SOLUTION to question 8.
The area of the shaded portion is caculated by using the formula
![\text{Area of external circle- Area of the internal circle}](https://img.qammunity.org/2023/formulas/mathematics/college/48eyn7kw7c8wn02vy2tq2138b04izlxeo9.png)
Hence
Area of the bigger circle will be obtain using the formula
![\begin{gathered} \pi R^2 \\ \text{where} \\ \pi=3.14, \\ R=(6)/(2)=3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/b0ilck9906riid3nztw57z6skmv4iar5i1.png)
substitute into the formula, we have
![=3.14*3^2=3.14*9=28.26cm^2](https://img.qammunity.org/2023/formulas/mathematics/college/afmze64zea4sstgluh8p8kfi9e1hcqdjfq.png)
Hence
Area of the bigger circle is 28.26cm²
There area of the bigger circle will be
![\begin{gathered} \pi r^2 \\ \text{where r=}\frac{\text{2}}{2}=1 \\ \\ 3.14*1^2=3.14cm^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8u8vnivgv2i63tbnueattezo8xvxpgu4aw.png)
Hence
The ares of the internal circle is 3.14cm²
Hence
There area of the shaded portion is
![28-26-3.14=25.12cm^2](https://img.qammunity.org/2023/formulas/mathematics/college/7muj0ypka55f5t2299v0x9f3fbvmou2w1z.png)
Therefore
The area of the of the shaded portion is 25.12cm²