Răspuns:
a) Din formula volumului aflăm înălțimea:
[tex] \boxed{ \boldsymbol{ \mathcal {V} = \dfrac {h \cdot (L^2+\ell^2+L \cdot \ell)}{3} }}[/tex]
[tex]\Rightarrow \dfrac {h \cdot ((12\sqrt {3})^2+(6\sqrt {3})^2+(12\sqrt {3}) \cdot (6\sqrt {3}))}{3} = 1134 \\ [/tex]
[tex]\Rightarrow \dfrac {h \cdot 756}{3} = 1134 \Rightarrow \bf h = 4,5 \ cm \\ [/tex]
[tex]a_B = \dfrac{L}{2} = \dfrac{12 \sqrt{3} }{2} = 6 \sqrt{3} \ cm[/tex]
[tex]a_b = \dfrac{\ell}{2} = \dfrac{6 \sqrt{3} }{2} = 3 \sqrt{3} \ cm[/tex]
[tex] \boldsymbol{ {a_t}^{2} = h^2 + (a_B-a_b)^2}[/tex]
[tex]{a_t}^{2} = 4.5^2 + (6 \sqrt{3} -3 \sqrt{3} )^2 \Rightarrow {a_t}^{2} = 20.25 + (3 \sqrt{3} )^2 \\ [/tex]
[tex]{a_t}^{2} = 20.25 + 27\Rightarrow \bf a_t = \sqrt{47.25} = \dfrac{3 \sqrt{21} }{2} \\ [/tex]
b) Aria laterală are formula:
[tex]\boxed{ \boldsymbol{ \mathcal {A}_{\ell} = \dfrac {(P_B + P_b) \cdot a_t}{2}}}[/tex]
[tex]\mathcal {A}_{\ell} = \dfrac {4 \cdot(12 \sqrt{3} + 6\sqrt{3}) \cdot \dfrac{3 \sqrt{21} }{2}}{2} = \dfrac {2 \cdot(18 \sqrt{3} \cdot 3 \sqrt{21})}{2} = \\ [/tex]
[tex]= \bf 162 \sqrt{7} \ {cm}^{2} [/tex]
[tex]c) \ \dfrac {H-h}{h} = \dfrac {a_p-a_t}{a_t} = \dfrac {a_b}{a_B} = \dfrac {6\sqrt{3}-3\sqrt{3}}{6\sqrt{3}} = \dfrac{1}{2} \\ [/tex]
[tex]\dfrac {a_p-a_t}{a_t} = \dfrac {1}{2} \Rightarrow \dfrac {H-4,5}{4,5} = \dfrac {1}{2} \\ [/tex]
[tex]\Rightarrow \boldsymbol {H = 6.75 \ cm}[/tex]
[tex]\dfrac {a_p-a_t }{a_t} = \dfrac {1}{2} \Rightarrow a_p = \dfrac {3a_t}{2} = \dfrac {3}{2} \cdot \dfrac{3 \sqrt{21}}{2} \\ [/tex]
[tex]\Rightarrow \boldsymbol{ a_p = \dfrac{9 \sqrt{21}}{4}}[/tex]
