Question

The sector COB is cut from the circle with center O. The ratio of the area of the sector removed from the outer circle to the area of the sector removed from the inner circle is

Answer 1

Hello

Answer 2

Is there a picture or image relating to this problem?

Answer 3

What is your question?

Answer 4

very top

Answer 5

Answer choices:
A. R^2/r^2
B. R/r
C. (R^2 - r^2)/r^2
D. 1

Answer 6

can anyone help?

Answer 7

it's not explicitly stated but let's let the big radius equal R and the small radius equal r

let a = area of inner circle removed = (pi)r^2 * (theta/360) where theta is just a generic angle

A = area of the whole entire sector, which includes the area removed from the small circle + area removed from the big circle = [(piR^2) - (pir^2)] * (theta/360)

the theta/360 ends up cancelling out so it doesn't really matter

Answer 8

so area removed from outer/area removed from inner = A/a = [(piR^2) - (pir^2)] / (pi)r^2

then simplify, you will end up with one of the answer choices