+0

# Help!

0
157
1
+1110

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

#1
+5057
+3

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

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

.
Dec 28, 2018

#1
+5057
+3
$$A_2 = \pi(r_2^2-r_1^2)\\ r_2^2 =\dfrac{A_2 + \pi r_1^2}{\pi} = \dfrac{A_2 + A_1}{\pi}\\ A_3 = \pi(r_3^2 - r_2^2) \\ r_3^2 = \dfrac{A_3 + \pi r_2^2}{\pi} = \dfrac{A_3+A_2+A_1}{\pi}\\ r_4^2 = \dfrac{A_4+A_3+A_2+A_1}{\pi}$$
$$r_4^2 = \dfrac {(4+3+2+1)A_1}{\pi} = \dfrac{10}{\pi}\cdot \pi r_1^2 = 10 r_1^2\\ \left(\dfrac{r_4}{r_1}\right)^2 = 10\\ \dfrac{r_4}{r_1} = \sqrt{10}$$