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# C2 ques - radian measure watch

1. 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

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

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??

2. (Original post by kiiten)
...
Could you post the actual question? What's OEF? What's the radius? What are you talking about?

3. 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??
4. Thats the question. 3.a)

Sorry my diagram is kinda messy.Attachment 510537510539 Im guessing o is the centre and e and f are points on the circumference so 10cm is the radius.
Attached Images

5. (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.
6. (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?
7. (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
8. (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?
9. (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?
10. (Original post by kiiten)
But my question is that: is pi/6 180/6 = 30rads or is it 30 degrees?
. So is in radians which is the same thing as .
11. 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
12. (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
No problem.
13. (Original post by Zacken)
No problem.

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

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Attached Images

14. (Original post by kiiten)
8.b) ?
R^2 x theta / 2 = r^2 ??

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You know that perimeter of the sector is given by so some simplification gets you .

Then, area sector is (because ).
15. (Original post by Zacken)
You know that perimeter of the sector is given by so some simplification gets you .

Then, area sector is (because ).
Thank you!!
Is the diagram like this for question 2.b) ?

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Attached Images

16. (Original post by kiiten)
Thank you!!
Is the diagram like this for question 2.b) ?

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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.
17. (Original post by Zacken)
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.
Yeah i remember but i got 41.9 and the answer is 49.9 ? This is what i did:

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18. anyone? :'(

I got 41.9 but the answer is 49.9
19. (Original post by kiiten)
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

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

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??

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

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