Respuesta :

etanauriel

Answer:

B.

[tex] \frac{275\pi}{6} {m}^{2} [/tex]

Step-by-step explanation:

[tex]area \: of \: sector \: is \\ \frac{the \: angle}{360} \times \pi {r}^{2} \\ hence \: we \: have \\ \frac{165}{360} \times {(10m)}^{2} = \frac{275\pi}{6} {m}^{2} [/tex]

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