The sector COB is cut from the circle with center O. The ratio of the area of the sector removed from the whole circle to the area of the sector removed from the inner circle is R^2/r^2 .

Answer:

[tex]Ratio = \frac{R^2 - r^2 }{ r^2}[/tex]

Step-by-step explanation:

Given

See attachment for circles

Required

Ratio of the outer sector to inner sector

The area of a sector is:

[tex]Area = \frac{\theta}{360}\pi r^2[/tex]

For the inner circle

[tex]r \to radius[/tex]

The sector of the inner circle has the following area

[tex]A_1 = \frac{\theta}{360}\pi r^2[/tex]

For the whole circle

[tex]R \to Radius[/tex]

The sector of the outer sector has the following area

[tex]A_2 = \frac{\theta}{360}\pi (R^2 - r^2)[/tex]

So, the ratio of the outer sector to the inner sector is:

[tex]Ratio = A_2 : A_1[/tex]

[tex]Ratio = \frac{\theta}{360}\pi (R^2 - r^2) : \frac{\theta}{360}\pi r^2[/tex]

Cancel out common factor

[tex]Ratio = R^2 - r^2 : r^2[/tex]

Express as fraction

[tex]Ratio = \frac{R^2 - r^2 }{ r^2}[/tex]

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