Is Terry Lee right? (1 Viewer)

[mreditor16]

So, I was doing this Q and underneath the Q, I have copied Terry Lee's solution for part i).

I don't recall learning that theorem first of all and, second of all, are we even allowed to use that as justification?

Also, does anyone have another way of doing i)?

thanks!

 

[Kurosaki]

That is a legit theorem (I remember learning it in 2 unit maths).
If you have the Year 12 Pender book, you can find that theorem in there as well (page 332, of my copy at least).

 

[seventhroot]

ofc he's right. He is asian

 

[RealiseNothing]

It's a theorem you learn in 2U.

(hence proves the importance of 2U even if you do 4U)

 

[Immortality]

Think of it as similar triangles and matching sides would be in 1:2 ratio (midpoints)

 

[Kurosaki]

In regards to your other question, midpoint theorem is probably the fastest way, can't really think of any others. (Well, I could but the work required would probably be too tedious for the 1 mark)

 

[Blueblade11]

I've never learned of that theorem. And looking through my notes and textbook, I can'd find any info on it.

 

[Kurosaki]

What textbook?

 

[mreditor16]

Think of it as similar triangles and matching sides would be in 1:2 ratio (midpoints)

This was the method I was planning on using. Proving triangle AKM is similar to ABC, and then matching angles in the similar triangles would prove KM // BL. and matching sides in similar triangles are in ratio, thus KM = half of BC . and since BL = BC (i.e. half of BC is BL), KM = BL

therefore a pair of equal and parallel sides - thus, parallelogram.

but I don't know if BOSTES was expecting quite a long solution for the one mark.

It's a theorem you learn in 2U.

(hence proves the importance of 2U even if you do 4U)

That is a legit theorem (I remember learning it in 2 unit maths).
If you have the Year 12 Pender book, you can find that theorem in there as well (page 332, of my copy at least).

lol looked through my notes, we never proved or touched this in class :/

I wonder whether BOSTES was really expecting us to remember that theorem? or they just wanted us to use the method I explained above. but it is a one mark "explain" Q, so idk :/

 

[we_rhere]

It's a theorem you learn in 2U.

(hence proves the importance of 2U even if you do 4U)

Na. circle geo is useless.

 

[we_rhere]

This was the method I was planning on using. Proving triangle AKM is similar to ABC, and then matching angles in the similar triangles would prove KM // BL. and matching sides in similar triangles are in ratio, thus KM = half of BC . and since BL = BC (i.e. half of BC is BL), KM = BL

therefore a pair of equal and parallel sides - thus, parallelogram.

but I don't know if BOSTES was expecting quite a long solution for the one mark.





lol looked through my notes, we never proved or touched this in class :/

I wonder whether BOSTES was really expecting us to remember that theorem? or they just wanted us to use the method I explained above. but it is a one mark "explain" Q, so idk :/

 

[Kurosaki]

What year is this exam?

 

[mreditor16]

What year is this exam?

2006 HSC

Q4d)

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2006exams/pdf_doc/maths_ext2_06.pdf

Here is Terry Lee's solutions:

http://www.hsccoaching.com/Resources/2006 Ext2.pdf

 

[Kurosaki]

Right, cool.

Notes from marking centre:

"Of the few candidates who attempted this part, most gave unnecessarily lengthy arguments involving similar or congruent triangles. A few obtained the answer quickly by recognising the intercept property of lines through the mid-points of two sides of a triangle. The most common mistake, among those who attempted this part, was to argue only that one pair of sides was parallel, or that one pair of sides was equal. Of course, many responses correctly observed that one pair of sides was both parallel and equal."

 

[Carrotsticks]

This same property was used for also last year's rectangular hyperbola question.

 

[mreditor16]

Right, cool.

Notes from marking centre:

"Of the few candidates who attempted this part, most gave unnecessarily lengthy arguments involving similar or congruent triangles. A few obtained the answer quickly by recognising the intercept property of lines through the mid-points of two sides of a triangle.
The most common mistake, among those who attempted this part, was to argue only that one pair of sides was parallel, or that one pair of sides was equal. Of course, many responses correctly observed that one pair of sides was both parallel and equal."

hmm so they were looking for Terry Lee's method

should have taken a look at NFTMC myself haha

thanks man

btw you screwed up your latex. :/

 

[mreditor16]

This same property was used for also last year's rectangular hyperbola question.

really? just took a look at your solutions for the rectangular hyperbola 2013 hsc Q12 d) and it doesn't use it at all. :/

 

[Carrotsticks]

I did for the very last part I believe.

 

[mreditor16]

I did for the very last part I believe.

oh yes my bad.

thanks everyone for the help

 

[we_rhere]

This same property was used for also last year's rectangular hyperbola question.

I didnt use that.