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C2 Questions

With a homework titled 'Radians', I've got two annoying questions here which I could do with some help with - pointing me in the right direction would be greatly appreciated.

1.

An equilateral triangle is inscribed in a circle of radius 10 cm.

Spoiler

(ii) Find the area of the triangle
(iii)Find the area of the three segments surrounding the triangle. - I know this is relatively easy once I have the area of the triangle (Part ii).

2. Once upon a time a hermit found an island shaped like a triangle with straightshores of lengths 6 km, 8 km, and 10 km. Needing seclusion, he declared that noone should approach within 1 km of his shore. What was the area of his‘exclusion’ zone?

Really not sure where to go with the latter, I've drew the triangle and 'exclusion zone' (circle), but I can't think of anything related to radians bar area of a sector. I know to do this I'll need to work out the necessary angles (I'm assuming I use the cosine rule) - but once I have the angles of the triangle, where do I then go?
Reply 1
Original post by Kozmo
With a homework titled 'Radians', I've got two annoying questions here which I could do with some help with - pointing me in the right direction would be greatly appreciated.

1.

An equilateral triangle is inscribed in a circle of radius 10 cm.

Spoiler

(ii) Find the area of the triangle
(iii)Find the area of the three segments surrounding the triangle. - I know this is relatively easy once I have the area of the triangle (Part ii).

2. Once upon a time a hermit found an island shaped like a triangle with straightshores of lengths 6 km, 8 km, and 10 km. Needing seclusion, he declared that noone should approach within 1 km of his shore. What was the area of his‘exclusion’ zone?

Really not sure where to go with the latter, I've drew the triangle and 'exclusion zone' (circle), but I can't think of anything related to radians bar area of a sector. I know to do this I'll need to work out the necessary angles (I'm assuming I use the cosine rule) - but once I have the angles of the triangle, where do I then go?


no 2 is rather nice ... they should ask it in a Maths interview ...
Reply 2
Original post by Kozmo
With a homework titled 'Radians', I've got two annoying questions here which I could do with some help with - pointing me in the right direction would be greatly appreciated.

1.

An equilateral triangle is inscribed in a circle of radius 10 cm.

Spoiler

(ii) Find the area of the triangle


You know that the area of a triangle is given by 12ABsinc\frac{1}{2}AB \sin c

You know it's equilateral, so c=60c = 60^{\circ}, both side lengths would be the same as well.
Reply 3
Original post by TeeEm
no 2 is rather nice ... they should ask it in a Maths interview ...


It's not bad is it, made me think so far and I'm yet to even solve it! :')
Going to give it another attempt tomorrow after 40 winks and see how far I get. :tongue:

I'm assuming Cosine and trig is necessary?
Reply 4
Original post by Zacken
You know that the area of a triangle is given by 12ABsinc\frac{1}{2}AB \sin c

You know it's equilateral, so c=60c = 60^{\circ}, both side lengths would be the same as well.


Yep, I managed to get that far. The problem was with working out the side length. I got the side length to be 10√3 - but this required a very long, tedious process and I'm not too sure on it anyway.
Reply 5
Original post by Kozmo
It's not bad is it, made me think so far and I'm yet to even solve it! :':wink:
Going to give it another attempt tomorrow after 40 winks and see how far I get. :tongue:

I'm assuming Cosine and trig is necessary?


An able experienced student should be able to do it without writing much down
Reply 6
Original post by Kozmo
Yep, I managed to get that far. The problem was with working out the side length. I got the side length to be 10√3 - but this required a very long, tedious process and I'm not too sure on it anyway.




Does this help?
Reply 7
Original post by Zacken


Does this help?


Yep. Using that I got the same side value of 10√3 which I got using my very prolonged method! Hopefully this is right, would you possibly be able to support this answer? :tongue:

Also, any idea on no. 2?
Reply 8
Original post by TeeEm
An able experienced student should be able to do it without writing much down


I like to consider myself able, although experienced I am not >.>
I'm quite looking forward to seeing the method though.
Reply 9
Original post by Kozmo
Yep. Using that I got the same side value of 10√3 which I got using my very prolonged method! Hopefully this is right, would you possibly be able to support this answer? :tongue:


Yep, that's correct.

Also, any idea on no. 2?


I'll have a look now.
Reply 10
Original post by Kozmo


Really not sure where to go with the latter, I've drew the triangle and 'exclusion zone' (circle), but I can't think of anything related to radians bar area of a sector. I know to do this I'll need to work out the necessary angles (I'm assuming I use the cosine rule) - but once I have the angles of the triangle, where do I then go?


Don't you just need to find area circle - area triangle? Area triangle should be easy to work out using Heron's.
Reply 11
Original post by Kozmo
I like to consider myself able, although experienced I am not >.>
I'm quite looking forward to seeing the method though.


I personally can give you the answer without writing anything down ... maybe I have seen this type of problem before ....
c13.jpg
Reply 12
Original post by Zacken
Don't you just need to find area circle - area triangle? Area triangle should be easy to work out using Heron's.


You know what, I didn't even consider that - I'm too busy over-complicating it.
I'll give this a shot, although I have no idea what Heron's formula is! :smile:
Reply 13
Original post by TeeEm
I personally can give you the answer without writing anything down ... maybe I have seen this type of problem before ....
c13.jpg


I'm envious :cry2:
Reply 14
Original post by Kozmo
You know what, I didn't even consider that - I'm too busy over-complicating it.
I'll give this a shot, although I have no idea what Heron's formula is! :smile:


Don't bother using Heron, bit of a fancy shmoozle, stick with what you do know. :smile:
Reply 15
Original post by Kozmo
I'm envious :cry2:


if I understood the problem well i get 24 + pi
Reply 16
Original post by Zacken
Don't bother using Heron, bit of a fancy shmoozle, stick with what you do know. :smile:


I just encountered a problem: obviously I have to work out the area of the circle, but how do I go about this if I don't know the radius?
Reply 17
Original post by Kozmo
I just encountered a problem: obviously I have to work out the area of the circle, but how do I go about this if I don't know the radius?


Ah, nevermind. The method you're meant to use is to notice that the excluded region consists of a rectangular region (1 km wide) across each shore and a circular region at the vertices. Picture:



That should do the trick.
Reply 18
Original post by Zacken
Ah, nevermind. The method you're meant to use is to notice that the excluded region consists of a rectangular region (1 km wide) across each shore and a circular region at the vertices. Picture:



That should do the trick.


Ahh, I see. I'll give this a go tomorrow. Thank you very much for your help!

Oh and I just realised another, much easier way to solve the first one so thought I'd share:

If we split the equilateral up into a further three triangles the inside angles are all 120 degrees, and thus we can use the area of a segment formula = 0.5xr^2(theta-sintheta) to calculate the segment area. Multiply this by three and minus it from the area of the circle.
- Same result but much more efficient method in my opinion! :smile:

Anyhow, I'm going to head to sleep.
Thanks to anyone who helped!

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