Question

Numa piramide quadrangular regular,uma aresta da base mede 2 raiz de 2cm e uma aresta lateral mede raiz de 22 cm.O volume dessa piramide,em cm,é?

Answer 1

aresta da base = 2*\/2 cm

diagonal da base = ( 2*\/2 )*\/2 = 4 cm

aresta lateral = \/22 cm

então para o cálculo da altura:

(\/22)² = h² + 2²

22 = h² + 4 => h = 3\/2 cm

V = (1/3)*8*3\/2 = 8\/2 cm³

Answer 2

Caso esteja pelo app, e tenha problemas para visualizar esta resposta, experimente abrir pelo navegador https://brainly.com.br/tarefa/16973181

                                                   

$\sf\ell^2=a_p^2+\bigg(\dfrac{a}{2}\bigg)^2\\\sf(\sqrt{22})^2=a_p^2+\bigg(\dfrac{\diagdown\!\!\!2\sqrt{2}}{\diagdown\!\!\!2}\bigg)^2\\\sf a_p^2+2=22\\\sf a_p^2=22-2\\\sf a_p^2=20\\\sf \\\sf a_p^2=h^2+m^2\\\sf 20=h^2+(\sqrt{2})^2\\\sf h^2+2=20\\\sf h^2=20-2\\\sf h^2=18\\\sf h=\sqrt{18}=3\sqrt{2}~cm$

$\sf V=\dfrac{1}{3}\cdot B\cdot h\\\sf V=\dfrac{1}{\backslash\!\!\!3}\cdot(2\sqrt{2})^2\cdot\backslash\!\!\!3\sqrt{2}\\\sf V=4\cdot2\cdot\sqrt{2}\\\huge\boxed{\boxed{\boxed{\boxed{\sf V=8\sqrt{2}~cm^3}}}}\blue{\checkmark}$