An archery target consists of three concentric circles. The outer rim of the target has a radius of 1.2m. Find the radii of the two inner circles so that the three areas( central circle and two rings are equal)
total area = $\displaystyle \pi(1.2)^2$
inner circle ... $\displaystyle \pi r_1^2 = \frac{\pi(1.2)^2}{3}$
middle ring ... $\displaystyle \pi(r_2^2 - r_1^2) = \frac{\pi(1.2)^2}{3}$
outer ring ... $\displaystyle \pi[(1.2)^2 - r_2^2] = \frac{\pi(1.2)^2}{3}$
Hello, requal!
An archery target consists of three concentric circles.
The outer rim of the target has a radius of 1.2m.
Find the radii of the two inner circles so that the three areas
(central circle and two rings) are equal)
Let $\displaystyle a$ = smallest radius.
Let $\displaystyle b$ = next larger radius.
We know that $\displaystyle c = 1.2$ is the largest radius.
The area of the central circle is: .$\displaystyle \pi a^2$ .[1]
The area of the middle ring is: .$\displaystyle \pi b^2 - \pi a^2 \:=\:\pi(b^2-a^2)$ .[2]
The area of the outer ring is: .$\displaystyle \pi(1.2^2) - \pi b^2 \:=\:(1.44 - b^2)\pi$ .[3]
Since the areas are equal, [1] = [2] gives us:
. . $\displaystyle \pi a^2 \:=\:\pi(b^2-a^2) \quad\Rightarrow\quad b^2 - 2a^2 \:=\:0$ .[4]
And [2] = [3] gives us:
. . $\displaystyle \pi(b^2-a^2) \:=\:\pi(1.44 - b^2) \quad\Rightarrow\quad 2b^2 - a^2 \:=\:1.44$ .[5]
Solve the system: .$\displaystyle \begin{array}{cccc}b^2-2a^2 &=& 0 & {\color{blue}[4]} \\ 2b^2 - a^2 &=& 1.44 & {\color{blue}[5]} \end{array}$
. . and we get: .$\displaystyle \boxed{a \:=\:\frac{2\sqrt{3}}{5}} \quad\boxed{ b \:=\:\frac{2\sqrt{6}}{5}}$