Got a Maths question to ask... (Circle Theorem)

Mr Jones · 5 Jul 2010 at 21:12

A circle has centre O and radius 2. A, B and C are points on the circumference of the circle such that AB is the perpendicular bisector of OC.

Find the area of the segment of the circle bounded by the line segment AB and the minor arc ACB.

Give the area in an exact form in terms of surds and (Pi).

This is the diagram I've annotated

It's been awhile since I've done circles but as I remember, there 2 equations to work them out is:

Area of segment AB
r(squared)/2 * ( (Pi)/180 * AngleC - Sin AngleC )

Length of arc ACB
r ( 2 (Pi) AngleC / 360 )

What else can I use to work out what AngleC is, or is my equation going to give the answer as
Area AB = ( (Pi)/90 * AngleC - 2 * Sin AngleC )
Arc ACB = ( (Pi)AngleC )/90

Help is much appreciated.

Mr Jones · 5 Jul 2010 at 22:42

Where are all you maths dudes!
I would like some confirmation as it's something I need to show tomorrow morning!

bennyB · 5 Jul 2010 at 22:48

You could use cDonald's theorem?

D D Danneh · 5 Jul 2010 at 22:57

I'm assuming OD == DC?

If so, that means OD is 1. Since OA is the radius, it is 2.

That means angle AOD is cos^-1(1 / 2), or 60 degrees.

Since triangle AOB is isosceles, angle C is equal to 60 x 2, or 120 degrees.

Inquisitor · 5 Jul 2010 at 23:02

bennyB said:
You could use cDonald's theorem?

n² + 9 + 9, I believe. Clever guy, that cDonald.

bennyB · 5 Jul 2010 at 23:05

yeah it reduces everything down to a simple quadratic, clever stuff I say

Mr Jones · 5 Jul 2010 at 23:07

D D Danneh said:
I'm assuming OD == DC?

If so, that means OD is 1. Since OA is the radius, it is 2.

That means angle AOD is cos^-1(1 / 2), or 60 degrees.

Since triangle AOB is isosceles, angle C is equal to 60 x 2, or 120 degrees.

I'm not sure if OD=DC unless there's something I'm missing? What theorem would suggest that?
The only equal distance that's firm is AD=DB.

D D Danneh · 5 Jul 2010 at 23:10

Mr Jones said:
I'm not sure if OD=DC unless there's something I'm missing? What theorem would suggest that?
The only equal distance that's firm is AD=DB.

AB bisects OC.

Mr Jones · 5 Jul 2010 at 23:14

D D Danneh said:
AB bisects OC.

BAH!
I can't believe this didn't dawn on me, AB bisects OC means it's not just AD=DB but OD = DC too! So it is indeed 1!