Question

áreas de regiones circulares
ayuda porfas en estos ejercicios​

Answer 1

Verified answer

Respuesta:

Explicación paso a paso:

2 )

[tex]Radio: r =\sqrt{5} m[/tex]

[tex]\pi =3.14[/tex]

[tex]Area: A = ?[/tex]

[tex]Formula:[/tex]

[tex]A = \pi r^{2}[/tex]

[tex]A = \pi (\sqrt{5} m)^{2} = \pi (5m^{2} )[/tex]

[tex]A = 5\pi m^{2}[/tex]

[tex]A = 5(3.14) m^{2}[/tex]

[tex]A = 15.7m^{2}[/tex]

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3 )

[tex]Area: A = 16m^{2}[/tex]

[tex]\pi = 3.14[/tex]

[tex]Radio:R =?[/tex]

[tex]Formula:[/tex]

[tex]R = \sqrt{\frac{A}{\pi } }[/tex]

[tex]R = \sqrt{\frac{16m^{2} }{3.14} }[/tex]

[tex]R = \sqrt{5.1m^{2} }[/tex]

[tex]R = 2.26m[/tex]

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4 )

[tex]Radio: R = 6m[/tex]

[tex]\pi =3.14[/tex]

[tex]Area -de-la -region -sombreada: A s = ?[/tex]

[tex]Formula:[/tex]

[tex]As = \frac{\pi R^{2} }{2}[/tex]

[tex]As = \frac{(3.14)(6m)^{2} }{2} =\frac{(3.14)(36m^{2}) }{2} =\frac{113.04m^{2} }{2}[/tex]

[tex]As = 56.52m^{2}[/tex]

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7 )

[tex]Radio: R = 4u[/tex]

[tex]\pi =3.14[/tex]

[tex]Area -de-la-region -sombreada: As = ?[/tex]

Fórmula:

[tex]As = \frac{\pi R^{2} }{4}[/tex]

[tex]As = \frac{\pi (4u)^{2} }{4} =\frac{(3.14)(16u^{2}) }{4} =(3.14)(4u^{2} )[/tex]

[tex]As= 12.56u^{2}[/tex]