Answer:
[tex]r = 4m[/tex] --- small circle
[tex]R =8m[/tex] --- big circle
Step-by-step explanation:
Given
[tex]Area = 80\pi\ m^2[/tex] -- sum of areas
[tex]R = 2r[/tex]
Required
The radius of the larger circle
Area is calculated as;
[tex]Area = \pi r^2[/tex]
For the smaller circle, we have:
[tex]A_1 = \pi r^2[/tex]
For the big, we have
[tex]A_2 = \pi R^2[/tex]
The sum of both is:
[tex]Area = A_1 + A_2[/tex]
[tex]Area = \pi r^2 + \pi R^2[/tex]
Substitute: [tex]R = 2r[/tex]
[tex]Area = \pi r^2 + \pi (2r)^2[/tex]
[tex]Area = \pi r^2 + \pi *4r^2[/tex]
Substitute [tex]Area = 80\pi\ m^2[/tex]
[tex]80\pi = \pi r^2 + \pi *4r^2[/tex]
Factorize
[tex]80\pi = \pi[ r^2 + 4r^2][/tex]
[tex]80\pi = \pi[ 5r^2][/tex]
Divide both sides by [tex]\pi[/tex]
[tex]80 = 5r^2[/tex]
Divide both sides by 5
[tex]16 = r^2[/tex]
Take square roots of both sides
[tex]4 = r[/tex]
[tex]r = 4m[/tex]
The radius of the larger circle is:
[tex]R = 2r[/tex]
[tex]R =2 * 4[/tex]
[tex]R =8m[/tex]
