L=AB=18cm (latura bazei mari)
l=A'B'=12cm (latura bazei mici)
h=OO'=3cm (inaltimea)
a)
Formula volumului trunchiului de piramida:
[tex]\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\\
[/tex]
ABCA'B'C' trunchi de piramida triunghiulara regulata (1)
Din (1) =>ΔABC, ΔA'B'C' echilaterale
=>
[tex]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})}[/tex]
Calculam volumul:
[tex]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})[/tex]
b)
Formula apotemei trunchiului de piramida:
[tex]\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)\\
\\ [/tex]
[tex]\\
\\ 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)
[/tex]
c)
Formula pentru aria laterala:
[tex]\boxed{A_{l}= \frac{(P_{B}+P_{b})\cdot a_{t}}{2}}[/tex]
Calculam perimetrele:
[tex]P_{B}=3\cdot L=3\cdot18=54(cm) \\
\\ P_{b}=3\cdot l=3\cdot 12=36(cm)\\
\\[/tex]
Calculam aria laterala folosind formula:
[tex]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})[/tex]
