Question

a quinta pfvr, !!!!!

Answer 1

Sabe-se que:

$(\alpha,\alpha*q,\alpha*q^2)$

São as dimensões do paralelepípedo

$4\alpha+4\alpha*q+4\alpha*q^2=104$

$4\alpha+4\alpha*3+4\alpha*3^2=104$

$4\alpha+12\alpha+36\alpha=104$

$52\alpha=104$

$\boxed{\boxed{\alpha=2}}$

$(2,6,18)$

Então:

$V=2*6*18$

$\boxed{\boxed{V=216~cm^3}}$