Cheeky c2 question, I wonder who'll get it??

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Terminatoring
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Struggling here a bit boys& girls, help me out please


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BioStudentx
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what year and exam board is it?
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lizard54142
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(Original post by BioStudentx)
what year and exam board is it?
Someone's jumping straight to the mark scheme
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Terminatoring
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I can send over the mark scheme if it'll help you guys explain it


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BioStudentx
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(Original post by lizard54142)
Someone's jumping straight to the mark scheme
Yep. But seriously if it's not AQA i dont care.
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Terminatoring
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Just wanted someone to point me in the right direction


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lizard54142
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(Original post by Terminatoring)
I can send over the mark scheme if it'll help you guys explain it


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No don't! I'll try and have a look at it in a minute.
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Terminatoring
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(Original post by BioStudentx)
Yep. But seriously if it's not AQA i dont care.
If may not be AQA but that doesn't mean it's any less important or valuable.
The thing about maths is we're all equal, regardless of the exam board the maths remains the same.

That's why i think you should help me




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Terminatoring
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(Original post by lizard54142)
No don't! I'll try and have a look at it in a minute.
Thank you very much


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Prince Philip
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(Original post by Terminatoring)
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Struggling here a bit boys& girls, help me out please


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Consider the unshaded region: ABO

Drawing a line from O to B gives a sector AOB : find the area of this in the normal way.

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Straight lines from A to B to O give a triangle AOB. What are the angles in this triangle?


You're left with a region (segment) enclosed by line OB and arc OB. Draw a line from A to B and notice that this segment is the area of another sector minus the area of the triangle AOB.
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AG98
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AO = AB = r, as they are both radii of the circle with centre A.
OB = OC r, as it is the radius of circle centre O.
Therefore, AOB = equliateral traingle.
Similarly, COD = equilateral triangle.
Therefore, angle DOC = 60 and BOA = 60.
Therefore angle BOC = 60 (angles on straight line sum to 180)

OB = OC. Therefore, angle BCO and CBO = 60.

Therefore BOC = equilateral triangle, side length of r.

Area of full circle centre O = (r^2)pi
Area of sector AOB = Area of sector COD = 1/6 (r^2)pi
Therefore, shaded area = 1/2(r^2)pi -2/6 (r^2)pi= 1/6 (r^2)pi

Where have I gone wrong? :/
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lizard54142
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(Original post by Terminatoring)
Thank you very much


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Consider the line OC, this gives a sector of the circle, you can work out this area easily. You're then left with the segment of arc OC and line OC. Then draw the line OD, notice you have an equilateral triangle OCD. You can work out this area, and then the area of the segment from this, etc... repeat similarly for AOB!

EDIT: Beaten to it by two people! Gosh I should I typed quicker
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Louisb19
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Did this question earlier today. Here is my working, if you still don't understand tell me.

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Prince Philip
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(Original post by AG98)
AO = AB = r, as they are both radii of the circle with centre A.
OB = OC r, as it is the radius of circle centre O.
Therefore, AOB = equliateral traingle.
Similarly, COD = equilateral triangle.
Therefore, angle DOC = 60 and BOA = 60.
Therefore angle BOC = 60 (angles on straight line sum to 180)

OB = OC. Therefore, angle BCO and CBO = 60.

Therefore BOC = equilateral triangle, side length of r.

Area of full circle centre O = (r^2)pi
Area of sector AOB = Area of sector COD = 1/6 (r^2)pi
Therefore, shaded area = 1/2(r^2)pi -2/6 (r^2)pi= 1/6 (r^2)pi

Where have I gone wrong? :/
Firstly, AB=r is incorrect.

EDIT: Ignore this. I made a mistake!

I think you're assuming that unshaded region ABO is a sector when it isn't. A sector is enlosed by one arc and two radii. This region is enclosed by one radius and two arcs.
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Aladino
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ez question
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Louisb19
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(Original post by Aladino)
ez question
Maybe for you but it is supposed to be one of the harder C2 questions being that it is on a Soloman paper.
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Terminatoring
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Not sure I'm seeing how you guys dealt with the segments

Thanks for the help!


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nats927
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damn maths looks another language to me. glad i'm not studying it.. btw answer is BOC= 3xy
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AG98
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(Original post by notnek)
Firstly, AB=r is incorrect.

I think you're assuming that unshaded region ABO is a sector when it isn't. A sector is enlosed by one arc and two radii. This region is enclosed by one radius and two arcs.
But OB is the arc of circle radius r centre A, so AB is a radius = r?
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Prince Philip
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(Original post by Louisb19)
Maybe for you but it is supposed to be one of the harder C2 questions being that it is on a Soloman paper.
It's unusual but I wouldn't say it's hard. It could be a question in a GCSE textbook.
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