You are Here: Home >< Maths

1. Hello I'm having huge problems with 2a even after looking at the mark scheme. I have no idea how using area of a sector = half radius squared times theter comes to this. Where on earth had the 8 pi come from?

Thanks
Posted from TSR Mobile
Attached Images

2. (Original post by ThrashMetal)
Hello I'm having huge problems with 2a even after looking at the mark scheme. I have no idea how using area of a sector = half radius squared times theter comes to this. Where on earth had the 8 pi come from?

Thanks
Posted from TSR Mobile

So, there's where pi comes in.

Post your initial equation, and we can check.
3. (Original post by ThrashMetal)
Hello I'm having huge problems with 2a even after looking at the mark scheme. I have no idea how using area of a sector = half radius squared times theter comes to this. Where on earth had the 8 pi come from?

Thanks
Posted from TSR Mobile
Well, usually, angles on a straight line add to 180, but 180 degrees is radians.

You know the radius and the relationship between the two sectors (in terms of area)

Does that help?
4. I think I've done it. I forgot half a circle was pi radians lol.

Posted from TSR Mobile
Attached Images

5. (Original post by ThrashMetal)
I think I've done it. I forgot half a circle was pi radians lol.

Posted from TSR Mobile
Looking good

A bit of a mess on the page, though
6. (Original post by Indeterminate)
Looking good

A bit of a mess on the page, though
Haha yeah I always answer the question roughly then write it out neatly
7. (Original post by ThrashMetal)
Hello I'm having huge problems with 2a even after looking at the mark scheme. I have no idea how using area of a sector = half radius squared times theter comes to this. Where on earth had the 8 pi come from?

Thanks
Posted from TSR Mobile
AB is the diameter of the circle and so bisects the circle (perhaps the point you are missing).
The area of the whole circle is pi x r^2 = pi x 4^2 = 16pi
Half the area of the whole circle is (1/2) x 16pi = 8pi
So we have (Area of AOC Sector) + (Area of BOC Sector) = 8pi
But (Area of AOC Sector) + 5 = (Area of BOC Sector)
So (Area of AOC Sector) + ( (Area of AOC Sector) + 5) = 8 pi
So 2 x (Area of AOC Sector) = 8pi - 5.
But: (Area of AOC Sector) = (1/2) (r^2)(theta)=(1/2)(4^2)(theta)= 8 x (theta)

So we have 2 x (8 x (theta)) = 16 (theta) = 8pi -5
Thus (theta) = (8pi - 5)/16.

Hope this helps
8. Thanks for the help guys. Got there in the end. Plus repped

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: April 8, 2013
Today on TSR

### How to stand out in an Oxbridge interview

What makes you memorable?

### A pregnant woman visits me every night

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE