Question 25 (1 Viewer)

Rednut

New Member
I know that for 25. a (ii) (the one about the similar triangles) the answer was d/200=20/x, but was there a way of symplifying that further?

Also, with the Annulus formula - was the answer R=Root(0.75squared +6.79/pi)??

Marty

Member
A = pi(R^2 - r^2)

A = pi R^2 - pi r^2

A + pi r^2 = pi R^2

(A + pi r^2) / pi = R^2

R = root [(A + pi r^2)/pi]

yeh or something like that

banx006

New Member
i put that marty, but im pretty sure its wrong. i think you only square root the r because its the only squared thing. could be wrong, but i worked it out after the exam, but i cant find it now!!

Untitled

New Member
no, when u take the r^2 to the other side u square root everything,

the answer for the question is

A= pie (R^2 - r^2)

A/pie = R^2 - r^2

A/pie + r^2 = R^2

root ( A/pie + r^2 ) = R

i am pretty sure thats the answer.
let me know if not

Eeeek

New Member
HAHA, all you smart ppl, i just wrote next to the question...i have no idea and aint gunna pretend i do! Do u think they will give me a mark??????

tammer

Remembering P.I.G.
sorry to break it to you Eeeek but as you may or may not have heard writing things like that may actually lose you marks as they consider it as an unserious attempt

Marty

Member
so where does losing marks come into it?

bugger

New Member
They could say it was a non-serious attemp and give 0 for the whole exam

tammer

Remembering P.I.G.
thats what i meant bugger

i was with marty on the solution tho

ie. R = sqroot (A + pi r^2)/pi