Question

Urgent!!!Am nevoie de ajutor!!​

Answer 1

Explicație pas cu pas:

a)

V = AB×AA'×BC

54 = 3√3 × 2√3 × BC

BC = 54 : 18 => BC = 3 cm

$d = \sqrt{ {(3 \sqrt{3} )}^{2} + {(2 \sqrt{3} )}^{2} + {3}^{2}} = \sqrt{48} = 4 \sqrt{3} \ cm$

$\sqrt{48} < \sqrt{49} \implies d < 7 \ cm$

b)

CM = ⅓×DC = ⅓×3√3 = √3 cm

MD = DC+CM => MD = 4√3 cm

BD² = AB²+AD² = (3√3)²+3² = 27+9 = 36 = 6²

=> BD = 6 cm

ducem MN ⊥ BD, N∈BD

(BDD'B') ⊥ (ABCD) => d(M,(BDD')) = MN

ΔBCD ~ ΔMND (dreptunghice, unghiul D comun)

$\dfrac{BC}{MN} = \dfrac{BD}{MD} \iff \dfrac{3}{MN} = \dfrac{6}{4 \sqrt{3} } \\ MN = \dfrac{12 \sqrt{3} }{6} \implies \bf MN = 2 \sqrt{3} \ cm$