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?

jonathanxu999 Feb 2, 2018

#2**+3 **

*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 r^{2} = π * 12^{2} , where r is the radius of the small circle.

3π r^{2} = 144π

Divide both sides by 3π .

r^{2} = 48

Take the positive square root of both sides.

r = 4√3

hectictar Feb 2, 2018

#2**+3 **

Best Answer

*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 r^{2} = π * 12^{2} , where r is the radius of the small circle.

3π r^{2} = 144π

Divide both sides by 3π .

r^{2} = 48

Take the positive square root of both sides.

r = 4√3

hectictar Feb 2, 2018

#3**+1 **

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

Good job, hectictar !!!

CPhill Feb 2, 2018

#4**+3 **

**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?**

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

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

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

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

heureka Feb 2, 2018