Respuesta :

[tex]\begin{gathered} \frac{AB}{r}=\frac{x}{10} \\ \text{Area of the circle is} \\ A=\pi r^2 \\ \text{Area of the circular sector is} \\ a=\frac{AB\cdot r}{2} \\ \text{Where AB is the arc-lenght} \\ \text{hence, the ratios is} \\ \frac{a}{A}=\frac{\frac{AB\cdot r}{2}}{\pi r^2} \\ \frac{a}{A}=\frac{AB\cdot r}{2\pi r^2} \\ \frac{a}{A}=\frac{AB}{2\pi\text{ r}} \end{gathered}[/tex][tex]\begin{gathered} \text{SINCE} \\ \frac{AB}{r}=\frac{x}{10}\Rightarrow AB\questeq\frac{r\cdot x}{10} \\ \text{THEN,} \\ \frac{a}{A}=\frac{r\cdot x}{(10)(2\pi r)} \\ \frac{a}{A}=\frac{x}{(10)(2\pi)} \\ \text{hence,} \\ \frac{a}{A}=\frac{x}{20\pi} \end{gathered}[/tex]

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