Question

Va rog frumos sa ma ajutati si pe mn...rezolvati va rog problema asta

Answer 1

17 a) Desenul trunchiului de piramida triunghiulara il gasesti in fisierul atasat.

17 b)

$\displaystyle\\ \text{Volumul trunchiului de piramida este:}\\\\ V = \frac{h\Big(A_B + A_b + \sqrt{A_B\cdot A_b }\Big)}{3}\\\\ A_B = \text{aria bazei mari}\\ A_b = \text{aria bazei mici}\\ A_B = \frac{L^2\sqrt{3}}{4}=\frac{6^2\sqrt{3}}{4}=\frac{36\sqrt{3}}{4}=9\sqrt{3}~cm^2\\\\ A_b = \frac{l^2\sqrt{3}}{4}=\frac{4^2\sqrt{3}}{4}=\frac{16\sqrt{3}}{4}=4\sqrt{3}~cm^2\\\\ h = 9~cm$

$\displaystyle\\ V = \frac{h\Big(A_B + A_b + \sqrt{A_B\cdot A_b }\Big)}{3}=\\\\ =\frac{9\Big(9\sqrt{3} + 4\sqrt{3} + \sqrt{9\sqrt{3}\cdot 4\sqrt{3} }\Big)}{3}=\\\\ =3\Big(9\sqrt{3} + 4\sqrt{3} + \sqrt{9\sqrt{3}\cdot 4\sqrt{3} }\Big)=\\\\ =3\Big(13\sqrt{3}+\sqrt{36\cdot 3}\Big)=\\\\ =3\Big(13\sqrt{3}+6\sqrt{3}\Big)=3\cdot 19\sqrt{3} = \boxed{57\sqrt{3}~cm^3}$