Three concentric circles

*edit again* I just thought of this....

area of small circle  +  area of middle ring + area of outer ring   =   area of big circle

area of small circle + area of small circle + area of outer ring   =   area of big circle

area of small circle + area of small circle + area of small circle   =   area of big circle

3(area of small circle)   =   area of big circle

3π r²  =  π * 12²      ,  where  r  is the radius of the small circle.

3π r²  =  144π

                               Divide both sides by  3π .

r²  =  48

                               Take the positive square root of both sides.

r  =  4√3

Mmmm....that problem seems familiar!!!!  [inside joke ]

Good job, hectictar  !!!

Three concentric circles are drawn such that the area of the smallest circle is equal to the area of each of the rings.
If the radius of the largest circle is 12 cm,
what is the radius of the smallest circle?

\(\begin{array}{rcll} \text{Let smallest circle radius $=r_s$ } \\ \text{Let middle circle radius $=r_m$ }\\ \text{Let largest circle radius $=r_l$ }\\ \end{array} \)

\(\begin{array}{|lrcll|} \hline & \pi r_s^2 = \pi r_m^2-\pi r_s^2 = \pi r_l^2 - \pi r_m^2 \quad & | \quad : \pi \\ & r_s^2 = r_m^2- r_s^2 = r_l^2- r_m^2 \\ \hline \end{array} \)

\(\begin{array}{lrcll} (1) & 2r_s^2 &=& r_m^2 \\ \hline (2) & 2r_m^2 &=& r_l^2 + r_s^2 \quad & | \quad r_m^2 = 2r_s^2 \\ & 2\cdot (2r_s^2) &=& r_l^2 + r_s^2 \\ & 4r_s^2 &=& r_l^2 + r_s^2 \\ & 3r_s^2 &=& r_l^2 \quad & | \quad \sqrt{ }\\ & \sqrt{3}r_s &=& r_l \quad & | \quad r_l = 12 \\ & \sqrt{3}r_s &=& 12 \\ & r_s &=& \dfrac{12}{\sqrt{3}} \\ & r_s &=& \dfrac{12 \sqrt{3}}{\sqrt{3}\sqrt{3}} \\ & r_s &=& \dfrac{12 \sqrt{3}}{3} \\ & \mathbf{r_s} &\mathbf{=}& \mathbf{4 \sqrt{3} } \\ \end{array}\)

\(\text{The radius of the smallest circle is $\mathbf{4 \sqrt{3} }$ cm $\approx \mathbf{6.9} $ cm } \)