Question

hei !este urgent!ma ajuta cineva#!

Answer 1

L=AB=18cm (latura bazei mari)
l=A'B'=12cm (latura bazei mici)
h=OO'=3cm (inaltimea)

a)
Formula volumului trunchiului de piramida:

$\boxed{V_{tr}=\frac{h_{tr}}{3}\cdot(A_{B}+A_{b}+\sqrt{A_{B} \cdot A_{b}})}\\ \\ Unde:\\ \\ A_{B} \Rightarrow aria~bazei~mari\\ \\ A_{b} \Rightarrow aria~bazei~mici$

ABCA'B'C' trunchi de piramida triunghiulara regulata (1)

Din (1) =>ΔABC, ΔA'B'C' echilaterale
=>
$A_{[ABC]}=\frac{L^{2}\sqrt{3}}{4}\\ \\A_{[ABC]}=\frac{18^{2}\sqrt{3}}{4}\\ \\ A_{[ABC]}=\frac{324\sqrt{3}}{4}\\ \\ \boxed{A_{[ABC]}=81\sqrt{3}(cm^{2})}\\ \\ \\ \\ \\ \\ A_{[A'B'C']}=\frac{l^{2}\sqrt{3}}{4}\\ \\A_{[A'B'C']}=\frac{12^{2}\sqrt{3}}{4}\\ \\ A_{[A'B'C']}=\frac{144\sqrt{3}}{4}\\ \\ \boxed{A_{[A'B'C']}=36\sqrt{3}(cm^{2})}$


Calculam volumul:

$V_{tr}=\frac{h_{tr}}{3}\cdot(A_{B}+A_{b}+\sqrt{A_{B} \cdot A_{b}})\\ \\ V_{tr}=\frac{3}{3}\cdot(81\sqrt{3}+36\sqrt{3}+\sqrt{81\sqrt{3} \cdot 36\sqrt{3}})\\ \\V_{tr}=117\sqrt{3}+9\cdot6\cdot\sqrt{3}=171\sqrt{3} (cm^{3})$

b)
Formula apotemei trunchiului de piramida:

$\boxed{a_{t}^{2}=h^{2}+(a_{B}-a_{b})^2}\\ \\ \\ \\a_{B}=\frac{L\sqrt{3}}{6}=\frac{18\sqrt{3}}{6}=3\sqrt{3}(cm)\\ \\ a_{b}=\frac{l\sqrt{3}}{6}=\frac{12\sqrt{3}}{6}=2\sqrt{3}(cm)$
$\\ \\ a_{t}^{2}=3^{2}+(3\sqrt{3}-2\sqrt{3})^{2}\\ \\ a_{t}^{2}=9+(\sqrt{3})^{2}\\ \\ a_{t}^{2}=9+3=12\\ \\ \Rightarrow a_{t}=\sqrt{12}=2\sqrt{3}(cm)$


c)
Formula pentru aria laterala:

$\boxed{A_{l}= \frac{(P_{B}+P_{b})\cdot a_{t}}{2}}$

Calculam perimetrele:

$P_{B}=3\cdot L=3\cdot18=54(cm) \\ \\ P_{b}=3\cdot l=3\cdot 12=36(cm)$

Calculam aria laterala folosind formula:

$A_{l}= \frac{(P_{B}+P_{b})\cdot a_{t}}{2}\\ \\ A_{l}= \frac{(54+36)\cdot 2\sqrt{3}}{2}\\ \\ A_{l}=\frac{180\sqrt{3}}{2}\\ \\ A_{l}=90\sqrt{3}(cm^{2})$