Author Topic: Mathematics Question Thread (Read 1625946 times) Tweet Share

ilikeapples · « **Reply #3780 on:** October 05, 2018, 09:07:29 am »

Hey could someone explain something to me in regards to the last question from the 2017 exam... why in the solutions attached for part (i) does BD=DE? Is this some ancient rule from year ten that I have forgotten? Could someone just explain what it means by equal intercept. thanks!

RuiAce · « **Reply #3781 on:** October 05, 2018, 09:38:53 am »

Quote from: ilikeapples on October 05, 2018, 09:07:29 am

Hey could someone explain something to me in regards to the last question from the 2017 exam... why in the solutions attached for part (i) does BD=DE? Is this some ancient rule from year ten that I have forgotten? Could someone just explain what it means by equal intercept. thanks!

It's one of those obscure theorems that some schools teach whilst other schools take for granted. I feel it's slightly too rushed, but the following explanation is still a bit rushed.
$\text{It is clear from the diagram that}\\ \triangle MBD ||| \triangle CBE.$
$\text{So by the proportional sides, }\frac{BM}{BC} = \frac{BD}{BE}\\ \text{however since }M\text{ is the midpoint of }BC,\\ \frac{BM}{BC} = \frac12.$
$\therefore \frac{1}{2} = \frac{BD}{BE} \implies BE = 2BD\\ \text{so since }BE = BD + DE...$

Mate2425 · « **Reply #3782 on:** October 05, 2018, 08:07:23 pm »

Hi guys this question is from HSC 2011, was wondering if someone could please help me understand how to work out part Q9(ii).

http://educationstandards.nsw.edu.au/wps/wcm/connect/f718de94-0e72-4393-bb72-714be7283934/maths-hsc-exam-2011.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-f718de94-0e72-4393-bb72-714be7283934-lGhOrSw

Thanks all!!

kauac · « **Reply #3783 on:** October 05, 2018, 08:17:09 pm »

Quote from: Mate2425 on October 05, 2018, 08:07:23 pm

Hi guys this question is from HSC 2011, was wondering if someone could please help me understand how to work out part Q9(ii).

Which part of question 9 was it?

Mate2425 · « **Reply #3784 on:** October 05, 2018, 09:02:56 pm »

Quote from: kauac on October 05, 2018, 08:17:09 pm

Which part of question 9 was it?

Q9a (ii) please

RuiAce · « **Reply #3785 on:** October 05, 2018, 09:16:47 pm »

Quote from: Mate2425 on October 05, 2018, 09:02:56 pm

Q9a (ii) please

Basically the similar triangles in part i) were proven using "two sides in proportion, included angle equal". But now that we have them, we can then use the other property, i.e. equiangular.
$\text{Since }\triangle ABC ||| \triangle ADE\text{, we observe that }\angle ACB = \angle AED\\ \text{by matching angles on similar triangles.}$
Of course, could also use the other ones.
$\text{But this proves to us that }BC \parallel DE\text{ since a pair of}\\ \text{corresponding angles are equal.}$
$\text{Then using alternate angles on parallel lines it becomes immediately obvious that}\\ \text{in fact, we have a second pair of similar triangles: }\boxed{\triangle BCF ||| \triangle EDF}\\ \text{using the equiangular test.}$
_________________________________________
$\text{From }\triangle ABC ||| \triangle ADE\text{, by the proportional sides we have}\\ BC:DE = AB:AD \implies \boxed{BC:DE = 1:2}$
$\text{But then from }\triangle BCF||| \triangle EDF,\\ BC:ED = BF:EF \implies \boxed{BF:FE = 1:2}\text{ as required.}$

Mate2425 · « **Reply #3786 on:** October 05, 2018, 09:51:51 pm »

Quote from: RuiAce on October 05, 2018, 09:16:47 pm

Basically the similar triangles in part i) were proven using "two sides in proportion, included angle equal". But now that we have them, we can then use the other property, i.e. equiangular.
$\text{Since }\triangle ABC ||| \triangle ADE\text{, we observe that }\angle ACB = \angle AED\\ \text{by matching angles on similar triangles.}$
Of course, could also use the other ones.
$\text{But this proves to us that }BC \parallel DE\text{ since a pair of}\\ \text{corresponding angles are equal.}$
$\text{Then using alternate angles on parallel lines it becomes immediately obvious that}\\ \text{in fact, we have a second pair of similar triangles: }\boxed{\triangle BCF ||| \triangle EDF}\\ \text{using the equiangular test.}$
_________________________________________
$\text{From }\triangle ABC ||| \triangle ADE\text{, by the proportional sides we have}\\ BC:DE = AB:AD \implies \boxed{BC:DE = 1:2}$
$\text{But then from }\triangle BCF||| \triangle EDF,\\ BC:ED = BF:EF \implies \boxed{BF:FE = 1:2}\text{ as required.}$

Thanks RuiAce

With the numbers 1:2 are these just figures that represent they are similar but different in size?

Could you please also explain why in question HSC 2010 Q6b (iii) and (iv) it was a necessity to change angle to degrees rather than keep it RAD. I tried my method and still don't understand why it is necessary and how to recognise when to do this.

Thankkkksss!!!

http://educationstandards.nsw.edu.au/wps/wcm/connect/e8e5dd58-ab1f-482d-b849-221495c629c6/maths-hsc-exam-2010.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-e8e5dd58-ab1f-482d-b849-221495c629c6-lGhPb9T

RuiAce · « **Reply #3787 on:** October 05, 2018, 10:13:26 pm »

Quote from: Mate2425 on October 05, 2018, 09:51:51 pm

Thanks RuiAce
With the numbers 1:2 are these just figures that represent they are similar but different in size?

Could you please also explain why in question HSC 2010 Q6b (iii) and (iv) it was a necessity to change angle to degrees rather than keep it RAD. I tried my method and still don't understand why it is necessary and how to recognise when to do this.

Thankkkksss!!!
http://educationstandards.nsw.edu.au/wps/wcm/connect/e8e5dd58-ab1f-482d-b849-221495c629c6/maths-hsc-exam-2010.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-e8e5dd58-ab1f-482d-b849-221495c629c6-lGhPb9T

I'm looking at your iv) and your "area of sector" is definitely correct. Also your angle in iii) makes sense.

On the other hand, your "area of total" though - I'm not too sure how you got that? Also I'm not sure why you used Pythagoras on 9^2 when 9 is the length of the arc? Have a look again, because I believe it's the other stuff giving you a wrong answer.

Mate2425 · « **Reply #3788 on:** October 05, 2018, 11:39:53 pm »

Quote from: RuiAce on October 05, 2018, 10:13:26 pm

I'm looking at your iv) and your "area of sector" is definitely correct. Also your angle in iii) makes sense.

On the other hand, your "area of total" though - I'm not too sure how you got that? Also I'm not sure why you used Pythagoras on 9^2 when 9 is the length of the arc? Have a look again, because I believe it's the other stuff giving you a wrong answer.

Ahhhhhhh, yes that makes sense thank you!

Just in regards to previous question - With the numbers 1:2 are these just figures that represent they are similar but different in size? (where do they come from? how do we know it is 1:2 specifically?)
Thanks

RuiAce · « **Reply #3789 on:** October 06, 2018, 09:57:37 am »

Quote from: Mate2425 on October 05, 2018, 11:39:53 pm

Ahhhhhhh, yes that makes sense thank you!
Just in regards to previous question - With the numbers 1:2 are these just figures that represent they are similar but different in size? (where do they come from? how do we know it is 1:2 specifically?)
Thanks

$AB:AD = 1:2\text{ because the length of }AD\\ \text{is double the length of }AB.\\ \text{This is the case because }B\text{ is the }\textbf{midpoint}\text{ of }AD\\ \text{so essentially }B\text{ 'chops up' }AD\text{ into two equal segments.}$

Mate2425 · « **Reply #3790 on:** October 06, 2018, 12:42:44 pm »

Quote from: RuiAce on October 06, 2018, 09:57:37 am

$AB:AD = 1:2\text{ because the length of }AD\\ \text{is double the length of }AB.\\ \text{This is the case because }B\text{ is the }\textbf{midpoint}\text{ of }AD\\ \text{so essentially }B\text{ 'chops up' }AD\text{ into two equal segments.}$

Many thanks Rui, true Ace!!

kauac · « **Reply #3791 on:** October 06, 2018, 01:16:34 pm »

Hi...
Could someone please explain the answer for HSC 2015 MC Q9?

RuiAce · « **Reply #3792 on:** October 06, 2018, 01:17:25 pm »

Quote from: kauac on October 06, 2018, 01:16:34 pm

Hi...
Could someone please explain the answer for HSC 2015 MC Q9?

Already addressed in the compilation.

kauac · « **Reply #3793 on:** October 06, 2018, 01:19:03 pm »

Quote from: RuiAce on October 06, 2018, 01:17:25 pm

Already addressed in the compilation.

Sweet, thanks for that.

Mate2425 · « **Reply #3794 on:** October 06, 2018, 03:45:25 pm »

Hi is there a process in order to derive the formulas for surface area or volume in an array of shapes in case you forget them in the test!
And also could someone please provide a really simple explanation to the topic of prob. when do you add and when do you times, what do each represent?
Thanks.

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