To solve this question we will use the following diagram:
Therefore:
![\begin{gathered} 2r_1+r_1=9in, \\ 2r_2+r_2=2r_(1.) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/on1w841hhb1mo99otffq1qheayga6tw9gx.png)
Adding like terms in the above equations we get:
![\begin{gathered} 3r_1=9in, \\ 3r_2=2r_1\text{.} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/z9jc3isj07cnj19tgfsp4y96bkik9bil45.png)
Therefore:
![\begin{gathered} r_1=(9in)/(3)=3in, \\ r_2=(6in)/(3)=2in\text{.} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/61sk055kpvfivltr54cmfelm1nzmgav4ce.png)
Now, notice that the orange region is formed by 2 semicircles of radius 3in and 2 semicircles of radius 1in, then, the area is:
![\begin{gathered} A=\pi(3in)^2+\pi(2in)^2 \\ =9\pi in^2+4\pi in^2 \\ =13\pi in^2\approx40.84in^2. \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/pdlj5e6ox4hdyrehmvhxf0twmhp354zxgm.png)
Answer: 40.84in².