Respuesta :

Yuzhu

Answer:

[tex]d = \sqrt{ {22}^{2} - {18}^{2} } = \sqrt{484 - 324} = \sqrt{160} = 4 \sqrt{10} [/tex]

[tex]circle \: area = ({4 \sqrt{10} \div 2 })^{2} \times \pi = ( {2 \sqrt{10} })^{2} \times \pi =40\pi [/tex]

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