C2 ques - radian measure

Im really confused with this c2 ques. I dont know what its asking me to do. Ive tried many methods - once i got the answer but it didnt make sense. I havent listed all of the because i cant remember what i was trying to do.

The question is:

Find the area of the major segment, OEF

A) OE = 10cm, theta = pi / 6 rads
Answer: 313cm^2

First i did pi r^2 - (r^2 x theta) / 2
= 100 pi - 1500
= -ve answer WRONG

then i tried area of major SECTOR + area of minor SECTOR
11/6 pi x 100 all divided by 2 + 100 x pi/6 all divided by 2
= 313.1799... CORRECT?!

BUT, the question asks for the major SEGMENT so shouldnt it be pi r^2 - 1/2 ab sin c (sin is 30?)
When i do that i get 338??

Help please :smile:

(edited 8 years ago)

Reply 1

Zacken

Original post by kiiten

...

Could you post the actual question? What's OEF? What's the radius? What are you talking about?

Reply 2

Zacken

Does this picture help? You can find the area of the minor sector and then subtract that from the area of the full circle to get the area of the major segment.

Original post by kiiten

Find the area of the major segment, OEF

BUT, the question asks for the minor SEGMENT so shouldnt it be pi r^2 - 1/2 ab sin c (sin is 30?)
When i do that i get 338??

Reply 3

kiiten

Thats the question. 3.a)

Sorry my diagram is kinda messy.

Attachment not found

Im guessing o is the centre and e and f are points on the circumference so 10cm is the radius.

image.jpg370.6KB

Reply 4

Zacken

Original post by kiiten

Im guessing o is the centre and e and f are points on the circumference so 10cm is the radius.

Huh? It sounds like they're talking about sectors instead of segments or have I forgotten my geometry?

Edit: lol my bad, half me forgetting geometry half the textbook not bothering to define stuff properly.

(edited 8 years ago)

Reply 5

kiiten

Original post by Zacken

Does this picture help? You can find the area of the minor sector and then subtract that from the area of the full circle to get the area of the major segment.

Doesnt that give you the area of the white part which is the major SECTOR?

Reply 6

kiiten

Original post by Zacken

Huh? It sounds like they're talking about sectors instead of segments or have I forgotten my geometry?

Thats just what i guessed it would look like but i have no idea where OEF are. 10 must definitely be the radius though because pi r^2 using 10 gives you 314

Reply 7

Zacken

Original post by kiiten

Thats just what i guessed it would look like but i have no idea where OEF are. 10 must definitely be the radius though because pi r^2 using 10 gives you 314

Okay, s'all fine. I understand it now. Look at my picture again, can you find the area of the minor segment?

Reply 8

kiiten

Original post by Zacken

Okay, s'all fine. I understand it now. Look at my picture again, can you find the area of the minor segment?

Theta is pi/6 = 30 rads? (Not sure if/when you convert it)
R^2 x theta all divided by 2
= 100 x (pi/6) / 2
= 25/3 pi

Area of triangle
1/2 a b sin c
1/2 x 100 x sin pi/6
= 25

Minus both to get 1.179999...
Ah so for the major segment you do pi r^2 = 100pi
100pi - 1.17999... = 312.979
= 313 (3s.f.)

But my question is that: is pi/6 180/6 = 30rads or is it 30 degrees?

Reply 9

Zacken

Original post by kiiten

But my question is that: is pi/6 180/6 = 30rads or is it 30 degrees?

\frac{\pi}{6} \text{rad} = 30^{\circ}

. So

\frac{\pi}{6}

is in radians which is the same thing as

30^{\circ}

Reply 10

kiiten

Sorry, 360 degrees is 2pi - i just read it a few pages before that question. Yeah, you're right 30 degrees is pi/6 rads
Thanks for your help :smile:

Reply 11

Zacken

Original post by kiiten

Sorry, 360 degrees is 2pi - i just read it a few pages before that question. Yeah, you're right 30 degrees is pi/6 rads
Thanks for your help :smile:

No problem.

Reply 12

kiiten

Original post by Zacken

No problem.

8.b) ?
R^2 x theta / 2 = r^2 ??

Posted from TSR Mobile

1457280641751.jpg47.4KB

Reply 13

Zacken

Original post by kiiten

8.b) ?
R^2 x theta / 2 = r^2 ??

Posted from TSR Mobile

You know that perimeter of the sector is given by

2r + r\theta = 4r

so some simplification gets you

\theta=2

.

Then, area sector is

\frac{1}{2} \times r^2 \times 2 = r^2

(because

\theta = 2

Reply 14

kiiten

Original post by Zacken

You know that perimeter of the sector is given by

2r + r\theta = 4r

so some simplification gets you

\theta=2

.

Then, area sector is

\frac{1}{2} \times r^2 \times 2 = r^2

(because

\theta = 2

Thank you!!
Is the diagram like this for question 2.b) ?

Posted from TSR Mobile

1457337954558.jpg44.6KB

1457338189729.jpg28.2KB

Reply 15

Zacken

Original post by kiiten

Thank you!!
Is the diagram like this for question 2.b) ?

Posted from TSR Mobile

The diagram is right, yeah. But I'm hoping that where you've written 8 you've written it to mean the side lengths of the equilateral triangle OAB and not APB, yeah?

BTW, remember your GCSE circle theorems? If the angle at the centre is something (AOB) then the angle at the cirumference (APB) is half of that.

Reply 16

kiiten

Original post by Zacken

Yeah i remember but i got 41.9 and the answer is 49.9 ? This is what i did:

Posted from TSR Mobile

(edited 8 years ago)

1457374282517.jpg31.5KB

Reply 17

kiiten

anyone? :'(

I got 41.9 but the answer is 49.9

Reply 18

Dr Cobbold

Original post by kiiten

I was wondering since our upcoming chemistry exams are coming up if people wanted to be part of a maths whatsapp revision group. I guessing most of us use whatsapp so I guess it will be more suitable for everyone then. If you are interested either pm me or responded to this message

Posted from TSR Mobile

Posted from TSR Mobile