Question

se poderem responder!

Answer 1

resposta da pergunta e d)

Answer 2

Teorema de Pitágoras:

$\textsf{H}^2 = \textsf{CO}^2 + \textsf{CA}^2\\16^2 = x^2 + (4\sqrt{7})^2\\x^2 = 16^2 - (4\sqrt{7})^2\\x^2 = 256 - 112\\x^2 = 144\\x = \sqrt{144}\\\\\boxed{x = 12}$

O circulo maior tem diâmetro 12cm (raio 6cm).

Para resolver a questão, vamos chamar o circulo maior de "C1", o segundo maior de "C2", o terceiro maior de "C3" e o menor de todos de "C4".

A àrea hachurada é C2 - C3 + C4

Descobrindo o valor das áreas:

$A_{\textsf{C2}} = \pi (\frac{3\times6}{4})^2\\\\A_{\textsf{C2}} = \pi (\frac{9}{2})^2\\\\\boxed{A_{\textsf{C2}} = \frac{81\pi}{4}}$

$A_{\textsf{C3}} = \pi (\frac{6}{2})^2\\\\A_{\textsf{C3}} = \pi 3^2\\\\\boxed{A_{\textsf{C3}} = 9\pi}$

$A_{\textsf{C4}} = \pi (\frac{6}{4})^2\\\\A_{\textsf{C4}} = \pi (\frac{3}{2})^2\\\\\boxed{A_{\textsf{C4}} = \frac{9\pi}{4}}$

Portanto, a área hachurada é:

$\textsf{C2} - \textsf{C3} + \textsf{C4}\\\\\frac{81\pi}{4} - 9\pi + \frac{9\pi}{4}\\\\\frac{81\pi}{4} -\frac{36\pi}{4} + \frac{9\pi}{4}\\\\\frac{81\pi-36\pi+9\pi}{4}\\\\\frac{54\pi}{4}\\\\\boxed{\boxed{\frac{27\pi}{2}}}$

Letra C)