hand, I'll think you're mega-wicked-awesome. XD Ahahaha.
Number Four:
Okie dokies. British, "a" is opposite from 60-degrees, making them vertical angles. You should be able to solve "b" and "c" with that information.
65-degrees is opposite from "e", making them vertical angles. You should be able to solve "d" and "f" with that information.
55-degrees is opposite from "h", making them vertical angles. XD Aaand, you should be able to solve "i" and "g" with that information.
Number Five:
I'm sure you've already solved "a", so I'll skip that. 163-degrees and "b" are a pair of vertical angles. After you solve "b", "b" and "c" are a pair of linear angles. 70-degrees and "e" are a pair of vertical angles, so you should be able to solve "d".
Number Nine:
Yoshi is going to make a right-angled corner. Ninety-degrees. I know this is going to sound a bit vague but, what and what equates to ninety-degrees?
Sincerely,
--Mary.
hello class! its robyn! i just got home...i had step practice and we didnt got out until 6:30...so sorry that why it took me so long...but like mostly everyone eslse has figured...i got the same thing
*The centroid divides the Euler segment into two parts so that the smallest part is half the larger part.
`robyn!
Weeeell Is Good To Know Everyone Got The Same Thinggg And I Did Too! Yesterday I Had To Go Over The Last Couple Of Sections So I Could Get It All Right Because I Wasnt Getting It Completely But Now I Think I Do =) Yey
I need help.... On 4.1 could someone please explaim question 9 in simpler terms. And can someone explain horizontal and vertical lines of symmetry.......... For C 14 i got:The centroid divides the Euler segment into two parts so that the smallest part is half the larger part.
O Yeah.... Andres' dog sadly passed away.
C-14: The centroid divides the Euler segment into two parts so that the smaller part is half as long as the larger part.
At least, thats what I got, correct me if I'm wrong.
Homework Was So Easyyy I Wish They Were All Like This ONeeee
yeah...ummm....i dont know...
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