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(D) Let the radius of the inner circle be $r$ and the radius of the outer circle be $R$.Given that the sum of their areas is $116 \, \pi \, cm^2$,we h

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$4 \, cm$ and $10 \, cm$

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(D) Let the radius of the inner circle be $r$ and the radius of the outer circle be $R$.Given that the sum of their areas is $116 \, \pi \, cm^2$,we have $\pi R^2 + \pi r^2 = 116 \, \pi$,which simplifies to $R^2 + r^2 = 116$.Since the circles touch internally,the distance between their centers is $R - r = 6 \, cm$.We know the identity $(R + r)^2 + (R - r)^2 = 2(R^2 + r^2)$.Substituting the known values: $(R + r)^2 + (6)^2 = 2(116)$.$(R + r)^2 + 36 = 232$.$(R + r)^2 = 232 - 36 = 196$.Taking the square root,$R + r = 14$.Now we have a system of linear equations: $R + r = 14$ and $R - r = 6$.Adding the two equations: $2R = 20 \Rightarrow R = 10 \, cm$.Subtracting the equations: $2r = 8 \Rightarrow r = 4 \, cm$.Thus,the radii are $10 \, cm$ and $4 \, cm$.

Two circles touch internally. The sum of their areas is $116\, \pi \, cm^2$ and the distance between their centers is $6 \, cm$. Then,the radii of the circles are?

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