question (1 Viewer)

[Pace_T]

JR q10 not sure how to do it
answer to II is 1339
thanks

 

[richz]

just find the area of the entire segment then minus the triangle

ie A (segment) = (r^2)@/2
where r = 505 and @ is 2*tan-1 (100/495)

then minus triangle A = 1/2*200*495

hope it works

 

[rama_v]

Draw up the whole circle and use the cosine rule; this was a beast of a question lol

 

[Pace_T]

im confused
please explain?
i cant do either i or ii
thanks

 

[mattchan]

EDIT; Basically, if you look at my dodgy diagram
Find the angle indicated on the diagram and find X through basical trigonometry.
Then use the cosine rule

where:
R^2 = R^2 + x^2 - 2 * R * x * CosTheta

That should find you R

 

[chin music]

na i thought the same thing mattchan but theres a way easier way. Make the centre of the circle R. Therefore From R to any point of the circle is equal. Let the radius = x(BR=x and AR =x). join D to R and then join A to R. u should get a right angled triangle (DRA) with right angle ADR. And as pythagoras theorem says (AD) squared +(DR) squared = (AD) squared. That is 100 squared (1m =100 cm) + (x- 10) squared = x squared. Ye and u just simplify that and get 505 . For the next part i just found the area of the sector and minussed the big triangles. hopefully i explained the first part well enough for ya. Its a pretty hard question i doubt itd be in the exam but its always good to do em.