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

The area of sector COB = ∅/360 π R²

The area of sector OEF = ∅/360 π r²


The area of the sector removed from the outer circle

=  ∅/360 π R² - ∅/360 π r²

= ∅/360 π (R² - r²)


The area of the sector removed from the inner circle = ∅/360 π r²


The ratio of the area of the sector removed from the outer circle to the area of the sector removed from the inner circle

=   ∅/360 π (R² - r²) :    ∅/360 π r²

=    R² - r² :   r²                           

Answer 2

the picture in the attached figure

we know that

area of a sector=(∅/2)*r²--------> when ∅ is in radians

area of sector inner circle=(∅/2)*r²

area of sector outer circle=(∅/2)*R²-(∅/2)*r²

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

[(∅/2)*R²-(∅/2)*r²]/[(∅/2)*r²]-----> [R²-r²]/[r²]

therefore

the answer is

(R^2 - r^2)/r^2