Answer:
[tex]\begin{gathered} a.\text{ The area is }\frac{25}{2}\pi\text{ cm}^2 \\ b.\text{ The area is }\frac{5}{9}\pi\text{ cm}^2 \end{gathered}[/tex]
Step-by-step explanation:
The area of a circle is represented by the following equation;
[tex]\begin{gathered} A_o=\pi *r^2 \\ \text{ Then, for a semicircle:} \\ A_o=\frac{\pi *r^2}{2} \end{gathered}[/tex]
Therefore, for a semicircle with a radius of 5cm:
[tex]\begin{gathered} A_o=\frac{\pi *5^2}{2} \\ A_o=\frac{25}{2}\pi \end{gathered}[/tex]
The area of the sector is given as:
[tex]\begin{gathered} A_{\text{ sector}}=(\frac{\theta}{360})*\pi *r^2 \\ \text{ arc length=r*}\theta \end{gathered}[/tex]
Given the arc length and the radius, solve for the angle:
[tex]\begin{gathered} 20=10\theta \\ \theta=\frac{20}{10} \\ \theta=2\text{ degrees} \end{gathered}[/tex]
Now, for the area of the sector:
[tex]\begin{gathered} Area_{sector}=(\frac{2}{360})*\pi *10^2 \\ Area_{sector}=\frac{5}{9}\pi \end{gathered}[/tex]
