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

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I did it by getting the equation for the segment via the equation and taking the angle of the sector as pi/3, I then took 2(segment) and 2(sector) away from the area of the semi circle to find the shaded area. Hope this helps.

Reply 21

lizard54142

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.

Surely AB is equal to r? The circle centre A has the same radius as the circle centre O, and B is a point on the circle centre A?

Reply 22

Notnek

Original post by AG98

But OB is the arc of circle radius r centre A, so AB is a radius = r?

Sorry - my mistake.

Reply 23

Louisb19

Original post by AG98

But OB is the arc of circle radius r centre A, so AB is a radius = r?

You don't need to know AB for any of the calculations though. The area of the segments and sectors can be found with the radius R and the angle pi/3.

Reply 24

AG98

Original post by lizard54142

Surely AB is equal to r? The circle centre A has the same radius as the circle centre O, and B is a point on the circle centre A?

That's what I thought :confused:

But I've clearly gone wrong somewhere

Reply 25

ComputerMaths97

Original post by Terminatoring

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 :smile:

Posted from TSR Mobile

Nah, MEI OCR is 10x harder than Edexcel, other than that I don't know enough to comment. But Edexcel is so boringly easy it's unfair.

Reply 26

lizard54142

Original post by AG98

That's what I thought :confused:

But I've clearly gone wrong somewhere

You're correct, no worries :smile:

Reply 27

rubberduckies321

Original post by lizard54142

Someone's jumping straight to the mark scheme :wink:

Hahaha

Reply 28

Louisb19

Original post by joe12345marc

Nah, MEI OCR is 10x harder than Edexcel, other than that I don't know enough to comment. But Edexcel is so boringly easy it's unfair.

It's easy to say that when you aren't studying it.

Reply 29

ComputerMaths97

Original post by Louisb19

It's easy to say that when you aren't studying it.

Do a June 2014 M2 paper in both, mark it and come back to me and try persuade me that Edexcel is harder :P
Nah seriously, MEI OCR just tricks people, it's horrible. Edexcel is more focused on your understanding, which is way easier than understanding attempts to catch you out.

Reply 30

AG98

Original post by lizard54142

You're correct, no worries :smile:

But I still don't get why there's a factor of (3sqrt3 - pi) instead of just (pi), as I wrote in my attempted solution

Reply 31

Louisb19

Original post by AG98

But I still don't get why there's a factor of (3sqrt3 - pi) instead of just (pi), as I wrote in my attempted solution

Scroll up and look at the working out that I linked. Pi is not a factor of everything because in order to find the area of the segment you need to find the area of a triangle (which does not contain pi in the calculation).

Reply 32

lizard54142

Original post by AG98

But I still don't get why there's a factor of (3sqrt3 - pi) instead of just (pi), as I wrote in my attempted solution

Original post by notnek

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.

What notnek said here! AOB as it is, is not a sector. It is the sector described by the lines OA and OB, and the segment described by the line OB and the arc OB.

Reply 33

Terminatoring

Original post by Louisb19

Did this question earlier today. Here is my working, if you still don't understand tell me.

This was very helpful, thanks a lot :smile:

Btw - is that a different equation for the segment? (As opposed to the theta - sintheta one)

Posted from TSR Mobile

Reply 34

Terminatoring

Original post by notnek

Sorry - my mistake.

Thanks for the help

Posted from TSR Mobile

Reply 35

Terminatoring

Original post by lizard54142

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 :wink:

Thanks for the help :smile:

Posted from TSR Mobile

Reply 36

lizard54142

Original post by Terminatoring

Thanks for the help :smile:

Posted from TSR Mobile

No problem. Did you get there in the end?

Reply 37

Louisb19

Original post by Terminatoring

This was very helpful, thanks a lot :smile:

Btw - is that a different equation for the segment? (As opposed to the theta - sintheta one)

Posted from TSR Mobile

The equation I used was Segment = Sector - Triangle (inside the sector)

Reply 38

AG98

Original post by lizard54142

What notnek said here! AOB as it is, is not a sector. It is the sector described by the lines OA and OB, and the segment described by the line OB and the arc OB.

I'm probably being extremely dumb here, so sorry for that :tongue:

But I still don't get why you can't just get the area of half the circle and take away the areas outlined in red here:

I understand that you could work out the area of the triangle OCB and take away the unshaded segments, but that just seems like a more complicated way of doing it, and while it does work, I still don't understand why the original way I did it was wrong