5.1-5.3 Connections
Feb 1st, 2012 by corricelli
Hello,
Please use this spot to post real connections that you have seen/found online about Triangles and their Bisectors, Medians, Altitudes, and Perpendiculars.
Alternatively, maybe you found a great video or site online that might help your teammates to learn this material. Post that here!
Be sure to explain where the site goes and what it will do/has done to help you.
Happy connecting,
Mrs. Corricelli
I am having trouble doing number 6 on the homework. I know that the distance from N to P is 11, but does the medians theorem apply to that part of the triangle? I am confused at what median actually is because NP doesn’t look shorter than PO, in fact, it looks about the same. Can anyone help me? Thanks.
Alyson,
Aargh. I do not have the actual document. Ok – let me try to tell you what I know and I hope it helps… A median has special properties which are: 1.) crossing a side at its midpoint, 2.) the centroid is the point of concurrency of 3 medians, 3.) The centroid location is 2/3 of the way down the median away from the vertex. The centroid is always closer to the side than the vertex. (which kinda makes sense b/c it is a balancing point).
I hope this answers your question. If not, can anyone who has the diagram help?
Mrs. Corricelli
Mrs.Corricelli, I was wondering on the homework of 2/7/11 (5.3) what does it mean by “Find the length of the median?” I kind of figured it meant the middle of that line but for number 16 and 18 there are two midpoints for the segments the homework is talking about.
Matt,
The median, of any triangle, is the segment that connects the midpoint of a segment to its opposite vertex. To find the length of this segment, you need to employ the distance formula: Distance = sqrt( (x1 – x2)^2 + (y1 – y2)^2). You are given two ordered pairs – a vertex and a midpoint (which you calculated) – use the distance formula to get the distance!
Hope this helps!
(Next time, try to phrase your questions to all teammates – not just me!)
Happy calculations,
Mrs. Corricelli
I found a great video that can help with any remaining questions people still have, or just some review before the quiz. Try taking notes when watching the video. The more I write, the more the information sinks in.
http://www.youtube.com/khanacademy#p/c/26812DF9846578C3/81/GiGLhXFBtRg
Hope this helps!
Amalia
Awesome post Amalia! Thanks for giving blogs a chance! You helped a lot of people tonight!
Sincerely,
Mrs. Corricelli
Hi, Can someone please further explain what exactly is an altitude of a Triangle and explain the difference between a median and an altitude?
Hey An,
Well to the best of my understanding an altitude is a height that is the perpendicular segm. from the vertex to the opp side of the triangle… For example, the perp. bisector of the hypotnuse of a right triangle would be an altitude. A median on the other hand is a segm. whose endpts are a vertex of the triangle, and the mdpt of the opp side. 3 medians would be concurrent and their pt of concurrency would be called the centroid. A median is split into 2 sections by a centroid, 1 part of it is 2/3 of the segm and the other part is 1/3 of the median. The 1/3 part is closer to the side of the triangle, while the 2/3 part is closer to an angle of the triangle. Hope this helped! Good luck!
Mohammed,
PERFECT explanation! Way to help a teammate out! Maybe the only thing that I would add is that the centroid is 2/3 of the way down the median away from the vertex. In other words, the centroid is always farther from the vertex than it is from the base that it intersects. In fact, it is always twice as far!
Hope this helps!
Happy Wednesday,
Mrs. Corricelli
An,
It’s the perpendicular segment from the vertex to opposite sides of the triangle (or the line containing the opposite side. An altitude can lie inside, on, or outside the triangle.
Hope this helps a little,
Amalia
Im sort of confused on # 18 on the homework because during class, oh wait. connection. Do we use the distance formula from the point A and E to find the length of the median AE?
And for # 19, how can I construct an altitude. We haven’t learned this yet. Knowing that the altitude is the height of the triangle, is it just a # that i need to put there?
Laura,
Yea! The power of the blog… In writing the question, sometimes the answer just pops out! Yes – to find the length of the median, use the distance formula.
Do not worry about 19 or 20.
Hope this helps,
Mrs. Corricelli
Hey Mrs Corricell, i have question on the homework for 5.3. since i missed class the other day I’m having a hard time with some of it. most of my trouble is with the graphing part 15-18? how do find the length go median?
Emma,
Ok – check out this video; should help!
http://www.khanacademy.org/video/triangle-medians-and-centroids?topic=core-geometry
(Or alternatively, check out the posts in the 5.1-5.3 hw).
We can talk more tomorrow, but I do think the video should support you a great deal.
Next time (if you feel this way again) I am going to challenge you to find the video! Remember – lots of people have learned/taught this information. More perspectives enrich our conversation – consider reaching out to find one or two (the Khan Academy is always awesome) and then you help you and your teammates! (A win-win!)
Post here and let me know how the video worked.
Happy 5.3-ing,
Mrs. Corricelli
Mrs.corricelli will we need to know altitudes for the quiz. If we do what does a right triangles altitudes look like. is their anything else important going to be on the quiz that i need to look at.
Fernando,
Please feel free to check out the description on the main page: http://blog.whps.org/corricelli/2012/02/08/geometry-videos-on-triangle-topics/.
Thanks so much for inspiring this post! I am sure it will benefit many of your teammates!
Happy studying,
Mrs. Corricelli
Mrs Corricelli,
I’m having trouble with question 18 on the practice B you gave us for the quiz review. It is the problem that asks to find the coordinates of E which is the midpoint of CB. Then it says to show the quotient AG/AE= 2/3
I’m not sure how to even get to the point so it is root 32 over root 72 so it can be reduced to 2/3?
Cierra,
Did you get E’s coordinates? If so, what are they? (Midpoint formula = ( (x1 + x2)/2, (y1 + y2)/2 ).
Then, do you know A and G?
Once you have these things, then it is the distance formula to find AG and AE.
Distance formula: Distance formula: distance from (x1, y1) to (x2, y2) is sqrt((x1-x2)^ + (y1-y2)^2) or go to http://www.purplemath.com/modules/distform.htm.
I’ll be here; keep me posted – hee hee!
Mrs. Corricelli
im still confused on how to get the altitude of a triangle.
and i forget what the distance formula is
Nick,
Altitude = height. Start at a vertex and then connect to the opposite side so that you are perpendicular to the opposite side. To do this, sometimes you have to extend the side.
Distance formula: distance from (x1, y1) to (x2, y2) is sqrt((x1-x2)^ + (y1-y2)^2) or go to http://www.purplemath.com/modules/distform.htm.
Hope this helps,
Mrs. Corricelli
Hi! I’m trying to do problem 17 on the 5.3 Practice B homework. It says to find the centroid coordinates. How do I do that? I mean a centroid is the point of concurrency for the median, right? I drew the medians in the triangle, but I don’t know what to do next. I mean is there a mathematical way of doing it or do i just have to say where the centroid is based on the graph?
Brittany,
The centroid is the point of concurrency for the three medians. The vocabulary, “point of concurrency” implies three lines are intersecting at this point – not just one. So to find the centroid coordinates you need a nice median in coordinate geometry. So look for one that is a vertical line or a horizontal line (that will make finding the centroid easier). Once you have that move 2/3 of the median’s length away from the vertex. You will always be 2/3 of the way down the median away from the vertex when you are on the centroid. So, for example, say you have a median that is 12 units in length. Then the centroid for the triangle will be 2/3 (12) = 8 units away from the vertex. To be 8 units to the right, you add 8 to the x. To be 8 units up you add 8 units to the y. Similarly, to be 8 units down you subtract 8 units from the y-coordinate and to be 8 units left, you subtract 8 units from the x-coordinate.
This is absolutely on the quiz.
Thanks for posting a great question – hope this helped!
Mrs. Corricelli
hi, i had a question on number 8, 5.2, practice A…i dont undersetand how <GEC can be congruent to <GEA…i know it's probably not drawn to scale, but i just don't get how you would determine this…
thanks
-amy
For the Euler Line Lab, question number 7, do the Centroid, Circumcenter, and orthocenter all lay on the same point? I drew each line, and they all connected at the same points. However, I know that a Euler line is a line which contains all of the points (circumcenter, centroid, and orthocenter.) Is it possible that there are three different Euler Lines? Thanks
Alyson,
They are close to each other, but not on the same point b/c the triangle is not equilateral (see the centroid demo). All three points are collinear, though, which is kinda cool.
A few things you will need to know is how to find the equation for a line given the slope and a point. So slope = (y- y_pt)/(x – x_pt)
Moreover, the slope of any perpendicular to a line is a negative reciprocal of the slope of a line.
Thanks,
Mrs. Corricelli
For the Euler Line Lab, how to i find the centroid, circumcenter, and orthocenter? (#4-6)
Brittany,
Consider looking in the book – remind yourself – what is a centroid? What is a circumcenter? What is an orthocenter? Can anyone help a little more? Could you be specific about what you have done so far?
Thank you,
Mrs. Corricelli
Mrs. Corricelli,
Ok so i tried finding the circumcenter, but i’m still confused. I’ve found out the the equations for both AB and BC. For AB I have y=500/0x-75000/0 and for the BC i’ve gotten y=8/5+617/5. I’m not sure if i’m right, but i don’t think so, but then i tried equation and i’m stuck and not sure what to do(if what i’m doing is right).
Hello Brittany,
First please try to address posts to more than just me – the idea is this is a window into a community of learners – not just me!!! Although I love being a part of the community, I do not want to be the center!
I am wondering about your equation. You wrote that for segment AB you have two numbers dividing by 0. You cannot divide any number by 0 so right at the start I am a little concerned. A = ? B = ? The slope between A and B is (y1 – y2) / (x1 – x2). Now, you are correct that this slope has a 0 in the denominator, but what does this mean? Which type of line has an undefined slope? For B to C, the slope should be negative (because the line is downward facing). The slope between B and C should be -5/8. How did you get your solution? Finally, why do you need the equation for this segment?
Please address to all of your teammates, not just me. Please also clarify your work – what did you do and why are you doing it – so others can help you.
Thanks,
Mrs. Corricelli
Hey,
If anyone is having trouble on the Euler Line Lab, go to the link that Mrs. C posted…it really helped me better understand because I could visualize exactly what I needed to draw. You can make the triangle show only the circumcenter or the centroid or all of them together which helps to see what needs to be done.
Hope this helps.
-aMy
I wasn’t sure where to post this, but can anyone help me with numbers 11 and 12 on the homework? It tells you what yz is so how would you have to solve for it?
First you have to solve for x. when you solve for X substitute the variable with your answer . Solve for YZ and your answer should be 4 units. same for 12. MN should be 8 meaning the midsegment is four units bc i\the midsegment is YZ
http://www.winpossible.com/lessons/Geometry_Medians_and_Midsegment.html This is a good site to better understand what midsegment and the medians are from chapter 5 && its easy to comprehend, it helped me with last night’s homework
For the worksheet that was given in class today, when we are finding the value of x on the triangles, do we label the lengths as a line segment and put a line over the two letters or do we label it as a distance and leave the two letters side by side?
Carlos,
It depends… Is the x labeling a side? If yes, then it is a distance and its solution should have units. If no, then the x would have no units but the length that you are being asked to find (is a distance) would have no “segment” on top and it would have units.
I hope this helps!
Mrs. Corricelli
Thanks, Mrs. C
Turns out I was doing it wrong and putting the little “segment” thing on top :p
Carlos,
Teamwork!
Mrs. Corricelli