Sectors

Hi,

Please can I have some help with this question?

I know:

S = r theta
6 = r theta

24 = (1/2) x a x b x sin theta
48 = ab sin theta

I was thinking next that:

Since OA = OB which is also the radius
48 = r^2 theta
which is the same as
48 = 2r x theta

Not sure how i would use this?

Then to find the area of the shaded segment
A = (1/2) r^2 theta

24 is actually the area of the sector

So 24 = (1/2) r^2 theta
48 = r^2 theta

Reply 3

ghostwalker

Original post by M.Johnson2111

24 is actually the area of the sector

So 24 = (1/2) r^2 theta
48 = r^2 theta

Yes.

And

6=r\theta

So, what's r?

And hence

\theta

Reply 4

M.Johnson2111

Original post by M.Johnson2111

24 is actually the area of the sector

So 24 = (1/2) r^2 theta
48 = r^2 theta

Then Im sure that area of the sector would be area of the shaded segment + area of the triangle?

So i need to find area of the triangle, (1/2) ab sin theta and subtract this from 24 (area of sector) ?

Reply 5

ghostwalker

Original post by M.Johnson2111

Yes.

See my previous post too.

Reply 6

M.Johnson2111

Original post by ghostwalker

Yes.

And

6=r\theta

So, what's r?

And hence

\theta

Well r is the radius and then to find theta I would 6/r

Would theta be 90, using the circle theorem then convert this to radians?

Reply 7

ghostwalker

Original post by M.Johnson2111

Well r is the radius and then to find theta I would 6/r

Would theta be 90, using the circle theorem then convert this to radians?

No, it's not 90.

Use the two equations you have:

6=r\theta

48=r^2\theta

Reply 8

M.Johnson2111

So would I let r = 6/theta

then sub this into 48 = r^2 theta?

So 36/theta^2 = 48

36 = 48 x theta^2

3/4 = theta^2

Then squareroot 3/4 = theta?

Reply 9

ghostwalker

Original post by M.Johnson2111

So would I let r = 6/theta

then sub this into 48 = r^2 theta?

So 36/theta^2 = 48

36 = 48 x theta^2

3/4 = theta^2

Then squareroot 3/4 = theta?

Your substituted equation, in red, is incorrect. Can you correct it? You missed a bit out.

Reply 10

M.Johnson2111

Original post by M.Johnson2111

So would I let r = 6/theta

then sub this into 48 = r^2 theta?

So 36/theta^2 = 48

36 = 48 x theta^2

3/4 = theta^2

Then squareroot 3/4 = theta?

Ignore ^

48 = (6/theta) x theta

48 = (36/theta^2) x theta

48 = 36 theta

theta = 4/3?

Reply 11

M.Johnson2111

Original post by ghostwalker

Your substituted equation, in red, is incorrect. Can you correct it? You missed a bit out.

We noticed this at same time perhaps, thank you

Reply 12

ghostwalker

Original post by M.Johnson2111

Ignore ^

48 = (6/theta) x theta

48 = (36/theta^2) x theta

48 = 36 theta

theta = 4/3?

Agreed - don't forget that's in radians.

You should also be able to get r now, and then proceed to work out the area of the triangle, etc.

Reply 13

M.Johnson2111

Original post by M.Johnson2111

We noticed this at same time perhaps, thank you

So r = 6/ theta = 6 / (4/3) = 9/2

Reply 14

M.Johnson2111

Original post by ghostwalker

Agreed - don't forget that's in radians.

You should also be able to get r now, and then proceed to work out the area of the triangle, etc.

thank you!

Original post by M.Johnson2111

So r = 6/ theta = 6 / (4/3) = 9/2

Using A = 1/2 x (9/2)^2 x sin (4/3)
A = 9.84...
So area of segment = 24 - 9.84... = 14.15912.... = 14.2cm^2 to 3 sigfig

Reply 15

ghostwalker

Original post by M.Johnson2111

Ignore ^

48 = (6/theta) x theta

48 = (36/theta^2) x theta

48 = 36 theta

theta = 4/3?

SORRY!

I shouldn't have agreed with you there. Theta doesn't = 4/3.

If you cancel out the theta's you get.

48 = 36/theta

So, theta = 3/4.

Then proceed as before.

Reply 16

M.Johnson2111

Original post by ghostwalker

SORRY!

I shouldn't have agreed with you there. Theta doesn't = 4/3.

If you cancel out the theta's you get.

48 = 36/theta

So, theta = 3/4.

Then proceed as before.

Ah yes thank you