Question

prisma regulata ABCA'B'C' are aria laterala de 54cm² si aria totala de 6(9+√3)cm². aflați: a) volumul prismei; b) masura unghiului planelor (ABC) si (A'BC).
Ajutor,dau coroana!:(((((

Answer 1

P*h=54 cm²
2Ab+Al=6(9+√3)cm²⇒a²√3/2+54=54+6√3⇒a²√3/2=6√3⇒a²=12⇒a=2√3 cm
3*2√3*h=54⇒6√3h=54⇒h=3√3 cm
V=a²√3/4*h=3√3*3√3=27 cm³

Answer 2

$a) A_{t}=A_{l}+2A_{b} , deci \,\,A_{b}=(A_{t}-A_{l})/2=3 \sqrt{3} \\ A_{b}=\frac{l^{2}\sqrt{3}}{4}=3\sqrt{3}, deci\, l=2\sqrt{3}. \\A_{l}=P_{b}h, deci \, h=\frac{A_{l}}{P_{b}}=3\sqrt{3}.\\V=A_{b}h=27 \\ b) unghiul\,este A`MA, unde\, M\, este\, mijlocul \, laturii\, BC. \\ Aflam\, AM=3 \,cu\, teorema\, lui \,Pitagora\, in\, \Delta ABM.\\ Inaltimea\, AA`=3\sqrt{3}. \\Aplicam\,tangenta\, in \,\Delta A`MA \, dreptunghic: tg(A`MA)=\frac{3\sqrt{3}}{3}=\sqrt{3}.\\ Deci \, unghiul\, este \, de\, 60 \,grade.$