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In the figure with four circles below, let A1 be the area of the smallest circle, let A2 be the area of the region inside the second-smallest circle but outside the smallest circle, and so on. If A1 : A2 : A3 : A4 = 1 : 2 : 3 : 4, then find the ratio of the largest radius to the smallest radius.

 Dec 27, 2018

Best Answer 

 #1
avatar+4394 
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\(A_2 = \pi(r_2^2-r_1^2)\\ r_2^2 =\dfrac{A_2 + \pi r_1^2}{\pi} = \dfrac{A_2 + A_1}{\pi}\\ A_3 = \pi(r_3^2 - r_2^2) \\ r_3^2 = \dfrac{A_3 + \pi r_2^2}{\pi} = \dfrac{A_3+A_2+A_1}{\pi}\\ r_4^2 = \dfrac{A_4+A_3+A_2+A_1}{\pi}\)

 

\(r_4^2 = \dfrac {(4+3+2+1)A_1}{\pi} = \dfrac{10}{\pi}\cdot \pi r_1^2 = 10 r_1^2\\ \left(\dfrac{r_4}{r_1}\right)^2 = 10\\ \dfrac{r_4}{r_1} = \sqrt{10}\)

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 Dec 28, 2018
 #1
avatar+4394 
+3
Best Answer

\(A_2 = \pi(r_2^2-r_1^2)\\ r_2^2 =\dfrac{A_2 + \pi r_1^2}{\pi} = \dfrac{A_2 + A_1}{\pi}\\ A_3 = \pi(r_3^2 - r_2^2) \\ r_3^2 = \dfrac{A_3 + \pi r_2^2}{\pi} = \dfrac{A_3+A_2+A_1}{\pi}\\ r_4^2 = \dfrac{A_4+A_3+A_2+A_1}{\pi}\)

 

\(r_4^2 = \dfrac {(4+3+2+1)A_1}{\pi} = \dfrac{10}{\pi}\cdot \pi r_1^2 = 10 r_1^2\\ \left(\dfrac{r_4}{r_1}\right)^2 = 10\\ \dfrac{r_4}{r_1} = \sqrt{10}\)

Rom Dec 28, 2018

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